problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
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1 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAJMO Problem 3 | Let $\alpha = \angle BAZ$, $\beta = \angle ABX$.
Since $XY$ is a chord of the circle with diameter $AB$,
$\angle XAY = \angle XBY = \gamma$. From the chord $YZ$,
we conclude $\angle YAZ = \angle YBZ = \delta$.
Triangles $BQY$ and $APY$ are both right-triangles, and share the
angle $\gamma$, therefore they are simi... | import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label("$O$", O, plain.S); pair A = r * plain.W; dot(A); label("$A$", A, unit(A)); pair B = r * plain.E; dot(B); label("$B$", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y... | [] |
1 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAJMO Problem 3 | Let $T$ be the foot of the perpendicular from $Y$ to $\overline{AB}$, let $O$ be the center of the semi-circle.
Since we have a semi-circle, if we were to reflect it over $\overline {AB}$, we would have a full circle, with $\triangle{AXB}$ and $\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that f... | currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P)... | [] |
2 | Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$
and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and
$CE$ meet at $I$. Determine whether or not it is possible for
segments $AB, AC, BI, ID, CI, IE$ to all have integer len... | 2010 USAJMO Problem 6 | We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to $45^{\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,
$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and tha... | import olympiad; // Scale unitsize(1inch); // Shape real h = 1.75; real w = 2.5; // Points void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); } pair A = origin; ldot(A, "$A$", plain.SW); pair B = w * plain.E; ldot(B, "$B$", plain.SE); pair C = h * plain.N; ldot(C, "$C$", plain.NW); pair D = ex... | [] |
3 | For a point $P = (a, a^2)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2a$. Consider the set of triangles with vertices of the form $P_1 = (a_1, a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$, such that the intersections of the lines $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$... | 2011 USAJMO Problem 3 | Solution 1
Note that the lines $l(P_1), l(P_2), l(P_3)$ are \[y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,\] respectively. It is easy to deduce that the three points of intersection are \[\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).\] The slopes of ... | Label f; f.p=fontsize(6); xaxis(-2,2); yaxis(-2,2); real f(real x) { return x^2; } draw(graph(f,-sqrt(2),sqrt(2))); real f(real x) { return (2*sqrt(3)/3)*x-1/3; } draw(graph(f,-5*sqrt(3)/6,2)); real f(real x) { return (-sqrt(3)/9)*x-1/108; } draw(graph(f,-2,2)); real f(real x) { return (-5*sqrt(3)/3)*x-25/12; } dra... | [] |
4 | Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P$, $A$, $C$ are collinear, and (iii) $\overline{DE} \parallel \overline{AC}$. Prove that $\overline{BE}$ bisects $\overline{AC}$. | 2011 USAJMO Problem 5 | This is the solution from EGMO Problem 1.43 page 242
Let $O$ be the center of the circle, and let $M$ be the midpoint of $AC$. Let $\theta$ denote the circle with diameter $OP$. Since $\angle OBP = \angle OMP = \angle ODP = 90^\circ$, $B$, $D$, and $M$ all lie on $\theta$.
Since quadrilateral $BOMP$ is cyclic, $\... | // Block 1
import graph;
unitsize(2 cm);
pair A, B, C, D, E, M, O, P;
path circ;
O = (0,0);
circ = Circle(O,1);
B = dir(100);
D = dir(240);
P = extension(B, B + rotate(90)*(B), D, D + rotate(90)*(D));
C = dir(-40);
A = intersectionpoint((P--(P + 0.9*(C - P))),circ);
E = intersectionpoint((D + 0.1*(C - A))--(D + C - ... | [] |
5 | Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle{BPS} = \angle{PRS}$, and $\angle{CQR} = \angle{QSR}$. Prove that $P$, $Q$,... | 2012 USAJMO Problem 1 | Since $\angle BPS = \angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.
For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle.... | // Block 1
import markers;
unitsize(0.5 cm);
pair A, B, C, O, P, Q, R, S;
A = (2,12);
B = (0,0);
C = (14,0);
P = intersectionpoint(A--B,Circle(A,8));
Q = intersectionpoint(A--C,Circle(A,8));
O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q));
S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270));... | [] |
6 | Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear. | 2012 USAJMO Problem 6 | By the sine law on triangle $AB'P$,
\[\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},\]
so
\[AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.\]
Similarly,
\begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'... | import graph; import geometry; unitsize(0.5 cm); pair[] A, B, C; pair P, R; A[0] = (2,12); B[0] = (0,0); C[0] = (14,0); P = (4,5); R = 5*dir(70); A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[... | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAJMO Problem 3 | In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that $\omega_A, \omega_B, \omega_C$ concur at a point $M$ (the Miquel point). Let $XM$ meet $\omega_B, \omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have \[XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM... | /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43... | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAJMO Problem 3 | We will use some construction arguments to solve the problem. Let $\angle BAC=\alpha,$ $\angle ABC=\beta,$ $\angle ACB=\gamma,$ and let $\angle APB=\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T... | // Block 1
import graph; size(12cm); real labelscalefactor = 1.9; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.6988888888888977,6.4266666666666... | [] |
7 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAJMO Problem 3 | We begin again by noting that the three circumcircles intersect at point $M$ by Miquel's theorem. In addition, we state that the angle $\angle MQC = \alpha$, hence $\angle MPC = \angle MZC = 180 - \alpha$, as well as $\angle AQM$, from which follows that $\angle ARM = \alpha$, so $\angle BRM = 180 - \alpha$, and $\angl... | /* DRAGON 0.0.9.6 */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43122416534181074); pair R = (-1... | [] |
8 | Let $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^{\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$, respectively.
(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.
(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respecti... | 2014 USAJMO Problem 2 | Lemma: $H$ is the reflection of $O$ over the angle bisector of $\angle BAC$ (henceforth 'the' reflection)
Proof: Let $H'$ be the reflection of $O$, and let $B'$ be the reflection of $B$.
Then reflection takes $\angle ABH'$ to $\angle AB'O$.
$\Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of... | // Block 1
import olympiad;
unitsize(1inch);
pair A,B,C,O,H,P,Q,i1,i2,i3,i4;
//define dots
A=3*dir(50);
B=(0,0);
C=right*2.76481496;
O=circumcenter(A,B,C);
H=orthocenter(A,B,C);
i1=2*O-H;
i2=2*i1-O;
i3=2*H-O;
i4=2*i3-H;
//These points are for extending PQ. DO NOT DELETE!
P=intersectionpoint(i2--i4,A--B);
Q=intersec... | [] |
9 | Let $ABC$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M,N,P$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $E,F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$, respectively. Let $U,V$ be the intersections of lin... | 2014 USAJMO Problem 6 | (a)
Solution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$, with $R$ and $Q$ not equal to $V$. Let $x = \angle A/2 = \angle IAE$ and $y = \angle C/2 = \angle ICA$. We know that $\overline{MP} \parallel \overline{AC}$ because $MP$ is a midsegment of tri... | // Block 1
unitsize(5cm);
import olympiad;
pair A, B, C, I, M, N, P, E, F, U, V, X, R;
A = dir(190);
B = dir(120);
C = dir(350);
I = incenter(A, B, C);
label("$A$", A, W);
label("$B$", B, dir(90));
label("$C$", C, dir(0));
dot(I); label("$I$", I, SSE);
draw(A--B--C--cycle);
real r, R;
r = inradius(A, B, C);
R = circ... | [] |
10 | The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the... | 2016 USAJMO Problem 1 | We claim that $M$ (midpoint of arc $BC$) is the fixed point.
We would like to show that $M$, $P$, $I_B$, $I_C$ are cyclic.
We extend $PI_B$ to intersect $\omega$ again at R.
We extend $PI_C$ to intersect $\omega$ again at S.
We invert around a circle centered at $P$ with radius $1$ (for convenience).
(I will denote X... | // Block 1
size(8cm);
pair A = dir(90);
pair B = dir(-10);
pair C = dir(190);
pair P = dir(-70);
pair U = incenter(A,B,P);
pair V = incenter(A,C,P);
pair M = dir(-90);
draw(circle((0,0),1));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$I_B$", U, NE);
dot("$I_C$", V,... | [] |
10 | The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the... | 2016 USAJMO Problem 1 | Let $M$ be the midpoint of arc $BC$. Let $D$ be the midpoint of arc $AB$. Let $E$ be the midpoint of arc $AC$.
Then, $P, I_B$, and $D$ are collinear and $P, I_C$, and $E$ are collinear.
We'll prove $MPI_B I_C$ is cyclic. (Intuition: we'll show that $M$ is the Miquel's point of quadrilateral $DE I_C I_B$.
$D$ is the c... | // Block 1
size(8cm);
pair A = dir(90);
pair B = dir(-10);
pair C = dir(190);
pair P = dir(-70);
pair U = incenter(A,B,P);
pair V = incenter(A,C,P);
pair M = dir(-90);
pair D = dir(40);
pair E = dir(140);
draw(circle((0,0),1));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D))... | [] |
11 | Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.
Given tha... | 2016 USAJMO Problem 5 | It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point,
\[AP\cdot AB=AH^2=AQ\cdot AC\] Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$-angle bisect... | // Block 1
size(8cm);
pair O=(0,0);
pair A=dir(110);
pair B=dir(-29);
pair C=dir(209);
pair H=foot(A,B,C);
pair P=foot(H,A,B);
pair Q=foot(H,A,C);
draw(A--B--C--A--H--P);
draw(circle(O,1));
draw(Q--H);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$H$", H, S);
dot("$P$", P, NE);
dot("$Q$", ... | [] |
12 | ($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$. | 2017 USAJMO Problem 3 | We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\sqrt{3})$. Using the facts $\triangle{CBP} \sim \triangle{CFB}$ and $\triangle{BCP} \sim \triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \frac{1}{2x}$, and $E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x... | // Block 1
import cse5;
import graph;
import olympiad;
size(3inch);
pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));
pair P = (-1, -sqrt(11)+sqrt(3));
path circle = Circle(O, 2sqrt(3));
pair D = extension(A,P,B,C);
pair E1 = extension(A,C,B,P);
pair F=extension(A... | [] |
13 | Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$. | 2017 USAJMO Problem 5 | (original diagram by integralarefun)
It's well known that the reflection of $H$ across $\overline{BC}$, $H'$, lies on $(ABC)$. Then $(BHC)$ is just the reflection of $(BH'C)$ across $\overline{BC}$, which is equivalent to the reflection of $(ABC)$ across $\overline{BC}$. Reflect points $A$ and $N$ across $\overline{BC... | // Block 1
import olympiad;
unitsize(100);
pair pA = dir(120);
pair pB = dir(225);
pair pC = dir(315);
pair pO = origin;
pair pH = orthocenter(pA, pB, pC);
pair pM = midpoint(pB--pC);
pair dD = bisectorpoint(pB, pA, pC);
pair pD = extension(pA, dD, pB, pC);
pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(p... | [] |
13 | Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$. | 2017 USAJMO Problem 5 | Suppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$, and let the foot of the A-altitude of $ABC$ be $E$. Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$. Likewise, $\angle CHE=\angle B$. So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$.
$BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=18... | // Block 1
size(9cm);
pair A = dir(130);
pair B = dir(220);
pair C = dir(320);
draw(unitcircle, lightblue);
pair P = dir(-90);
pair Q = dir(90);
pair D = extension(A, P, B, C);
pair O = origin;
pair M = extension(B, C, O, P);
pair N = 2*M-P;
draw(A--B--C--cycle, lightblue);
draw(A--P--Q, lightblue);
draw(A--... | [] |
14 | $(*)$ Let $ABC$ be a triangle with $\angle ABC$ obtuse. The $A$-excircle is a circle in the exterior of $\triangle ABC$ that is tangent to side $\overline{BC}$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respect... | 2019 USAJMO Problem 4 | We claim that the answer is no. We proceed with contradiction. Suppose that $EF$ is indeed tangent to the a-excenter. Define the point of tangency to be $X$. Let $G$ to be the intersection of the a-excircle with the extension of $AB$ and $H$ to be the intersection of the a-excircle with the extension of $AC$. Define $I... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black... | [] |
15 | Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ the... | 2023 USAJMO Problem 3 | To start off, we put the initial non-covered square in a corner (marked by the shaded square). Let's consider what happens when our first domino slides over the empty square. We will call such a move where we slide a domino over the uncovered square a "step":
When the vertically-oriented domino above the shaded squa... | // Block 1
size(5cm);
draw((0,0)--(0,3.2));
draw((1,0)--(1,3.2));
draw((0,0)--(1.2,0));
draw((0,1)--(1.2,1));
draw((0,2)--(1.2,2));
draw((0,3)--(1.2,3));
fill(origin--(1,0)--(1,1)--(0,1)--cycle, grey);
draw((0.1,1.1)--(0.1,2.9)--(0.9,2.9)--(0.9,1.1)--cycle,dashed);
draw((0.1,0.1)--(0.1,1.9)--(0.9,1.9)--(0.9,0.1)--cyc... | [] |
16 | Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Points $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral. | 2024 USAJMO Problem 1 | First, let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. It is clear that $AE=BE=3.5$, $PE=QE=0.5$, $DF=CF=4$, and $SF=RF=2$. Also, let $O$ be the circumcenter of $ABCD$.
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' ... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/wiki/User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* p... | [] |
17 | A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the a... | 1972 USAMO Problem 5 | Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).
Proof: For the "only if" direction, since $[A_0A_1A_2]... | // Block 1
size(120);
defaultpen(fontsize(10));
pathpen = black;
pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72));
D(A--B--C--D--E--cycle);
D(A--C--E--B--D--cycle);
pair Ap = MP("A'",IP(B--D,C--... | [] |
18 | Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $x=u+v+w$. | 1974 USAMO Problem 5 | Solution 1
We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$:
Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then
\[[ABC] + \frac... | // Block 1
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE);
// Block 2
size(20... | [] |
19 | Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that
\[AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.\] | 1975 USAMO Problem 2 | Solution 1
If we project points $A,B,C,D$ onto the plane parallel to $\overline{AB}$ and $\overline{CD}$, $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ are coplanar:
Let $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$. We... | // Block 1
defaultpen(fontsize(8)); pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); label("A",A,(0,1));label("D",D,(1,0));label("B",B,(-1,-1));label("C",C,(1,-1)); axialshade(A--C--D--cycle, lightgray, A, gray, D); draw(A--B--C--A--D--C);draw(B--D, linetype("8 8")); label("$m$",(A+B)/2,(-1,1));label("$n$",(C+D)/2,(1,0)); labe... | [] |
20 | (a) Suppose that each square of a $4\times 7$ chessboard, as shown above, is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are ... | 1976 USAMO Problem 1 | There are many ways to prove the first part, we will show one.
Consider the first row. It contains at least four cells of the same color, say white. Pick any four such cells. Let's now consider these four columns only. If any of the remaining three rows contains two white cells in some of these columns, we are done. O... | // Block 1
void fillsq(int x, int y){
fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray);
}
int i;
fillsq(0,0);fillsq(0,1);
fillsq(1,0);fillsq(1,2);
fillsq(2,0);fillsq(2,3);
fillsq(3,1);fillsq(3,2);
fillsq(4,1);fillsq(4,3);
fillsq(5,2);fillsq(5,3);
for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);}
for(i=0;... | [] |
20 | (a) Suppose that each square of a $4\times 7$ chessboard, as shown above, is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are ... | 1976 USAMO Problem 1 | We will prove the first part by the pigeonhole principle. To form a rectangle, two squares have in one column have to be colored the same color as the corresponding squares in another column. In the example given, the two second and third squares of the fourth and seventh columns are all white. When coloring this recta... | // Block 1
void fillsq(int x, int y){
fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray);
}
int i;
fillsq(0,3);fillsq(0,2);
fillsq(1,3);fillsq(1,1);
fillsq(2,3);fillsq(2,0);
fillsq(3,2);fillsq(3,1);
fillsq(4,2);fillsq(4,0);
fillsq(5,1);fillsq(5,0);
for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);}
for(i=0;... | [] |
21 | $P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible. | 1979 USAMO Problem 4 | Let $r = OP, x = \angle OPR, a = \angle POR,$ and $b = \angle POQ.$ Then $\angle ORP = \pi - x - a$ and $\angle OQP = x - b.$ Using the Law of Sines on $\triangle OPR$ gives
\[PR = \sin a * \frac{r}{\sin(\pi - x - a)} = \sin a * \frac{r}{\sin(x + a)},\]
and using the Law of Sines on $\triangle OPQ$ gives
\[PQ = \sin b ... | // Block 1
pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16);
dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P);
label("O", O, S);
label("A", A, W);
label("Q", Q, W);
label("B", B, S);
label("R", R, S);
label("P", P, E);
draw(O--R--Q--O); draw(O--P);
label("r", O--P, N);
// Block 2
pair O = (0,... | [] |
22 | In triangle $ABC$, angle $A$ is twice angle $B$, angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter. | 1991 USAMO Problem 1 | Solution 1
(diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of $\angle A$, and let it intersect $\overline{BC}$ at $D$. Notice that $\triangle ADC\sim \triangle BAC$, and let $AD=x$. Now from similarity,
\[x=\frac{bc}{a}\]
However, from the angle bisector theorem, we have
\[BD=\fr... | import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label... | [] |
23 | Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that t... | 1991 USAMO Problem 5 | Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$, respectively; let these circles touch $CD$ at $C_a$, $C_b$, respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$, respectively, as shown in the diagram below.
We note that
\[CE = CC_a - EC... | // Block 1
size(220);
defaultpen(1);
pair A=(0,0), B=(220,0), C=(18.7723,118.523);
pair D=(72.6,0);
pair Ia=incenter(A,D,C), Ib=incenter(B,D,C);
pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129);
pair E=IntersectionPoint((Ta--Tb),(C--D));
path Oa=circle(Ia,inradius(A,D,C));
path Ob=circle(Ib,inradius(B,D,C));
pair Da=IP(... | [] |
24 | Prove
\[\frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.\] | 1992 USAMO Problem 2 | Solution 1
Consider the points $M_k = (1, \tan k^\circ)$ in the coordinate plane with origin $O=(0,0)$, for integers $0 \le k \le 89$.
Evidently, the angle between segments $OM_a$ and $OM_b$ is $(b-a)^\circ$, and the length of segment $OM_a$ is $1/\cos a^\circ$. It then follows that the area of triangle $M_aOM_b$... | size(200); defaultpen(1); pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30); pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=y/y.x, Z=z/z.x, E=(D+Y)/2; pair X=(O+E)/2; draw(O--A--Z); draw(B--O--C--D--O--Y--Z--O); label("\(O\)",O,SW); label("\(M_0\)",A,ESE); label("\(M_1\)",B,ESE); la... | [] |
25 | Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. | 1993 USAMO Problem 2 | Diagram
Work
Let $X$, $Y$, $Z$, $W$ be the foot of the altitude from point $E$ of $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$.
Note that reflection of $E$ over all the points of $XYZW$ is similar to $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections a... | import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('$E$',E,N); pair A,B,C,D; A=(9,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('$A$',A,E); label('$B$',B,N); label('$C$',C,W); label('$D$',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=refl... | [] |
26 | A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE$, and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $\frac{CP}{PE}=(\frac{AC}{CE})^2$. | 1994 USAMO Problem 3 | Let the diagonals $AD$, $BE$, $CF$ meet at $Q$.
First, let's show that the triangles $\triangle AEC$ and $\triangle QED$ are similar.
$\angle ACE=\angle ADE$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\angle ADE=\angle QDE$. $\angle AEB=\angle CED$ because $AB=CD$, and $A$,$B$,$C$,$D$ and $E$ all lie... | // Block 1
pair A,B,C,D,E,F,P,Q;
A=(-0.96,0.28);
B=(-0.352,0.936);
C=(0,1);
D=(4/5,3/5);
E=(4/5,-3/5);
F=(0,-1);
P=IntersectionPoint(A--D,C--E);
Q=IntersectionPoint(A--D,C--F);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E,green);
draw(E--F);
draw(F--A);
draw(A--C,red);
draw(A--Q);
draw(A--E,red);
draw(B--Q);
draw(C-... | [] |
27 | Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$ , $\angle MAC= 40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. | 1996 USAMO Problem 5 | Solution 1
Clearly, $\angle AMB = 150^\circ$ and $\angle AMC = 110^\circ$. Now by the Law of Sines on triangles $ABM$ and $ACM$, we have \[\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}\] and \[\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.\] Combining these equations gives us \[\frac{AB}{AC} = \frac{\sin ... | pair A,B,C,M; A=(0,0); B=(1,2); C=(2,0); M=(0.8,1.1); draw(A--B); draw(B--C); draw(C--A); draw(A--M); draw(B--M); draw(C--M); label("\(A\)",A,SW); label("\(B\)",B,N); label("\(C\)",C,SE); label("\(M\)",M,NE); | [] |
28 | To clip a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon $P_6$ of area $1$ i... | 1997 USAMO Problem 4 | $\textbf{Claim:}$ It is impossible to choose two non-adjacent sides and clip a whole part of it off.
$\textbf{Proof:}$ If you clip adjacent sides, you can cut off at most up to the blue lines; Clipping more is impossible due to the degrees getting larger and larger and more and more circular.
Thus, after infinitely... | // Block 1
size(200);
draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0));
draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue);
draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue);
// Block 2
size(200); draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--... | [] |
29 | Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line... | 1998 USAMO Problem 2 | First, $AD=\frac{AB}{2}=\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC$. Therefore, $\triangle ACF \sim \triangle AED$, so $\angle ACF = \angle AED$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, w... | // Block 1
pair O,A,B,C,D,E,F,DEb,CFb,Fo,M;
O=(0,0);
A=(1.732,1);
B=(0,1);
C=(-1.732,1);
D=(0.866,1);
Fo=(-1,-0.5);
path AC,AF,DE,CF,DEbM,CFbM,C1,C2;
C1=circle(O,2);
C2=circle(O,1);
E=intersectionpoints(A--Fo,C2)[0];
F=intersectionpoints(A--Fo,C2)[1];
DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0);
CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.... | [] |
30 | Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.) | 1998 USAMO Problem 6 | Lemma: If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ in an equiangular $n$-gon are tangential, and $A_iA_{i+3}$ is the longest side quadrilateral $A_iA_{i+1}A_{i+2}A_{i+3}$ for all $i$, then quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential.
Proof:
If quadrilaterals $A... | // Block 1
import geometry;
size(10cm);
pair A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U;
A = (-1,0);
B = (1,0);
draw(Circle(A,1)^^Circle(B,1));
C = (sqrt(2)/2-1,sqrt(2)/2);
D = (-sqrt(3)/2 - 1, .5);
E = (-sqrt(3)/2 - 1, -.5);
F = (-1,-1);
G = (1,-1);
H = (sqrt(3)/2 + 1, -.5);
I = (sqrt(3)/2 + 1, .5);... | [] |
31 | Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | 2000 USAMO Problem 4 | We claim that $n = 1999$ is the smallest such number. For $n \le 1998$, we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$. We call a row or column fill... | // Block 1
for(int i = 0; i < 10; ++i){
for(int j = 0; j < 10; ++j){
if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j)*unitsquare,rgb(0.3,0.3,0.3));
else draw(shift(i,j)*unitsquare);
}
}
// Block 2
for(int i = 0; i < 10; ++i){ for(int j = 0; j < 10; ++j){ if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,... | [] |
32 | Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k - 1}$ and passes through $A_k$ and $A_{k + 1},$ where $A_{n + 3... | 2000 USAMO Problem 5 | Solution 1
Let the circumcenter of $\triangle ABC$ be $O$, and let the center of $\omega_k$ be $O_k$. $\omega_k$ and $\omega_{k-1}$ are externally tangent at the point $A_k$, so $O_k, A_k, O_{k-1}$ are collinear.
$O$ is the intersection of the perpendicular bisectors of $\overline{A_1A_2}, \overline{A_2A_3}, \overlin... | size(300); pathpen = linewidth(0.7); pen t = linetype("2 2"); pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("... | [] |
33 | Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection... | 2001 USAMO Problem 2 | Solution 1
It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = ... | pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca... | [] |
34 | Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$. | 2003 USAMO Problem 4 | Solution 1
Extend segment $DM$ through $M$ to $G$ such that $FG\parallel CD$.
Then $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\parallel CG$. Hence $MC = MF$ if and only if $\angle GCD = \angle FDA$, that is, $\angle FDA + \angle CGF = 180^\circ$.
Because quadrilateral $ABED$ is cyclic, $... | // Block 1
defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c... | [] |
35 | At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert... | 2003 USAMO Problem 6 | Assume the original numbers are $a,b,c,d,e,f$. Since $a+b+c+d+e+f$ is odd, either $a+c+e$ or $b+d+f$ must be odd. WLOG let $a+c+e$ be odd and $a\ge c\ge e \ge 0$.
Case 1
$a,c,e>0$. Define Operation A as the sequence of moves from Step 1 to Step 3, shown below:
Notice that Operation A changes the numbers $a,c,e$ to $... | // Block 1
size(300); defaultpen(fontsize(9)); label("$d$",expi(0),(0,0)); label("$c$",expi(pi/3),(0,0),red); label("$b$",expi(2*pi/3),(0,0)); label("$a$",expi(pi),(0,0),red); label("$f$",expi(4*pi/3),(0,0)); label("$e$",expi(5*pi/3),(0,0),red); label("Step 1",(0,-2),(0,0)); label("$c-e$",(5,0)+expi(0),(0,0)); label("... | [] |
36 | (Zuming Feng) Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex quadrilateral ... | 2005 USAMO Problem 3 | Solution 1
Let $B_1'$ be the second intersection of the line $C_1A$ with the circumcircle of $APC$, and let $Q'$ be the second intersection of the circumcircle of $B_1' C_1P$ and line $BC$. It is enough to show that $B_1'=B_1$ and $Q' =Q$. All our angles will be directed, and measured mod $\pi$.
Since points $C_1,... | size(300); defaultpen(1); pair A=(2,5), B=(-1,0), C=(5,0); pair C1=(.5,5.7); path O1=circumcircle(A,B,C1); pair P=IntersectionPoint(O1,B--C,1); path O2=circumcircle(A,P,C); pair B1=IntersectionPoint(O2,C1--5A-4C1,0); path O=circumcircle(B1,C1,P); pair Q=IntersectionPoint(O,B--C,1); draw(C1--P--A--B--C--A); draw(P--B1... | [] |
37 | (Kiran Kedlaya, Sungyoon Kim) Let $ABC$ be an acute triangle with $\omega$, $\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively. The circle $\omega_A$ is tangent internally to $\Omega$ at $A$ and externally tangent to $\omega$. Circle $\Omega_A$ is internally tangent to $\Omega$ at $A$ ... | 2007 USAMO Problem 6 | Solution 1
Lemma.
\[P_{A}Q_{A}=\frac{ 4R^{2}(s-a)^{2}(s-b)(s-c)}{rb^{2}c^{2}}\]
Proof. Note $P_{A}$ and $Q_{A}$ lie on $AO$ since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let $\omega$ touch $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Note $AE=AF=s-a$. Cons... | size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a=x*(O-A)+A; q_a=... | [] |
38 | (Zuming Feng) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $... | 2008 USAMO Problem 2 | Solution 1 (synthetic)
Without Loss of Generality, assume $AB >AC$. It is sufficient to prove that $\angle OFA = 90^{\circ}$, where $O$ is the circumcenter of $\triangle{ABC},$ as this would immediately prove that $A,P,O,F,N$ are concyclic.
By applying the Menelaus' Theorem in the Triangle $\triangle BFC$ for the trans... | // Block 1
/* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F... | ["https://artofproblemsolving.com/wiki/images/0/05/2008usamo2-sol8.png", "https://artofproblemsolving.com/wiki/images/d/df/2008usamo2-sol9.png"] |
39 | (Gabriel Carroll) Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$ with integer coordinates such that
\[\left|x\right| + \left|y + \frac {1}{2}\right| < n\]
A path is a sequence of distinct points $(x_1 , y_1 ), (x_2 , y_2 ), \ldots , (x_\ell, y_\ell)$ in $S_n$ such that, for $i = 2, \ldots , \... | 2008 USAMO Problem 3 | Solution 1
Color all the points in $S_{n}$ red or black such that each row of points starts and ends with a red point, and alternates between red and black for each point in the row. This creates a checkerboard pattern, except that the middle two rows are identical.
Examples for $n=4$
Suppose there is a par... | // Block 1
/* variable declarations */ int n = 4; /* code */ pen r = red, b = black; int absol(int k){if(k >= 0) return k; else return -1*k-1;} pen col(bool k){if(k) return r; else return b;} for(int y = -n; y < n; ++y){ bool b = true; for(int x = absol(y)-n+1; x < n-absol(y); ++x){ dot((x,y),col(b)); b = !b... | [] |
40 | (Gregory Galparin) Let $\mathcal{P}$ be a convex polygon with $n$ sides, $n\ge3$. Any set of $n - 3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\mathcal{P}$ into $n - 2$ triangles. If $\mathcal{P}$ is regular and there is a triangulation of $\mathcal{P}... | 2008 USAMO Problem 4 | We label the vertices of $\mathcal{P}$ as $P_0, P_1, P_2, \ldots, P_n$. Consider a diagonal $d = \overline{P_a\,P_{a+k}},\,k \le n/2$ in the triangulation. We show that $k$ must have the form $2^{m}$ for some nonnegative integer $m$.
This diagonal partitions $\mathcal{P}$ into two regions $\mathcal{Q},\, \mathcal{R}$,... | // Block 1
size(200); defaultpen(linewidth(0.7)+fontsize(10));
int n = 17; real r = 1; real rad = pi/2;
pair pt(real k=0) {
return (r*expi(rad-2*pi*k/n));
}
for(int i=0; i<n; ++i){
dot(pt(i));
draw(pt(i)--pt(i+1));
}
draw(pt()--pt(8));
draw(pt()--pt(4)--pt(8),linewidth(0.7)+linetype("4 4"));
draw(pt()--pt(2)--pt(... | [] |
41 | Trapezoid $ABCD$, with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ ... | 2009 USAMO Problem 5 | We will use directed angles in this solution. Extend $QR$ to $T$ as follows:
If:
Note that \begin{align*}\measuredangle GBT+\measuredangle TRG&=\frac{m\widehat{TQ}}{2}+\measuredangle TRB+\measuredangle BRG\\ &=\frac{m\widehat{TQ}+m\widehat{DQ}+m\widehat{CB}+m\widehat{BT}}{2}.\\ \end{align*}
Thus, $BTRG$ is cyclic.
... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
draw(circle);
pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
draw(A--B--C--D--cycle); draw(B--D);
dot("$A$", A, NW); dot("$B$", B, NE); ... | [] |
41 | Trapezoid $ABCD$, with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ ... | 2009 USAMO Problem 5 | Extend $QR$ to $T$, and let line $l \parallel AB$ intersect $\omega$ at $K$ and another point $V$, as shown:
If:
Suppose that $VP \cap CB = S'$, and $AC \cap QV = R'$. Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that the points $S'$, $R'$, and $G = PA \cap BQ$ are collinear. However, $AC$ and $BD$ are... | // Block 1
import cse5;
import graph;
import olympiad;
dotfactor = 3;
unitsize(1.5inch);
path circle = Circle(origin, 1);
draw(circle);
pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
draw(A--B--C--D--cycle); draw(B--D);
dot("$A$", A, NW); dot("$B$", B, NE); ... | [] |
42 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAMO Problem 1 | Let $\alpha = \angle BAZ$, $\beta = \angle ABX$.
Since $XY$ is a chord of the circle with diameter $AB$,
$\angle XAY = \angle XBY = \gamma$. From the chord $YZ$,
we conclude $\angle YAZ = \angle YBZ = \delta$.
Triangles $BQY$ and $APY$ are both right-triangles, and share the
angle $\gamma$, therefore they are simi... | // Block 1
import olympiad;
// Scale
unitsize(1inch);
real r = 1.75;
// Semi-circle: centre O, radius r, diameter A--B.
pair O = (0,0); dot(O); label("$O$", O, plain.S);
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));
draw(arc(O, r, 0, 180)--cycle);
// ... | [] |
42 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter
$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto
lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle
formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where
$O$ is the midpoint of segment $AB$. | 2010 USAMO Problem 1 | Let $T$ be the foot of the perpendicular from $Y$ to $\overline{AB}$, let $O$ be the center of the semi-circle.
Since we have a semi-circle, if we were to reflect it over $\overline {AB}$, we would have a full circle, with $\triangle{AXB}$ and $\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that f... | // Block 1
currentpicture=new picture;
size(12cm);
pair O, A, B, X, Y, Z, P, Q, R, SS, T;
O=(0, 0);
A=(-1, 0);
B=(1, 0);
X=(Cos(144), Sin(144));
Y=(Cos(105), Sin(105));
Z=(Cos(27), Sin(27));
P=foot(Y, A, X);
Q=foot(Y, B, X);
R=foot(Y, A, Z);
SS=foot(Y, B, Z);
T=foot(Y, A, B);
dot(O); dot(A); dot(B); dot(X); dot(Y); dot... | [] |
43 | Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$
and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and
$CE$ meet at $I$. Determine whether or not it is possible for
segments $AB, AC, BI, ID, CI, IE$ to all have integer len... | 2010 USAMO Problem 4 | We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to $45^{\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,
$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and tha... | // Block 1
import olympiad;
// Scale
unitsize(1inch);
// Shape
real h = 1.75;
real w = 2.5;
// Points
void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); }
pair A = origin; ldot(A, "$A$", plain.SW);
pair B = w * plain.E; ldot(B, "$B$", plain.SE);
pair C = h * plain.N; ldot(C, "$C$", plain.NW);
... | [] |
44 | Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear. | 2012 USAMO Problem 5 | By the sine law on triangle $AB'P$,
\[\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},\]
so
\[AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.\]
Similarly,
\begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'... | // Block 1
import graph;
import geometry;
unitsize(0.5 cm);
pair[] A, B, C;
pair P, R;
A[0] = (2,12);
B[0] = (0,0);
C[0] = (14,0);
P = (4,5);
R = 5*dir(70);
A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0]));
B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0]));
C[1] = extension(A[0],B[0],P,reflect(P + R... | [] |
45 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAMO Problem 1 | In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that $\omega_A, \omega_B, \omega_C$ concur at a point $M$ (the Miquel point). Let $XM$ meet $\omega_B, \omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have \[XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM... | // Block 1
/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad;
import cse5;
size(11cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair A = (-1.0, 3.0);
pair B = (-3.0, -3.0);
pair C = (4.0, -3.0);
pair P = (-0.6698198198198195, -3.0);
pair Q = (1.14064652888... | [] |
45 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAMO Problem 1 | We will use some construction arguments to solve the problem. Let $\angle BAC=\alpha,$ $\angle ABC=\beta,$ $\angle ACB=\gamma,$ and let $\angle APB=\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T... | // Block 1
import graph; size(12cm);
real labelscalefactor = 1.9;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
draw((-3.6988888888888977,6.4266666666666... | [] |
45 | In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$... | 2013 USAMO Problem 1 | We begin again by noting that the three circumcircles intersect at point $M$ by Miquel's theorem. In addition, we state that the angle $\angle MQC = \alpha$, hence $\angle MPC = \angle MZC = 180 - \alpha$, as well as $\angle AQM$, from which follows that $\angle ARM = \alpha$, so $\angle BRM = 180 - \alpha$, and $\angl... | // Block 1
/* DRAGON 0.0.9.6 */
import olympiad;
import cse5;
size(11cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair A = (-1.0, 3.0);
pair B = (-3.0, -3.0);
pair C = (4.0, -3.0);
pair P = (-0.6698198198198195, -3.0);
pair Q = (1.1406465288818244, 0.43122416534181074);
p... | [] |
46 | For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. ... | 2013 USAMO Problem 2 | We label the points in clockwise order as $1,2,3,\ldots,n$, where point $A$ is the same as point $1$. We start and end at point $1$, and we must cross over it, either by visiting it again, or else by making the move from point $n$ to point $2$. We interpret each of these cases in terms of tiling. In each move, we eith... | // Block 1
unitsize(10);
draw((0,0)--(3,0)--(3,2)--(0,2)--cycle^^(0,1)--(3,1),linewidth(2));
draw((1,0)--(1,1)^^(2,1)--(2,2));
draw(shift((0,-2.5))*((0,0)--(3,0)--(3,2)--(0,2)--cycle^^(0,1)--(3,1)),linewidth(2));
draw(shift((0,-2.5))*((1,0)--(1,1)^^(2,1)--(2,2)));
draw(shift((7,0))*((0,0)--(4,0)--(4,2)--(0,2)--cycle^^(... | [] |
47 | In convex cyclic quadrilateral $ABCD,$ we know that lines $AC$ and $BD$ intersect at $E,$ lines $AB$ and $CD$ intersect at $F,$ and lines $BC$ and $DA$ intersect at $G.$ Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at ... | 2018 USAMO Problem 5 | \begin{align*} &\mathrel{\phantom{=}}\angle DEQ+\angle AED+\angle AEP\\ &=\angle DAQ+\angle AQD+\angle AEP\\ &=180-\angle ADC+\angle AEP\\ &=180-\angle ADC+\angle ABP\\ &=\angle ABP+\angle ABC\\ &=180 \end{align*}
so $P,E,Q$ are collinear. Furthermore, note that $DQBP$ is cyclic because:
\[\angle EDQ = \angle BAE = BPE... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black... | [] |
48 | Let ABC be a triangle with incenter $I$ and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively. Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$ that does not lie on any of the lines $II_{a}$, $I_{b}I_{c}$, or $BC$, suppose the circumcircles of $\triangle DIIa$ and $\triangle DI_b... | 2023 USAMO Problem 6 | Consider points $G,H,J,K,P,$ and $Q$ such that the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DII_A$ are $D$ and $G$, the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DI_BI_C$ are $D$ and $H$, the intersections of the circumci... | // Block 1
size(500);
pair A,B,C,D,E,F,G,H,I,J,K,IA,IB,IC,P,Q;
B=(0,0);
C=(8,0);
A=intersectionpoint(Circle(B,6),Circle(C,9));
I=incenter(A,B,C);
path c=circumcircle(A,B,C);
J=intersectionpoint(I--(4*I-3*A),c);
IA=2*J-I;
IB=2*intersectionpoint(I--(4*I-3*B),c)-I;
IC=2*intersectionpoint(I--(4*I-3*C),c)-I;
K=intersectionp... | [] |
49 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 2025 USAMO Problem 3 | Alice has a winning strategy: Choose $\mathcal S$ to be the set of of points strictly outside the disk with diameter $PQ$. Then two cities $A$ and $B$ have a road between them if and only if every other city is outside the closed disk with diameter $AB$. Note that this condition is equivalent to the angle $\angle ACB$ ... | // Block 1
import markers;
size(12cm);
pair A = (-1,0);
pair B = (1,0);
pair O = (0,0);
// Point C in the upper-left of the disk
pair C = dir(120) * 0.8;
draw(circle(O,1));
draw(A--B, dashed);
draw(A--C);
draw(C--B);
label("less than $\sqrt{n+1}$", A--B, S);
markangle(radius=10, A, C, B);
dot(A); dot(B); dot(C);
... | [] |
50 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problem 4 | Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$.
Applying the Pythagorean Theorem, $OA^2 = OD^... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(D--O--F--B,dashed);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);
label("$A$... | [] |
50 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problem 4 | We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\cos \angle OAB$, we'll be in good shape.
We can find $\cos \angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras,
\[AC = \sqrt{AB^2 + BC^2} = \sqrt{... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(O--B); draw(A--C);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);
label("$A$"... | [] |
50 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problem 4 | Mark the midpoint $M$ of $AC$. Then, drop perpendiculars from $O$ to $AB$ (with foot $T_1$), $M$ to $OT_1$ (with foot $T_2$), and $M$ to $AB$ (with foot $T_3$).
First notice that by computation, $OAC$ is a $\sqrt {50} - \sqrt {40} - \sqrt {50}$ isosceles triangle, so $AC = MO$.
Then, notice that $\angle MOT_2 = \angl... | // Block 1
size(200);
pair dl(string name, pair loc, pair offset) {
dot(loc);
label(name,loc,offset);
return loc;
};
pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)};
string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};
for(int i=0;i<a.length;++i) {
dl(n[i],a[i],dir(degrees(a[i],fal... | [] |
50 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. | 1983 AIME Problem 4 | I will use the law of cosines in triangle $\triangle OAC$ and $\triangle OBC$.
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}$
$\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}$
$\cos \angle ACO = \frac{AC^2+OC^2-OA^2}{2 \cdot AC \cdot OC} = \frac{(2\sqrt{10})^2+(\sqrt{50})^2-(\sqrt{50})^2}{2 ... | // Block 1
size(150); defaultpen(linewidth(0.6)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r);
pair D=(A.x,0),F=(0,B.y);
path P=circle(O,r);
pair C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P);
draw(C--B--O--A--B);
draw(O--B); draw(A--C); draw(O--C);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);... | [] |
51 | The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid? | 1983 AIME Problem 11 | Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$.
Next, we complete t... | // Block 1
size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s... | [] |
52 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the central angle o... | 1983 AIME Problem 15 | Solution 1
As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\ell$? The answer to this question is: a line $m$ parallel to $\ell$, such that $m$ and $P$ ... | // Block 1
size(170); pair P = (0,0), L1 = (-3, 0), L2 = (1.5, 1.5), M1 = (-3, 1), M2 = (1.5, 2.5); pair X = .6*M1 + .4*M2, M = (P+X)/2; draw(L1--L2); draw(M1--M2); draw(P--X, dotted); dot(M); dot("$P$", P, S); dot("$X$", X, N); label("$\ell$", L2, E); label("$m$", M2, E);
// Block 2
size(170); pair O = (0,0), D = (0, ... | ["https://artofproblemsolving.com/wiki/images/thumb/b/b7/Aime1983p15s2.png/500px-Aime1983p15s2.png", "https://artofproblemsolving.com/wiki/images/thumb/3/3c/Dgram.png/800px-Dgram.png"] |
53 | In tetrahedron $ABCD$, edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$. | 1984 AIME Problem 9 | Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron. So, $h = ... | // Block 1
/* modified version of olympiad modules */
import three;
real markscalefactor = 0.03;
path3 rightanglemark(triple A, triple B, triple C, real s=8)
{
triple P,Q,R;
P=s*markscalefactor*unit(A-B)+B;
R=s*markscalefactor*unit(C-B)+B;
Q=P+R-B;
return P--Q--R;
}
path3 anglemark(triple A, triple B, triple C, re... | [] |
54 | In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator... | 1985 AIME Problem 9 | All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle.
This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 ... | // Block 1
size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP("2",(rotate(a/... | [] |
54 | In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator... | 1985 AIME Problem 9 | It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$, and using the Law of Cosines, we get:
\[2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\]
Which, rearranges to:
\[21 = 24\cos\frac{\alpha}{2}\]
And, that gets us:
\[\cos\frac{\alpha}{2} = 7/8\]
U... | // Block 1
size(200);
pointpen = black; pathpen = black + linewidth(0.8);
real r = 8/15^0.5, a = 57.91, b = 93.135;
pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A;
D(CR(O,r));
D(O--A1--A2--cycle);
D(O--A2--A3--cycle);
D(O--A1--A3--cycle);
MP("2",(A1+A2)/2,NE);
MP("3",(... | [] |
55 | An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis? | 1985 AIME Problem 11 | An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$, $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$-axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is m... | // Block 1
size(200);
pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10);
pair F1=(9,20),F2=(49,55);
D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle);
D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60));
path p = F1--(49,-55);
pair X = IP(p,(0,0)--(80,0));
D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((... | [] |
56 | In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$. | 1986 AIME Problem 9 | Solution 1
Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$). The remaining three se... | // Block 1
size(200); pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C... | [] |
57 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | 1986 AIME Problem 15 | Let $\theta_1$ be the angle that the median through $A$ makes with the positive $y$-axis, and let $\theta_2$ be the angle that the median through $B$ makes with the positive $x$-axis. The tangents of these two angles are the slopes of the respective medians; in other words, $\tan \theta_1 = 1$, and $\tan \theta_2 =2$.... | // Block 1
size(170);
pair A = (0,0), B = (3, 2), C = (3, 0);
pair M = (B+C)/2, NN = (A+C)/2, G = (A+B+C)/3;
draw(A--B--C--cycle);
draw(A--M);
draw(B--NN);
label("$A$", A, S);
label("$C$", C, S);
label("$B$", B, N);
label("$M$", M, NW);
label("$N$", NN, NW);
label("$a$", M, E);
label("$b$", NN, S);
label("$G$", G, NW);... | [] |
57 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | 1986 AIME Problem 15 | We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$.
Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$. Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$. Note that the acute angle between the r... | // Block 1
import graph;
size(150);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 2.0));
yaxis(-8,8,Ticks(f, 2.0));
real f(real x)
{
return (x + 3);
}
real g( real x){
return (2x + 4);
}
draw(graph(f,-8,5),red+linewidth(1));
draw(graph(g,-6,2),blue+linewidth(1));
// Block 2
import graph; size(4cm);
real label... | [] |
58 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problem 6 | Label the point of intersection as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the law of cosines,
\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{1... | // Block 1
pointpen=black; pathpen=black+linewidth(0.7);
pair A=(0,0),B=(10,0),C=16*expi(pi/3);
D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t",(B+C)/2,NE);
// Block 2
pointpen=black; pathpen=black+linewidth(0.7); pair... | [] |
58 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problem 6 | Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$. Letting this line be the $x$-axis, we can reflect $B$ over the $x$-axis to get $B'$. As reflections preserve length, $B'X = XB$.
We then draw lines $BB'$ and $PB'$. W... | // Block 1
draw((0,0)--(11,0)--(7,14)--cycle);
draw((7,14)--(11,28));
draw((11,28)--(11,0));
label("$A$",(-1,-1),N);
label("$B$",(12,-1),N);
label("$P$",(6,15),N);
label("$B'$",(12,29),N);
draw((-10,14)--(20,14));
label("$X$",(12.5,15),N);
draw((7,0)--(7,14),dashed);
label("$Y$",(7,-4),N);
draw((6,0)--(6,1));
draw((6,1... | [] |
58 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problem 6 | We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram:
Now, if we drop an altitude from point$M$, we get :
We know this from the $30-60-90$ triangle that is form... | // Block 1
draw((0,0)--(100,0)--(80,139)--cycle);
label("8x",(0,0)--(80,139),NW);
label("7x",(100,0)--(80,139),NE);
label("100",(0,0)--(100,0),S);
dot((0,0));
label("A",(0,0),S);
dot((100,0));
label("B",(100,0),S);
dot((80,139));
label("M",(80,139),N);
// Block 2
size(300); draw((0,0)--(100,0)--(80,139)--cycle);
label(... | [] |
58 | Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 1989 AIME Problem 6 | Drop the altitude from $M$ to $AB$, and call it $P$. $\triangle AMP$ is a $30-60-90$ triangle, so $AP = 4t$ and $MP = 4\sqrt{3}t$, and by the Pythagorean theorem on $\triangle MPB$, $PB = t$. $AP + BP = AB$, so $t=20$. Therefore, $8t = \boxed{160}$.
~~Disphenoid_lover | // Block 1
draw((0,0)--(100,0)--(80,139)--cycle);
label("8x",(0,0)--(80,139),NW);
label("7x",(100,0)--(80,139),NE);
label("100",(0,0)--(100,0),S);
dot((0,0));
label("A",(0,0),S);
dot((100,0));
label("B",(100,0),S);
dot((80,139));
label("M",(80,139),N);
draw((80,139)--(80,0),dashed); // Altitude MP
label("$P$",(80,0),S)... | [] |
59 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that
\begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*}
Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | 1989 AIME Problem 8 | Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:
Label equations $(1),(2),(3),$ and $(4)$ as Solution 2 does. Since the coefficients of $x_1,x_2,x_3,x_4,x_5,x_6,x_7,$ or $(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36... | // Block 1
/* Made by MRENTHUSIASM */
size(20cm);
for (real i=1; i<=10; ++i) {
label("\boldmath{$"+string(i^2)+"$}",(i-1,0));
}
for (real i=1; i<=9; ++i) {
label("$"+string(1+2*i)+"$",(i-0.5,-0.75));
}
for (real i=1; i<=8; ++i) {
label("$2$",(i,-1.5));
}
for (real i=1; i<=9; ++i) {
draw((0.1+(i-1),-0.15... | [] |
60 | Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find
$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5
2.6 Solution 6
2.7 Sol... | 1989 AIME Problem 10 | Solution 1
We draw the altitude $h$ to $c$, to get two right triangles.
Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$, from the definition of the cotangent.
Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$, so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$.
By identical logic, we can find similar expressions ... | // Block 1
size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$a$", (B+C)/2, NE); label("$b$", (A+C)/2, NW); label("$c$", (A+B)/2, S); label("$h... | [] |
61 | A triangle has vertices $P_{}^{}=(-8,5)$, $Q_{}^{}=(-15,-19)$, and $R_{}^{}=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$. Find $a+c_{}^{}$.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Trigonometry)
3 ... | 1990 AIME Problem 7 | Use the distance formula to determine the lengths of each of the sides of the triangle. We find that it has lengths of side $15,\ 20,\ 25$, indicating that it is a $3-4-5$ right triangle. At this point, we just need to find another point that lies on the bisector of $\angle P$.
Solution 1
Use the angle bisector theor... | // Block 1
import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
// B... | [] |
62 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$. If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segmen... | 1990 AIME Problem 14 | Solution 1(Synthetic)
Our triangular pyramid has base $12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}$.
... | // Block 1
import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple ... | [] |
63 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} ... | 1991 AIME Problem 2 | The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$. Thus, its length is $5 \cdot \frac{168-k}{168}$. Let $a_k=\frac{5(168-k)}{168}$. We want to... | // Block 1
real r = 0.35; size(220);
pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8);
pair A=(0,0),B=(4,0),C=(4,3),D=(0,3);
D(A--B--C--D--cycle);
pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0);
pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r);
D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4);
MP("... | [] |
64 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$, $Q^{}_{}$, $R^{}_{}$, and $S^{}_{}$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB^{}_{}=15$, $BQ^{}_{}=20$, $PR^{}_{}=30$, and $QS^{}_{}=40$. ... | 1991 AIME Problem 12 | Note that this is a modified version of the original diagram. By the Pythagorean theorem, the side length of the rhombus is $25$. Since a rhombus is a parallelogram, its diagonals bisect each other. Thus, $\triangle PBQ \cong \triangle RDS$ and $\triangle QCR \cong \triangle SAP$ by symmetry.
Let $RC=x$ and $QC=y$. Ap... | // Block 1
defaultpen(fontsize(12)+linewidth(1.3));
pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20);
pair F=(0, Q.y);
O=intersectionpoint(A--C,B--D);
draw(A--B--C--D--cycle);
draw(P--R..Q--S);
draw(P--Q--R--S--cycle);
draw(Q--F, dashed);
draw(rightanglemark(Q... | [] |
65 | A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$. | 1991 AIME Problem 14 | Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$.
Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$.
Subtracting these equations give $y^2-81y-112\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into th... | // Block 1
defaultpen(fontsize(9));
pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));
draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--... | [] |
66 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 1993 AIME Problem 13 | Solution 1
Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$
The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t... | // Block 1
size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,5... | [] |
66 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 1993 AIME Problem 13 | Consider our diagram here, where Jenny goes from $A$ to $D$ and Kenny goes from $B$ to $C$. Really what we are asking is the length of $AD$, knowing that $AB$ and $CD$ are tangents and $AB$ is perpendicular to the parallel lines. Draw lines $EF$ and $GH$ as shown such that they are parallel to $BC$ and $AD$ and are tan... | // Block 1
import math; import geometry; import olympiad;
point A,B,C,D,F,G,H,I,M,O,P; A=(0,0); B=(0,200); C=(160,200); D=(160/3,0); F=(0,150); G=(400/3,150); H=(80,50); I=(0,50); M=(0,100); O=(320/3,100); P=(80,150);
draw(I--H--D--A--B--C--G--F); draw(M--O--H); draw(circle((50,100),50)); draw(G--H--P);
label("A",A,SW)... | [] |
67 | Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$. If $AB = 1995\,$, $AC = 1994\,$, and $BC = 1993\,$, then $RS\,$ can be expressed as $m/n\,$, where $m\,$ and $n\,$ are relatively ... | 1993 AIME Problem 15 | From the Pythagorean Theorem, $AH^2+CH^2=1994^2$, and $(1995-AH)^2+CH^2=1993^2$.
Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$.
After simplification, we see that $2*1995AH-1995^2=3987$, or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$.
Note that $AH+BH=1995$.
Therefore we have that $BH=\frac{1995}{2}-... | // Block 1
unitsize(48);
pair A,B,C,H;
A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);
label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE);
draw(circle((2,1),1));
pair [] x=intersectionpoints(C--H,circle((2,1),1));
dot(x[0]); label("$S$",x[0],SW);
draw(circle((4.29843788128,... | [] |
68 | The graphs of the equations
$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | 1994 AIME Problem 6 | Solution 1
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.
Solving the above equations for $k=\pm 10$, we see that the hexagon in question is regular, with side length $\frac{20}{... | // Block 1
size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);} else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));} } picb = rotate(120,origin)*pica; picc = rotat... | [] |
69 | For certain ordered pairs $(a,b)\,$ of real numbers, the system of equations
$ax+by=1\,$
$x^2+y^2=50\,$
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there? | 1994 AIME Problem 7 | The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$, centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$, $(\pm5,\pm5)$, and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cdo... | // Block 1
size(150); draw(circle((0,0),sqrt(50)));
draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green);
dot((0,0));
dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue);
dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue);
dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); ... | [] |
70 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Supp... | 1994 AIME Problem 15 | Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$. According to the problem statement, the circumcenters ... | // Block 1
pair project(pair X, pair Y, real r){return X+r*(Y-X);}
path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);}
pathpen = linewidth(1); size(250); pen dots = linetype("2 3") + linewidth(0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewid... | [] |
71 | Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$. The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. | 1995 AIME Problem 4 | We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\overline{O_3A_3... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7); size(150);
pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9));
D(CR(D(MP("O_9",A)),9)); D(CR(D(MP("O_3",B)),3)); D(CR(D(MP("O_6",C)),6)); D(MP("P",P,NW)--... | [] |
71 | Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$. The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. | 1995 AIME Problem 4 | Let $A$ be defined as the origin of a coordinate plane with the $y$-axis running across the chord and $C(6\sqrt{2},0)$ by the Pythagorean Theorem. Then we have $D(0,-6)$ and $F(6\sqrt{2},-3)$, and since $\frac{DE}{DF}=\frac{1}{3}$, the point $E$ is one-third of the way from $D$ to $F$, so point $E$ has coordinates $(2\... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7); size(150);
pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9));
D(CR(D(MP("E",A)),9)); D(CR(D(MP("F",B)),3)); D(CR(D(MP("D",C)),6)); D((-9,0)--(9,0)); D(MP... | [] |
72 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n... | 1995 AIME Problem 12 | Solution 1 (trigonometry)
The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\triangle... | // Block 1
import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple proje... | [] |
73 | In a circle of radius $42$, two chords of length $78$ intersect at a point whose distance from the center is $18$. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form ... | 1995 AIME Problem 14 | Let the center of the circle be $O$, and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$, such that $AE = CE < BE = DE$. Let $F$ be the midpoint of $\overline{AB}$. Then $\overline{OF} \perp \overline{AB}$.
By the Pythagorean Theorem, $OF = \sqrt{OB^2 - BF^2} = \sqrt{42^2 - 39^2} = 9\sqrt{3}... | // Block 1
size(200); pathpen = black + linewidth(0.7); pen d = dashed+linewidth(0.7);
pair O = (0,0), E=(0,18), B=E+48*expi(11*pi/6), D=E+48*expi(7*pi/6), A=E+30*expi(5*pi/6), C=E+30*expi(pi/6), F=foot(O,B,A);
D(CR(D(MP("O",O)),42)); D(MP("A",A,NW)--MP("B",B,SE)); D(MP("C",C,NE)--MP("D",D,SW)); D(MP("E",E,N)); D(C--B-... | [] |
74 | A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centi... | 1996 AIME Problem 4 | (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$, and so the sides of the shadow are $7$. Using the similar triangles in blue,
$\frac {x}{1} = \frac {1}{6}$, and $\left\lfloor 1000x \right\rfloor = \boxed{166}$. | // Block 1
import three;
size(250);defaultpen(0.7+fontsize(9));
real unit = 0.5;
real r = 2.8;
triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P);
draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit));
draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit));... | [] |
75 | Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? | 1996 AIME Problem 7 | There are ${49 \choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.
For most pairs, there will be
thre... | // Block 1
pathpen = black; pair O = (3.5,3.5); D(O);
for(int i=0;i<7;++i)
for(int j=0;j<7;++j)
D(shift(i,j)*unitsquare);
fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4));
fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));
fill(shift(2,3)*unitsquare,rg... | [] |
76 | In triangle $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$... | 1996 AIME Problem 13 | Let $E$ be the midpoint of $\overline{BC}$. Since $BE = EC$, then $\triangle ABE$ and $\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$, so $\frac{[BDE]}{[BAE]} = \frac{DE}{AE}$ (see ... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);
D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E));
MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(rightanglema... | [] |
76 | In triangle $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$... | 1996 AIME Problem 13 | First, extend side \(AC\) to a point \(F\) such that \(AF \perp BF\).
We begin by using the Law of Cosines to find \(\cos \angle ACB\):
\[(\sqrt{6})^2 + (\sqrt{15})^2 - 2(\sqrt{6})(\sqrt{15})\cos \angle ACB = (\sqrt{30})^2.\]
Simplifying,
\[6 + 15 - 2\sqrt{90}\cos \angle ACB = 30,\]
which gives
\[-6\sqrt{10}\cos \angl... | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);
pair F = foot(B, A, C);
D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B);
D(MP("E",E));
D(C--MP("F",F,E)--B, dashed); // Draw C-F-B, dashed to show it's an extension
... | [] |
77 | In parallelogram $ABCD$, let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$. Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$, and angle $ACB$ is $r$ times as large as angle $AOB$. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Solution
2.1 Solution 1 (trigonometry)
2.2 Sol... | 1996 AIME Problem 15 | Solution 1 (trigonometry)
Let $\theta = \angle DBA$. Then $\angle CAB = \angle DBC = 2\theta$, $\angle AOB = 180 - 3\theta$, and $\angle ACB = 180 - 5\theta$. Since $ABCD$ is a parallelogram, it follows that $OA = OC$. By the Law of Sines on $\triangle ABO,\, \triangle BCO$,
$\frac{\sin \angle CBO}{OC} = \frac{\sin \... | size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP("A",A,N)--MP("B",B)--MP("C",C)--MP("D",D,N)--cycle); D(B--D); D(A--C); D(MP("O",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(anglemark(C,B,D,3.... | [] |
78 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$, the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the st... | 1997 AIME Problem 7 | We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\left(\frac{1}{6}, \frac{1}{2}\right)$. Labeling the car's starting position A, the storm center B, and the right triangle for... | // Block 1
size(200,200);
draw((0,0)--(0,110));
label("A",(0,0),S);
dot((0,0));
dot((0,110));
label("B",(0,110),NE);
draw(circle((0,110),51));
draw((0,0)--(161/3,161.0),EndArrow);
draw((0,110)--(110/3,110.0));
label("C",(110/3,110.0),SE);
dot((110/3,110.0));
label("D",(33,99),SE);
dot((33,99));
draw((0,110)--(33,99));
... | [] |
79 | Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ minutes. The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m = a - b\sqrt {c},$ where $a, b,$ and $c$ are... | 1998 AIME Problem 9 | Solution 1
Let the two mathematicians be $M_1$ and $M_2$. Consider plotting the times that they are on break on a coordinate plane with one axis being the time $M_1$ arrives and the second axis being the time $M_2$ arrives (in minutes past 9 a.m.). The two mathematicians meet each other when $|M_1-M_2| \leq m$. Also b... | import graph; size(180); real m=60-12*sqrt(15); draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray); draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle); xaxis("$M_1$",-10,80); yaxis("$M_2$",-10,80); label(rotate(45)*"$M_1-M_2\le m$",((m+60)... | ["https://artofproblemsolving.com/wiki/images/8/86/AIME_1998-9.png", "https://artofproblemsolving.com/wiki/images/7/7a/AIME_1998-9b.png"] |
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