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=... | [] |
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