Found problems: 126
2014 ELMO Shortlist, 2
$ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent.
[i]Proposed by Yang Liu[/i]
2008 Paraguay Mathematical Olympiad, 4
Let $\Gamma$ be a circumference and $A$ a point outside it. Let $B$ and $C$ be points in $\Gamma$ such that $AB$ and $AC$ are tangent to $\Gamma$. Let $P$ be a point in $\Gamma$. Let $D$, $E$ and $F$ be points in $BC$, $AC$ and $AB$ respectively, such that $PD \perp BC$, $PE \perp AC$, and $PF \perp AB$.
Show that $PD^2 = PE \cdot PF$
2013 AMC 12/AHSME, 11
Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$?
[asy]
size(180);
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
real s=1/2,m=5/6,l=1;
pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;
draw(A--B--C--cycle^^D--E^^F--G);
dot(A^^B^^C^^D^^E^^F^^G);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,S);
label("$E$",E,NW);
label("$F$",F,S);
label("$G$",G,NW);
[/asy]
$\textbf{(A) }1\qquad
\textbf{(B) }\dfrac{3}{2}\qquad
\textbf{(C) }\dfrac{21}{13}\qquad
\textbf{(D) }\dfrac{13}{8}\qquad
\textbf{(E) }\dfrac{5}{3}\qquad$
2014 Postal Coaching, 4
Let $ABC$ and $PQR$ be two triangles such that
[list]
[b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$.
[b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$
[/list]
Prove that $AB+AC=PQ+PR$.
2002 India IMO Training Camp, 13
Let $ABC$ and $PQR$ be two triangles such that
[list]
[b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$.
[b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$
[/list]
Prove that $AB+AC=PQ+PR$.
2014 Contests, 2
Let $ AB$ be the diameter of semicircle $O$ ,
$C, D $ be points on the arc $AB$,
$P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ .
Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy]
import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black;
real h=sqrt(55/64);
pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B);
D(arc(O,1,0,180),darkgreen);
D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue);
D(O);
[/asy]
1995 AIME Problems, 14
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 lenghts, and the area of either of them can be expressed uniquley in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime number. Find $m+n+d.$
2012 APMO, 4
Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
(Here we denote $XY$ the length of the line segment $XY$.)
2007 Princeton University Math Competition, 1
Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$.
[asy]
/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 2.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */
/* draw figures */
draw(circle((1.37,2.54), 5.17));
draw((-2.62,-0.76)--(-3.53,4.2));
draw((-3.53,4.2)--(5.6,-0.44));
draw((5.6,-0.44)--(-2.62,-0.76));
draw(circle((-0.9,0.48), 2.12));
/* dots and labels */
dot((-2.62,-0.76),dotstyle);
label("$C$", (-2.46,-0.51), SW * labelscalefactor);
dot((-3.53,4.2),dotstyle);
label("$A$", (-3.36,4.46), NW * labelscalefactor);
dot((5.6,-0.44),dotstyle);
label("$B$", (5.77,-0.17), SE * labelscalefactor);
dot((0.08,2.37),dotstyle);
label("$D$", (0.24,2.61), SW * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor);
label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor);
label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor);
/* end of picture */
[/asy]
2013 Sharygin Geometry Olympiad, 18
Let $AD$ be a bisector of triangle $ABC$. Points $M$ and $N$ are projections of $B$ and $C$ respectively to $AD$. The circle with diameter $MN$ intersects $BC$ at points $X$ and $Y$. Prove that $\angle BAX = \angle CAY$.
2014 USA Team Selection Test, 1
Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.)
[i]Robert Simson et al.[/i]
1963 AMC 12/AHSME, 38
Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals:
[asy]
size(7cm);
pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5);
pair G = intersectionpoints(B--F, D--C)[0];
pair E = intersectionpoints(A--C, B--F)[0];
draw(A--D--C--B--cycle);
draw(A--C);
draw(D--F--B);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$F$", F, N);
label("$G$", G, NE);
label("$E$", E, SE);
//Credit to MSTang for the asymptote
[/asy]
$\textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 8\qquad
\textbf{(C)}\ 10 \qquad
\textbf{(D)}\ 12 \qquad
\textbf{(E)}\ 16$
2012 Purple Comet Problems, 12
Pentagon $ABCDE$ consists of a square $ACDE$ and an equilateral triangle $ABC$ that share the side $\overline{AC}$. A circle centered at $C$ has area 24. The intersection of the circle and the pentagon has half the area of the pentagon. Find the area of the pentagon.
[asy]/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(4.26cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -1.52, xmax = 2.74, ymin = -2.18, ymax = 6.72; /* image dimensions */
draw((0,1)--(2,1)--(2,3)--(0,3)--cycle);
draw((0,3)--(2,3)--(1,4.73)--cycle);
/* draw figures */
draw((0,1)--(2,1));
draw((2,1)--(2,3));
draw((2,3)--(0,3));
draw((0,3)--(0,1));
draw((0,3)--(2,3));
draw((2,3)--(1,4.73));
draw((1,4.73)--(0,3));
draw(circle((0,3), 1.44));
label("$C$",(-0.4,3.14),SE*labelscalefactor);
label("$A$",(2.1,3.1),SE*labelscalefactor);
label("$B$",(0.86,5.18),SE*labelscalefactor);
label("$D$",(-0.28,0.88),SE*labelscalefactor);
label("$E$",(2.1,0.8),SE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]
2014 Contests, 1
Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.)
[i]Robert Simson et al.[/i]
2014 Online Math Open Problems, 21
Consider a sequence $x_1,x_2,\cdots x_{12}$ of real numbers such that $x_1=1$ and for $n=1,2,\dots,10$ let \[ x_{n+2}=\frac{(x_{n+1}+1)(x_{n+1}-1)}{x_n}. \] Suppose $x_n>0$ for $n=1,2,\dots,11$ and $x_{12}=0$. Then the value of $x_2$ can be written as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a,b,c$ with $a>b$ and no square dividing $a$ or $b$. Find $100a+10b+c$.
[i]Proposed by Michael Kural[/i]
1963 AMC 12/AHSME, 18
Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, triangle $EUM$ is similar to triangle:
[asy]
pair B = (-0.866, -0.5);
pair C = (0.866, -0.5);
pair E = (0, -1);
pair F = (0, 1);
pair M = midpoint(B--C);
pair A = (-0.99, -0.141);
pair U = intersectionpoints(A--E, B--C)[0];
draw(B--C);
draw(F--E--A);
draw(unitcircle);
label("$B$", B, SW);
label("$C$", C, SE);
label("$A$", A, W);
label("$E$", E, S);
label("$U$", U, NE);
label("$M$", M, NE);
label("$F$", F, N);
//Credit to MSTang for the asymptote
[/asy]
$\textbf{(A)}\ EFA \qquad
\textbf{(B)}\ EFC \qquad
\textbf{(C)}\ ABM \qquad
\textbf{(D)}\ ABU \qquad
\textbf{(E)}\ FMC$
2007 All-Russian Olympiad, 3
$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear.
[i]V. Astakhov[/i]
1985 ITAMO, 6
As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$.
[asy]
size(200);
pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C);
draw(F--C--A--B--C^^A--D^^B--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("84", centroid(H, C, E), fontsize(9.5));
label("35", centroid(H, B, D), fontsize(9.5));
label("30", centroid(H, F, B), fontsize(9.5));
label("40", centroid(H, A, F), fontsize(9.5));[/asy]
2012 Purple Comet Problems, 30
The diagram below shows four regular hexagons each with side length $1$ meter attached to the sides of a square. This figure is drawn onto a thin sheet of metal and cut out. The hexagons are then bent upward along the sides of the square so that $A_1$ meets $A_2$, $B_1$ meets $B_2$, $C_1$ meets $C_2$, and $D_1$ meets $D_2$. If the resulting dish is filled with water, the water will rise to the height of the corner where the $A_1$ and $A_2$ meet. there are relatively prime positive integers $m$ and $n$ so that the number of cubic meters of water the dish will hold is $\sqrt{\frac{m}{n}}$. Find $m+n$.
[asy]
/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 14.52, ymin = -8.3, ymax = 6.3; /* image dimensions */
draw((0,1)--(0,0)--(1,0)--(1,1)--cycle);
draw((1,1)--(1,0)--(1.87,-0.5)--(2.73,0)--(2.73,1)--(1.87,1.5)--cycle);
draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle);
draw((0,0)--(1,0)--(1.5,-0.87)--(1,-1.73)--(0,-1.73)--(-0.5,-0.87)--cycle);
draw((0,1)--(0,0)--(-0.87,-0.5)--(-1.73,0)--(-1.73,1)--(-0.87,1.5)--cycle);
/* draw figures */
draw((0,1)--(0,0));
draw((0,0)--(1,0));
draw((1,0)--(1,1));
draw((1,1)--(0,1));
draw((1,1)--(1,0));
draw((1,0)--(1.87,-0.5));
draw((1.87,-0.5)--(2.73,0));
draw((2.73,0)--(2.73,1));
draw((2.73,1)--(1.87,1.5));
draw((1.87,1.5)--(1,1));
draw((0,1)--(1,1));
draw((1,1)--(1.5,1.87));
draw((1.5,1.87)--(1,2.73));
draw((1,2.73)--(0,2.73));
draw((0,2.73)--(-0.5,1.87));
draw((-0.5,1.87)--(0,1));
/* dots and labels */
dot((1.87,-0.5),dotstyle);
label("$C_1$", (1.72,-0.1), NE * labelscalefactor);
dot((1.87,1.5),dotstyle);
label("$B_2$", (1.76,1.04), NE * labelscalefactor);
dot((1.5,1.87),dotstyle);
label("$B_1$", (0.96,1.8), NE * labelscalefactor);
dot((-0.5,1.87),dotstyle);
label("$A_2$", (-0.26,1.78), NE * labelscalefactor);
dot((-0.87,1.5),dotstyle);
label("$A_1$", (-0.96,1.08), NE * labelscalefactor);
dot((-0.87,-0.5),dotstyle);
label("$D_2$", (-1.02,-0.18), NE * labelscalefactor);
dot((-0.5,-0.87),dotstyle);
label("$D_1$", (-0.22,-0.96), NE * labelscalefactor);
dot((1.5,-0.87),dotstyle);
label("$C_2$", (0.9,-0.94), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
1997 Putnam, 6
The dissection of the $3-4-5$ triangle shown below has diameter $5/2$.
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(23cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 1.42, xmax = 24.42, ymin = 3.8, ymax = 15.54; /* image dimensions */
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true);
yaxis(ymin, ymax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */
/* draw figures */
draw((9.44,8.52)--(12.44,8.52));
draw((9.44,12.52)--(9.44,8.52));
draw((9.44,12.52)--(12.44,8.52));
draw((9.44,10.52)--(10.94,10.52));
draw((10.94,10.52)--(10.94,8.52));
draw((9.44,8.52)--(10.94,10.52));
/* dots and labels */
dot((9.44,8.52),dotstyle);
label("$A$", (9.52,8.64), NE * labelscalefactor);
dot((12.44,8.52),dotstyle);
label("$B$", (12.52,8.64), NE * labelscalefactor);
dot((9.44,12.52),dotstyle);
label("$C$", (9.52,12.64), NE * labelscalefactor);
dot((10.94,8.52),dotstyle);
label("$D$", (11.02,8.64), NE * labelscalefactor);
dot((9.44,10.52),dotstyle);
label("$E$", (9.52,10.64), NE * labelscalefactor);
dot((10.94,10.52),dotstyle);
label("$F$", (11.02,10.64), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]
Find the least diameter of this triangle into four parts. (The diameter of a dissection is the least upper bound of the distances between pairs of points belonging to the same part.)
2008 Paraguay Mathematical Olympiad, 3
Let $ABC$ be a triangle, where $AB = AC$ and $BC = 12$. Let $D$ be the midpoint of $BC$. Let $E$ be a point in $AC$ such that $DE \perp AC$. Let $F$ be a point in $AB$ such that $EF \parallel BC$. If $EC = 4$, determine the length of $EF$.
2008 AIME Problems, 15
A square piece of paper has sides of length $ 100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance $ \sqrt {17}$ from the corner, and they meet on the diagonal at an angle of $ 60^\circ$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $ \sqrt [n]{m}$, where $ m$ and $ n$ are positive integers, $ m < 1000$, and $ m$ is not divisible by the $ n$th power of any prime. Find $ m \plus{} n$.
[asy]import math;
unitsize(5mm);
defaultpen(fontsize(9pt)+Helvetica()+linewidth(0.7));
pair O=(0,0);
pair A=(0,sqrt(17));
pair B=(sqrt(17),0);
pair C=shift(sqrt(17),0)*(sqrt(34)*dir(75));
pair D=(xpart(C),8);
pair E=(8,ypart(C));
draw(O--(0,8));
draw(O--(8,0));
draw(O--C);
draw(A--C--B);
draw(D--C--E);
label("$\sqrt{17}$",(0,2),W);
label("$\sqrt{17}$",(2,0),S);
label("cut",midpoint(A--C),NNW);
label("cut",midpoint(B--C),ESE);
label("fold",midpoint(C--D),W);
label("fold",midpoint(C--E),S);
label("$30^\circ$",shift(-0.6,-0.6)*C,WSW);
label("$30^\circ$",shift(-1.2,-1.2)*C,SSE);[/asy]
2012 NIMO Problems, 8
Points $A$, $B$, and $O$ lie in the plane such that $\measuredangle AOB = 120^\circ$. Circle $\omega_0$ with radius $6$ is constructed tangent to both $\overrightarrow{OA}$ and $\overrightarrow{OB}$. For all $i \ge 1$, circle $\omega_i$ with radius $r_i$ is constructed such that $r_i < r_{i - 1}$ and $\omega_i$ is tangent to $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\omega_{i - 1}$. If
\[
S = \sum_{i = 1}^\infty r_i,
\]
then $S$ can be expressed as $a\sqrt{b} + c$, where $a, b, c$ are integers and $b$ is not divisible by the square of any prime. Compute $100a + 10b + c$.
[i]Proposed by Aaron Lin[/i]
2006 Pan African, 6
Let $ABC$ be a right angled triangle at $A$. Denote $D$ the foot of the altitude through $A$ and $O_1, O_2$ the incentres of triangles $ADB$ and $ADC$. The circle with centre $A$ and radius $AD$ cuts $AB$ in $K$ and $AC$ in $L$. Show that $O_1, O_2, K$ and $L$ are on a line.
2020 BAMO, D/2
Here’s a screenshot of the problem. If someone could LaTEX a diagram, that would be great!