This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 126

2012 Canadian Mathematical Olympiad Qualification Repechage, 8

Suppose circles $\mathit{W}_1$ and $\mathit{W}2$, with centres $\mathit{O}_1$ and $\mathit{O}_2$ respectively, intersect at points $\mathit{M}$ and $\mathit{N}$. Let the tangent on $\mathit{W}_2$ at point $\mathit{N}$ intersect $\mathit{W}_1$ for the second time at $\mathit{B}_1$. Similarly, let the tangent on $\mathit{W}_1$ at point $\mathit{N}$ intersect $\mathit{W}_2$ for the second time at $\mathit{B}_2$. Let $\mathit{A}_1$ be a point on $\mathit{W}_1$ which is on arc $\mathit{B}_1\mathit{N}$ not containing $\mathit{M}$ and suppose line $\mathit{A}_1\mathit{N}$ intersects $\mathit{W}_2$ at point $\mathit{A}_2$. Denote the incentres of triangles $\mathit{B}_1\mathit{A}_1\mathit{N}$ and $\mathit{B}_2\mathit{A}_2\mathit{N}$ by $\mathit{I}_1$ and $\mathit{I}_2$, respectively.* [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.1cm); 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 = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; /* image dimensions */ pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773); /* draw figures */ draw(circle(O_2, 4.000000000000000)); draw(circle(O_1, 3.000000000000000)); draw((4.048892687647541,4.413249028538064)--B_2); draw(B_2--A_2); draw(A_2--(4.048892687647541,4.413249028538064)); draw((4.048892687647541,4.413249028538064)--B_1); draw(B_1--A_1); draw(A_1--(4.048892687647541,4.413249028538064)); /* dots and labels */ dot(O_2,dotstyle); label("$O_2$", (7.788512439159622,6.243082420501817), NE * labelscalefactor); dot(O_1,dotstyle); label("$O_1$", (2.298205165350667,6.929370829727937), NE * labelscalefactor); dot(M,dotstyle); label("$M$", (4.383466101076183,8.935444641311980), NE * labelscalefactor); dot((4.048892687647541,4.413249028538064),dotstyle); label("$N$", (3.855551940133015,3.761885864068922), NE * labelscalefactor); dot(B_2,dotstyle); label("$B_2$", (10.19052187145104,9.463358802255147), NE * labelscalefactor); dot(A_2,dotstyle); label("$A_2$", (11.80066006232771,4.659339937672310), NE * labelscalefactor); dot(B_1,dotstyle); label("$B_1$", (1.981456668784765,10.09685579538695), NE * labelscalefactor); dot(A_1,dotstyle); label("$A_1$", (0.08096568938935705,3.973051528446190), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Show that \[\angle\mathit{I}_1\mathit{MI}_2=\angle\mathit{O}_1\mathit{MO}_2.\] *[size=80]Given a triangle ABC, the incentre of the triangle is defined to be the intersection of the angle bisectors of A, B, and C. To avoid cluttering, the incentre is omitted in the provided diagram. Note also that the diagram serves only as an aid and is not necessarily drawn to scale.[/size]

2009 China Team Selection Test, 1

Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$

2012 NIMO Summer Contest, 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]

1998 AMC 8, 24

A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? [asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy] $\text{(A)}\ 36 \qquad \text{(B)}\ 64 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 120$

2009 ELMO Problems, 2

Let $ABC$ be a triangle such that $AB < AC$. Let $P$ lie on a line through $A$ parallel to line $BC$ such that $C$ and $P$ are on the same side of line $AB$. Let $M$ be the midpoint of segment $BC$. Define $D$ on segment $BC$ such that $\angle BAD = \angle CAM$, and define $T$ on the extension of ray $CB$ beyond $B$ so that $\angle BAT = \angle CAP$. Given that lines $PC$ and $AD$ intersect at $Q$, that lines $PD$ and $AB$ intersect at $R$, and that $S$ is the midpoint of segment $DT$, prove that if $A$,$P$,$Q$, and $R$ lie on a circle, then $Q$, $R$, and $S$ are collinear. [i]David Rush[/i]

2009 Purple Comet Problems, 3

In the diagram $ABCDEFG$ is a regular heptagon (a $7$ sided polygon). Shown is the star $AEBFCGD$. The degree measure of the obtuse angle formed by $AE$ and $CG$ is $\dfrac{m}{n}$ where m and n are relatively prime positive integers. Find $m + n$. [asy] size(150); defaultpen(linewidth(1)); string lab[]={"A","B","C","D","E","F","G"}; real r = 360/7; pair A=dir(90-r),B=dir(90),C=dir(90+r),D=dir(90+2*r),E=dir(90+3*r),F=dir(90+4*r),G=dir(90+5*r); draw(A--E--B--F--C--G--D--cycle); for(int k = -1;k <= 5;++k) { label("$"+lab[k+1]+"$",dir(90+k*r),dir(90+k*r)); } [/asy]

2013 ELMO Shortlist, 13

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

2014 India Regional Mathematical Olympiad, 5

Let $ABC$ be a triangle and let $X$ be on $BC$ such that $AX=AB$. let $AX$ meet circumcircle $\omega$ of triangle $ABC$ again at $D$. prove that circumcentre of triangle $BDX$ lies on $\omega$.

2013 ELMO Shortlist, 13

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

2007 F = Ma, 19

A non-Hookian spring has force $F = -kx^2$ where $k$ is the spring constant and $x$ is the displacement from its unstretched position. For the system shown of a mass $m$ connected to an unstretched spring initially at rest, how far does the spring extend before the system momentarily comes to rest? Assume that all surfaces are frictionless and that the pulley is frictionless as well. [asy] size(250); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,0)--(0,-1)--(2,-1)--(2+sqrt(3),-2)); draw((2.5,-2)--(4.5,-2),dashed); draw(circle((2.2,-0.8),0.2)); draw((2.2,-0.8)--(1.8,-1.2)); draw((0,-0.6)--(0.6,-0.6)--(0.75,-0.4)--(0.9,-0.8)--(1.05,-0.4)--(1.2,-0.8)--(1.35,-0.4)--(1.5,-0.8)--(1.65,-0.4)--(1.8,-0.8)--(1.95,-0.6)--(2.2,-0.6)); draw((2+0.3*sqrt(3),-1.3)--(2+0.3*sqrt(3)+0.6/2,-1.3+sqrt(3)*0.6/2)--(2+0.3*sqrt(3)+0.6/2+0.2*sqrt(3),-1.3+sqrt(3)*0.6/2-0.2)--(2+0.3*sqrt(3)+0.2*sqrt(3),-1.3-0.2)); //super complex Asymptote code gg draw((2+0.3*sqrt(3)+0.3/2,-1.3+sqrt(3)*0.3/2)--(2.35,-0.6677)); draw(anglemark((2,-1),(2+sqrt(3),-2),(2.5,-2))); label("$30^\circ$",(3.5,-2),NW); [/asy] $ \textbf{(A)}\ \left(\frac{3mg}{2k}\right)^{1/2} $ $ \textbf{(B)}\ \left(\frac{mg}{k}\right)^{1/2} $ $ \textbf{(C)}\ \left(\frac{2mg}{k}\right)^{1/2} $ $ \textbf{(D)}\ \left(\frac{\sqrt{3}mg}{k}\right)^{1/3} $ $ \textbf{(E)}\ \left(\frac{3\sqrt{3}mg}{2k}\right)^{1/3} $

1979 AMC 12/AHSME, 10

If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$, and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$, then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is $\textbf{(A) }6\qquad\textbf{(B) }2\sqrt{6}\qquad\textbf{(C) }\frac{8\sqrt{3}}{3}\qquad\textbf{(D) }3\sqrt{3}\qquad\textbf{(E) }4\sqrt{3}$

2014 AIME Problems, 8

Circle $C$ with radius $2$ has diameter $\overline{AB}$. Circle $D$ is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C,$ externally tangent to circle $D,$ and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$ and can be written in the form $\sqrt{m} - n,$ where $m$ and $n$ are positive integers. Find $m+n$.

1984 AIME Problems, 3

A point $P$ is chosen in the interior of $\triangle ABC$ so that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles, $t_1$, $t_2$, and $t_3$ in the figure, have areas 4, 9, and 49, respectively. Find the area of $\triangle ABC$. [asy] size(200); pathpen=black+linewidth(0.65);pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);[/asy]

1964 AMC 12/AHSME, 39

Tags: Asymptote , AMC
The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than: $\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$ [asy] import math; defaultpen(fontsize(11pt)); pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); //Credit to bobthesmartypants for the diagram [/asy]

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 $AB+AC=PQ+PR$.

2013 Sharygin Geometry Olympiad, 14

Let $M$, $N$ be the midpoints of diagonals $AC$, $BD$ of a right-angled trapezoid $ABCD$ ($\measuredangle A=\measuredangle D = 90^\circ$). The circumcircles of triangles $ABN$, $CDM$ meet the line $BC$ in the points $Q$, $R$. Prove that the distances from $Q$, $R$ to the midpoint of $MN$ are equal.

2012 Purple Comet Problems, 9

Points $E$ and $F$ lie inside rectangle $ABCD$ with $AE=DE=BF=CF=EF$. If $AB=11$ and $BC=8$, find the area of the quadrilateral $AEFB$.

2013 USAMO, 6

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]

2014 India Regional Mathematical Olympiad, 1

Let $ABC$ be a triangle with $\angle ABC $ as the largest angle. Let $R$ be its circumcenter. Let the circumcircle of triangle $ARB$ cut $AC$ again at $X$. Prove that $RX$ is perpendicular to $BC$.

2004 AMC 12/AHSME, 24

A plane contains points $ A$ and $ B$ with $ AB \equal{} 1$. Let $ S$ be the union of all disks of radius $ 1$ in the plane that cover $ \overline{AB}$. What is the area of $ S$? $ \textbf{(A)}\ 2\pi \plus{} \sqrt3 \qquad \textbf{(B)}\ \frac {8\pi}{3} \qquad \textbf{(C)}\ 3\pi \minus{} \frac {\sqrt3}{2} \qquad \textbf{(D)}\ \frac {10\pi}{3} \minus{} \sqrt3 \qquad \textbf{(E)}\ 4\pi \minus{} 2\sqrt3$

2010 ELMO Shortlist, 1

Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. [i]Carl Lian.[/i]

2013 Canadian Mathematical Olympiad Qualification Repechage, 7

Consider the following layouts of nine triangles with the letters $A, B, C, D, E, F, G, H, I$ in its interior. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(200); 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 = 1.740000000000003, xmax = 8.400000000000013, ymin = 3.500000000000005, ymax = 9.360000000000012; /* image dimensions */ draw((5.020000000000005,8.820000000000011)--(2.560000000000003,4.580000000000005)--(7.461947712046029,4.569577506690286)--cycle); /* draw figures */ draw((5.020000000000005,8.820000000000011)--(2.560000000000003,4.580000000000005)); draw((2.560000000000003,4.580000000000005)--(7.461947712046029,4.569577506690286)); draw((7.461947712046029,4.569577506690286)--(5.020000000000005,8.820000000000011)); draw((3.382989341689345,5.990838871467448)--(4.193333333333338,4.580000000000005)); draw((4.202511849578174,7.405966442513598)--(5.828619600041468,4.573707435672692)); draw((5.841878190157451,7.408513542990484)--(4.193333333333338,4.580000000000005)); draw((6.656214943659867,5.990342259816768)--(5.828619600041468,4.573707435672692)); draw((4.202511849578174,7.405966442513598)--(5.841878190157451,7.408513542990484)); draw((3.382989341689345,5.990838871467448)--(6.656214943659867,5.990342259816768)); label("\textbf{A}",(4.840000000000007,8.020000000000010),SE*labelscalefactor,fontsize(22)); label("\textbf{B}",(3.980000000000006,6.640000000000009),SE*labelscalefactor,fontsize(22)); label("\textbf{C}",(4.820000000000007,7.000000000000010),SE*labelscalefactor,fontsize(22)); label("\textbf{D}",(5.660000000000008,6.580000000000008),SE*labelscalefactor,fontsize(22)); label("\textbf{E}",(3.160000000000005,5.180000000000006),SE*labelscalefactor,fontsize(22)); label("\textbf{F}",(4.020000000000006,5.600000000000008),SE*labelscalefactor,fontsize(22)); label("\textbf{G}",(4.800000000000007,5.200000000000007),SE*labelscalefactor,fontsize(22)); label("\textbf{H}",(5.680000000000009,5.620000000000007),SE*labelscalefactor,fontsize(22)); label("\textbf{I}",(6.460000000000010,5.140000000000006),SE*labelscalefactor,fontsize(22)); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] A sequence of letters, each letter chosen from$ A, B, C, D, E, F, G, H, I$ is said to be [i]triangle-friendly[/i] if the first and last letter of the sequence is $C$, and for every letter except the first letter, the triangle containing this letter shares an edge with the triangle containing the previous letter in the sequence. For example, the letter after $C$ must be either $A, B$, or $D$. For example, $CBF BC$ is triangle-friendly, but $CBF GH$ and $CBBHC$ are not. [list] [*] (a) Determine the number of triangle-friendly sequences with $2012$ letters. [*] (b) Determine the number of triangle-friendly sequences with exactly $2013$ letters.[/list]

2013 Sharygin Geometry Olympiad, 9

Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$.

2014 Saudi Arabia BMO TST, 2

Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\omega_1$ such that $P Y \parallel BD$. Point $Z$ lies on $\omega_2$ such that $P Z \parallel AC$. Prove that points $Q,~ X,~ Y,~ Z$ are collinear.

2003 Bulgaria Team Selection Test, 5

Let $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its in center on $AC$. Prove that $\angle {APB}=\angle {APD}$