Found problems: 127
$BB_{1}$ 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]
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]
Here’s a screenshot of the problem. If someone could LaTEX a diagram, that would be great!
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 $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear.
Find $\frac{area(CDF)}{area(CEF)}$ in the figure.
[asy]
/* File unicodetex not found. */
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(5.75cm);
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 = -2, xmax = 21, ymin = -2, ymax = 16; /* image dimensions */
/* draw figures */
draw((0,0)--(20,0));
draw((13.48,14.62)--(7,0));
draw((0,0)--(15.93,9.12));
draw((13.48,14.62)--(20,0));
draw((13.48,14.62)--(0,0));
label("6",(15.16,12.72),SE*labelscalefactor);
label("10",(18.56,5.1),SE*labelscalefactor);
label("7",(3.26,-0.6),SE*labelscalefactor);
label("13",(13.18,-0.71),SE*labelscalefactor);
label("20",(5.07,8.33),SE*labelscalefactor);
/* dots and labels */
dot((0,0),dotstyle);
label("$B$", (-1.23,-1.48), NE * labelscalefactor);
dot((20,0),dotstyle);
label("$C$", (19.71,-1.59), NE * labelscalefactor);
dot((7,0),dotstyle);
label("$D$", (6.77,-1.64), NE * labelscalefactor);
dot((13.48,14.62),dotstyle);
label("$A$", (12.36,14.91), NE * labelscalefactor);
dot((15.93,9.12),dotstyle);
label("$E$", (16.42,9.21), NE * labelscalefactor);
dot((9.38,5.37),dotstyle);
label("$F$", (9.68,4.5), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy]
size(10cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));
draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));
draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));
draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));
draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));
draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));
draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));
draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));
draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));
dot((0,0));
dot((22,0));
label("$A$",(0,0),WNW);
label("$B$",(22,0),E);
filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);
filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);
filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);
filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);
filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);
filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);
filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);
filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);
filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);
filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);
filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);
filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);
filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);
filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);
filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);
filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);
filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);
filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);
filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);
filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);
filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);
filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);
filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$ \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 $
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]
Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.
In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.
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 $AB+AC=PQ+PR$.
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 $XY$.)
In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.
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 $\angle BAX = \angle CAY$.
Find the area of the region bounded by the graphs $y=x^2$, $y=x$, and $x=2$.
Let $ \overline{AB}$ be a diameter of circle $ \omega$. Extend $ \overline{AB}$ through $ A$ to $ C$. Point $ T$ lies on $ \omega$ so that line $ CT$ is tangent to $ \omega$. Point $ P$ is the foot of the perpendicular from $ A$ to line $ CT$. Suppose $ AB \equal{} 18$, and let $ m$ denote the maximum possible length of segment $ BP$. Find $ m^{2}$.
The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:
${{ \textbf{(A)}\ \text{I and II} \qquad\textbf{(B)}\ \text{II and III} \qquad\textbf{(C)}\ \text{I and III} \qquad\textbf{(D)}\ \text{III and IV} }\qquad\textbf{(E)}\ \text{I and IV} } $
The sum of the distances from one vertex of a square with sides of length two to the midpoints of each of the sides of the square is
$\textbf{(A) }2\sqrt{5}\qquad\textbf{(B) }2+\sqrt{3}\qquad\textbf{(C) }2+2\sqrt{3}\qquad\textbf{(D) }2+\sqrt{5}\qquad \textbf{(E) }2+2\sqrt{5}$
Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic.
[i]Proposed by Eric Chen[/i]
Let $ K $ be the midpoint of the side $ AB $ of a given triangle $ ABC $. Let $ L $ and $ M$ be points on the sides $ AC $ and $ BC$, respectively, such that $ \angle CLK = \angle KMC $. Prove that the perpendiculars to the sides $ AB, AC, $ and $ BC $ passing through $ K,L, $ and $M$, respectively, are concurrent.
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]
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 $\omega$.
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 $m+n$.
Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic.
[i]Proposed by Eric Chen[/i]
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]
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]