Source: 2018 Canadian Open Math Challenge Part C Problem 1 ----- At Math-$e^e$-Mart, cans of cat food are arranged in an pentagonal pyramid of 15 layers high, with 1 can in the top layer, 5 cans in the second layer, 12 cans in the third layer, 22 cans in the fourth layer etc, so that the $k^{\text{th}}$ layer is a pentagon with $k$ cans on each side. [center][img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9lLzA0NTc0MmM2OGUzMWIyYmE1OGJmZWQzMGNjMGY1NTVmNDExZjU2LnBuZw==&rn=YzFhLlBORw==[/img][img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9hLzA1YWJlYmE1ODBjMzYwZDFkYWQyOWQ1YTFhOTkzN2IyNzJlN2NmLnBuZw==&rn=YzFiLlBORw==[/img][/center] $\text{(a)}$ How many cans are on the bottom, $15^{\text{th}}$, [color=transparent](A.)[/color]layer of this pyramid? $\text{(b)}$ The pentagonal pyramid is rearranged into a prism consisting of 15 identical layers. [color=transparent](B.)[/color]How many cans are on the bottom layer of the prism? $\text{(c)}$ A triangular prism consist of indentical layers, each of which has a shape of a triangle. [color=transparent](C.)[/color](the number of cans in a triangular layer is one of the triangular numbers: 1,3,6,10,...) [color=transparent](C.)[/color]For example, a prism could be composed of the following layers: [center][img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi85L2NlZmE2M2Y3ODhiN2UzMTRkYzIxY2MzNjFmMDJkYmE0ZTJhMTcwLnBuZw==&rn=YzFjLlBORw==[/img][/center] Prove that a pentagonal pyramid of cans with any number of layers $l\ge 2$ can be rearranged (without a deficit or leftover) into a triangluar prism of cans with the same number of layers $l$.

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]

Let $a,b,c$ be the answers to problems $4$, $5$, and $6$, respectively. In $\triangle ABC$, the measures of $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, $c$ in degrees, respectively. Let $D$ and $E$ be points on segments $AB$ and $AC$ with $\frac{AD}{BD} = \frac{AE}{CE} = 2013$. A point $P$ is selected in the interior of $\triangle ADE$, with barycentric coordinates $(x,y,z)$ with respect to $\triangle ABC$ (here $x+y+z=1$). Lines $BP$ and $CP$ meet line $DE$ at $B_1$ and $C_1$, respectively. Suppose that the radical axis of the circumcircles of $\triangle PDC_1$ and $\triangle PEB_1$ pass through point $A$. Find $100x$. [i]Proposed by Evan Chen[/i]

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]

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]

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 $\angle BAX = \angle CAY$.

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

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]

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]

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]

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]

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} $

$20$ points with no three collinear are given. How many obtuse triangles can be formed by these points? $ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 2{{10}\choose{3}} \qquad \textbf{(D)}\ 3{{10}\choose{3}} \qquad \textbf{(E)}\ {{20}\choose{3}}$

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 be a twice-differentiable function $ f:(0,\infty )\longrightarrow\mathbb{R} $ that admits a polynomial function of degree $ 1 $ or $ 2, $ namely, $ \alpha :(0,\infty )\longrightarrow\mathbb{R} $ as its asymptote. Prove the following propositions: [b]a)[/b] $ f''>0\implies f-\alpha >0 $ [b]b)[/b] $ \text{supp} f''=(0,\infty )\wedge f-\alpha >0\implies f''=0 $

$BB_{1}$ 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. 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 $XY$.)

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 $\textbf{(A) }6\qquad\textbf{(B) }2\sqrt{6}\qquad\textbf{(C) }\frac{8\sqrt{3}}{3}\qquad\textbf{(D) }3\sqrt{3}\qquad\textbf{(E) }4\sqrt{3}$

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 $AB+AC=PQ+PR$.

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); 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 = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

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 $AB+AC=PQ+PR$.

In $\bigtriangleup ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$? [asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35)); [/asy] $\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$

The figure $ABCD$ is bounded by a semicircle $CDA$ and a quarter circle $ABC$. Given that the distance from $A$ to $C$ is $18$, find the area of the figure. [asy] size(200); defaultpen(linewidth(0.8)); pair A=(-9,0),B=(0,9*sqrt(2)-9),C=(9,0),D=(0,9); dot(A^^B^^C^^D); draw(arc(origin,9,0,180)^^arc((0,-9),9*sqrt(2),45,135)); label("$A$",A,S); label("$B$",B,N); label("$C$",C,S); label("$D$",D,N); [/asy]

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]

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}$.