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 $l$.

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]

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]

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 triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic. [i]Proposed by Mahan Malihi[/i]

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]

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

Let $ABC$ be a triangle with symmedian point $K$. Select a point $A_1$ on line $BC$ such that the lines $AB$, $AC$, $A_1K$ and $BC$ are the sides of a cyclic quadrilateral. Define $B_1$ and $C_1$ similarly. Prove that $A_1$, $B_1$, and $C_1$ are collinear. [i]Proposed by Sammy Luo[/i]

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]

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 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.$

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 $\angle BCA$.

In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$? $\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$

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] $\text{(A)}\ 36 \qquad \text{(B)}\ 64 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 120$

$BB_{1}$ 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]