Let $ABC$ be a triangle and $I$ its incentre. Let $\varrho_1$ and $\varrho_2$ be the inradii of triangles $IAB$ and $IAC$ respectively. (a) Show that there exists a function $f: ( 0, \pi ) \mapsto \mathbb{R}$ such that \[ \frac{ \varrho_1}{ \varrho_2} = \frac{f(C)}{f(B)} \] where $B = \angle ABC$ and $C = \angle BCA$ (b) Prove that \[ 2 ( \sqrt{2} -1 ) < \frac{ \varrho_1} { \varrho_2} < \frac{ 1 + \sqrt{2}}{2} \]

Let $ABCDE$ be a right quadrangular pyramid with vertex $E$ and height $EO$. Point $S$ divides this height in the ratio $ES: SO=m$. In which ratio does the plane $(ABC)$ divide the lateral area of the pyramid.

Prove that if $\frac{AC}{BC}=\frac{AB + BC}{AC}$ in a triangle $ABC$ , then $\angle B = 2 \angle A$ .

In a triangle $ABC$, the bisector of the angle at $A$ of a triangle $ABC$ intersects the segment $BC$ and the circumcircle of $ABC$ at points $A_1$ and $A_2$, respectively. Points $B_1,B_2,C_1,C_2$ are analogously defined. Prove that $$\frac{A_1A_2}{BA_2+CA_2}+\frac{B_1B_2}{CB_2+AB_2}+\frac{C_1C_2}{AC_2+BC_2}\ge\frac34.$$

Let $ABC$ be a triangle and let $P$ be a point in the plane of $ABC$ that is inside the region of the angle $BAC$ but outside triangle $ABC$. [b](a)[/b] Prove that any two of the following statements imply the third. [list] [b](i)[/b] the circumcentre of triangle $PBC$ lies on the ray $\stackrel{\to}{PA}$. [b](ii)[/b] the circumcentre of triangle $CPA$ lies on the ray $\stackrel{\to}{PB}$. [b](iii)[/b] the circumcentre of triangle $APB$ lies on the ray $\stackrel{\to}{PC}$.[/list] [b](b)[/b] Prove that if the conditions in (a) hold, then the circumcentres of triangles $BPC,CPA$ and $APB$ lie on the circumcircle of triangle $ABC$.

Let $AD$ be a diameter of a circle of radius $r,$ and let $B,C$ be points on the circle such that $AB=BC=\frac r2$ and $A\neq C.$ Find the ratio $\frac{CD}{r}.$

In the figure, $ ABCD$ is a $ 2\times 2$ square, $ E$ is the midpoint of $ \overline{AD}$, and $ F$ is on $ \overline{BE}$. If $ \overline{CF}$ is perpendicular to $ \overline{BE}$, then the area of quadrilateral $ CDEF$ is [asy]defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair D = (2,2); pair E = midpoint(A--D); pair F = foot(C,B,E); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,N);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$E$",E,N);label("$F$",F,NW); draw(A--B--C--D--cycle); draw(B--E); draw(C--F); draw(rightanglemark(B,F,C,4));[/asy]$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \minus{} \frac {\sqrt {3}}{2}\qquad \textbf{(C)}\ \frac {11}{5}\qquad \textbf{(D)}\ \sqrt {5}\qquad \textbf{(E)}\ \frac {9}{4}$

The square $ABCD$ is inscribed in a circle. Points $E$ and $F$ are located on the side of the square, and points $G$ and $H$ are located on the smaller arc $AB$ of the circle so that the $EFGH$ is a square. Find the area ratio of these squares.

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$, then a possible value for the ratio $\frac{a}{b}$, to the nearest integer, is $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ \text{none of these} $

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]

In a triangle $ABC$, the bisector of the angle at $A$ of a triangle $ABC$ intersects the segment $BC$ and the circumcircle of $ABC$ at points $A_1$ and $A_2$, respectively. Points $B_1,B_2,C_1,C_2$ are analogously defined. Prove that $$\frac{A_1A_2}{BA_2+CA_2}+\frac{B_1B_2}{CB_2+AB_2}+\frac{C_1C_2}{AC_2+BC_2}\ge\frac34.$$

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]

$\triangle ABC$ has area $240$. Points $X, Y, Z$ lie on sides $AB$, $BC$, and $CA$, respectively. Given that $\frac{AX}{BX} = 3$, $\frac{BY}{CY} = 4$, and $\frac{CZ}{AZ} = 5$, find the area of $\triangle XYZ$. [asy] size(175); defaultpen(linewidth(0.8)); pair A=(0,15),B=(0,-5),C=(25,0.5),X=origin,Y=(4C+B)/5,Z=(5A+C)/6; draw(A--B--C--cycle^^X--Y--Z--cycle); label("$A$",A,N); label("$B$",B,S); label("$C$",C,E); label("$X$",X,W); label("$Y$",Y,S); label("$Z$",Z,NE);[/asy]

Akello divides a square up into finitely many white and red rectangles, each (rectangle) with sides parallel to the sides of the parent square. Within each white rectangle, she writes down the value of its width divided by its height, while within each red rectangle, she writes down the value of its height divided by its width. Finally, she calculates $x$, the sum of these numbers. If the total area of the white rectangles equals the total area of the red rectangles, what is the least possible value of $x$ she can get?

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1));[/asy]$ \textbf{(A)}\ 3\minus{}2\sqrt2 \qquad \textbf{(B)}\ 2\minus{}\sqrt2 \qquad \textbf{(C)}\ 4(3\minus{}2\sqrt2) \qquad \textbf{(D)}\ \frac12(3\minus{}\sqrt2)$ $ \textbf{(E)}\ 2\sqrt2\minus{}2$

At Frank's Fruit Market, $ 3$ bananas cost as much as $ 2$ apples, and $ 6$ apples cost as much as $ 4$ oranges. How many oranges cost as much as $ 18$ bananas? $ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Given a parallelogram $ABCD$, join $A$ to the midpoints $E$ and $F$ of the opposite sides $BC$ and $CD$. $AE$ and $AF$ intersect the diagonal $BD$ in $M$ and $N$. Prove that $M$ and $N$ divide $BD$ into three equal parts.

A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?

Points $A,B,C,D$ are on a circle with radius $R$. $|AC|=|AB|=500$, while the ratio between $|DC|, |DA|, |DB|$ is $1,5,7$. Find $R$.

Let $r\geq1$ be areal number that holds with the property that for each pair of positive integer numbers $m$ and $n$, with $n$ a multiple of $m$, it is true that $\lfloor{nr}\rfloor$ is multiple of $\lfloor{mr}\rfloor$. Show that $r$ has to be an integer number. [b]Note: [/b][i]If $x$ is a real number, $\lfloor{x}\rfloor$ is the greatest integer lower than or equal to $x$}.[/i]

Let $ABC$ be an acute-angled triangle with $AB \ne BC$ and radius $k$. Let $P$ and $Q$ be the points of intersection of $k$ with the internal bisector and the external bisector of $\angle CBA$ respectively. Let $D$ be the intersection of $AC$ and $PQ$. Find the ratio $AD: DC$.

Let $A,B,C$ be fixed points in the plane , and $D$ be a variable point on the circle $ABC$, distinct from $A,B,C$ . Let $I_{A},I_{B},I_{C},I_{D}$ be the Simson lines of $A,B,C,D$ with respect to triangles $BCD,ACD,ABD,ABC$ respectively. Find the locus of the intersection points of the four lines $I_{A},I_{B},I_{C},I_{D}$ when point $D$ varies.

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $ m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is $ \textbf{(A)}\ \frac {1}{2m \plus{} 1} \qquad \textbf{(B)}\ m \qquad \textbf{(C)}\ 1 \minus{} m \qquad \textbf{(D)}\ \frac {1}{4m} \qquad \textbf{(E)}\ \frac {1}{8m^2}$

Given distinct straight lines $ OA$ and $ OB$. From a point in $ OA$ a perpendicular is drawn to $ OB$; from the foot of this perpendicular a line is drawn perpendicular to $ OA$. From the foot of this second perpendicular a line is drawn perpendicular to $ OB$; and so on indefinitely. The lengths of the first and second perpendiculars are $ a$ and $ b$, respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is: $ \textbf{(A)}\ \frac {b}{a \minus{} b} \qquad \textbf{(B)}\ \frac {a}{a \minus{} b} \qquad \textbf{(C)}\ \frac {ab}{a \minus{} b} \qquad \textbf{(D)}\ \frac {b^2}{a \minus{} b} \qquad \textbf{(E)}\ \frac {a^2}{a \minus{} b}$

The sides of a triangle are in the ratio $ 6: 8: 9$. Then: $ \textbf{(A)}\ \text{the triangle is obtuse} \qquad\textbf{(B)}\ \text{the angles are in the ratio } 6: 8: 9$ $ \textbf{(C)}\ \text{the triangle is acute}$ $ \textbf{(D)}\ \text{the angle opposite the largest side is double the angle opposite the smallest side}$ $ \textbf{(E)}\ \text{none of these}$