Alex the Kat has written $61$ problems for a math contest, and there are a total of $187$ problems submitted. How many more problems does he need to write (and submit) before he has written half of the total problems?

Two different ellipses are given. One focus of the first ellipse coincides with one focus of the second ellipse. Prove that the ellipses have at most two points in common.

Fixed points $B$ and $C$ are on a fixed circle $\omega$ and point $A$ varies on this circle. We call the midpoint of arc $BC$ (not containing $A$) $D$ and the orthocenter of the triangle $ABC$, $H$. Line $DH$ intersects circle $\omega$ again in $K$. Tangent in $A$ to circumcircle of triangle $AKH$ intersects line $DH$ and circle $\omega$ again in $L$ and $M$ respectively. Prove that the value of $\frac{AL}{AM}$ is constant. [i]Proposed by Mehdi E'tesami Fard[/i]

Find all possible values of $\dfrac{d}{a}$ where $a^2-6ad+8d^2=0$, $a\neq 0$.

Given a parallelogram $ABCD$ with angle $A$ equal to $60^o$. Point $O$ is the the center of a circle circumscribed around triangle $ABD$. Line $AO$ intersects the bisector of the exterior angle $C$ at point $K$. Find the ratio $AO/OK$.

Let $A$ be a finite set of positive reals, let $B = \{x/y\mid x,y\in A\}$ and let $C = \{xy\mid x,y\in A\}$. Show that $|A|\cdot|B|\le|C|^2$. [i](Proposed by Gerhard Woeginger, Austria)[/i]

Let $\omega$ be a circle with center $O$ and diameter $AB$. A circle with center at $B$ intersects $\omega$ at C and $AB$ at $D$. The line $CD$ intersects $\omega$ at a point $E$ ($E\ne C$). The intersection of lines $OE$ and $BC$ is $F$. (a) Prove that triangle $OBF$ is isosceles. (b) If $D$ is the midpoint of $OB$, find the value of the ratio $\frac{FB}{BD}$.

Let $ABCD$ be a square of side length $13$. Let $E$ and $F$ be points on rays $AB$ and $AD$ respectively, so that the area of square $ABCD$ equals the area of triangle $AEF$. If $EF$ intersects $BC$ at $X$ and $BX = 6$, determine $DF$.

Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

Let $ABC$ be an acute-angled triangle in which $D,E,F$ are points on $BC,CA,AB$ respectively such that $AD \perp BC$;$AE = BC$; and $CF$ bisects $\angle C$ internally, Suppose $CF$ meets $AD$ and $DE$ in $M$ and $N$ respectively. If $FM$$= 2$, $MN =1$, $NC=3$, find the perimeter of $\Delta ABC$.

In $ \triangle ABC$ points $ D$ and $ E$ lie on $ \overline{BC}$ and $ \overline{AC}$, respectively. If $ \overline{AD}$ and $ \overline{BE}$ intersect at $ T$ so that $ AT/DT \equal{} 3$ and $ BT/ET \equal{} 4$, what is $ CD/BD$? [asy]unitsize(2cm); defaultpen(linewidth(.8pt)); pair A = (0,0); pair C = (2,0); pair B = dir(57.5)*2; pair E = waypoint(C--A,0.25); pair D = waypoint(C--B,0.25); pair T = intersectionpoint(D--A,E--B); label("$B$",B,NW);label("$A$",A,SW);label("$C$",C,SE);label("$D$",D,NE);label("$E$",E,S);label("$T$",T,2*W+N); draw(A--B--C--cycle); draw(A--D); draw(B--E);[/asy]$ \textbf{(A)}\ \frac {1}{8}\qquad \textbf{(B)}\ \frac {2}{9}\qquad \textbf{(C)}\ \frac {3}{10}\qquad \textbf{(D)}\ \frac {4}{11}\qquad \textbf{(E)}\ \frac {5}{12}$

Consider a shape obtained from two equal squares with the same center. Prove that the ratio of the area of this shape to the perimeter does not change when the squares are rotated around their center. [img]http://4.bp.blogspot.com/-1AI4FxsNSr4/XovZWkvAwiI/AAAAAAAALvY/-kIzOgXB5rk3iIqGbpoKRCW9rwJPcZ3uQCK4BGAYYCw/s400/estonia%2B2000%2Bo.j.1.3.png[/img]

Let $ABCD$ be a parallelogram with $AB=8$, $AD=11$, and $\angle BAD=60^\circ$. Let $X$ be on segment $CD$ with $CX/XD=1/3$ and $Y$ be on segment $AD$ with $AY/YD=1/2$. Let $Z$ be on segment $AB$ such that $AX$, $BY$, and $DZ$ are concurrent. Determine the area of triangle $XYZ$.

Given is a triangle $ABC$ and points $D$ and $E$, respectively on $] BC [$ and $] AB [$. $F$ it is intersection of lines $AD$ and $CE$. We denote as $| CD | = a, | BD | = b, | DF | = c$ and $| AF | = d$. Determine the ratio $\frac{| BE |}{|AE |}$ in terms of $a, b, c$ and $d$ [img]https://cdn.artofproblemsolving.com/attachments/5/7/856c97045db2d9a26841ad00996a2b809addaa.png[/img]

Let $A_1,B_1,C_1$ be the midpoints of the sides $BC,CA,AB$ respectively of a triangle $ABC$. Points $K$ on segment $C_1A_1$ and $L$ on segment $A_1B_1$ are taken such that $$\frac{C_1K}{KA_1}=\frac{BC+AC}{AC+AB}\enspace\enspace\text{and}\enspace\enspace\frac{A_1L}{LB_1}=\frac{AC+AB}{BC+AB}.$$If $BK$ and $CL$ meet at $S$, prove that $\angle C_1A_1S=\angle B_1A_1S$.

Let be a sequence $ \left( s_n\right)_{n\geqslant 0} $ of positive real numbers, with $ s_0 $ being the golden ratio, and defined as $$ s_{n+2}=\frac{1+s_{n+1}}{s_n} . $$ Establish the necessary and sufficient condition under which $ \left( s_n\right)_{n\geqslant 0} $ is convergent.