Let $\vartriangle ABC$ be an equilateral triangle with side length $M$ such that points $E_1$ and $E_2$ lie on side $AB$, $F_1$ and $F_2$ lie on side $BC$, and $G1$ and $G2$ lie on side $AC$, such that $$m = \overline{AE_1} = \overline{BE_2} = \overline{BF_1} = \overline{CF_2} = \overline{CG_1} = \overline{AG_2}$$ and the area of polygon $E_1E_2F_1F_2G_1G_2$ equals the combined areas of $\vartriangle AE_1G_2$, $\vartriangle BF_1E_2$, and $\vartriangle CG_1F_2$. Find the ratio $\frac{m}{M}$. [img]https://cdn.artofproblemsolving.com/attachments/a/0/88b36c6550c42d913cdddd4486a3dde251327b.png[/img]

The triangle $ ABC$ has vertices in such manner that $ AB \equal{} 3, BC \equal{} 4,$ and $ AC \equal{} 5$. The inscribed circle is tangent to $ AB$ in $ C'$, $ BC$ in $ A'$ and $ AC$ in $ B'.$ What is the ratio between the area of the triangles $ A'B'C'$ and $ ABC$? A. 1/4 B. 1/5 C. 2/9 D. 4/21 E. 5/24

Find $\lim_{n\to\infty} \left(\frac{_{3n}C_n}{_{2n}C_n}\right)^{\frac{1}{n}}$ where $_iC_j$ is a binominal coefficient which means $\frac{i\cdot (i-1)\cdots(i-j+1)}{j\cdot (j-1)\cdots 2\cdot 1}$.

Let $AB=AC$ in $\triangle ABC$, and let $D$ be a point on segment $AB$. The tangent at $D$ to the circumcircle $\omega$ of $BCD$ hits $AC$ at $E$. The other tangent from $E$ to $\omega$ touches it at $F$, and $G=BF \cap CD$, $H=AG \cap BC$. Prove that $BH=2HC$. [i]Proposed by David Stoner[/i]

$ABC$ is a triangle with $\angle A = 90^\circ$, $\angle B = 60^\circ$. The points $A_1$, $B_1$, $C_1$ on $BC$, $CA$, $AB$ respectively are such that $A_1B_1C_1$ is equilateral and the perpendiculars (to $BC$ at $A_1$, to $CA$ at $B_1$ and to $AB$ at $C_1$) meet at a point $P$ inside the triangle. Find the ratios $PA_1:PB_1:PC_1$.

Given triangle $ABC$. Suppose $P$ and $P_1$ are points on $BC, Q$ lies on $CA, R$ lies on $AB$, such that $$\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP_1}{P_1B}$$ Let $G$ be the centroid of triangle $ABC$ and $K = AP_1 \cap RQ$. Prove that points $P,G$, and $K$ are collinear.

Prove that when three circles share the same chord $AB$, every line through $A$ different from $AB$ determines the same ratio $X Y : Y Z$, where $X$ is an arbitrary point different from $B$ on the first circle while $Y$ and $Z$ are the points where AX intersects the other two circles (labeled so that $Y$ is between $X$ and $Z$).

In right triangle $ ABC$ with right angle at $ C$, $ \angle BAC < 45$ degrees and $ AB \equal{} 4$. Point $ P$ on $ AB$ is chosen such that $ \angle APC \equal{} 2\angle ACP$ and $ CP \equal{} 1$. The ratio $ \frac{AP}{BP}$ can be represented in the form $ p \plus{} q\sqrt{r}$, where $ p,q,r$ are positive integers and $ r$ is not divisible by the square of any prime. Find $ p\plus{}q\plus{}r$.

A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains $N$ astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength $g$. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength $g'$ . What is the ratio $g'/g$ in terms of $N$?[asy] import roundedpath; size(300); path a = roundedpath((0,-0.3)--(4,-0.3)--(4,-1)--(5,-1)--(5,0),0.1); draw(scale(+1,-1)*a); draw(scale(+1,+1)*a); draw(scale(-1,-1)*a); draw(scale(-1,+1)*a); filldraw(circle((0,0),1),white,black); filldraw(box((-2,-0.27),(2,0.27)),white,white); draw(arc((0,0),1.5,+35,+150),dashed,Arrow); draw(arc((0,0),1.5,-150,-35),dashed,Arrow);[/asy] $ \textbf{(A)}\ 2N/(N-1) $ $ \textbf{(B)}\ N/(N-1) $ $ \textbf{(C)}\ \sqrt{(N-1)/N} $ $ \textbf{(D)}\ \sqrt{N/(N-1)} $ $ \textbf{(E)}\ \text{none of the above} $

(F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A \equal{} 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC \equal{} \angle LAB \equal{} 10^{\circ}$. Determine the ratio $ CK/LB$.

Let $M$, $N$ and $P$ be midpoints of sides $BC, AC$ and $AB$, respectively, and let $O$ be circumcenter of acute-angled triangle $ABC$. Circumcircles of triangles $BOC$ and $MNP$ intersect at two different points $X$ and $Y$ inside of triangle $ABC$. Prove that \[\angle BAX=\angle CAY.\]

Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

Given a convex figure in the Cartesian plane that is symmetric with respect of both axis, we construct a rectangle $A$ inside it with maximum area (over all posible rectangles). Then we enlarge it with center in the center of the rectangle and ratio lamda such that is covers the convex figure. Find the smallest lamda such that it works for all convex figures.

Consider two circles $k_{1},k_{2}$ touching externally at point $T$. a line touches $k_{2}$ at point $X$ and intersects $k_{1}$ at points $A$ and $B$. Let $S$ be the second intersection point of $k_{1}$ with the line $XT$ . On the arc $\widehat{TS}$ not containing $A$ and $B$ is chosen a point $C$ . Let $\ CY$ be the tangent line to $k_{2}$ with $Y\in k_{2}$ , such that the segment $CY$ does not intersect the segment $ST$ . If $I=XY\cap SC$ . Prove that : (a) the points $C,T,Y,I$ are concyclic. (b) $I$ is the excenter of triangle $ABC$ with respect to the side $BC$.