AoPS - Problem

Source: IMO 1980 Luxembourg, problem 2

Tags: geometry, function, area, Triangle, IMO Shortlist

orl

06.05.2004 19:01

Three points $A,B,C$ are such that $B \in ]AC[$. On the side of $AC$ we draw the three semicircles with diameters $[AB], [BC]$ and $[AC]$. The common interior tangent at $B$ to the first two semi-circles meets the third circle in $E$. Let $U$ and $V$ be the points of contact of the common exterior tangent to the first two semi-circles. Denote the area of the triangle $ABC$ as $S(ABC)$. Evaluate the ratio $R=\frac{S(EUV)}{S(EAC)}$ as a function of $r_1 = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$.

$X=EA\cap (AB),\ Y=EC\cap (BC)$, where, in general, $(PQ)$ is the semicircle with diameter $(AB)$. It's obvious tht $X,\ Y$ are the projections of $B$ on $EA,\ EB$ respectively. At the same time, $E$ is on the radical axis of $(AB)$ and $(BC)$, so $EX*EA=EY*EC$, so $\triangle EYX \sim \triangle EAC$. This means that $\angle BYX=\angle EXY=\angle ECA=\angle BCY$, so $XY$ is tangent to $(BC)$. In the same way we show that $XY$ is tangent to $(AB)$. This means that $X=U,\ Y=V$. $\frac{S(EUV)}{S(EAC)}=(\frac{UV}{AC})^2=(\frac{EB}{AC})^2=\frac{AB*BC}{(AB+BC)^2}=\frac{r_1*r_2}{(r_1+r_2)^2}$ (because triangles $EVU$ and $EAB$ are similar and $EUBV$ is a rectangle).