Let $ ABC$ be an acute triangle. Points $ D$, $ E$, and $ F$ lie on segments $ BC$, $ CA$, and $ AB$, respectively, and each of the three segments $ AD$, $ BE$, and $ CF$ contains the circumcenter of $ ABC$. Prove that if any two of the ratios $ \frac{BD}{DC}$, $ \frac{CE}{EA}$, $ \frac{AF}{FB}$, $ \frac{BF}{FA}$, $ \frac{AE}{EC}$, $ \frac{CD}{DB}$ are integers, then triangle $ ABC$ is isosceles.
Problem
Source: CGMO 2007 P2
Tags: geometry, circumcircle, ratio, inequalities, geometry proposed
oty
23.05.2012 03:01
No solution for this problem ?
TripteshBiswas
12.01.2015 18:49
If i have understood the problem, $AD,BE,CF$ are concurrent and the point of concurrency is the circumcenter of $\triangle ABC.$ If we consider four following cases then we are done.(Other cases are symmetric) $\boxed {case (i):-}$ $\frac {BD}{CD}$ and $\frac {CD}{BD}$ are both integers then $BD=CD \implies AB=AC$ $\boxed {case (ii):-}$ $\frac {AF}{BF}$ and $\frac {AE}{CE}$ are both integers. $\implies \frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}$ and $\frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}$ are integers. $\implies \frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}- \frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}=\frac {b^2-c^2}{a^2}$ is an integer. Since $\triangle ABC$ is acute, So $-1 < \frac {b^2-c^2}{a^2} < 1$ hence $AB=AC$ $\boxed {case (iii):-}$ $\frac {AF}{BF}$ and $\frac {CE}{AE}$ are integers. $\implies \frac {CD}{BD}$ is an integer. Now similarly as previous we get $CA=CB$ $\boxed {case (iv):-}$ $\frac {BF}{AF}$ and $\frac{CE}{AE}$ are integers.
Someone for this case please ???
MariusBocanu
16.02.2015 10:33
Yeah, I can help with the last case ($\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)}$ and $\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)}$ are integers). Denote $a^2=x,b^2=y, c^2=z$ and $x,y,z$ are the sidelenghts of a triangle, so we can write $x=v+w, y=u+w, z=u+v$. Now, we know that $\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)} \ge 2$ and $\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)} \ge 2$ (in fact, if we have both of them $2$, we are done since $D$ is the midpoint of $BC$, case already treated). If you are patient enough to multiply the paranthesis, you will get $xz+2y^2 \ge x^2+xy+2yz, xy+2z^2 \ge x^2+2yz+xz$. We have 3 cases: 1. $x \ge y \ge z$ , when $xy+2z^2 <x^2+2yz+xz$, impossible (the equations are symmetric in $y$ and $z$). 2. $y \ge z \ge x$ which implies $ u \ge w \ge v$ and the two inequlities become $uw \ge uv+2vw ; uv \ge uw+2vw$, obviously impossible. 3. $y \ge x \ge z$ which implies $ w \ge u \ge v$ and the two inequlities become $uw \ge uv+2vw ; uv \ge uw+2vw$, obviously impossible. So, at least one of the ratios is $1$ and we already solved this case.