On sides $ AB$ and $ AC$ of triangle $ ABC$ there are given points $ D,E$ such that $ DE$ is tangent of circle inscribed in triangle $ ABC$ and $ DE \parallel BC$. Prove $ AB+BC+CA\geq 8DE$
Problem
Source: Middle Europe Mathematical Olympiad 2009 TST First day, Third problem
Tags: geometry unsolved, geometry
18.06.2009 18:52
Matematika wrote: On sides $ AB$ and $ AC$ of triangle $ ABC$ there are given points $ D,E$ such that $ DE$ is tangent of circle inscribed in triangle $ ABC$ and $ DE \parallel BC$. Prove $ AB + BC + CA\geq 8DE$ Quite easy as a TST. \[ \frac{DE}{a}=\frac{h_a-2r}{h_a}\iff DE=a-\frac{2ar}{h_a}=a-\frac{2a^2sr}{ash_a}=a-\frac{2a^2\Delta}{2\Delta s} =a-\frac{a^2}{s}\] We need to prove that, \[ 2s \geq 8a-\frac{8a^2}{s} \iff 4a^2-4as+s^2 \geq 0 \iff (2a-s)^2 \geq 0\] Done!
18.06.2009 20:47
Wow! Now I find that it is Italy 1999.
19.06.2009 02:24
Don't make mistake this isn't IMO TST this is MEMO tst.
19.06.2009 04:50
Hmm...I know. BTW how did Adrian do?
22.06.2009 20:01
He didn't write he's IMO contestant
06.05.2022 10:29
Let $r$ and $r_a$ denote the inradius and $A-$exradius, respectively. It is standard to see that $rs=[ABC]=r_a(s-a)$, where $s$ is the semiperimeter. Then, since $\triangle ADE\sim\triangle ABC$ and the $A-$excircle of $\triangle ADE$ is the incircle of $\triangle ABC$, we find that $\frac{DE}{BC}=\frac{r}{r_a}=\frac{s-a}{s}=\frac{b+c-a}{a+b+c}$. Thus, $DE=\frac{a(b+c-a)}{a+b+c}$. Finally, \begin{align*} (b+c-3a)^2&\ge0\\ 9a^2+b^2+c^2-6ab-6ca+2bc&\ge0\\ a^2+b^2+c^2+2ab+2ca+2bc&\ge8ab+8ca-8a^2\\ (a+b+c)^2&\ge8a(b+c-a)\\ a+b+c&\ge\frac{8a(b+c-a)}{a+b+c}\\ AB+BC+CA\ge8DE.\,\blacksquare \end{align*}