Let $\Delta ABC$ be an acute triangle. $D,E,F$ are the touch points of incircle with $BC,CA,AB$ respectively. $AD,BE,CF$ intersect incircle at $K,L,M$ respectively. If,$$\sigma = \frac{AK}{KD} + \frac{BL}{LE} + \frac{CM}{MF}$$$$\tau = \frac{AK}{KD}.\frac{BL}{LE}.\frac{CM}{MF}$$Then prove that $\tau = \frac{R}{16r}$. Also prove that there exists integers $u,v,w$ such that, $uvw \neq 0$, $u\sigma + v\tau +w=0$.
Problem
Source: IMOTC Practice Test 1 2018 P1, India
Tags: geometry
18.07.2018 21:53
$\frac{AK}{AD} = \frac{AK \cdot AD}{AD^2} = \frac{(s-a)^2}{AD^2}$. Using Stewart's theorem, we have $(d^2 + (s-b)(s-c)) a = b^2(s-b) + c^2(s-c)$. So we have $\frac{AK}{KD} = \frac{a(s-a)}{4(s-b)(s-c)}$, so using $R = \frac{abc}{4 \Delta}, \Delta = rs$ and Heron's formula, we are done for the first part. Note that $\sigma = \frac{\sum_{cyc} a(s-a)^2}{4(s-a)(s-b)(s-c)}$. For the sum of $a(s-a)^2$, let $s-a=x$, and define $y,z$ similarly. We wish to calculate $\sum_{sym}xy^2$, which equals $\prod_{cyc} (x+y) - 2xyz$, so we're done. Edit : thanks @below, I totally forgot to correct that.
22.07.2018 11:09
WizardMath wrote: So we have $\frac{AK}{KD} = \frac{a(s-a)}{(s-b)(s-c)}$ I think $\frac{AK}{KD} = \frac{a(s-a)}{4(s-b)(s-c)}$
07.08.2018 15:49
Dear Mathlinkers, for the sommation (part a), I have found sigma = R/r - 1/2... Sincerely Jean-Louis