Track / Stage 4 / 69 of 340 #809 of 2444

Problem 809

AMC 12 late, AIME early

Geometry Difficulty 4.4 Prove it Junior Turkish Mathematical Olympiad · Turkey

In an equilateral triangle $ABC$ , the point $X$ on the segment $[BC]$ and the points $Y$ , $Z$ on the rays $[BA$ and $[CA$ , respectively, are given such that $AX$ , $BZ$ , $CY$ are parallel. Let $XY$ intersect $AC$ at $M$ , and let $XZ$ intersect $AB$ at $N$ . Show that $MN$ is tangent to the incircle of $ABC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$AX \parallel BZ$ implies $\frac{AN}{NB} = \frac{AX}{BZ} = \frac{CX}{CB}$ , and $AX \parallel CY$ implies $\frac{AM}{MC} = \frac{AX}{CY} = \frac{BX}{BC}$ . Hence,
$\frac{AN}{NB} + \frac{AM}{MC} = \frac{CX}{CB} + \frac{BX}{BC} = 1. \quad (1)$
Now let $a$ denote the sidelength of $ABC$ . Using (1), one finds
$\frac{a^2}{NB \cdot MC} = \left(1 + \frac{AN}{NB}\right) \left(1 + \frac{AM}{MC}\right) = 2 + \frac{AN \cdot AM}{NB \cdot MC},$
Therefore,
$AN \cdot AM = a^2 - 2 \cdot NB \cdot MC. \quad (2)$
Now the Law of Cosines implies $MN^2 = AM^2 + AN^2 - AM \cdot AN$ which together with (2) gives
$\begin{aligned} MN^2 &= AM^2 + AN^2 - a^2 + 2 \cdot NB \cdot MC \\ &= (a - NB)^2 + (a - MC)^2 - a^2 + 2 \cdot NB \cdot MC \\ &= (NB + MC - a)^2. \end{aligned}$
Hence, $MN = NB + MC - a$ (it can be easily seen by (2) that $NB + MC - a > 0$ ). Finally $MN + BC = NB + MC$ implies that the quadrilateral $MNBC$ has an incircle, thus the incircle of $ABC$ is tangent to the segment $MN$ . Done.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.