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Taiwan IMO Selection Camp

Geometry Difficulty 4.4 AIME Prove it Taiwan

$\triangle ABC$ 中, $BC$ 邊的中點為 $M$ , $AM$ 再交 $\triangle ABC$ 的外接圓 $\Gamma$ 於 $R$ , 過 $R$ 且與 $BC$ 平行的直線再交 $\Gamma$ 於 $S$ 。自 $R$ 至 $BC$ 的垂線的垂足為 $U$ , $T$ 為 $U$ 對 $R$ 的對稱點。 $D$ 是 $BC$ 上的一點使得 $AD$ 為 $\triangle ABC$ 的高, $N$ 為 $AD$ 中點。最後令 $AS$ , $MN$ 交於 $K$ 。證明: $AT$ 平分 $MK$ 。

In $\triangle ABC$ , let the midpoint of side $BC$ be $M$ , and let $AM$ meet the circumcircle $\Gamma$ of $\triangle ABC$ again at $R$ . The line through $R$ parallel to $BC$ meets $\Gamma$ again at $S$ . Let the foot of the perpendicular from $R$ to $BC$ be $U$ , and let $T$ be the reflection of $U$ over $R$ . Let $D$ be a point on $BC$ such that $AD$ is an altitude of $\triangle ABC$ , and let $N$ be the midpoint of $AD$ . Finally, let $AS$ and $MN$ meet at $K$ . Prove that: $AT$ bisects $MK$ .

Solution

(a) Let $L$ be the midpoint of $MK$ . The perpendicular to $BC$ through $L$ meets $BC$ , $AM$ at $E$ , $F$ respectively.
Since $AD \parallel EF$ , $L$ is the midpoint of $EF$ . Therefore $\triangle LEM \cong \triangle LFK$ , and we obtain
$\angle LKF = 90^\circ$ , $KF \parallel BC$ .

(b) Let $MN$ meet $RU$ at $W$ . Since $AD \parallel RU$ , and $N$ is the midpoint of $AD$ , hence $W$ is the midpoint of $RU$ .

(c) Let the foot of the perpendicular from $S$ to $BC$ be $V$ . Clearly $S, R$ are symmetric with respect to $OM$ ( $O$ is the circumcenter of $\triangle ABC$ ), hence $M$ is the midpoint of $UV$ . By (b) we get $NMW \parallel VR$ . Also, since $VS = RU = TR$ , and $TRU \parallel SV$ , we know $SVRT$ is a parallelogram, so $ST \parallel VR \parallel MN$ .

(d) By (a) and (c), we know that $\triangle LFK$ , $\triangle TRS$ have corresponding sides parallel, and since $KS, RF$ meet at $A$ , by the parallel case of Desargues' theorem (or by computing ratios), we get $A, L, T$ are collinear. This completes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.