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Problem 1767

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.2 Prove it IMO Team Selection Test 1 · Netherlands

Let $\triangle ABC$ be an acute triangle such that $|AB| + |BC| = 4|AC|$ and $|AB| < |BC|$ . Let $D$ be the intersection of the bisector of $\angle ABC$ with the side $AC$ . Points $P$ and $Q$ lie on segment $BD$ such that $|BP| = 2|DQ|$ . Let $\ell$ be the line through $P$ parallel to $AC$ . The line through $Q$ perpendicular to $BD$ intersects the segments $AB$ and $BC$ in $X$ and $Y$ , respectively. Suppose that both $X$ and $Y$ lie on the opposite side of $\ell$ to $B$ .
An ant starts their journey in $X$ and goes from there to a point on $AC$ , then a point on $\ell$ , then back to a (possibly different) point on $AC$ and finally to $Y$ . Prove that the length of the ant's shortest possible route is equal to $4|XY|$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $R, S$ and $T$ be respectively the points where the ant is on $AC$ the first time, is on $\ell$ and is on $AC$ the second time. Let $\ell'$ and $S'$ be the reflections of $\ell$ and $S$ in $AC$ , and let $Y'$ be the reflection of $Y$ in $\ell'$ . Then, because of the triangle inequality for the total length of the ant's path, we have
$\begin{aligned} |XR| + |RS| + |ST| + |TY| &= |XR| + |RS'| + |S'T| + |TY| \\ &\geq |XS'| + |S'Y| = |XS'| + |S'Y'| \\ &\geq |XY'|, \end{aligned}$
with equality if $S'$ lies on $XY'$ , $R$ lies on $XS'$ and $T$ lies on $S'Y$ . So now we are left to prove that $|XY'| = 4|XY|$ .

We redefine $S'$ as the intersection point of $XY'$ and $\ell'$ . Then, by definition, $\ell'$ is the external bisector of $\angle XS'Y$ . We also define $B'$ as the intersection point of $BQ$ with $\ell'$ . Then $B'$ also lies on the perpendicular bisector of $XY$ . Using the incentre-excentre lemma with respect to $\triangle XYS'$ , we find that $B'$ is the midpoint of the arc $XY$ of the circumcircle of $\triangle XYS'$ that does contain $S'$ . (Note that $BQ$ is not parallel to $\ell'$ or perpendicular to it, because $\ell' \parallel AC$ and $\triangle BAC$ is not isosceles. So $B'$ is well-defined and unequal to $S'$ .) In particular, $XYB'S'$ is a cyclic quadrilateral. Because of our definition of $\ell'$ , we have that $B'$ is also the reflection of $P$ in $D$ . So, together with the condition that $2|QD| = |BP|$ , we calculate that
$\begin{aligned} |QB'| &= |QD| + |DB'| = |QD| + |PD| \\ &= |QD| + |PQ| + |QD| = |BP| + |PQ| \\ &= |BQ|. \end{aligned}$

Since $Q$ is also the midpoint of $XY$ and the diagonals are perpendicular to each other, we conclude that $BXB'Y$ is a rhombus. We now angle chase
$\begin{align*} \angle S'XY &= 180^\circ - \angle S'B'Y = 180^\circ - \angle S'B'B - \angle BB'Y \\ &= \angle BDC - \angle YBB' = \angle DBA + \angle BAD - \angle BDA \\ &= \angle BAC. \end{align*}$
Analogously, it holds that $\angle S'YX = \angle BCA$ . So $\triangle XYS' \sim \triangle ACB$ , so the condition $|AB| + |BC| = 4|AC|$ also holds in $\triangle XYS'$ . We conclude that $|XY'| = |XS'| + |S'Y| = 4|XY|$ , which is exactly what we wanted to show. $\square$

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.