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2004 IMO Shortlist Problems/G3

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Problem

(South Korea) Let $\displaystyle O$ be the circumcenter of an acute-angled triangle $\displaystyle ABC$ with $\angle B < \angle C$ . The line $\displaystyle AO$ meets the side $\displaystyle BC$ at $\displaystyle D$ . The circumcenters of the triangles $\displaystyle ABD$ and $\displaystyle ACD$ are $\displaystyle E$ and $\displaystyle F$ , respectively. Extend the sides $\displaystyle BA$ and $\displaystyle CA$ beyond $\displaystyle A$ , and choose, on the respective extensions points $\displaystyle G$ and $\displaystyle H$ such that $\displaystyle AG= AC$ and $\displaystyle AH = AB$ . Prove that the quadrilateral $\displaystyle EFGH$ is a rectangle if and only if $\angle ACB - \angle ABC = 60^{\circ}$ .

(This was also Problem 2 of the 2005 3rd German TST; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.)

Solution

Lemma. In any triangle $\displaystyle ABC$ with circumcenter $\displaystyle O$ , the altitude from $\displaystyle A$ is the reflection of $\displaystyle AO$ over the angle bisector of $\displaystyle A$ .

Proof. This is well-known, but we prove it anyway. Let $\displaystyle AO, BO, CO$ meet sides $\displaystyle a,b,c$ at $\displaystyle A', B', C'$ , and let $\displaystyle H_a$ be the foot of the altitude from $\displaystyle A$ . Let us denote $\angle BOA' = \angle B'OA = x$ , $\angle COB' = \angle C'OB = y$ , $\angle AOC' = \angle A'OC = z$ , and let us use the notation $\displaystyle \alpha, \beta, \gamma$ for the angles of triangle $\displaystyle ABC$ . By virtue of inscribed arcs in the circumcircle of $\displaystyle ABC$ , we know $\displaystyle x+y = 2\beta$ , $\displaystyle y+z = 2\gamma$ , $\displaystyle z+x = 2\alpha$ , so $\displaystyle x = \beta + \alpha - \gamma = \pi - 2\gamma$ , and again by inscribed arcs, $\angle BAO = \frac{1}{2}x = \frac{\pi}{2}-\gamma = \angle H_aAC$ . The lemma follows. ∎

We first note that $\displaystyle AGH$ is the reflection of $\displaystyle ACB$ over the exterior angle bisector of $\displaystyle B$ . It follows that line $\displaystyle AO$ is the altitude from $\displaystyle A$ in triangle $\displaystyle AGH$ , i.e., $\displaystyle AO \perp GH$ . Since both $\displaystyle E$ and $\displaystyle F$ line on the perpendicular bisector of $\displaystyle AD$ , it follows that $\displaystyle GH$ and $\displaystyle FE$ are always parallel.

We extend $\displaystyle HG$ and $\displaystyle BC$ to meet a point $\displaystyle P$ . Since $\angle B < \angle C$ , $\displaystyle ACPG$ is a convex quadrilateral. In particular, if we use the notation $\alpha= \angle BAC$ , $\beta = \angle CBA$ , $\gamma = \angle ACB$ , then $\angle PCA = \alpha + \beta$ , $\angle CAG = \beta + \gamma$ , $\angle AGP = \angle PCA = \alpha + \beta$ , so $\angle GPC = \gamma - \beta$ . It follows that line $\displaystyle EF$ makes an angle of $\displaystyle \gamma - \beta$ with $\displaystyle AC$ . Now, if $\displaystyle E'$ is the midpoint of $\displaystyle AD$ and $\displaystyle F'$ is the midpoint of $\displaystyle DC$ , we note that $\displaystyle EE'$ and $\displaystyle FF'$ are perpendicular to $\displaystyle AC$ . Hence $\displaystyle E'F' = EF \cos (\gamma- \beta)$ . But if $\displaystyle EFGH$ is a rectangle, then $\displaystyle EF = GH = AC = 2E'F'$ , so $\cos(\gamma -\beta) = \frac{1}{2}$ and $\angle C - \angle B = \frac{\pi}{3}$ . Thus the condition $\angle C - \angle B = \frac{\pi}{3}$ is necessary for $\displaystyle EFGH$ to be a rectangle.

We now prove that it is sufficient. From the previous paragraph, we know that if $\angle C - \angle B = \frac{\pi}{3}$ , then $\displaystyle EFGH$ is a parallelogram. It is sufficient to show that if $\displaystyle Q$ is the intersection of line $\displaystyle AO$ with $\displaystyle GH$ and $\displaystyle R$ is the intersection of line $\displaystyle AO$ and $\displaystyle EF$ , then $\displaystyle GQ = FR$ , since $\displaystyle AO$ is perpendicular to $\displaystyle EF$ and $\displaystyle HG$ . Indeed, since the cosine of the angle between lines $\displaystyle EF$ and $\displaystyle AC$ is $\frac{1}{2}$ , it is sufficient to show that if $\displaystyle R'$ is the projection of $\displaystyle R$ onto $\displaystyle AC$ , then $F'R' = \frac{1}{2}GQ$ . Let $\displaystyle H_a$ be the projection of $\displaystyle A$ onto $\displaystyle AC$ . Since $\displaystyle ABC, AHG$ are congruent, $\displaystyle CH_a = GQ$ . On the other hand, since $\displaystyle R$ is the midpoint of $\displaystyle AD$ , $\displaystyle R'$ is the midpoint of the projection of $\displaystyle AD$ onto $\displaystyle AC$ , namely, $\displaystyle H_aD$ , so $\displaystyle F'R' = F'D - R'D = \frac{1}{2}CD - \frac{1}{2}\left( CD- H_aD \right) = \frac{1}{2}H_aD$ , as desired. Thus $\displaystyle EFGH$ is a rectangle if and only if $\displaystyle \angle C - \angle B = \frac{\pi}{3}$ , as desired.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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