2004 IMO Shortlist Problems/G3

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(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.)


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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