2004 IMO Shortlist Problems/G2

Problem

(Kazakhstan) The circle $\displaystyle \Gamma$ and the line $\ell$ do not intersect. Let $\displaystyle AB$ be the diameter of $\displaystyle\Gamma$ perpendicular to $\ell$ , with $\displaystyle B$ closer to $\ell$ than $\displaystyle A$ . An arbitrary point $C \neq A,B$ is chosen on $\displaystyle \Gamma$ . The line $\displaystyle AC$ intersects $\ell$ at $\displaystyle D$ . The line $\displaystyle DE$ is tangent to $\displaystyle \Gamma$ at $\displaystyle E$ , with $\displaystyle B$ and $\displaystyle E$ on the same side of $\displaystyle AC$ . Let $\displaystyle BE$ intersect $\ell$ at $\displaystyle F$ , and let $\displaystyle AF$ intersect $\displaystyle \Gamma$ at $G \neq A$ . Prove that the reflection of $\displaystyle G$ in $\displaystyle AB$ lies on the line $\displaystyle CF$ .

This was also a Problem 2 on the 2005 Greece TST, Problem 1 of Day 1 of the 2005 Moldova TST, and Problem 2 of the final exam of the 3rd 2005 Taiwan TST.

Solution

Solution 1

We use directed angles mod $\displaystyle \pi$ .

Let $\displaystyle CF$ meet $\displaystyle \Gamma$ at $\displaystyle H$ . The problem is equivalent to showing that lines $\displaystyle GH$ and $\ell$ are parallel, which happens if and only if $\angle AGH \equiv \angle AFD$ . But by cyclic quadrilaterals and vertical angles, $\angle AGH \equiv \angle ACH \equiv \angle DCF$ . To prove $\angle DCF \equiv \angle DFA$ , it suffices to show that triangles $\displaystyle DFC, DAF$ are similar. Since these triangles share a common angle, it then suffices to show $\frac{DC}{DF} = \frac{DF}{DA}$ , or $DF^2 = DC\cdot DA$ .

By considering the power of the point $\displaystyle D$ with respect to $\displaystyle \Gamma$ , we see $\displaystyle DE^2 = DC \cdot DA$ . Hence it suffices to show that $\displaystyle DE \equiv DF$ . Let $\displaystyle O$ be the center of $\displaystyle \Gamma$ . Since $AO \perp DF, AE \perp EF, OE \perp ED$ , it follows that there is a spiral similarity mapping $\displaystyle AOE$ to $\displaystyle FDE$ , i.e., these triangles are similar. Since $\displaystyle OA = OE$ , it follows that $\displaystyle DF = DE$ . Q.E.D.

Solution 2

We use directed angles mod $\displaystyle \pi$ .

Lemma. Let $\displaystyle \omega_1 , \omega_2$ be two circles with centers $\displaystyle O_1, O_2$ , and common points $\displaystyle M, N$ . Let $\displaystyle A$ be a point on $\displaystyle \omega_1$ , and let $\displaystyle A'$ be the second intersection of line $\displaystyle AM$ and $\displaystyle \omega_2$ . Then $\angle AO_1N \equiv \angle A'O_2N$ .

Proof. Since $\displaystyle A, M, A'$ are collinear, $\angle AO_1N \equiv 2\angle AMN \equiv 2\angle A'MN$ . But since $\displaystyle A', M, N$ lie on a circle with center $\displaystyle O_2$ , $2\angle A'MN \equiv \angle A'ON$ , as desired.

Let the center of $\displaystyle \Gamma$ be $\displaystyle O$ . Let $\displaystyle G'$ be the reflection of $\displaystyle G$ across $\displaystyle AB$ . It is sufficient to show that $\angle ACG' \equiv \angle DCF$ , since $\displaystyle A,C,D$ are collinear.

Since $\displaystyle \Gamma$ is symmetric about its diameter $\displaystyle AB$ , $\displaystyle G'$ lies on $\displaystyle \Gamma$ , and ${\rm m}\widehat{AG} = {\rm m}\widehat{AG'}$ , so

$\angle ACG' \equiv \angle GCA$ .

If we consider the line which passes through $\displaystyle B$ and is parallel to $\ell$ , we see ${\rm m}\angle GFD = \frac{1}{2}{\rm m}(\widehat{AB} - \widehat{BG}) = \frac{1}{2}{\rm m}[\widehat{AB} - (\widehat{BA} - \widehat{GA})] = \frac{1}{2}{\rm m}\widehat{GA} = {\rm m}\angle GCA$ , since $\widehat{AB} , \widehat{BA}$ are semicircles. Thus

$\angle GCA \equiv \angle GFD$ ,

or $\angle GCA \equiv \angle GFD \equiv \angle AFD$ . Since we also have $\angle FAD \equiv -\angle CAG$ , it follows that triangles $\displaystyle AFD, ACG$ are similar, with opposite orientation. In particular, $\frac{AF}{AC} = \frac{AD}{AG}$ , or $\displaystyle AF \cdot AG = AC \cdot AD$ , so $\displaystyle F,G, D,C$ are concyclic.

At this point, we note that $OE \perp EP$ , $AE \perp EF$ (since $\angle AEF$ is inscribed in a semicircle), and $AO \perp DF$ . It follows that there is a spiral similarity centered at $\displaystyle E$ with rotation of ${} \frac{\pi}{2}$ mapping triangle $\displaystyle AOE$ to triangle $\displaystyle FDE$ and $\displaystyle \Gamma$ to a circle $\displaystyle \Gamma'$ centered at $\displaystyle D$ with radius $\displaystyle DE = DF$ . Let $\displaystyle G''$ be the second intersection of $\displaystyle \Gamma$ and $\displaystyle \Gamma'$ . We note that $\displaystyle AG''$ must intersect $\ell$ at some point $\displaystyle F'$ on the same side of $\displaystyle D$ as $\displaystyle F$ , since $\displaystyle G', A$ must be on the same side of $\displaystyle DO$ as $\displaystyle F$ . By the lemma, $\angle F'DE \equiv \angle AOE \equiv \angle FDE$ , and since $\displaystyle F'$ is on $\displaystyle \Gamma'$ , $\displaystyle DF' = DE = DF$ . It follows that $\displaystyle F = F'$ . Since $\displaystyle G''$ is the intersection point of $\displaystyle AF' = AF$ and $\displaystyle \Gamma$ , $\displaystyle G'' = G$ , and $\displaystyle G$ lies on $\displaystyle \Gamma'$ . In particular, $\displaystyle DG = DF$ , and

$\angle GFD \equiv \angle DGF$ .

Now, since $\displaystyle F,G, D, C$ are concyclic, as we noted above,

$\angle DGF \equiv \angle DCF$ .

To summarize,

$\angle ACG' \equiv \angle GCA \equiv \angle GFD \equiv \angle DGF \equiv \angle DCF$ ,

as desired. Q.E.D.

Solution 3

We use projective geometry. Let $\displaystyle G'$ be the reflection of $\displaystyle G$ over $\displaystyle AB$ . Let $\displaystyle \Phi$ be the intersection of two distinct parallels of $\ell$ ; let $\displaystyle Y$ be the intersection of $\displaystyle AE$ and $\displaystyle BG$ ; let $\displaystyle S$ be the intersection of $\displaystyle AB$ and $\displaystyle G'G$ ; and let $\displaystyle F'$ be the intersection of $\displaystyle BE$ and $\displaystyle G'C$ . It is sufficient to show that $\displaystyle F'$ lies on $\ell$ .

We apply Pascal's Theorem for cyclic hexagons several times. By applying it to the degenerate hexagon $\displaystyle AAEBBG$ , we see that $\displaystyle \Phi, Y, F$ are collinear, i.e., $\displaystyle Y$ lies on $\ell$ . By applying the theorem to the hexagon $\displaystyle ABGG'EA$ , we see that $\displaystyle S, Y, \Phi$ are collinear, i.e., $\displaystyle S$ also lies on $\ell$ . Finally, by appling the theorem to $\displaystyle G'EEBAC$ , we see that $\displaystyle S, D, F'$ are collinear, so $\displaystyle F'$ lies on $\ell$ , as desired.

Solution 4

Again, we use projective geometry. We send $\ell$ to infinity, so that $\displaystyle \Gamma$ becomes an ellpise with axis $\displaystyle AB$ . A distortion then makes $\displaystyle \Gamma$ a circle again. We shall now refer to objects as their images under these transformations.

We know that $\displaystyle AC$ is parallel to the tangent to $\displaystyle \Gamma$ at $\displaystyle E$ , and $\displaystyle \displaystyle AG$ is parallel to $\displaystyle BE$ . Now, since $\displaystyle AB$ is a diameter, $GA \equiv BE$ . Also, $\angle BED \equiv \angle GAC$ . Since equal angles inscribe equal arcs, we have $\displaystyle GC \equiv GA$ . Since $\displaystyle G'$ is the reflection of $\displaystyle G$ over $\displaystyle AB$ , we have $GC \equiv GA \equiv GA'$ . This implies that $\displaystyle GA$ and $\displaystyle CG'$ are parallel, i.e., they pass through the same point on the line at infinity as $\displaystyle GA$ , i.e., $\displaystyle F$ . Q.E.D.