2004 IMO Shortlist Problems/G2


(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 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.

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