IMO 2010: Problem 2

In this year’s IMO, there were two geometry problems. Problem 4 is generally considered the easier one of the two. Problem 2 is the other problem, being a bit harder. Of the 517 contestants, 366 of them received a full mark on problem 4; only 162 students got a full mark for problem 2.

The Problem

Let $I$ be the incentre of triangle $ABC$ and let $\Gamma$ be its circumcircle. Let the line $AI$ intersect $\Gamma$ again at $D$ . Let $E$ be a point on the arc $\widehat{BDC}$ and $F$ a point on the side $BC$ such that

$\angle BAF = \angle CAE < \frac{1}{2} \angle BAC$ .

Finally, let $G$ be the midpoint of the segment $IF$ . Prove that the lines $DG$ and $EI$ intersect on $\Gamma$ .

Solution

It turns out that several solutions exist. Several of the solutions use Menelaus’s theorem, which I’ve never even heard of; others rely on properties of excircles. I’ll give the solution that I think is the most elegant (although it doesn’t seem to be the standard solution).

We start by working backwards.

Let us denote $K$ to be the point of intersection between $DG$ and $EI$ ; we wish to prove that $K$ lies on $\Gamma$ . If $K$ lies on $\Gamma$ , then quadrilateral $AKDE$ is cyclic, and $\triangle IKD \sim \triangle IAE$ .

Obviously $\angle KID = \angle AIE$ . If we can show that $\angle GDI = \angle AEI$ , then triangles $IKD$ and $IAE$ are similar and our result follows.

Let $M$ be the midpoint of $AI$ .

We will prove that $\angle GDI = \angle AEI$ by proving that $\triangle MGD$ and $\triangle AIE$ are similar (these are the triangles highlighted in red).

It is given that $\angle CAE = \angle BAF$ , so angles $\angle DAE$ and $\angle DAF$ are also equal (since $AD$ passes through the incenter $I$ and thus bisects $\angle A$ ).

It is also given that $FG = GI$ ; now $IM = MA$ by definition so triangles $IFA$ and $IGM$ are similar. Then $\angle GMD = \angle IAE$ .

All that remains to prove the similarity of $\triangle MGD$ and $\triangle AIE$ (and thus the result) is to prove that

$\frac{MG}{MD} = \frac{AI}{AE}$

or equivalently

$MG \cdot AE = AI \cdot MD$ .

As $\triangle IFA \sim \triangle IGM$ , and $IG = \frac{1}{2}IF$ , it follows that $MG = \frac{1}{2}AF$ . By substitution,

$\frac{1}{2}AF \cdot AE = AI \cdot MD$ .

It can be shown that the product $AF \cdot AE$ is constant no matter where $E$ is on the circle:

Draw the line $EC$ ; the triangle formed, $\triangle ACE$ , is similar to $\triangle AFB$ since $\angle CAE = \angle CAF$ and $\angle ABF$ and $\angle AEC$ subtend the same arc.

Then

$\frac{AF}{AB} = \frac{AC}{AE}$

$AE \cdot AF = AB \cdot AC$ .

Since $AB \cdot AC$ is constant with respect to $E$ (and $F$ ), so is $AE \cdot AF$ .

Consequentially we can prove that $\frac{1}{2}AE \cdot AF = AI \cdot MD$ for all values of $E$ and $F$ by proving it for one value of $E$ and $F$ , since $\frac{1}{2} AE \cdot AF$ is constant and obviously $AI \cdot MD$ is constant.

We prove the case for $\angle CAE = 0$ , or in other words when $E$ coincides with $C$ and $F$ coincides with $B$ :

So here we need to prove that

$\frac{1}{2} AB \cdot AC = AI \cdot MD$ .

This is equivalent to proving that $K$ lies on $\Gamma$ in this instance, as that would prove the above equation too for this instance.

Obviously $\angle ACK = \angle ADK$ , also $\angle BCK = \angle BDK$ . As $\angle ACK = \angle BCK$ since $CK$ passes through the incircle and is an angle bisector, it follows that $\angle ADK = \angle BDK$ .

Given $BG=GI$ , this is sufficient to prove $\triangle BDI$ to be isosceles.

So $\angle BDG = \angle GDI$ , and $DG$ meets $\Gamma$ at the midpoint of minor ark $\widehat{AB}$ . Obviously $CI$ meets $\Gamma$ at the same point, and we are done. QED.

Checking the solution

Just to make sure, we can trace out steps back to get from the equation $\frac{1}{2} AE \cdot AF = AI \cdot MD$ to our result that $K$ lies on $\Gamma$ . I’m going to go through it very briefly:

From the equation we get $\frac{MG}{MD} = \frac{AI}{AE}$ , proving triangles $MGD$ and $AIE$ to be similar.

Then $\angle GDI = \angle AEI$ and triangles $KDI$ and $AEI$ are similar. Finally $AKDE$ is cyclic, leading to the result.

It is also valid to, starting with the result, arrive at the equation, which is what we implicitly did near the end of the solution.

This entry was posted on Friday, July 16th, 2010 at 5:00 pm and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to IMO 2010: Problem 2

lightest says:

March 25, 2011 at 11:41 am

I think the first official solution is elegant, too.

http://imo2010org.kz/download/sol_eng.pdf

It just uses a classical (though I haven’t ever heard of, too!) property of incenter: AI/IL=AD/DI. This property is so nice that all segments involved in this ratio lie just on the same line.

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xxtlc13 says:

September 23, 2011 at 12:57 pm

a nice problem to solve by euclidiean geo.the solution is the same as mine.

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