# 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

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

or

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

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.

## 2 thoughts on “IMO 2010: Problem 2”

1. lightest says:

I think the first official solution is elegant, too.

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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2. xxtlc13 says:

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

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