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.


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.


\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.

2 thoughts on “IMO 2010: Problem 2

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