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 be the incentre of triangle and let be its circumcircle. Let the line intersect again at . Let be a point on the arc and a point on the side such that

.

Finally, let be the midpoint of the segment . Prove that the lines and intersect on .

### 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 to be the point of intersection between and ; we wish to prove that lies on . If lies on , then quadrilateral is cyclic, and .

Obviously . If we can show that , then triangles and are similar and our result follows.

Let be the midpoint of .

We will prove that by proving that and are similar (these are the triangles highlighted in red).

It is given that , so angles and are also equal (since passes through the incenter and thus bisects ).

It is also given that ; now by definition so triangles and are similar. Then .

All that remains to prove the similarity of and (and thus the result) is to prove that

or equivalently

.

As , and , it follows that . By substitution,

.

It can be shown that the product is **constant **no matter where is on the circle:

Draw the line ; the triangle formed, , is similar to since and and subtend the same arc.

Then

or

.

Since is constant with respect to (and ), so is .

Consequentially we can prove that for **all** values of and by proving it for **one** value of and , since is constant and obviously is constant.

We prove the case for , or in other words when coincides with and coincides with :

So here we need to prove that

.

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

Obviously , also . As since passes through the incircle and is an angle bisector, it follows that .

Given , this is sufficient to prove to be isosceles.

So , and meets at the midpoint of minor ark . Obviously meets 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 to our result that lies on . I’m going to go through it very briefly:

From the equation we get , proving triangles and to be similar.

Then and triangles and are similar. Finally 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.

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

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