# IMO 2010: Problem 4

The International Olympiad of Mathematics took place on July 8 and 9, and the six questions were made available the second day. This was a two day contest, three questions per day, and four and a half hours per day, giving a total of nine hours for the six questions. The international contest included questions on geometry, algebra, combinatorics, and number theory.

This year, it seems that problems 1 and 4 were the easiest (by comparison). I’ve only looked at problem 4 in detail (probably because it’s geometry).

### Problem

Let P be a point inside the triangle $ABC$. The lines $AP$, $BP$ and $CP$ intersect the circumcircle $\Gamma$ of triangle $ABC$ again at the points K, L and M respectively. The tangent to $\Gamma$ at C intersects the line $AB$ at S. Suppose that $SC = SP$. Prove that $MK = ML$.

### Solution

As often with geometry problems, a variety of different solutions are possible. Some are simple, while others rely on homothetic transformations and inversions. I’m going to go with the method I came up with (albeit with a few peeks at the Art of Problem Solving forums). From the Power of a Point theorem, we have $SC^2 = SA \cdot SB$

But it’s given that $SP = SC$, so $SP^2 = SA \cdot SB$

Or, $\frac{SP}{SA} = \frac{SB}{SP}$

This is enough to show that triangles $SAP$ and $SPB$ are similar, since they also share an angle.

Therefore, $\frac{SA}{AP} = \frac{SP}{BP}$

Or, $\frac{AP}{BP} = \frac{SA}{SP}$

Since $SP = SC$, $\frac{AP}{BP} = \frac{SA}{SC} \; \; \; (1)$

Triangles $SCA$ and $SBC$ are similar, since they share an angle and $\angle SCA = \angle ABC$. Thus, $\frac{SA}{AC} = \frac{SC}{BC}$

Or, $\frac{SA}{SC} = \frac{AC}{BC}$

From (1), $\frac{AP}{BP} = \frac{AC}{BC}$

Or equivalently, $\frac{AC}{AP} = \frac{BC}{BP} \; \; \; (2)$

Notice that triangles $APC$ and $MPK$ are similar ( $\angle CMK$ and $\angle CAK$ subtending the same arc); so, $\frac{AC}{AP} = \frac{MK}{MP}$

Similarly, triangles $MPL$ and $BPC$ are also similar, giving $\frac{BC}{BP} = \frac{ML}{MP}$

From (2), $\frac{MK}{MP} = \frac{ML}{MP}$

This shows that $MK = ML$, as desired. QED.

## 2 thoughts on “IMO 2010: Problem 4”

1. trax 0101 says:

hello,,, can you help me,,, do you have any books recommendation in preparing for the international mathematical olympiad,,,, maybe looking at your specialty,, books that are about imo level geometry?

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1. hehe says:

check imo copendium

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