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.

Contest questions may be downloaded here.

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. 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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