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”

trax 0101 says:

August 30, 2010 at 8:09 am Reply

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?
1. hehe says:
  
  January 29, 2012 at 4:41 am Reply
  
  check imo copendium

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IMO 2010: Problem 4

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