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).
Let P be a point inside the triangle . The lines , and intersect the circumcircle of triangle again at the points K, L and M respectively. The tangent to at C intersects the line at S. Suppose that . Prove that .
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
But it’s given that , so
This is enough to show that triangles and are similar, since they also share an angle.
Triangles and are similar, since they share an angle and . Thus,
Notice that triangles and are similar ( and subtending the same arc); so,
Similarly, triangles and are also similar, giving
This shows that , as desired. QED.