Problem 4 from the IMO 2009
Here is Problem 4 from the IMO 2009:
Let ABC be a triangle with AB = AC. The angle bisectors of ∠CAB and ∠ABC meet the sides BC and CA at D and E, respectively. Let K be the incentre of triangle ADC. Suppose that |
Solution by Vo Duc Dien
(Saturday July 18, 2009)
Dedicated to my teacher Nguyen Hoi
Extend CK to meet BE at H. H is the center of the incircle of triangle ABC.
Since CH bisects ∠ACB, the two right triangles HDC and HMC are equal. Therefore
There are two possibilities: either M and E coincide or they are different. In the former case, BM is both the median and the altitude from B, implying BC = AB. Since also, AB = AC, all three sides in ΔABC are equal and the triangle is equilateral. Thus
If M and E are distinct, we recall that ∠HEK = ∠BEK = 45°. Thus the quadrilateral HMEK is cyclic and, since
Further, ∠BHC = 180° - ∠KHE = 135°, 2b = 45°.
In triangle ABC we have ∠CAB = 180° - 4b = 90° which is the second possible value for ∠CAB.