# BMO 2007 Problem 2

Problem 2 of the British Mathematical Olympiad 2007

Let triangle ABC have incentre I and circumcentre O. Suppose that ∠AIO = 90° and ∠CIO = 45°. Find the ratio AB : BC : CA.

Solution by Vo Duc Dien Let the incircle tangent AB at D and r be its radius. We have

```		½ ∠A + ½ ∠C + ∠B = ∠AIC = 90° + 45° = 135°
```

and ∠A + ∠B + ∠C = 180°

or ∠B = 90° and the circumcenter O of triangle ABC is the midpoint of AC. Now let ½AC = OA = OC = OB = R, the radius of the circumcircle, OI = b and ∠AOB = ε

Applying the law of the sine function for triangle OIC, we obtain

R/sin45° = b/sin½∠C but in triangle AOI, R = b/sin1/2∠A and the previous expression becomes R/sin45° = R sin½∠A / sin½∠C or sin½∠A = 2 sin½∠C (*)

We also have ½∠A + ½∠C = 45° or sin(½∠A + ½∠C) = sin45° Now expand it sin½∠A cos½∠C + cos½∠A sin½∠C = 2 /2 or

Now substituting sin½∠C from (*), we have sin½∠A cos(45° - ½∠A) + cos½∠A sin½∠A 2 /2 = 2 /2 or

sin½∠A (cos½∠A + sin½∠A ) + cos½∠A sin½∠A = 1 or

2sin½∠A cos½∠A = cos² ½∠A or 2sin½∠A = cos½∠A or tan½∠A = 0.5 = DI/AD or DI = r, AD = 2r and AI = r 5

and cos∠A = cos² ½∠A - sin² ½∠A = (AD/AI)² - (DI/AI)² = 3/5 = 0.6

But cos∠A = AB/CA = 0.6

Now applying the Pythagorean formula CA² = AB² + BC², we have

CA² = 0.36 CA² + BC² or BC/CA = 0.8

Finally, AB : BC : CA = 3 : 4 : 5