**Problem 2 of the Canadian Mathematical Olympiad 2010**

Let A, B, P be three points on a circle. Prove that if a and b are the distances from P to the tangents at A and B and c is the distance from P to the chord AB, then c² = ab.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Cathy Diễm and Huy)**

Let O be the center of the circumcircle Г of triangle ABP, I be the foot of P to AB, PI = c. Extend PI to meet Г at K. Now let E and F be the midpoints of AK and BP, respectively; E and F are also the feet of O to AK and BP, respectively. From P draw the perpendicular lines to meet the tangents at A and at B of circumcircle Г at C and D, respectively, PC = a, PD = b. Let G be the foot of O to PC and r the radius of circle Г.

Since AB ┴ PK, angles subtending arcs AK plus BP is 90° or

∠AOE (½ ∠AOK) + ∠POF (½ ∠POB) = 90°

but ∠AOE + ∠EAO = 90° or ∠EAO = ∠POF

Combined with OA = OP = r, the triangles AOE and OPF are congruent and AE = OF, and OE = PF.

We now have AK² + PB² = (2AE)² + (2PF)² = 4AE² + 4OE² = 4r². Moreover,

AK² + PB² = AI² + KI² + PI² + BI² or 4r² = c² + AI² + BI² + KI² (*)

Now the right triangle GOP gives us OP² = OG² + GP² or r² = AC² + (a – r)² or a² = 2ar - AC² or 2ar = AP² Similarly, on the right side of the configuration 2br = BP²

Multiply the previous two equations side by side 4abr² = AP² × BP² or

4abr² = (AI² + c²)(BI² + c²) = c^4 + (AI² + BI²)c² + AI² × BI²

but AI × BI = c × KI and now 4abr² = c^4 + (AI² + BI² + KI²)c²

Multiply both side of (*) by c², we have 4c^2r^2 = c^4 + (AI² + BI² + KI²)c²

or c² = ab

**Remark**: It is a curious observation that this problem made its appearance at different times and in a little different guises in three mathematical olympiads. Besides the 2010 Canadian Olympiad, it showed up also at the the 2008 Australian Mathematical Olympiad and even much earlier in a published form.