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 c = 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 ( ∠AOK) + ∠POF ( ∠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 AK + PB = (2AE) + (2PF) = 4AE + 4OE = 4r. Moreover,

AK + PB = AI + KI + PI + BI or 4r = c + AI + BI + KI (*)

Now the right triangle GOP gives us OP = OG + GP or r = AC + (a r) or a = 2ar - AC or 2ar = AP Similarly, on the right side of the configuration 2br = BP

Multiply the previous two equations side by side 4abr = AP × BP or

4abr = (AI + c)(BI + c) = c^4 + (AI + BI)c + AI × BI

but AI × BI = c × KI and now 4abr = c^4 + (AI + BI + KI)c

Multiply both side of (*) by c, we have 4c^2r^2 = c^4 + (AI + BI + KI)c

or c = 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.