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Bài thi giỏi toán Canada 2010

[ Wednesday, 31 March, 2010 ]

Bài thi này các bạn sẽ không tìm đâu ra, ngay cả trên mạng. Một thí sinh gởi cho mình chiều nay 03/30/2010.
This problem and solution you will not find anywhere yet in the web. A contestant sent it to me this afternoon for solution.

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 Vo Duc Dien

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 x BP² or

4abr² = (AI² + c²)(BI² + c²) = c**4 + (AI² + BI²)c² + AI² x BI²

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

Multiply both side of (*) by c², we have 4c²r² = c**4 + (AI² + BI² + KI²)c²

or c² = ab

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