Saturday, December 4th, 2021

## Bài thi giỏi toán Ái Nhĩ Lan 2007

[ Saturday, 13 March, 2010 ]

Nhân tiện St. Patrick’s day sắp đến mình đăng một bài thi học sinh giỏi toán Ái Nhĩ Lan 2007.

Problem 3 of Irish Mathematical Olympiad 2007

The point P is a fixed point on a circle and Q is a fixed point on a line. The
point R is a variable point on the circle such that P, Q and R are not collinear. The circle through P, Q and R meets the line again at V. Show that the line VR passes through a fixed point.

Solution by Vo Duc Dien

Let C be the circle where the fixed point P is on. Link QR and VR and extend
them to intercept C at U and S, respectively. Let I and IN be the center and the diameter of the circumcircle of triangle PQR, respectively.

Now let ε = ∠SPU. We also have ε = ∠SRU = ∠VRQ = ∠VPQ and let θ = ∠PRQ = ∠PVQ = ∠PNQ.

We have ∠SPQ = ∠UPQ + ∠SPU (*)

But O and I are centers of the two circles and P and R are their intersections, we then have

∠IOP = ½ ∠ROP = ∠PUQ and similarly ∠OIP = ∠PQU. The two triangles OPI and UPQ are then similar and ∠OPI = ∠UPQ

Equation (*) can now be written as
∠SPQ = ∠OPI + ε = ∠OPI + ∠VPQ = ∠OPI + 180° – ∠PQV – ∠PVQ = ∠OPI + 180° – ∠PQV – ∠PNQ = ∠OPI + 180° – ∠PQV – (90° – ∠NPQ ) = ∠OPI + 180° – ∠PQV – 90° + ∠NPQ = ∠OPQ + 90° – ∠PQV

Since both angles ∠OPQ and ∠PQV are constants, ∠SPQ is then constant and
VR passes through a fixed point.

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