Circle through the Circumcenter

The applet below illustrates problem 3 from the 1996 British Mathematical Olympiad.

Let O be the circumcenter of an acute triangle ABC. Let Γ be the circle passing through B, O and C. Suppose AB and AC intersect Γ the second time in P and Q respectively. Prove that OA ⊥ PQ.

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Solution

Let O be the circumcenter of an acute triangle ABC. Let Γ be the circle passing through B, O and C. Suppose AB and AC intersect Γ the second time in P and Q respectively. Prove that OA ⊥ PQ.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Inscribed in Γ angles BPQ and BCQ are equal as subtended by the same arc BQ. Let X denote the intersection of OA and PQ. The condition that OA is perpendicular to PQ is equivalent to

∠APX + ∠PAX = 90°.

On the other hand,

∠APX = ∠APQ = ∠BPQ = ∠BCQ = ∠BCA.

And also

∠PAX = ∠BAO.

It follows that the orthogonality condition is equivalent to

∠BCA + ∠BAO = 90°.

But this is indeed so. In Γ, BCQ is an inscribed circle subtended by arc BQ; the same of course holds for ∠BCQ = ∠BCA. In the circumcircle, the inscribed ∠BCA is subtended by arc AB. So that the central angle AOB = 2∠BCA. Then in the isosceles ΔAOB, ∠BAO = (180° - ∠AOB)/2 = 90° - ∠BCA. Which is exactly what we need.

Observe that the acuteness of ΔABC played no role in the solution. In fact, AO ⊥ PQ for any triangle ABC.

References

1. T. Gardiner, The Mathematical Olympiad Handbook, Oxford University Press, 1997.