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.

What if applet does not run?

Solution

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Copyright © 1996-2018 Alexander Bogomolny

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.

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.

What Is Red Herring

• On the Difference of Areas
• Area of the Union of Two Squares
• Circle through the Incenter
• Circle through the Incenter And Antiparallels
• Circle through the Circumcenter
• Inequality with Logarithms
• Breaking Chocolate Bars
• Circles through the Orthocenter
• 100 Grasshoppers on a Triangular Board
• Simultaneous Diameters in Concurrent Circles
• An Inequality from the 2015 Romanian TST
• Schur's Inequality
• Further Properties of Peculiar Circles
• Inequality with Csc And Sin
• Area Inequality in Trapezoid
• Triangles on HO
• From Angle Bisector to 120 degrees Angle
• A Case of Divergence
• An Inequality for the Cevians through Spieker Point via Brocard Angle
• An Inequality In Triangle and Without
• Problem 3 from the EGMO2017
• Mickey Might Be a Red Herring in the Mickey Mouse Theorem
• A Cyclic Inequality from the 6th IMO, 1964
• Three Complex Numbers Satisfy Fermat's Identity For Prime Powers
• Probability of Random Lines Crossing
• Planting Trees in a Row
• Two Colors - Three Points

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Copyright © 1996-2018 Alexander Bogomolny