# 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

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Copyright © 1996-2017 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

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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

Observe that the acuteness of ΔABC played no role in the solution. In fact,

### References

- 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

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Copyright © 1996-2017 Alexander Bogomolny

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