# Angles Inscribed in an Absent Circle

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to suggest the following problem [Math Circles, p. 230] from the 1999 Bay Area Mathematical Olympiad:

Let *k* be a circle in the *xy*-plane with center on the *y*-axis and passing through points *a*)*b*)*a* < *b*.*x*-axis, and let

The problem admits a 1-step solution if we observe that the angles BQP and BOP have same end points and if they are indeed equal the four points B, O, P, Q must be concyclic. But the points O, B, Q form a right triangle whose circumscribed circle has BQ as a diameter. ∠BPQ is subtended by that diameter and should therefore be right. But it is in fact right because it is subtended by the diameter AB in circle *k*.

Thus the solution is this. Angles BOQ and BPQ (= ∠BPA) are right, making the four points B, O, P, Q concyclic. The two angles BQP and BOP are inscribed into the circumcircle of BQOP and are subtended by the same arc BP, therefore ∠BQP = ∠BOP.

### References

*A Decade of the Berkeley Mathematical Circle, The American Experience, Volume I*, Z. Stankova, Tom Rike (eds), AMS/MSRI, 2008

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny