The point P lies on side CD of rectangle ABCD, with CD = 20 and AD = 10. The
circumcircle ω of △ABP re-intersects CD at Q. Given that the radius of ω is 11, find the distance
PQ.
As circle omega goes through A and B, its intersections with CD will be symmetrically placed on that segment, so that DP = CQ and CP = DQ.
CQ will be twice the distance of P from the center of CD.
The center of omega will lie 10*tan(arccos(10/11)) below AB (assuming AB is above CD); that's 4.58257569495584, making the distance of the circle's center 10 - 4.58257569495584 = 5.41742430504416 above CD.
Then
(PQ/2)^2 = 11^2 - 5.41742430504416 ^ 2
PQ then comes out to 19.1469594347632.
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Posted by Charlie
on 2025-04-23 14:24:50 |
solution
