# Three Tangents, Three Secants: What is this about? A Mathematical Droodle

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Explanation The applet purports to suggest the following problem by Peter Y. Woo (#1557, Math Magazine, 1998):

Let PQ be a diameter of a circle, with A and B two distinct points on the circle on the same side of PQ. Let C be the intersection of the tangents to the circle at A and B. Let the tangent to the circle at Q meet PA, PB, and PC at A', B', and C', respectively. Prove that C' is the midpoint of A'B'.

### 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.

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Two solutions have been published Math Magazine, Vol. 72, No. 4, October 1999. The one by Richard E. Pfiefer is outright beautiful in its simplicity.

The key is to make an inversion with center P and radius PQ. Point Q remains invariant; the given circle inverts into its tangent at Q. The circle with center at C and radius AC, which is orthogonal to the given circle, inverts into a circle orthogonal to that tangent. Under the inversion, A' and B' are the images of A and B, respectively. The only circle through A', B' orthogonal to the tangent has its center at the midpoint of A'B'. Therefore A'C' = B'C'. ### Inversion - Introduction 