A Geometry Problem from the
2008 Pan African Mathematical Olympiad

Steven R. Dunbar
Department of Mathematics
University of Nebraska-Lincoln
Lincoln, Nebraska
www.math.unl.edu/ sdunbar1
sdunbar1 at unl dot edu

April 21, 2009


In circle C1 with center O, chord AB¯ is not a diameter. Point M is the midpoint of AB¯. We choose any point T on the circle C2 with OM¯ as diameter. The tangent to C2 at T intersects C1 at 2 points. Choose either one of them and denote it by P. Show that PA2 + PB2 = 4PT2.



Let C be the midpoint of OM¯, and therefore the center of C2. Extend OM¯ to meet circle C2 at F. Let G be the foot of the perpendicular from P to OF¯.

Let R be the radius of the larger circle C1 and let r be the radius of the smaller circle C2. Let x = FG, so that

CG = R r x, OG = R x PG2 = PO2 OG2 = R2 (R x)2 = 2xR x2.


PC2 = PG2 + CG2 = (R x r)2 + (2xR x2) = R2 2Rr + r2 2xr.


PT2 = PC2 CT2 = R2 2Rr + r2 + 2xr r2 = R2 2Rr + 2xr.

Let H be the foot of the perpendicular from P to AB¯.

Note that BM2 = R2 4r2. Note that

PH = GM = R 2r x.

Note that

MH = PG = 2xR x2.


BH = BM + MH = R2 4r2 + 2xR x2.


PB2 = BH2 + PH2 = (R2 4r2 + 2xR x2)2 + (R 2r x)2.


AH = AM MH = R2 4r2 2xR x2.


PA2 = AH2 + PH2 = (R2 4r2 2xR x2)2 + (R 2r x)2.

Combining and expanding,

PA2 + PB2 = 4R2 8rR + 8rx.

Then it is easy to see that PA2 + PB2 = 4PT2.


Note that if AB¯ is a diameter, then M is the center of C1, the circle C2 shrinks to a point so T = M, and the tangent segment PT becomes a radius, and PT = AM = MB. Also APB = 90 since it subtends a diameter. Then by the Pythagorean Theorem, PA2 + PB2 = AB2 = (2 AM)2 = 4AM2 = 4PT2. In this way, the problem is a generalization of the Pythagorean Theorem.

As a second special case, note that if the point T is taken to be M, the “top” of the circle C2, then P = A (or P = B). Then PA = 0 and PB = 2 PT and PA2 + PB2 = 4PT2.

A third special case occurs when PT is “vertical”. For reference, set the diagram in a coordinate plane with the origin at O and OM¯ oriented along the y-axis. Take the special case when the point T = (r,r) or symmetrically T = (r,r), so that the tangent line is perpendicular to the x-axis at either the right side or the left side of the smaller circle C2. Then PA2 + PB2 = 4R2 8rR2 r2. On the other hand PT = R2 r2 r. Then 4PT2 = 4R2 8rR2 r2, so that in this special case again PA2 + PB2 = 4PT2.

A fourth special case occurs when P is at the midpoint G of the arc AB. Then the points C, M and P are collinear on a line which is a radius of the larger circle. In this case the tangent point T is on the shoulder of the smaller circle, but from experimentation with GeoGebra, the point does not seem to have any special properties such as angle, or distance. The distance PC = R r. Then because of the tangency, the angle PTC is a right angle and so

PT2 + r2 = (R r)2

so PT2 = R2 2Rr. On the other hand AM = R R (2r)2 and MP = R 2r and AMP is a right angle. Therefore

PA2 = AM2 + MP2 = (R (2r)2)2 + (R 2r)2 = 2R2 4Rr

Then PA2 + PB2 = 4R2 8Rr. Hence PA2 + PB2 = 4PT2.

In each of these special cases, I was able to explicitly calculate the chord-quantity PA2 + PB2 and I have been able to explicitly calculate the tangent length PT.

I was able to finally solve this problem by thinking about what was common to the special cases. I decided that what enabled me most was to be able to calculate a vertical distance equivalent to what I have in the diagram as PH. Also, the special case when P is the midpoint of the arc AB finally suggested to me that I should take something like point P, or the foot of the perpendicular G as my “independent” variable, not the point T. Freeing my thinking in terms of G, not T, was the key was the key to finding all the right triangles in the problem in order to use the Pythagorean Theorem to find the distances. This problem ultimately was simple, depending only on multiple use of the Pythagorean Theorem. However, the problem statement has a lot of misdirection in it, causing me to focus on the small circle C2 and the tangent point T.