A Geometry Problem from the
2007 Thai Mathematical Olympiad

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

January 23, 2009

Problem: Problem 9 from the Selected Problems from the Thai Mathematical Olympiad 2007. Collected at the 2008 International Mathematical Olympiad, Madrid, Spain, July 2008.

Let PA¯ and PB¯ be the tangents to circle ω from an external point P. Let M and N be the midpoints of AP¯ and AB¯, respectively. Extend MN¯ to meet ω at C, where N is between M and C. The line PC¯ meets ω at D and extend ND¯ to intersect PB¯ at Q. Show that MNQP is a rhombus.

Solution: The strategy is to take the midpoint Q of PB¯ symmetric to M, extend it to determine point F symmetric to C and a point E symmetric to D. Then the proof will show that Q is identical to Q.

Draw the symmetry line NP¯. Because N is the midpoint of the chord AB¯ between the symmetric points of tangency A and B, NP¯ extends through the center of the circle ω. Now symmetrize the problem by letting Q be the midpoint of PB¯. Extend QN¯ to meet ω at F, where N is between Q and F. Then across the symmetry line NP¯, the points C and F are symmetric. Let D be the intersection of FN¯ with ω where D is between N and Q. Likewise, let E be the intersection point of CM¯ with ω where E is between N and M.

The CND is congruent to FNE. (Reason: FNE = CND and by symmetry FN = CN and NE = ND, so side-angle-side shows congruence.)

Point D is the intersection of CP¯ with ω. Likewise by symmetry, and also the congruence of CND and FNE, E is the intersection of FP¯ with ω. Now, using the symmetry of FE¯ and CD¯ (because of the symmetry of F and C across NP¯ and because of the congruence of FNE to CND, then CP¯ must intersect ω at E which is symmetric to FP¯ which intersects ω at E.) Hence D is identical with D. Then since E is determined by the midpoint M of AP¯, Q which determines D must also be identical with Q. That is, Q is the midpoint of BP¯.

Since N is the midpoint of the chord AB¯, ANP is a right triangle. Then NM¯ is the median to the hypotenuse of a right triangle, so NM = MP = AM, so MNP is isosceles, and MNP = MPN = α. Then NAP = π2 α. Also, ANM is isosceles, so ANM = π2 α. Because APB is isosceles, NBP = π2 α. Since NBP is a right triangle, NPB = α. Therefore, line CM is parallel to BP¯. Likewise, NQ¯ is parallel to AP¯ and NQ = QP.

Therefore, quadrilateral MNQP is a rhombus.

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Official Solution from the Selected Problems from the Thai Mathematical Olympiad 2007 ( with additional explanation and commentary in italics).

As in the previous proof above, observe that AB¯ NP¯. Thus, M is the circumcenter of ANP and hence MN = MP. As in the previous proof, it can also be seen that MN¯ PQ¯. From the power of the point M, PM2 = MA2 = ME MC. So


and hence PME CMP. (Note that PME and CMP share CMP.) Thus MPE = MCP.

Since O, A, P, B are cyclic (because of the right angles at A and B, the diameter is OP¯), the power of the point N tells us that


(Note that this measures the power of the point N first with respect to the circle ω through C, A, E, B, and then with respect to the circle through O, A, C, B using the common product AN NB.)

Thus C, P, E, O are also cyclic, and hence EPN = NCO. (Another way is to note that CNNP = ONNE and ONC = ENP, so ONC ENP. Then EPN = OCN.) Since PAN POA (both have a right angle and share OPA),


Thus PN PO = PA2 = PD PC. (This occurs first from the similarity of the triangles, then because of the power of point P with respect to the circle ω.) So C, D, N O are cyclic and hence QNP = PCO. (Another way is to note that PNPD = PCPO and PDN and POC share NPD, so PDN POC. Then QNP = DNP = PCO.) We can now see that


Thus MP¯ NQ¯. Therefore, MNPQ is a rhombus.

Commentary: I found this problem to be surprisingly difficult. One reason is that a lot of information is given about the points in ANP, since M is a midpoint of the hypotenuse. However the point Q on the corresponding tangent PB¯ is constructed in an entirely different fashion, and we have practically no symmetric information about D. On the other hand, the problem has tangents, circles, and many triangles. We have so much information, and so many possible relationships that I got distracted chasing many irrelevant relations.

Another reason this problem seemed to be difficult is that my proof is somewhat indirect. By that, I mean that I construct symmetric points in BNP and then prove that those symmetric points are identical with the original constructed points. I have not seen many indirect proofs in geometry, although indirect proofs occur in other areas of mathematics.

I was once told that solving a geometry problem depends on drawing in the correct auxiliary lines or points. In this case, it seems natural to start by drawing in the line of symmetry NP, which should be the diagonal of the presumed rhombus. The line also passes through the center of the circle, providing the line of symmetry that is the key to my solution.

Note that in the official solution, the orientation of PME and CMP are reversed, which makes them hard to see as similar. Likewise for PDN and POC. Also it is hard to see that O, A, P, B are cyclic since the circle is not drawn. It is even harder for me to see that C, D, N, O are cyclic since the large circle holding them is not drawn. The official proof depends on three “invisible” circles. Even harder, the official proof asserts that since four points have a “cross-ratio” the four points must be the vertices of a cyclic quadrilateral, and hence there is a circumscribing circle. Although this fact may be true, I could not find a simple statement of it as a theorem or even a remark in any of three sources I consulted. I found it easier to find similar triangles with the points as vertices, then to get the angle equalities from the similar triangles. That is, the official proof relies on an obscure fact when the result can more easily be seen with more elementary methods.