## A Geometry Problem from the 2007 Thai Mathematical Olympiad

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

Solution: The strategy is to take the midpoint ${Q}^{\prime }$ of $\overline{PB}$ 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 ${Q}^{\prime }$ is identical to $Q$.

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

The $△CN{D}^{\prime }$ is congruent to $△FNE$. (Reason: $\angle FNE=\angle CN{D}^{\prime }$ and by symmetry $FN=CN$ and $NE=N{D}^{\prime }$, so side-angle-side shows congruence.)

Point $D$ is the intersection of $\overline{CP}$ with $\omega$. Likewise by symmetry, and also the congruence of $△CN{D}^{\prime }$ and $△FNE$, $E$ is the intersection of $\overline{FP}$ with $\omega$. Now, using the symmetry of $\overline{FE}$ and $\overline{C{D}^{\prime }}$ (because of the symmetry of $F$ and $C$ across $\overline{NP}$ and because of the congruence of $△FNE$ to $△CN{D}^{\prime }$, then $\overline{CP}$ must intersect $\omega$ at $E$ which is symmetric to $\overline{FP}$ which intersects $\omega$ at $E$.) Hence $D$ is identical with ${D}^{\prime }$. Then since $E$ is determined by the midpoint $M$ of $\overline{AP}$, $Q$ which determines $D$ must also be identical with ${Q}^{\prime }$. That is, $Q$ is the midpoint of $\overline{BP}$.

Since $N$ is the midpoint of the chord $\overline{AB}$, $△ANP$ is a right triangle. Then $\overline{NM}$ is the median to the hypotenuse of a right triangle, so $NM=MP=AM$, so $△MNP$ is isosceles, and $\angle MNP=\angle MPN=\alpha$. Then $\angle NAP=\pi ∕2-\alpha$. Also, $△ANM$ is isosceles, so $\angle ANM=\pi ∕2-\alpha$. Because $△APB$ is isosceles, $\angle NBP=\pi ∕2-\alpha$. Since $△NBP$ is a right triangle, $\angle NPB=\alpha$. Therefore, line $CM$ is parallel to $\overline{BP}$. Likewise, $\overline{NQ}$ is parallel to $\overline{AP}$ 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 $\overline{AB}\perp \overline{NP}$. Thus, $M$ is the circumcenter of $△ANP$ and hence $MN=MP$. As in the previous proof, it can also be seen that $\overline{MN}\parallel \overline{PQ}$. From the power of the point $M$, $P{M}^{2}=M{A}^{2}=ME\cdot MC$. So

$\frac{PM}{ME}=\frac{MC}{PM}.$

and hence $△PME\sim △CMP$. (Note that $△PME$ and $△CMP$ share $\angle CMP$.) Thus $\angle MPE=\angle MCP$.

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

$CN\cdot NE=AN\cdot NB=ON\cdot NP.$

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

Thus $C$, $P$, $E$, $O$ are also cyclic, and hence $\angle EPN=\angle NCO$. (Another way is to note that $CN∕NP=ON∕NE$ and $\angle ONC=\angle ENP$, so $△ONC\sim △ENP$. Then $\angle EPN=\angle OCN$.) Since $△PAN\sim △POA$ (both have a right angle and share $\angle OPA$),

$\frac{PA}{PN}=\frac{PO}{PA}.$

Thus $PN\cdot PO=P{A}^{2}=PD\cdot PC$. (This occurs first from the similarity of the triangles, then because of the power of point $P$ with respect to the circle $\omega$.) So $C$, $D$, $N$ $O$ are cyclic and hence $\angle QNP=\angle PCO$. (Another way is to note that $PN∕PD=PC∕PO$ and $△PDN$ and $△POC$ share $\angle NPD$, so $△PDN\sim △POC$. Then $\angle QNP=\angle DNP=\angle PCO$.) We can now see that

$\angle QNP=\angle PCO=\angle PCM+\angle MCO=\angle MPE+\angle EPN=\angle MPN.$

Thus $\overline{MP}\parallel \overline{NQ}$. 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 $\overline{PB}$ 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.