2008 Pan African Mathematical Olympiad

April 21, 2009

Problem:

In circle ${C}_{1}$ with center $O$, chord $\overline{AB}$ is not a diameter. Point $M$ is the midpoint of $\overline{AB}$. We choose any point $T$ on the circle ${C}_{2}$ with $\overline{OM}$ as diameter. The tangent to ${C}_{2}$ at $T$ intersects ${C}_{1}$ at 2 points. Choose either one of them and denote it by $P$. Show that $P{A}^{2}+P{B}^{2}=4P{T}^{2}$.

Solution:

Let $C$ be the midpoint of $\overline{OM}$, and therefore the center of ${C}_{2}$. Extend $\overline{OM}$ to meet circle ${C}_{2}$ at $F$. Let $G$ be the foot of the perpendicular from $P$ to $\overline{OF}$.

Let $R$ be the radius of the larger circle ${C}_{1}$ and let $r$ be the radius of the smaller circle ${C}_{2}$. Let $x=FG$, so that

$$\begin{array}{llll}\hfill CG& =R-r-x,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill OG& =R-x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P{G}^{2}& =P{O}^{2}-O{G}^{2}={R}^{2}-{\left(R-x\right)}^{2}=2xR-{x}^{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Then

$$P{C}^{2}=P{G}^{2}+C{G}^{2}={\left(R-x-r\right)}^{2}+\left(2xR-{x}^{2}\right)={R}^{2}-2Rr+{r}^{2}-2xr.$$

Then

$$P{T}^{2}=P{C}^{2}-C{T}^{2}={R}^{2}-2Rr+{r}^{2}+2xr-{r}^{2}={R}^{2}-2Rr+2xr.$$

Let $H$ be the foot of the perpendicular from $P$ to $\overline{AB}$.

Note that $B{M}^{2}={R}^{2}-4{r}^{2}$. Note that

$$PH=GM=R-2r-x.$$

Note that

$$MH=PG=\sqrt{2xR-{x}^{2}}.$$

Then

$$BH=BM+MH=\sqrt{{R}^{2}-4{r}^{2}}+\sqrt{2xR-{x}^{2}}.$$

Then

$$P{B}^{2}=B{H}^{2}+P{H}^{2}={\left(\sqrt{{R}^{2}-4{r}^{2}}+\sqrt{2xR-{x}^{2}}\right)}^{2}+{\left(R-2r-x\right)}^{2}.$$

Likewise

$$AH=AM-MH=\sqrt{{R}^{2}-4{r}^{2}}-\sqrt{2xR-{x}^{2}}.$$

Then

$$P{A}^{2}=A{H}^{2}+P{H}^{2}={\left(\sqrt{{R}^{2}-4{r}^{2}}-\sqrt{2xR-{x}^{2}}\right)}^{2}+{\left(R-2r-x\right)}^{2}.$$

Combining and expanding,

$$P{A}^{2}+P{B}^{2}=4{R}^{2}-8rR+8rx.$$

Then it is easy to see that $P{A}^{2}+P{B}^{2}=4P{T}^{2}$.

Commentary:

Note that if $\overline{AB}$ is a diameter, then $M$ is the center of ${C}_{1}$, the circle ${C}_{2}$ shrinks to a point so $T=M$, and the tangent segment $PT$ becomes a radius, and $PT=AM=MB$. Also $\angle APB=9{0}^{\circ}$ since it subtends a diameter. Then by the Pythagorean Theorem, $P{A}^{2}+P{B}^{2}=A{B}^{2}={\left(2\cdot AM\right)}^{2}=4A{M}^{2}=4P{T}^{2}$. 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 ${C}_{2}$, then $P=A$ (or $P=B$). Then $PA=0$ and $PB=2\cdot PT$ and $P{A}^{2}+P{B}^{2}=4P{T}^{2}$.

A third special case occurs when $PT$ is “vertical”. For reference, set the diagram in a coordinate plane with the origin at $O$ and $\overline{OM}$ oriented along the $y$-axis. Take the special case when the point $T=\left(r,r\right)$ or symmetrically $T=\left(-r,r\right)$, so that the tangent line is perpendicular to the $x$-axis at either the right side or the left side of the smaller circle ${C}_{2}$. Then $P{A}^{2}+P{B}^{2}=4\phantom{\rule{0em}{0ex}}{R}^{2}-8\phantom{\rule{0em}{0ex}}r\sqrt{{R}^{2}-{r}^{2}}$. On the other hand $PT=\sqrt{{R}^{2}-{r}^{2}}-r$. Then $4P{T}^{2}=4\phantom{\rule{0em}{0ex}}{R}^{2}-8\phantom{\rule{0em}{0ex}}r\sqrt{{R}^{2}-{r}^{2}}$, so that in this special case again $P{A}^{2}+P{B}^{2}=4P{T}^{2}$.

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

$$P{T}^{2}+{r}^{2}={\left(R-r\right)}^{2}$$

so $P{T}^{2}={R}^{2}-2Rr$. On the other hand $AM=R-\sqrt{{R}^{-}{\left(2r\right)}^{2}}$ and $MP=R-2r$ and $\angle AMP$ is a right angle. Therefore

$$P{A}^{2}=A{M}^{2}+M{P}^{2}={\left(\sqrt{{R}^{-}{\left(2r\right)}^{2}}\right)}^{2}+{\left(R-2r\right)}^{2}=2{R}^{2}-4Rr$$

Then $P{A}^{2}+P{B}^{2}=4{R}^{2}-8Rr$. Hence $P{A}^{2}+P{B}^{2}=4P{T}^{2}$.

In each of these special cases, I was able to explicitly calculate the chord-quantity $P{A}^{2}+P{B}^{2}$ 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 ${C}_{2}$ and the tangent point $T$.