Olympiad Geometry | Homothety 2

Question

Two noncongruent circles intersect at $X$ and $Y$. Their common (external) tangents intersect at $Z$. One of the common tangents touches the circles at $P$ and $Q$. Prove that $ZX$ is tangent to the circumcircle of triangle $PXQ$.

I let $M$ be a point that is the intersection of XZ and the smaller circle, connecting $Q$ and $M$ gives us trapezium $XMQP$ where $XP||MQ$. But now I'm stuck what should I do next? I know I can move forward with angle chasing but I don't see how.

Answer 1

By the power of the point $Z$: $$ZQ^2 = ZX\cdot ZM$$

As you mentioned we have $${ZM\over ZX} = {ZQ\over ZP}\implies ZM = ZX\cdot {ZQ\over ZP}$$

so $$ZQ^2 = ZX^2\cdot {ZQ\over ZP} \implies ZX^2 = ZP\cdot ZQ$$

and we are done.

Answer 2

Şolution with an inversion with pole at $Z$ and radius $r=\sqrt{ZP\cdot ZQ}$.

It swaps points $P$ and $Q$.

It takes smaller circle to new circle which is tangent to $PQ$ at $P$, so this new circle is actualy bigger circle. With the same reasoning it takes bigger circle to smaller. So their intersection points stay where they are so $r=ZX = ZY$.

Now we have $$ZX^2 = ZP\cdot ZQ$$ which means that line $ZX$ is tangent to circle $(XPQ)$.

Answer 3

Nice question, I've created a diagram for it using asymptote. Here's the proof;

Let the circles be $A$ and $B$. Without Loss Of Generality, we let $A$ is the smaller circle among $A$ and $B$. Note that in our diagram, where we situate $P$ and $Q$ is not important (by symmetry).

We also let $K$ be the leftmost intersection of $ZX$ with circle $A$. Consider the tangent $ZP$ to circle $A$, by the Power Of a Point theorem on point $Z$, we have; $$ZP^2=ZK\cdot ZX.\tag{Equation 1}$$

Consider $\triangle\mathit{ZKP}$ and $\triangle\mathit{ZPX}$, we know that these two triangles share an angle; $$\angle\mathit{PZK} = \angle\mathit{XZP}$.$

Furthermore, by Equation 1, we have; $$ \begin{align*} & ZP^2=ZK\cdot ZX\ \Rightarrow & \frac{ZP}{ZK}=\frac{ZX}{ZP}. \end{align*}

Hence, by the SAS similarity theorem, we have $\triangle\mathit{ZKP} \sim \triangle\mathit{ZPX}$. We have; $$\angle\mathit{ZKP} = \angle\mathit{ZPX}\tag{Equation 2}$$

Note that we have a homothety centered at point $Z$, that maps circle $B$ to circle $A$, denote it by $h$. Since $Z$, $P$, and $S$ are collinear, and so are $Z$, $K$, and $X$, we have; $$h(Q)=h(P) \text{ and } h(X)=h(K)$$ $$\Rightarrow h(\overline{XQ})=h(\overline{KP}).$$

Hence, $KP$ and $XQ$ must be parallel. Since $ZX$ is a transversal to these parallel lines; $$\angle\mathit{ZXQ} = \angle\mathit{ZKP}.$$

Substituting Equation 2, we have; $$\angle\mathit{ZPX} = \angle\mathit{ZXQ}.$$

Now, we consider $\triangle\mathit{ZPX}$ and $\triangle\mathit{ZXQ}$, by our above Equation, and the fact that $\triangle\mathit{ZPX}$ and $\triangle\mathit{ZXQ}$ share an angle at vertex $Z$, we can infer that by the $AA$ congruence theorem, $$ \begin{align*} \triangle\mathit{ZPX} &\sim \triangle\mathit{ZXQ}\\ \Rightarrow \frac{ZP}{ZX} &=\frac{ZX}{ZQ}\\ \Rightarrow ZX^2 &=ZP\cdot ZQ. \end{align*} $$

Note that by the converse of the Power Of a Point theorem, the above Equation could only be true if $ZX$ is tangent to the circle passing through points $P$, $X$, and $Q$. Hence, we know that $ZX$ is tangent to the circumcircle of $\triangle\mathit{PXQ}$.