Too many tangents

Question

Let $C_1$ and $C_2$ be two circles of unequal radius. Circles $C_1$ and $C_2$ intersect at points $A$ and $B$; let $L_1$ be the tangent line to $C_1$ at $A$, and let $L_2$ be the tangent line to $C_2$ at $B$, and let $P$ be the intersection of $L_1$ and $L_2$. Let $M$ and $N$ be points in $C_1$ and $C_2$, respectively, such that $PM$ is tangent to $C_1$, and $PN$ is tangent to $C_2$.

Let $AM$ and $BP$ intersect at $S$, and let $BN$ and $AP$ intersect at $T$. Show that $\square ATBS$ is cyclic.

It was from at list of Power of a Point, so there should be a solution using power of a point.

I tried to show that triangles $\triangle APM$ and $\triangle BPN$ are similar. Also, I tried a lot of things using power of a point, but I did not make any progress.

Answer 1

I tried to solve it with regular geometry and found out that $\overline{AQ}\parallel\overline{BR}$, where $Q=(PB\cap\circ{O1})\setminus{B}$ and $R=(PA\cap\circ{O2})\setminus{A}$.

Usually it is not a bad idea to try inversion, if you try to prove points are cyclic, hence I have performed an inversion with a center at $A$ and the following result is obtained:

I have tried to color-code each line, so it is easier to find the corresponding lines and/or circles, but in generall:

• circles going through $A$ become lines not passing $A$
• lines passing through $A$ do not change
• Lines not passing through $A$ become circles

Since points $A,T,B,S$ need to be on a circle it results into $T,B,S$ need to be colinear in the inverted problem. I have labeled most of the angles and the only thing to prove is $\alpha=\beta$ or $\epsilon=\varphi$. Most of the angles can be calculated by inscribed circles or angles between circle segments and tangents (if you have any questions why two angles are equal, feel free to ask, but by doing it on your own you learn a lot).

Now we have from triangles $\triangle SQB$: $$\delta+\gamma+\varphi+\beta=\pi$$ $$\frac{\sin\varphi}{\sin\gamma}=\frac{\overline{SM}}{\overline{QM}}=\frac{\overline{SM}}{\overline{MB}}=\frac{\sin\beta}{\sin\delta}\Rightarrow \frac{\sin\varphi}{\sin\beta}=\frac{\sin\gamma}{\sin\delta}$$

and from $\triangle PNT$ ($\measuredangle PNR = \epsilon$): $$\delta+\gamma+\epsilon+\alpha=\pi$$ $$\frac{\sin\epsilon}{\sin\alpha}=\frac{\overline{PR}}{\overline{NR}}=\frac{\overline{TR}}{\overline{NR}}=\frac{\sin\gamma}{\sin\delta}\Rightarrow \frac{\sin\epsilon}{\sin\alpha}=\frac{\sin\gamma}{\sin\delta}$$

Now we have: $$\delta+\gamma+\varphi+\beta=\pi=\delta+\gamma+\epsilon+\alpha\Rightarrow\varphi+\beta=\epsilon+\alpha$$ $$\frac{\sin\varphi}{\sin\beta}=\frac{\sin\gamma}{\sin\delta}=\frac{\sin\epsilon}{\sin\alpha}\Rightarrow\frac{\sin\varphi}{\sin\beta}=\frac{\sin\epsilon}{\sin\alpha}$$

Modifying the trigonometric equation gives us: $$\sin\alpha\cdot\sin\varphi=\sin\beta\cdot\sin\epsilon$$ $$\frac{1}{2}(\cos(\alpha-\varphi)+\cos(\alpha+\varphi))=\frac{1}{2}(\cos(\beta-\epsilon)+\cos(\beta+\epsilon))$$

Since $\varphi+\beta=\epsilon+\alpha\Rightarrow\varphi-\alpha=\epsilon-\beta\Rightarrow\cos(\varphi-\alpha)=\cos(\epsilon-\beta)$ we get that $$\cos(\alpha+\varphi)=\cos(\beta+\epsilon)$$ $$\cos(\alpha+\varphi)-\cos(\beta+\epsilon)=-2\sin(\frac{\alpha+\varphi+\beta+\epsilon}{2})\sin(\frac{\alpha+\varphi-\beta-\epsilon}{2})=0$$ Following solutions arise:

• $\alpha+\varphi-\beta-\epsilon = 0 \Rightarrow \alpha+\varphi=\beta+\epsilon$,
• $\alpha+\varphi-\beta-\epsilon = 2\pi$, not possible,
• $\alpha+\varphi+\beta+\epsilon = 0$, not possible,
• $\alpha+\varphi+\beta+\epsilon = 2\pi \Rightarrow \varphi+\beta=\epsilon+\alpha=\pi$, which is not always true.

Hence only the first solution is possible from which follows that $$\alpha=\beta \land \varphi=\epsilon$$

This implies that the points $S,B,T$ must lie on one line and will be on the same circle with the point $A$.

Answer 2

Let $O_1$ be the center of $C_1$, and $O_2$ be the center of $C_2$. Look at the kite-like quadrilateral $AO_1BO_2$. Let the point $O_3$ be the symmetric image of $O_1$ with resepct to the line $AB$. Since the line $O_1O_2$ is the orthogonal bisector of $AB$, point $O_3$ lies on $O_1O_2$ and $AO_1BO_3$ is a rombus because $O_1A=O_1B=O_3A=O_3B$. After a strightforward angle chasing, you can show that $\angle APB = \angle O_3BO_2$ as well as $\angle PBA = \angle BO_2O_3$. Therefore triangles $ABP$ and $O_3BO_2$ are similar and thus $$\frac{BP}{O_2B} = \frac{AP}{O_3B}$$ However, recall that $O_3B = O_1A$ so $$\frac{BP}{O_2B} = \frac{AP}{O_1A}$$ The latter fact, combined with the fact that $\angle \, PAO_1 = 90^{\circ} = \angle \, PBO_2$, becase $PA$ and $PB$ are tangents from point $P$ to the cricles $C_1$ and $C_2$ respectively, implies that the triangles $PAO_1$ and $PBO_2$ are similar, which yields $\angle\, PO_1A = \angle\, PO_2B = \alpha$. Since $PM$ and $PN$ are the other two tangents from $P$ to the circles $C_1$ and $C_2$ respectively, $$\angle PAM = \angle PO_1A = \alpha = \angle PO_2B = \angle PBN$$ Consequently, since $$\angle SAT + \angle SBT = \angle SAT + \angle PBN = \angle SAT + \alpha = 180^{\circ} - \angle PAM + \alpha = 180 - \alpha + \alpha = 180^{\circ}$$ the quadrilateral $ASBT$ is inscribed in a circle.