Internal common tangents and angle condition

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Source: China TST 3 2018 Day 1 Q1

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#1 h Mar 27, 2018, 1:19 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $\omega_1,\omega_2$ be two non-intersecting circles, with circumcenters $O_1,O_2$ respectively, and radii $r_1,r_2$ respectively where $r_1 < r_2$ . Let $AB,XY$ be the two internal common tangents of $\omega_1,\omega_2$ , where $A,X$ lie on $\omega_1$ , $B,Y$ lie on $\omega_2$ . The circle with diameter $AB$ meets $\omega_1,\omega_2$ at $P$ and $Q$ respectively. If $\angle AO_1P+\angle BO_2Q=180^{\circ},$ find the value of $\frac{PX}{QY}$ (in terms of $r_1,r_2$ ).

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ABCDE

#2 h Mar 27, 2018, 4:29 PM • 2 Y

Y by Adventure10, Mango247

The answer is $\frac{r_1}{r_2}$ . Let the common external tangents of $\omega_1$ and $\omega_2$ meet at $E$ and the common internal tangents of $\omega_1$ and $\omega_2$ meet at $I$ . Let $A'$ and $B'$ be the antipodes of $A$ and $B$ on $\omega_1$ and $\omega_2$ . Note that $AB'$ and $A'B$ meet at $E$ , and that $P$ lies on $A'B$ and $Q$ lies on $AB'$ because $AA'P$ and $BB'Q$ are right triangles. As $\angle AO_1P+\angle BO_2Q=180^\circ$ , we have that $\angle AA'P+\angle BB'Q=90^\circ$ , so $AA'P$ is similar to $B'BQ$ . Hence, $A'AB$ is similar to $ABB'$ . As $P$ and $Q$ are the feet from $A$ to $A'B$ and $B$ to $AB'$ , $APB$ and $AQB$ are congruent right triangles. As $\left(\frac{AP}{PB}\right)^2=\left(\frac{A'A}{AB}\right)^2=\frac{A'A}{B'B}=\frac{r_1}{r_2}$ , $PQ$ passes through $I$ . Hence, $P$ and $Q$ are images of each other under the homothety at $I$ swapping $\omega_1$ and $\omega_2$ . But this is also true for $X$ and $Y$ , from which it follows that $\frac{PX}{QY}=\frac{r_1}{r_2}$ .

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#3 h Mar 28, 2018, 1:38 PM • 3 Y

Y by msinghal, Adventure10, Mango247

ABCDE wrote:

The answer is $\frac{r_1}{r_2}$ . Let the common external tangents of $\omega_1$ and $\omega_2$ meet at $E$ and the common internal tangents of $\omega_1$ and $\omega_2$ meet at $I$ . Let $A'$ and $B'$ be the antipodes of $A$ and $B$ on $\omega_1$ and $\omega_2$ . Note that $AB'$ and $A'B$ meet at $E$ , and that $P$ lies on $A'B$ and $Q$ lies on $AB'$ because $AA'P$ and $BB'Q$ are right triangles. As $\angle AO_1P+\angle BO_2Q=180^\circ$ , we have that $\angle AA'P+\angle BB'Q=90^\circ$ , so $AA'P$ is similar to $B'BQ$ . Hence, $A'AB$ is similar to $ABB'$ . As $P$ and $Q$ are the feet from $A$ to $A'B$ and $B$ to $AB'$ , $APB$ and $AQB$ are congruent right triangles. As $\left(\frac{AP}{PB}\right)^2=\left(\frac{A'A}{AB}\right)^2=\frac{A'A}{B'B}=\frac{r_1}{r_2}$ , $PQ$ passes through $I$ . Hence, $P$ and $Q$ are images of each other under the homothety at $I$ swapping $\omega_1$ and $\omega_2$ . But this is also true for $X$ and $Y$ , from which it follows that $\frac{PX}{QY}=\frac{r_1}{r_2}$ .

Well, the answer may be $\sqrt{\frac{r_1}{r_2}}$ , according to a Chinese Geometry teacher.

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#4 h Apr 1, 2018, 8:33 PM • 1 Y

Y by Adventure10

I see, let me fix my solution.

Everything is the same until the last two sentences. But $P$ and $Q$ are not images of each other under the homothety as $IP=IQ$ . Instead, they are images under the inversion at $I$ swapping $\omega_1$ and $\omega_2$ . Thus, $PXQY$ is cyclic with $PQ$ and $XY$ meeting at $I$ , so $\frac{PX}{QY}=\frac{IX}{IQ}=\frac{IA}{IQ}=\sqrt{\frac{IA}{IB}}=\sqrt{\frac{r_1}{r_2}}$ .

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#5 h May 3, 2022, 11:00 PM

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Answer is $\frac{PX}{QY} = \sqrt{\frac{r_1}{r_2}}$ Let $\Omega$ be the circle with diameter $AB$ . Then $O_1A,O_2B$ are tangents to $\Omega$ , meaning $\Omega$ is orthogonal to both $\omega_2,\omega_2$ . More precisely, $O_1P$ and $O_2Q$ are also tangents to $\Omega$ . Let $AB \cap XY \cap O_1O_2 = T$ . Look at $\Omega$ . We know $AA \cap PP, BB \cap QQ, AB \cap PQ$ should be collinear, implying $T \in PQ$ .
$[asy] size(250); pair A=dir(130),B=-A,P=dir(185),Q=2*foot(P,A,B)-P,O1=2*A*P/(A+P),O2=2*B*Q/(B+Q),T=extension(A,B,O1,O2),X=2*foot(A,O1,O2)-A,Y=2*foot(B,O1,O2)-B; draw(unitcircle,red); dot("$A$",A,dir(90)); dot("$B$",B,dir(-90)); dot("$P$",P,dir(-130)); dot("$Q$",Q,dir(20)); dot("$O_1$",O1,dir(O1)); dot("$O_2$",O2,dir(O2)); dot("$T$",T,dir(90)); dot("$X$",X,dir(-70)); dot("$Y$",Y,dir(Y)); draw(X--Y^^A--B^^O1--O2,purple); draw(P--Q,green); draw(P--X^^Y--Q,brown); draw(A--O1--P^^Q--O2--B,blue); draw(circle(O1,abs(O1-A))); draw(circle(O2,abs(O2-B))); label("$\omega_1$",2.25*O1-1.25*A); label("$\omega_2$",2.1*O2-1.1*Y); label("$\Omega$",1.15*dir(-90)); [/asy]$ Now the condition $\angle AO_1P + \angle BO_2Q$ would force $AP = AQ$ , and hence $AB$ is the perpendicular bisector of segment $PQ$ . Particularly, $TP = TQ$ . Now,
$\begin{align*} \frac{PX}{QY} &= \frac{r_1 \sin \angle PAX}{r_2 \sin \angle QBY} = \frac{r_1}{r_2} \cdot \frac{\sin \angle TXP}{\sin \angle TYQ} \qquad \text{and} \\ \frac{PX}{QY} &= \frac{TP \cdot \frac{\sin \angle PTX}{\sin \angle TXP}}{TQ \cdot \frac{\sin \angle QTY}{\sin \angle TYQ}} = \frac{\sin \angle TYQ}{\sin \angle TXP} \\ \implies \left( \frac{PX}{QY} \right)^2 &= \frac{PX}{QY} \cdot \frac{PX}{QY} = \left( \frac{r_1}{r_2} \cdot \frac{\sin \angle TXP}{\sin \angle TYQ} \right) \cdot \frac{\sin \angle TYQ}{\sin \angle TXP} = \frac{r_1}{r_2} \\ \therefore ~ \frac{PX}{QY} &= \sqrt{\frac{r_1}{r_2}} \end{align*}$

This post has been edited 1 time. Last edited by guptaamitu1, May 3, 2022, 11:01 PM

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