# Geometry problem about two externally touching circles

Two circles of radius $$~12~$$ and $$~3~$$ touch externally. A line intersecting both of them intersects first circle at points $$P$$ and $$Q$$, second circle - at points $$R$$ and $$S$$. Three resulting line segments, two inside the circles and the one between them, are equal: $$PQ=QR=RS$$. Find their common length.

I have prepared a picture with Geogebra to illustrate

I was trying to solve this with no luck.

After formulating it as a system of equations based on coordinates, with the origin being circles' common point and $$X$$-axis on the line connecting their centers, Wolfram Alpha helped me to find that the answer should be $$\frac{3}{2}\sqrt{13}~$$.

Can you give any hints on how to solve this?

• Lovely problem! Commented Jul 20, 2019 at 4:08

Let $$PQ=QR=RS=2x$$, $$AM$$ and $$O_2N$$ be perpendiculars to $$PS$$

and $$O_2K$$ be a perpendicular to $$AM$$.

Thus, since $$KO_2=MN=x+2x+x=4x,$$ $$AK=\sqrt{12^2-x^2}-\sqrt{3^2-x^2}$$ and $$AO_2=12+3=15,$$ by the Pythagoras's theorem for $$\Delta AO_2K$$ we obtain: $$(4x)^2+\left(\sqrt{12^2-x^2}-\sqrt{3^2-x^2}\right)^2=15^2.$$ Can you end it now?

I got $$PQ=QR=RS=\frac{3\sqrt{13}}{2}.$$

• On which triangle did you apply Pythagoras's Theorem?
– eem
Commented Jul 20, 2019 at 5:28
• @Aditya Dutt I added something. See now. Commented Jul 20, 2019 at 5:31
• $PQ=QR=RS=6$ is not a plausible answer because then in $\triangle O_2NR$ you have $NR=3=\frac{1}{2}RS$ and $O_2R=3$ as well. So hypotenuse is same as base length, which is definitely not possible. I think your error is you have put $AK=9$ which is NOT correct. Commented Jul 20, 2019 at 6:03
• @Anurag A Thank you! I fixed. But the idea is the same. Commented Jul 20, 2019 at 6:16

Let $$PQ=QR=RS=2\ell$$.

Drop a perpendicular from center $$A$$ onto chord $$PQ$$, let that point be $$J$$. Similarly drop a perpendicular from the center $$O_2$$ onto chord $$RS$$ and call it $$K$$.

Consider the right $$\triangle AJP$$, we have $$AJ=\sqrt{144-\ell^2}.$$ Likewise in right $$\triangle O_2KR$$, we have $$O_2K=\sqrt{9-\ell^2}.$$ Now draw the line $$O_2T$$ which is parallel to $$JK$$ (same as saying parallel to the line intersecting the two circles). Then $$O_2KJT$$ forms a rectangle. Now $$AT=AJ-TJ=AJ-O_2K=\sqrt{144-\ell^2}-\sqrt{9-\ell^2}.$$ Consider the right $$\triangle ATO_2$$. Observe that $$AO_2=12+3=15$$ and $$O_2T=JQ+QR+RK=4\ell$$. So $$AT^2+O_2T^2=AO_2^2 \implies \left[\sqrt{144-\ell^2}-\sqrt{9-\ell^2}\right]^2+(4\ell)^2=15^2.$$ So we need to solve for $$\ell$$.

Once you simplify this equation by appropriately squaring etc. you end up getting $$\ell^2(48\ell^2-351)=0.$$ This implies $$\ell^2=\frac{117}{16}$$. Consequently, $$\color{red}{2\ell=\frac{3\sqrt{13}}{2}}.$$