Geometry problem 3 by Dhaval Furia

Geometry Level pending

Two circles, each of radius $4$ cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is $_____$

\sqrt{2}

\frac{1}{\sqrt{2}}

\frac{\pi}{3}

1

2 solutions

Shikhar Srivastava
May 14, 2020

Let the radius of smaller circle be $r$ . Then $AC = 4 +r$ and $BC = 4 - r$ . Also $\triangle ABC$ is right-angled triangle. So applying pythagorean theorem in $\triangle ABC$

$\hspace{10pt} AB^2 + BC^2 = AC^2\newline\Rightarrow 4^2 + (4 - r)^2 = (4 + r)^2\newline\Rightarrow 16 + 4^2 - 8r + r^2 = 4^2 + 8r + r^2 \newline\Rightarrow r = 1$

Thanks for posting the drawing. I've misunderstood the problem. Thank you again.

Robert Bommarito - 1 year ago

A Former Brilliant Member
May 14, 2020

Using Descartes's theorem, we get

$\dfrac{1}{\sqrt a}=\dfrac{1}{\sqrt 4}+\dfrac{1}{\sqrt 4}=1\implies a=\boxed 1$ , where $a$ is the radius asked for.

Geometry problem 3 by Dhaval Furia

2 solutions

1 pending report

Vote up reports you agree with