INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

March 7, 2020

Radius of a Circle - SMO 2013 - Problem 25

Try this beautiful problem from Geometry based on the radius and tangent of a circle.

SMO 2013 - Geometry (Problem 25)


As shown in the figure below ,circles $C_1 $and$ C_2$ of radius 360 are tangent to each other , and both tangent to the straight line l.if the circle$ C_3$ is tangent to $C_1$ ,$C_2$ and l ,and circle$ C_4 $is tangent to$ C_1$,$C_3$ and l ,find the radius of$ C_4$

radius of a circle

  • 30
  • 35
  • 40

Key Concepts


Geometry

Pythagoras theorm

Distance Formula

Check the Answer


Answer:40

SMO -Math Olympiad-2013

Pre College Mathematics

Try with Hints


Let R be the radius of $C_3$

$C_2E$ =360-R

$C_3E=360$

$C_2C_3$=360+R

Using pythagoras theorm ....

$ (360-R)^2+360^2=(360+R)^2$

i.e R=90

Can you now finish the problem ..........

Let the radius of$ C_4$ be r

then use the distacce formula and tangent property........

can you finish the problem........

Let r be the radius of $C_4$ (small triangle).

LO+OC=360

$\sqrt{(360+p)^2-(360-p)^2}+\sqrt{(90+r)^2-(90-r)^2}=360$

i.e r=40.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com