Consider sector of a circle $OAB$.
Circle with center $ M $ touches $OA$ at $P$, $OB$ a $Q$ and arc $AB$ at $N$.
Circle with center at $L$ touches $OA$ at $C$, $OB$ at $D$ and circle with center $M$ at $K$.
$OA = 98$. $MP = 21$.
Find $LC$.


I can't even begin. I think $ O , L , M $are colinear and then $\triangle OCL$ and $\triangle OPM$ are similar.I don't know what to do. Please give some starting hint.


enter image description here From the figure, by similar triangles, (x+r) : r = (x + 2r + R) : R

Then, $x = \frac {2r^2}{(R - r)}$

Therefore, $2R + 2r + \frac {2r^2} {(R - r)} = 98$ with R = 21

Solve the corresponding equation to find r. And r should be 12.

  • $\begingroup$ I got the answer as 12. Is it right? Also thanks a lot. $\endgroup$ – A Googler Apr 2 '14 at 17:22
  • $\begingroup$ Yes, as shown in my edited version. $\endgroup$ – Mick Apr 2 '14 at 17:25
  • $\begingroup$ Same way as mine. $\endgroup$ – Ajay Apr 3 '14 at 1:44
  • $\begingroup$ Q55 Mailed to you. $\endgroup$ – Ajay Apr 3 '14 at 1:45
  • $\begingroup$ @Ajay Thanks a lot! $\endgroup$ – A Googler Apr 3 '14 at 8:50

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