1962 IMO Problems/Problem 6

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Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle and $\rho$ the radius of its inscribed circle. Prove that the distance $d$ between the centers of these two circles is



<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> Instead of an isosceles triangle, let us consider an arbitrary triangle $ABC$. Let $ABC$ have circumcenter $O$ and incenter $I$. Extend $AI$ to meet the circumcircle again at $L$. Then extend $LO$ so it meets the circumcircle again at $M$. Consider the point where the incircle meets $AB$, and let this be point $D$. We have $\angle ADI = \angle MBL = 90^{\circ}, \angle IAD = \angle LMB$; thus, $\triangle ADI \sim \triangle MBL$, or $\frac {ID}{BL} = \frac {AI} {ML} \iff ID \cdot ML = 2r\rho = AI \cdot BL$. Now, drawing line $BI$, we see that $\angle BIL = \frac {1}{2}\angle A + \frac {1}{2}\angle ABC, \angle IBL = \frac {1}{2}\angle ABC + \angle CBL = \frac {1}{2}\angle ABC + \frac {1}{2}\angle A$. Therefore, $BIL$ is isosceles, and $IL = BL$. Substituting this back in, we have $2r\rho = AI\cdot IL$. Extending $OI$ to meet the circumcircle at $P,Q$, we see that $AI\cdot IL = PI\cdot QI$ by Power of a Point. Therefore, $2r\rho = PI \cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)$, and we have $2r\rho = r^2 - d^2 \iff d = \sqrt {r(r - 2\rho)}$, and we are done.

See Also

1962 IMO (Problems) • Resources
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Problem 7
All IMO Problems and Solutions