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Difference between revisions of "1962 IMO Problems/Problem 6"

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==Solution==
 
==Solution==
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<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra>
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Instead of an isosceles triangle, let us consider an arbitrary triangle <math>ABC</math>. Let <math>ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>. Extend <math>AI</math> to meet the circumcircle again at <math>L</math>. Then extend <math>LO</math> so it meets the circumcircle again at <math>M</math>.
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Consider the point where the incircle meets <math>AB</math>, and let this be point <math>D</math>. We have <math>\angle ADI = \angle MBL = 90^{\circ}, \angle IAD = \angle LMB</math>; thus, <math>\triangle ADI \sim \triangle MBL</math>, or <math>\frac {ID}{BL} = \frac {AI} {ML} \iff ID \cdot ML = 2r\rho = AI \cdot BL</math>.
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Now, drawing line <math>BI</math>, we see that <math>\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</math>. Therefore, <math>BIL</math> is isosceles, and <math>IL = BL</math>.
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Substituting this back in, we have <math>2r\rho = AI\cdot IL</math>. Extending <math>OI</math> to meet the circumcircle at <math>P,Q</math>, we see that <math>AI\cdot IL = PI\cdot QI</math> by Power of a Point. Therefore, <math>2r\rho = PI \cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)</math>, and we have <math>2r\rho = r^2 - d^2 \iff d = \sqrt {r(r - 2\rho)}</math>, and we are done.
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1962|num-b=5|num-a=7}}
 
{{IMO box|year=1962|num-b=5|num-a=7}}

Latest revision as of 05:29, 16 December 2023

Problem

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

$d=\sqrt{r(r-2\rho)}$.

Solution

<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
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