Difference between revisions of "1962 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | {{ | + | <geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> |
+ | 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>. | ||
+ | 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 = 2rp = AI \cdot BL</math>. | ||
+ | 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>. | ||
+ | Substituting this back in, we have <math>2rp = 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>2rp = PI \cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)</math>, and we have <math>2rp = r^2 - d^2 \iff d = \sqrt {r(r - 2p)}</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 17:34, 16 February 2009
Problem
Consider an isosceles triangle. Let be the radius of its circumscribed circle and the radius of its inscribed circle. Prove that the distance between the centers of these two circles is
Solution
<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> Instead of an isosceles triangle, let us consider an arbitrary triangle . Let have circumcenter and incenter . Extend to meet the circumcircle again at . Then extend so it meets the circumcircle again at . Consider the point where the incircle meets , and let this be point . We have ; thus, , or . Now, drawing line , we see that . Therefore, is isosceles, and . Substituting this back in, we have . Extending to meet the circumcircle at , we see that by Power of a Point. Therefore, , and we have , 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 |