Difference between revisions of "1970 IMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | + | Let <math>M</math> be a point on the side <math>AB</math> of <math>\triangle ABC</math>. Let <math>r_1, r_2</math>, and <math>r</math> be the inscribed circles of triangles <math>AMC, BMC</math>, and <math>ABC</math>. Let <math>q_1, q_2</math>, and <math>q</math> be the radii of the exscribed circles of the same triangles that lie in the angle <math>ACB</math>. Prove that | |
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<center> | <center> | ||
− | <math> | + | <math>\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}</math>. |
</center> | </center> | ||
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We use the conventional triangle notations. | We use the conventional triangle notations. | ||
− | Let <math> | + | Let <math>I</math> be the incenter of <math>ABC</math>, and let <math>I_{c}</math> be its excenter to side <math>c</math>. We observe that |
<center> | <center> | ||
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<center> | <center> | ||
<math> \begin{matrix} | <math> \begin{matrix} | ||
− | c & = & | + | c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\ |
− | & = & | + | & = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}</math> |
</center> | </center> | ||
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<center> | <center> | ||
− | <math> | + | <math>\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)</math>. |
</center> | </center> | ||
− | But this follows from the fact that the angles <math> | + | But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary. |
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | {{IMO box|year=1970|before=First question|num-a=2}} |
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 19:25, 25 October 2007
Problem
Let be a point on the side of . Let , and be the inscribed circles of triangles , and . Let , and be the radii of the exscribed circles of the same triangles that lie in the angle . Prove that
.
Solution
We use the conventional triangle notations.
Let be the incenter of , and let be its excenter to side . We observe that
,
and likewise,
Simplifying the quotient of these expressions, we obtain the result
.
Thus we wish to prove that
.
But this follows from the fact that the angles and are supplementary.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1970 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |