Difference between revisions of "1970 IMO Problems/Problem 1"
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<center> | <center> | ||
<math> \begin{matrix} | <math> \begin{matrix} | ||
− | c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\ | + | 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> | + | \ |
+ | & = & q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . | ||
+ | \end{matrix}</math> | ||
</center> | </center> | ||
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<cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath> | <cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath> | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | |||
+ | |||
+ | |||
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+ | [COMING SOON.] | ||
Revision as of 19:06, 13 November 2024
Contents
[hide]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 escribed 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.
Solution 2
By similar triangles and the fact that both centers lie on the angle bisector of , we have , where is the semi-perimeter of . Let have sides , and let . After simple computations, we see that the condition, whose equivalent form is is also equivalent to Stewart's Theorem (see Stewart's_theorem or https://en.wikipedia.org/wiki/Stewart's_theorem)
Solution 3
[COMING SOON.]
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 |