Difference between revisions of "1970 IMO Problems/Problem 1"
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==Solution 3== | ==Solution 3== | ||
+ | Let <math>I</math> be the incenter, and <math>E</math> be the excenter relative to C, | ||
+ | and let <math>B_1, B_2</math> the points where the incircle and the escircle | ||
+ | touch <math>AC</math>. | ||
+ | [[File:Prob_1970_1.png|400px]] | ||
+ | The triangles <math>\triangle CIB_1</math> and <math>\triangle CEB_2</math> are similar, | ||
+ | so <math>\frac{r}{q} = \frac{CI}{CE}.</math> | ||
− | [ | + | |
+ | |||
+ | |||
+ | [FINISHING SOON.] | ||
Revision as of 15:14, 14 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
Let be the incenter, and be the excenter relative to C, and let the points where the incircle and the escircle touch .
The triangles and are similar, so
[FINISHING 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 |