Difference between revisions of "1970 IMO Problems/Problem 1"
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The triangles <math>\triangle CIB_1</math> and <math>\triangle CEB_2</math> are similar, | The triangles <math>\triangle CIB_1</math> and <math>\triangle CEB_2</math> are similar, | ||
so <math>\frac{r}{q} = \frac{CI}{CE}.</math> | so <math>\frac{r}{q} = \frac{CI}{CE}.</math> | ||
+ | |||
+ | Let <math>I_1</math> be the incenter, and <math>E_1</math> be the excenter relative to C | ||
+ | of the triangle <math>\triangle ACM</math>, and <math>I_2</math> be the incenter, and | ||
+ | <math>E_2</math> be the excenter relative to C of the triangle <math>\triangle BCM</math> | ||
+ | (<math>E_2</math> is not shown on the picture). | ||
+ | |||
+ | To solve the problem, we need to prove that | ||
+ | <math>\frac{CI_1}{CE_1} \cdot \frac{CI_2}{CE_2} = \frac{CI}{CE}.</math> | ||
+ | |||
Revision as of 16:04, 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
Let be the incenter, and be the excenter relative to C of the triangle , and be the incenter, and be the excenter relative to C of the triangle ( is not shown on the picture).
To solve the problem, we need to prove that
[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 |