Difference between revisions of "1970 IMO Problems/Problem 1"
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− | But this follows from the fact that the angles <math>AMC</math> and <math> | + | But this follows from the fact that the angles <math>AMC</math> and <math>CMB</math> are supplementary. |
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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> be the points where the incircle and the excircle | ||
+ | relative to <math>C</math> 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> | ||
+ | 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> | ||
+ | |||
+ | Applying the law of sines (see [[Law of Sines]] or | ||
+ | https://en.wikipedia.org/wiki/Law_of_sines) | ||
+ | in triangle <math>\triangle ICA</math> we get | ||
+ | |||
+ | <math>\frac{IC}{\sin \angle CAI} = \frac{AC}{\sin \angle CIA}</math>, or | ||
+ | <math>IC = b\ \frac{\sin \frac{A}{2}}{\sin (\pi - (\frac{A}{2} + \frac{C}{2}))}</math>. | ||
+ | Since <math>C = \pi - A - B</math> this becomes | ||
+ | <math>IC = b\ \frac{\sin \frac{A}{2}}{\cos \frac{B}{2}}.</math> | ||
+ | |||
+ | Similarly, using the law of sines in triangle <math>\triangle ECA</math> and | ||
+ | replacing some angles, we get | ||
+ | <math>EC = b\ \frac{\cos \frac{A}{2}}{\sin \frac{B}{2}}.</math> | ||
+ | |||
+ | It follows that <math>\frac{r}{q} = \tan \frac{A}{2} \tan \frac{B}{2}.</math> | ||
+ | |||
+ | Now, we proceed like in the first solution: Apply the above to | ||
+ | triangles <math>\triangle AMC</math> and <math>\triangle MBC</math>. | ||
+ | |||
+ | We get that | ||
+ | <math>\frac{r_1}{q_1} = \tan \frac{A}{2} \tan \frac{\angle AMC}{2}</math> | ||
+ | and <math>\frac{r_2}{q_2} = \tan \frac{\angle BMC}{2} \tan \frac{B}{2}.</math> | ||
+ | |||
+ | The desired equality follows from | ||
+ | <math>\tan \frac{\angle AMC}{2} \cdot \tan \frac{\angle BMC}{2} = 1</math> | ||
+ | (since <math>\frac{\angle AMC}{2}</math> and <math>\frac{\angle BMC}{2}</math> are | ||
+ | complementary). | ||
+ | |||
+ | [Solution by pf02, November 2024] | ||
Latest revision as of 17:03, 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 be the points where the incircle and the excircle relative to 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
Applying the law of sines (see Law of Sines or https://en.wikipedia.org/wiki/Law_of_sines) in triangle we get
, or . Since this becomes
Similarly, using the law of sines in triangle and replacing some angles, we get
It follows that
Now, we proceed like in the first solution: Apply the above to triangles and .
We get that and
The desired equality follows from (since and are complementary).
[Solution by pf02, November 2024]
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 |