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

Revision as of 15:14, 14 November 2024

Problem

Let $M$ be a point on the side $AB$ of $\triangle ABC$. Let $r_1, r_2$, and $r$ be the inscribed circles of triangles $AMC, BMC$, and $ABC$. Let $q_1, q_2$, and $q$ be the radii of the escribed circles of the same triangles that lie in the angle $ACB$. Prove that

$\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$.


Solution

We use the conventional triangle notations.

Let $I$ be the incenter of $ABC$, and let $I_{c}$ be its excenter to side $c$. We observe that

$r \left[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) \right] = c$,

and likewise,

$\begin{matrix} c & = & q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right] \\ \\ & = & q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}$

Simplifying the quotient of these expressions, we obtain the result

$\frac{r}{q} = \tan (A/2) \tan (B/2)$.

Thus we wish to prove that

$\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)$.

But this follows from the fact that the angles $AMC$ and $CMB$ are supplementary.


Solution 2

By similar triangles and the fact that both centers lie on the angle bisector of $\angle{C}$, we have $\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}$, where $s$ is the semi-perimeter of $ABC$. Let $ABC$ have sides $a, b, c$, and let $AM = c_1, MB = c_2, MC = d$. After simple computations, we see that the condition, whose equivalent form is \[\frac{b + d - c_1}{b + d + c_1} \cdot \frac{a + d - c_2}{a + d + c_2} = \frac{a + b - c}{a + b + c},\] is also equivalent to Stewart's Theorem (see Stewart's_theorem or https://en.wikipedia.org/wiki/Stewart's_theorem)

\[d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.\]


Solution 3

Let $I$ be the incenter, and $E$ be the excenter relative to C, and let $B_1, B_2$ the points where the incircle and the escircle touch $AC$.

Prob 1970 1.png

The triangles $\triangle CIB_1$ and $\triangle CEB_2$ are similar, so $\frac{r}{q} = \frac{CI}{CE}.$



[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
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