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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>CBM</math> are supplementary.
 
But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary.
  
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==Solution 2==
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By similar triangles and the fact that both centers lie on the angle bisector of <math>\angle{C}</math>, we have <math>\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}</math>, where <math>s</math> is the semi-perimeter of <math>ABC</math>. Let <math>ABC</math> have sides <math>a, b, c</math>, and let <math>AM = c_1, MB = c_2, MC = d</math>. After simple computations, we see that the condition, whose equivalent form is
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<cmath>\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},</cmath>
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is also equivalent to Stewart's Inequality
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<cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath>
 
{{alternate solutions}}
 
{{alternate solutions}}
  

Revision as of 21:12, 19 May 2015

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 exscribed 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 $CBM$ 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 Inequality \[d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.\] 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
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First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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