1970 IMO Problems/Problem 1

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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.


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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