Difference between revisions of "1970 IMO Problems/Problem 1"

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* [[1970 IMO Problems]]
* [[1970 IMO Problems]]
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366686#p366686]Discussion on AoPS/Mathlinks]
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366686#p366686 Discussion on AoPS/Mathlinks]
[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Geometry Problems]]

Revision as of 18:21, 18 September 2006


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

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


We use the conventional triangle notations.

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

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

and likewise,

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

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

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

But this follows from the fact that the angles $\displaystyle AMC$ and $\displaystyle CBM$ are supplementary.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


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