1970 IMO Problems/Problem 1

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 $B_1, B_2$ be the points where the incircle and the excircle relative to $C$ touch $AC$ .

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

Let $I_1$ be the incenter, and $E_1$ be the excenter relative to C of the triangle $\triangle ACM$ , and $I_2$ be the incenter, and $E_2$ be the excenter relative to C of the triangle $\triangle BCM$ ( $E_2$ is not shown on the picture).

To solve the problem, we need to prove that $\frac{CI_1}{CE_1} \cdot \frac{CI_2}{CE_2} = \frac{CI}{CE}.$

Applying the law of sines (see Law of Sines or https://en.wikipedia.org/wiki/Law_of_sines) in triangle $\triangle ICA$ we get

$\frac{IC}{\sin \angle CAI} = \frac{AC}{\sin \angle CIA}$ , or $IC = b\ \frac{\sin \frac{A}{2}}{\sin (\pi - (\frac{A}{2} + \frac{C}{2}))}$ . Since $C = \pi - A - B$ this becomes $IC = b\ \frac{\sin \frac{A}{2}}{\cos \frac{B}{2}}.$

Similarly, using the law of sines in triangle $\triangle ECA$ and replacing some angles, we get $EC = b\ \frac{\cos \frac{A}{2}}{\sin \frac{B}{2}}.$

It follows that $\frac{r}{q} = \tan \frac{A}{2} \tan \frac{B}{2}.$

Now, we proceed like in the first solution: Apply the above to triangles $\triangle AMC$ and $\triangle MBC$ .

We get that $\frac{r_1}{q_1} = \tan \frac{A}{2} \tan \frac{\angle AMC}{2}$ and $\frac{r_2}{q_2} = \tan \frac{\angle BMC}{2} \tan \frac{B}{2}.$

The desired equality follows from $\tan \frac{\angle AMC}{2} \cdot \tan \frac{\angle BMC}{2} = 1$ (since $\frac{\angle AMC}{2}$ and $\frac{\angle BMC}{2}$ 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
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1970 IMO Problems/Problem 1

Contents

Problem

Solution

Solution 2

Solution 3