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Difference between revisions of "1970 IMO Problems/Problem 1"
 
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</center>
 
</center>
  
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But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary.
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But this follows from the fact that the angles <math>AMC</math> and <math>CMB</math> are supplementary.
  
  
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==Solution 3==
 
==Solution 3==
  
 +
Let <math>I</math> be the incenter, and <math>E</math> be the excenter relative to C,
 +
and let <math>B_1, B_2</math> be the points where the incircle and the excircle
 +
relative to <math>C</math> touch <math>AC</math>.
  
 +
[[File:Prob_1970_1.png|400px]]
  
 +
The triangles <math>\triangle CIB_1</math> and <math>\triangle CEB_2</math> are similar,
 +
so <math>\frac{r}{q} = \frac{CI}{CE}.</math>
  
 +
Let <math>I_1</math> be the incenter, and <math>E_1</math> be the excenter relative to C
 +
of the triangle <math>\triangle ACM</math>, and <math>I_2</math> be the incenter, and
 +
<math>E_2</math> be the excenter relative to C of the triangle <math>\triangle BCM</math>
 +
(<math>E_2</math> is not shown on the picture).
  
−
[COMING SOON.]
+
To solve the problem, we need to prove that
 +
<math>\frac{CI_1}{CE_1} \cdot \frac{CI_2}{CE_2} = \frac{CI}{CE}.</math>
 +
 
 +
Applying the law of sines (see [[Law of Sines]] or
 +
https://en.wikipedia.org/wiki/Law_of_sines)
 +
in triangle <math>\triangle ICA</math> we get
 +
 
 +
<math>\frac{IC}{\sin \angle CAI} = \frac{AC}{\sin \angle CIA}</math>, or
 +
<math>IC = b\ \frac{\sin \frac{A}{2}}{\sin (\pi - (\frac{A}{2} + \frac{C}{2}))}</math>.
 +
Since <math>C = \pi - A - B</math> this becomes
 +
<math>IC = b\ \frac{\sin \frac{A}{2}}{\cos \frac{B}{2}}.</math>
 +
 
 +
Similarly, using the law of sines in triangle <math>\triangle ECA</math> and
 +
replacing some angles, we get
 +
<math>EC = b\ \frac{\cos \frac{A}{2}}{\sin \frac{B}{2}}.</math>
 +
 
 +
It follows that <math>\frac{r}{q} = \tan \frac{A}{2} \tan \frac{B}{2}.</math>
 +
 
 +
Now, we proceed like in the first solution:  Apply the above to
 +
triangles <math>\triangle AMC</math> and <math>\triangle MBC</math>.
 +
 
 +
We get that
 +
<math>\frac{r_1}{q_1} = \tan \frac{A}{2} \tan \frac{\angle AMC}{2}</math>
 +
and <math>\frac{r_2}{q_2} = \tan \frac{\angle BMC}{2} \tan \frac{B}{2}.</math>
 +
 
 +
The desired equality follows from
 +
<math>\tan \frac{\angle AMC}{2} \cdot \tan \frac{\angle BMC}{2} = 1</math>
 +
(since <math>\frac{\angle AMC}{2}</math> and <math>\frac{\angle BMC}{2}</math> are
 +
complementary).
 +
 
 +
[Solution by pf02, November 2024]
  
  

Latest revision as of 17:03, 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 $B_1, B_2$ be the points where the incircle and the excircle relative to $C$ touch $AC$.

Prob 1970 1.png

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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