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

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 $AC$ .

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
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 1: / Line 1: @@
 == Problem ==
-( ''Proposed by Poland'' ) Let <math>\displaystyle M</math> be a point on the side <math>\displaystyle AB</math> of <math>\displaystyle \triangle ABC</math>.  Let <math>\displaystyle r_1, r_2</math>, and <math>\displaystyle r</math> be the inscribed circles of triangles <math>\displaystyle AMC, BMC</math>, and <math>\displaystyle ABC</math>.  Let <math>\displaystyle q_1, q_2</math>, and <math>\displaystyle q</math> be the radii of the exscribed circles of the same triangles that lie in the angle <math>\displaystyle ACB</math>.  Prove that
+Let <math>M</math> be a point on the side <math>AB</math> of <math>\triangle ABC</math>.  Let <math>r_1, r_2</math>, and <math>r</math> be the inscribed circles of triangles <math>AMC, BMC</math>, and <math>ABC</math>.  Let <math>q_1, q_2</math>, and <math>q</math> be the radii of the escribed circles of the same triangles that lie in the angle <math>ACB</math>.  Prove that
 <center>
-<math>\displaystyle \frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}</math>.
+<math>\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}</math>.
 </center>
 == Solution ==
@@ Line 11: / Line 12: @@
 We use the conventional triangle notations.
-Let <math>\displaystyle I</math> be the incenter of <math>\displaystyle ABC</math>, and let <math>\displaystyle I_{c}</math> be its excenter to side <math>\displaystyle c</math>.  We observe that
+Let <math>I</math> be the incenter of <math>ABC</math>, and let <math>I_{c}</math> be its excenter to side <math>c</math>.  We observe that
 <center>
@@ Line 21: / Line 22: @@
 <center>
 <math> \begin{matrix}
-c & = & \displaystyle q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\
+c & = & 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}</math>
+\\
+& = & q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; .
+\end{matrix}</math>
 </center>
@@ Line 34: / Line 37: @@
 <center>
-<math>\displaystyle \tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)</math>.
+<math>\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)</math>.
 </center>
-But this follows from the fact that the angles <math>\displaystyle AMC</math> and <math>\displaystyle CBM</math> are supplementary.
+But this follows from the fact that the angles <math>AMC</math> and <math>CMB</math> are supplementary.
+==Solution 2==
+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
+<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>
+is also equivalent to Stewart's Theorem (see [[Stewart's_theorem]] or https://en.wikipedia.org/wiki/Stewart's_theorem)
+<cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath>
+==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).
+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>.
-{{alternate solutions}}
+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]
-== Resources ==
-* [[1970 IMO Problems]]
+{{alternate solutions}}
-* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366686#p366686]Discussion on AoPS/Mathlinks]
+{{IMO box|year=1970|before=First question|num-a=2}}
 [[Category:Olympiad Geometry Problems]]

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