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Difference between revisions of "1974 IMO Problems/Problem 2"
 
Line 59: Line 59:
 
==Solution 2==
 
==Solution 2==
  
−
First, assume that <math>CD^2 = AD \cdot DB</math>.  We want to prove that
+
Consider the triangle <math>\triangle ABC</math> and <math>D \in AB</math>.  Let
−
<math>\sin{A}\sin{B} \le \sin^2 \frac{C}{2}</math>.
+
<math>C_1 = \angle ACD</math> and <math>C_2 = \angle DCB</math>.
  
−
Let <math>C_1 = \angle ACD</math> and <math>C_2 = \angle DCB</math>.
+
[[File:prob_1974_2_2.png|400px]]
  
−
[[File:prob_1974_2_2.png|800px]]
+
From the law of sines in <math>\triangle CAD</math> we have
 +
<math>\frac{CD}{\sin A} = \frac{AD}{\sin C_1}</math>.
 +
From the law of sines in <math>\triangle CDB</math> we have
 +
<math>\frac{CD}{\sin B} = \frac{BD}{\sin C_2}</math>.
  
 +
It follows <math>CD = AD \cdot \frac{\sin A}{\sin C_1}</math> and
 +
<math>CD = BD \cdot \frac{\sin B}{\sin C_2}</math>.  Multiplying,
 +
we get <math>CD^2 = AD \cdot BD \cdot \frac{\sin A \sin B}{\sin C_1 \sin C_2}</math>.
  
 +
So <math>CD^2 = AD \cdot BD</math> if and only if <math>\sin A \sin B = \sin C_1 \sin C_2</math>.
 +
In other words, <math>CD^2 = AD \cdot BD</math> if and only if <math>\angle C</math> can be split
 +
into <math>\angle C = \angle C_1 + \angle C_2</math> so that
 +
<math>\sin A \sin B = \sin C_1 \sin C_2</math>.
  
 +
First, assume that <math>D</math> exists such that <math>CD^2 = AD \cdot DB</math>.  We want to
 +
prove that <math>\sin{A} \sin{B} \le \sin^2 \frac{C}{2}</math>.
 +
 +
Because of the assumption, this is equivalent to
 +
<math>\sin{C_1} \sin{C_2} \le \sin^2 \frac{C}{2}</math>.  Using well known
 +
trigonometric identities, this reduces to
 +
 +
<math>\frac{\cos({C_1 - C_2}) - \cos({C_1 + C_2})}{2} \le \frac{1 - \cos C}{2}</math>.
 +
 +
This is true since <math>C = C_1 + C_2</math> and <math>\cos(C_1 - C_2) \le 1</math>.
 +
 +
Now assume that <math>\sin{A} \sin{B} \le \sin^2 \frac{C}{2}</math>.
 +
We want to prove that <math>D</math> exists such that <math>CD^2 = AD \cdot DB</math>,
 +
or, in other words, that <math>\angle C</math> can be split into
 +
<math>\angle C = \angle C_1 + \angle C_2</math> so that
 +
<math>\sin A \sin B = \sin C_1 \sin C_2</math>.
 +
 +
Consider the function
 +
<math>f(x) = \sin \left( \frac{C}{2} + x \right) \sin \left( \frac{C}{2} - x \right)</math>
 +
for <math>x \in \left[ 0, \frac{C}{2} \right]</math>.  This can be written as
 +
<math>f(x) = \frac{\cos (2x) - \cos C}{2}</math>.
 +
 +
This is just a segment of a cosine function.  Its values go from
 +
<math>\frac{1 - \cos C}{2} = \sin^2 \frac{C}{2}</math> (at <math>x = 0</math>) to <math>0</math>
 +
at <math>\left( x = \frac{C}{2} \right)</math>.  If <math>\sin{A} \sin{B} \le \sin^2 \frac{C}{2}</math>,
 +
then there must be an <math>x</math> such that
 +
<math>f(x) =
 +
\sin \left( \frac{C}{2} + x \right) \sin \left( \frac{C}{2} - x \right) =
 +
\sin{A} \sin{B}</math>.
 +
 +
This gives the splitting <math>\angle C = \angle C_1 + \angle C_2</math> we wanted.
 +
 +
[Solution by pf02, October 2025]
  
  

Latest revision as of 22:44, 1 October 2025

Problem

In the triangle $ABC$, prove that there is a point $D$ on side $AB$ such that $CD$ is the geometric mean of $AD$ and $DB$ if and only if $\sin{A}\sin{B} \leq \sin^2 \frac{C}{2}$.


Solution

Let a point $D$ on the side $AB$. Let $CF$ the altitude of the triangle $\triangle ABC$, and $C'$ the symmetric point of $C$ through $F$. We bring a parallel line $L$ from $C'$ to $AB$. This line intersects the ray $CD$ at the point $E$, and we know that $DE=DC$.

The distance $d(L,AB)$ between the parallel lines $L$ and $AB$ is $CF$.

Let $w = (O,R)$ the circumscribed circle of $\triangle ABC$, and $MM'$ the perpendicular diameter to $AB$, such that $M,C$ are on difererent sides of the line $AB$.

In fact, the problem asks when the line $L$ intersects the circumcircle. Indeed:

Suppose that $DC$ is the geometric mean of $DA,DB$.

$DA \cdot DB = DC^{2}\Rightarrow DA \cdot DB = DC \cdot DE$

Then, from the power of $D$ we can see that $E$ is also a point of the circle $w$. Or else, the line $L$ intersects $w \Leftrightarrow$

$d(L,AB)\leq d(M,AB) \Leftrightarrow$

$CF \leq MN,$ where $MN$ is the altitude of the isosceles $\triangle MAB$.

$\Leftrightarrow \frac{1}{2}CF \cdot AB \leq \frac{1}{2}MN \cdot AB \Leftrightarrow$ $(ABC) \leq (MAB) \Leftrightarrow$ $\frac{AB \cdot BC \cdot AC}{4R}\leq \frac{AB \cdot MA^{2}}{4R}\Leftrightarrow$

$BC \cdot AC \leq MA^{2}$

We use the formulas:

$BC = 2R \cdot \sin A$, $AC = 2R \cdot \sin B$

and $\angle CMA = \frac{C}{2}\Rightarrow MA = 2R \cdot \sin\frac{C}{2}$

So we have $(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow$ $\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

Prob 1974 2.png

For $(\Leftarrow)$

Suppose that $\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

Then we can go inversely and we find that $d(L,AB)\leq d(M,AB) \Leftrightarrow$ the line $L$ intersects the circle $w$ (without loss of generality; if $d(L,AB)=d(M,AB)$ then $L$ is tangent to $w$ at $M$)

So, if $E \in L \cap w$ then for the point $D = CE \cap AB$ we have $DC=DE$ and $AD \cdot DB = CD \cdot DE \Rightarrow$

$AD \cdot DB = CD^{2}$

The above solution was posted and copyrighted by pontios.


Solution 2

Consider the triangle $\triangle ABC$ and $D \in AB$. Let $C_1 = \angle ACD$ and $C_2 = \angle DCB$.

Prob 1974 2 2.png

From the law of sines in $\triangle CAD$ we have $\frac{CD}{\sin A} = \frac{AD}{\sin C_1}$. From the law of sines in $\triangle CDB$ we have $\frac{CD}{\sin B} = \frac{BD}{\sin C_2}$.

It follows $CD = AD \cdot \frac{\sin A}{\sin C_1}$ and $CD = BD \cdot \frac{\sin B}{\sin C_2}$. Multiplying, we get $CD^2 = AD \cdot BD \cdot \frac{\sin A \sin B}{\sin C_1 \sin C_2}$.

So $CD^2 = AD \cdot BD$ if and only if $\sin A \sin B = \sin C_1 \sin C_2$. In other words, $CD^2 = AD \cdot BD$ if and only if $\angle C$ can be split into $\angle C = \angle C_1 + \angle C_2$ so that $\sin A \sin B = \sin C_1 \sin C_2$.

First, assume that $D$ exists such that $CD^2 = AD \cdot DB$. We want to prove that $\sin{A} \sin{B} \le \sin^2 \frac{C}{2}$.

Because of the assumption, this is equivalent to $\sin{C_1} \sin{C_2} \le \sin^2 \frac{C}{2}$. Using well known trigonometric identities, this reduces to

$\frac{\cos({C_1 - C_2}) - \cos({C_1 + C_2})}{2} \le \frac{1 - \cos C}{2}$.

This is true since $C = C_1 + C_2$ and $\cos(C_1 - C_2) \le 1$.

Now assume that $\sin{A} \sin{B} \le \sin^2 \frac{C}{2}$. We want to prove that $D$ exists such that $CD^2 = AD \cdot DB$, or, in other words, that $\angle C$ can be split into $\angle C = \angle C_1 + \angle C_2$ so that $\sin A \sin B = \sin C_1 \sin C_2$.

Consider the function $f(x) = \sin \left( \frac{C}{2} + x \right) \sin \left( \frac{C}{2} - x \right)$ for $x \in \left[ 0, \frac{C}{2} \right]$. This can be written as $f(x) = \frac{\cos (2x) - \cos C}{2}$.

This is just a segment of a cosine function. Its values go from $\frac{1 - \cos C}{2} = \sin^2 \frac{C}{2}$ (at $x = 0$) to $0$ at $\left( x = \frac{C}{2} \right)$. If $\sin{A} \sin{B} \le \sin^2 \frac{C}{2}$, then there must be an $x$ such that $f(x) = \sin \left( \frac{C}{2} + x \right) \sin \left( \frac{C}{2} - x \right) = \sin{A} \sin{B}$.

This gives the splitting $\angle C = \angle C_1 + \angle C_2$ we wanted.

[Solution by pf02, October 2025]


The original thread for this problem can be found here: [1]


See Also

1974 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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