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1974 IMO Problems/Problem 2: Difference between revisions

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==Problem==
==Problem==


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




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The above solution was posted and copyrighted by pontios.
The above solution was posted and copyrighted by pontios.
==Solution 2==
First, assume 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>.
Let <math>C_1 = \angle ACD</math> and <math>C_2 = \angle DCB</math>.
[[File:prob_1974_2_2.png|800px]]




Revision as of 18:59, 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 $(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow$ $\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

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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

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

Let $C_1 = \angle ACD$ and $C_2 = \angle DCB$.



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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