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Difference between revisions of "1974 IMO Problems/Problem 2"

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In the triangle ABC; prove that there is a point D on side AB such that CD
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==Problem==
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is the geometric mean of AD and DB if and only if  
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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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<math>\sin{A}\sin{B} \leq  \sin^2 (\frac{C}{2})</math>.
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==Solution==
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Let a point <math>D</math> on the side <math>AB</math>.
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Let <math>CF</math> the altitude of the triangle <math>\triangle ABC</math>, and <math>C'</math> the symmetric point of <math>C</math> through <math>F</math>.
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We bring a parallel line <math>L</math> from <math>C'</math> to <math>AB</math>. This line intersects the ray <math>CD</math> at the point <math>E</math>, and we know that <math>DE=DC</math>.
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The distance <math>d(L,AB)</math> between the parallel lines <math>L</math> and <math>AB</math> is <math>CF</math>.
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Let <math>w = (O,R)</math> the circumscribed circle of <math>\triangle ABC</math>, and <math>MM'</math> the perpendicular diameter to <math>AB</math>, such that <math>M,C</math> are on difererent sides of the line <math>AB</math>.
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In fact, the problem asks when the line <math>L</math> intersects the circumcircle. Indeed:
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Suppose that <math>DC</math> is the geometric mean of <math>DA,DB</math>.
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<math>DA \cdot DB = DC^{2}\Rightarrow DA \cdot DB = DC \cdot DE</math>
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Then, from the power of <math>D</math> we can see that <math>E</math> is also a point of the circle <math>w</math>.
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Or else, the line <math>L</math> intersects <math>w \Leftrightarrow</math>
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<math>d(L,AB)\leq d(M,AB) \Leftrightarrow</math>
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<math>CF \leq MN,</math> where <math>MN</math> is the altitude of the isosceles <math>\triangle MAB</math>.
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<math>\Leftrightarrow \frac{1}{2}CF \cdot AB \leq \frac{1}{2}MN \cdot AB \Leftrightarrow</math> <math>(ABC) \leq (MAB) \Leftrightarrow</math> <math>\frac{AB \cdot BC \cdot AC}{4R}\leq \frac{AB \cdot MA^{2}}{4R}\Leftrightarrow</math>
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<math>BC \cdot AC \leq MA^{2}</math>
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We use the formulas:
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<math>BC = 2R \cdot \sin A</math>
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<math>AC = 2R \cdot \sin B</math>
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and <math>\angle CMA = \frac{C}{2}\Rightarrow MA = 2R \cdot \sin\frac{C}{2}</math>
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So we have
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<math>(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow</math>
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<math>\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}</math>
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For <math>(\Leftarrow)</math>
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Suppose that <math>\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}</math>
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Then we can go inversely and we find that <math>d(L,AB)\leq d(M,AB) \Leftrightarrow</math>
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the line <math>L</math> intersects the circle <math>w</math> (without loss of generality; if <math>d(L,AB)=d(M,AB)</math> then <math>L</math> is tangent to <math>w</math> at <math>M</math>)
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So, if <math>E \in L \cap w</math> then for the point <math>D = CE \cap AB</math> we have <math>DC=DE</math> and <math>AD \cdot AB = CD \cdot DE \Rightarrow</math>
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<math>AD \cdot AB = CD^{2}</math>
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The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [https://aops.com/community/p365059]

Revision as of 14:56, 29 January 2021

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



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 $AD \cdot AB = CD \cdot DE \Rightarrow$

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

The above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]

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