Let $ABC$ be a triangle. Prove that there exists a point $D$ on the side $AB$ of the triangle $ABC$, such that $CD$ is the geometric mean of $AD$ and $DB$, iff the triangle $ABC$ satisfies the inequality $\sin A\sin B\le\sin^2\frac{C}{2}$.
HIDE: Comment Alternative formulation, from IMO ShortList 1974, Finland 2: We consider a triangle $ABC$. Prove that: $\sin(A) \sin(B) \leq \sin^2 \left( \frac{C}{2} \right)$ is a necessary and sufficient condition for the existence of a point $D$ on the segment $AB$ so that $CD$ is the geometrical mean of $AD$ and $BD$.Problem
Source: IMO ShortList 1974, Finland 1, IMO 1973, Day 1, Problem 2
Tags: trigonometry, circumcircle, geometry, Triangle, Trigonometric inequality, IMO, IMO 1974
pontios
09.11.2005 20:35
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 $AD \cdot AB = CD \cdot DE \Rightarrow$ $AD \cdot AB = CD^{2}$
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