1974 IMO Problems/Problem 2

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

Since this is an "if and only if" statement, we will prove it in two parts.

Before we begin, note a few basic but important facts.

1. When two variables $x$ and $y$ are restricted by an equation $x+y=k$ for some constant $k$ , the maximum of their product occurs when $x=y=\frac{k}{2}$ .

2. The triangle inequality states that for a triangle with sides $a$ , $b$ , and $c$ fulfills $a + b > c$ , meaning that $a > \frac{c}{2}$ or $b > \frac{c}{2}$ , which is equivalent to saying that $a$ and $b$ both cannot be less than or equal to $\frac{c}{2}$ .

3. As point D moves along the base of the side AB, the locus of points where C can exist fills in a semicircle because C must be a distance $\sqrt{(AD)(DB)} = \sqrt{(\frac{AB}{2}-k)(\frac{AB}{2}+k)} = \sqrt{(\frac{AB}{2})^2-k^2}$

Part 1: If $\sin{A}\sin{B} \leq \sin^2{\frac{C}{2}}$ , then such a point D exists.

First note that in order for D to exist, the line segment CD must be able to reach from the vertex C to the side AB, meaning that

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1974 IMO Problems/Problem 2

Solution