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Geometry Difficulty 4.8 AIME Prove it Bulgaria

In a circle $k$ is inscribed $ABCD$ , for which $S_{ACB} = s$ , $S_{ACD} = t$ , and $s < t$ . Find the minimum of $A = \frac{4s^2 + t^2}{5st}$ , and determine when it is achieved.

Solution

Since $0 < s < t$ , it suffices to find the minimum of $B(y) = \frac{4}{5}y + \frac{1}{5y}$ , when $0 < y = \frac{s}{t} < 1$ . There are many different approaches for analyzing $B(y)$ in the unit interval that work (e.g., by differentiation). We deduce that the minima of $B(y)$ is $\frac{4}{5}$ , and is achieved for $y = \frac{1}{2}$ . Finally, the minimum of $A = \frac{4s^2 + t^2}{5st}$ is $\frac{4}{5}$ , achieved for $t = 2s$ . Obviously, there exist such geometric configurations (it is enough to choose $B$ and $D$ , such that $2d(B, AC) = d(D, AC)$ ).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.