Competition problem — Geometry — Maths Olympiad Prep

Library / /351 of 520

Geometry Difficulty 3.4 AMC 10/12 Find the answer

$\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$ .
Point $C$ does not coincide with either $A$ or $B$ . Let $s=AC+BC$ . Then, for all permissible positions of $C$ :
$\textbf{(A)}\ s^2\le8r^2$ $\textbf{(B)}\ s^2=8r^2$ $\textbf{(C)}\ s^2 \ge 8r^2$ $\\ \textbf{(D)}\ s^2\le4r^2$ $\textbf{(E)}\ s^2=4r^2$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

Since $s=AC+BC$ , $s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2$ . Since $\triangle ABC$ is inscribed and $AB$ is the diameter, $\triangle ABC$ is a right triangle, and by the Pythagorean Theorem, $AC^2 + BC^2 = AC^2 = (2r)^2$ . Thus, $s^2 = 4r^2 + 2 \cdot AC \cdot BC$ .

The area of $\triangle ABC$ is $\frac{AC \cdot BC}{2}$ , so $2 \cdot [ABC] = AC \cdot BC$ . That means $s^2 = 4r^2 + 4 \cdot [ABC]$ . The area of $\triangle ABC$ can also be calculated by using base $AB$ and the altitude from $C$ . The maximum possible value of the altitude is $r$ , so the maximum area of $\triangle ABC$ is $r^2$ .

Therefore, $s^2 \le 8r^2$ , so the answer is $\boxed{\textbf{(A)}}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.