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Geometry Difficulty 4.3 AIME Find the answer United States

Suppose $A$ , $B$ , and $C$ are points in the plane with $AB = 40$ and $AC = 42$ , and let $x$ be the length of the line segment from $A$ to the midpoint of $BC$ . Define a function $f$ by letting $f(x)$ be the area of $\triangle ABC$ . Then the domain of $f$ is an open interval $(p, q)$ , and the maximum value of $f(x)$ occurs at $x = s$ . What is $p + q + r + s$ ?

Pick one

Solution

By the Triangle Inequality, $BC$ is between $40 + 42 = 82$ and $42 - 40 = 2$ . The corresponding bounding values of $x$ are $p = 42 - \frac{40+42}{2} = 1$ and $q = 42 - \frac{42-40}{2} = 41$ .

The area of $\triangle ABC$ is maximized when $\angle BAC$ is a right angle, in which case the area is $r = \frac{1}{2} \cdot 40 \cdot 42 = 840$ , and the median to the hypotenuse has half the length of the hypotenuse, namely
$s = \frac{1}{2}\sqrt{40^2 + 42^2} = \frac{1}{2}\sqrt{3364} = \frac{1}{2} \cdot 58 = 29.$
(This right triangle is the double of the $20$ - $21$ - $29$ right triangle.) The requested sum is $1 + 41 + 840 + 29 = 911$ .

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