Question

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 $\overline{BC}.$ 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 $r$ of $f(x)$ occurs at $x = s.$ What is $p + q + r + s?$

$909$

$910$

$911$

$912$

$913$

Solution:

Let $a = BC.$ The median length gives $x^2 = \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4} = \dfrac{6728 - a^2}{4}.$ The triangle inequality requires $2 \lt a \lt 82,$ i.e. $4 \lt a^2 \lt 6724,$ which translates to $1 \lt x \lt 41.$ So $(p, q) = (1, 41).$

With $AB = 40$ and $AC = 42$ fixed, the area $\tfrac12\cdot 40\cdot 42\sin A$ is largest when $\angle A = 90^\circ,$ giving $r = 840.$ Then $a^2 = 40^2 + 42^2 = 3364,$ so $x^2 = \dfrac{6728 - 3364}{4} = 841,$ i.e. $s = 29.$

Thus $p + q + r + s = 1 + 41 + 840 + 29 = 911.$

Thus, the correct answer is C.

2024 AMC 12B Problem 20

Solution:

Problem 20 in Other Years