2024 AMC 12B Problems/Problem 20

Problem 20

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 $p+q+r+s$ ?

$\textbf{(A) }909\qquad \textbf{(B) }910\qquad \textbf{(C) }911\qquad \textbf{(D) }912\qquad \textbf{(E) }913\qquad$

Solution #1

Let midpoint of BC as M, extends AM to D and MD = x,

triangle ACD has 3 sides (40,42,2x) as such, 2< 2x < 82 1<= x <=41 so p = 1, q=41

2*Area(ABC) = 40 * 42 * sin(A) <= 2*840 so r = max(Area{ABC)) = 840 which is achieved when A = 90 degree , then angle ACD = 90 degree, $(2x)^2 = 40^2 + 42^2$ x = 29 s= 29 p+q+s+r = 1 + 41 + 29 + 840 = $\fbox{\textbf{(C) } 911}$

~luckuso

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2024 AMC 12B Problems/Problem 20

Problem 20

Solution #1

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