2024 AMC 12B Problems/Problem 20

Revision as of 02:51, 14 November 2024 by Cyantist (talk | contribs) (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


See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png