Difference between revisions of "2024 AMC 12B Problems/Problem 20"

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Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>,  
 
Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>,  
  

Revision as of 11:08, 14 November 2024

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

2024 amc 12B P20.PNG


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

triangle $ACD$ has $3$ sides $(40,42,2x)$ as such,\[2<  2x < 82\] \[1\le x \le41\] so \[p = 1, q=41\]

\[2\cdot f(x) =   40 \cdot 42 \cdot \sin(A) \le 2\cdot840\] so $r = 840$ which is achieved when $A = 90^\circ$ , then $\angle ACD = 90^\circ$ \[(2x)^2 = 40^2 + 42^2\] \[x = 29\] \[s= 29\] \[p+q+s+r = 1 + 41 + 29 + 840 = \fbox{\textbf{(C) } 911}\]

~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

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