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

Revision as of 12:08, 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

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

2024 AMC 12B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 ==Solution #1 ==
+[[Image:2024_amc_12B_P20.PNG|thumb|center|300px|]]
 Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>,

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

Revision as of 12:08, 14 November 2024

Problem 20

Solution #1

See also