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

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==Solution==
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==Solution 1==
  
 
Let the midpoint of <math>BC</math> be <math>M</math>, and let the length <math>BM = CM = a</math>. We know there are limits to the value of <math>x</math>, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length <math>BC</math> to <math>AC</math> and <math>AB</math>, and doesn't contain any information about the median. Therefore we're going to have to write the side <math>BC</math> in terms of <math>x</math> and then use the triangle inequality to find bounds on <math>x</math>.
 
Let the midpoint of <math>BC</math> be <math>M</math>, and let the length <math>BM = CM = a</math>. We know there are limits to the value of <math>x</math>, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length <math>BC</math> to <math>AC</math> and <math>AB</math>, and doesn't contain any information about the median. Therefore we're going to have to write the side <math>BC</math> in terms of <math>x</math> and then use the triangle inequality to find bounds on <math>x</math>.
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Our final answer is <math>1+41+840+29 = \boxed{\textbf{911 } (C)}</math>
 
Our final answer is <math>1+41+840+29 = \boxed{\textbf{911 } (C)}</math>
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~KingRavi
  
 
==Solution #1 ==
 
==Solution #1 ==

Revision as of 11:31, 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 the midpoint of $BC$ be $M$, and let the length $BM = CM = a$. We know there are limits to the value of $x$, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length $BC$ to $AC$ and $AB$, and doesn't contain any information about the median. Therefore we're going to have to write the side $BC$ in terms of $x$ and then use the triangle inequality to find bounds on $x$.

We use Stewart's theorem to relate $BC$ to the median $AM$: $man + dad = bmb + cnc$. In this case $m = a$, $n=a$, $a = m+n$, $d = x$, $b = 42$, $c = 40$.

Therefore we get the equation $2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2$

$2a^2 + 2x^2 = 42^2 + 40^2$.

Notice that since $20-21-29$ is a pythagorean triple, this means $2a^2 + 2x^2 = 58^2$.

\[\implies a^2 = \frac{58^2}{2}-x^2\] \[\implies a = \sqrt{\frac{58^2}{2}-x^2}\]

By triangle inequality, $2a+40>42 \implies a>1$ and $40+42>2a \implies a<41$

Let's tackle the first inequality: \[\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1\]

\[\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2\]

Here we use the property that $\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2$.

Therefore in this case, $x<41$.

For the second inequality, \[\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2\]

\[\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2\]

\[\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1\]

Therefore we have $1<x<41$, so the domain of $f(x)$ is $(1,41)$.


The area of this triangle is $\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)$. The maximum value of the area occurs when the triangle is right, i.e. $\theta = 90^{\circ}$. Then the area is $\frac{1}{2} \cdot 40 \cdot 42 = 840$. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is $40^2+42^2 = 58^2$. Thus the length of $x$ is $29$.

Our final answer is $1+41+840+29 = \boxed{\textbf{911 } (C)}$

~KingRavi

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 (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

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