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

Revision as of 11:19, 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

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 $2a^2 + 2x^2 = 58^2$ .

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

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

@@ Line 9: / Line 9: @@
 \textbf{(E) }913\qquad
 </math>
+==Solution==
+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>.
+We use Stewart's theorem to relate <math>BC</math> to the median <math>AM</math>: <math>man + dad = bmb + cnc</math>. In this case <math>m = a</math>, <math>n=a</math>, <math>a = m+n</math>, <math>d = x</math>, <math>b = 42</math>, <math>c = 40</math>.
+Therefore we get the equation <math>2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2</math>
+<math>2a^2 + 2x^2 = 42^2 + 40^2</math>.
+Notice that since <math>20-21-29</math> is a pythagorean triple, this means <math>2a^2 + 2x^2 = 58^2</math>.
+<cmath>\implies a^2 = \frac{58^2}{2}-x^2</cmath>
+<cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath>
 ==Solution #1 ==

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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Difference between revisions of "2024 AMC 12B Problems/Problem 20"

Revision as of 11:19, 14 November 2024

Contents

Problem 20

Solution

Solution #1

See also