2024 AMC 12B Problems/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 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 2 (Geometry)

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

triangle $ACD$ has $3$ sides $(40,42,2x)$ , based on triangle inequality, $42 - 40 < 2x < 42 + 40$ $1 < x < 41$ 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

Solution 3 (Trigonometry)

Let A = (0, 0) , B =(b, 0) , C= ( $c\cos\theta , c\sin\theta$ ) $M = \left(\frac{b + c\cos\theta}{2}, \frac{c\sin\theta}{2}\right).$

$AM = x = \sqrt{\left(\frac{b + c\cos\theta}{2}\right)^2 + \left(\frac{c\sin\theta}{2}\right)^2} = \frac{\sqrt{c^2 + 2bc\cos\theta+b^2}}{2}.$

When $\cos\theta = 1$ : $x = \frac{\sqrt{(c+b)^2}}{2} = \frac{c+b}{2} = \frac{42+40}{2} = 41.$

When $\cos\theta = -1$ : $x = \frac{\sqrt{(c-b)^2}}{2} = \frac{c-b}{2} = \frac{42-40}{2} = 1.$ The domain of $f(x)$ is the open interval: $\boxed{(1, 41)}.$

The rest follows Solution 2

Solution 4 (Apollonius)

Here's a faster way to solve this problem using Apollonius's Theorem (which is a special case of Stewart's Theorem for medians). In this case, ${x}^2= \frac{1}{4}*(2{AB}^2+2{AC}^2-{BC}^2)$ . So, $x^2=1682-\frac{1}{4}*{BC}^2.$

We know that, by the Triangle Inequality, $2<BC<82$ . Applying these to Apollonius, we have that the minimum value of $x$ is $x=1$ and the maximum value is $x=41$ (both cannot be reached, however).

The rest of the solution follows Solution 1.

~xHypotenuse

Solution 5 (AM-GM Inequality)

By letting BC equal ${2a}$ , we can use Heron's formula to calculate the area. Notice the semi-perimeter is just $\frac{40 + 42 + 2a}{2}$ which is just ${a + 41}$ . Next, by Heron's formula, the area of ABC is: $\sqrt{(a + 41)(a + 1)(a - 1)(41 - a)}$ which simplifies to the $\sqrt{(a^2 - 1)(41^2 - a^2)}$ . We now know that the domain of ${f(x)}$ is just the domain of $\sqrt{(a^2 - 1)(41^2 - a^2)}$ . This domain is very easy to calculate. We see that $a^{2} >$ 1 and $a^{2} <$ $41^{2}$ . Because ${a}$ is always positive, we see that ${a}$ is in the open interval ${(1, 41)}$ . Now, we find the maximum of ${f(x)}$ . By the AM-GM inequality, we have: $\frac{((a^2 - 1) + (41^2 - a^2))}{2}$ ≥ $\sqrt{(a^2 - 1)(41^2 - a^2)}$ . Simplifying and letting $\sqrt{(a^2 - 1)(41^2 - a^2)}$ = ${f(x)}$ , we get that ${f(x)}$ ≤ $\frac{41^2 - 1}{2}$ = ${840}$ . We know by AM-GM that ${f(x)}$ = ${840}$ if and only if $a^{2} -$ 1 = $41^{2} -$ $a^{2}$ . Solving, ${a}$ = ${29}$ . Therefore, we have found the domain of ${f}$ is the open interval ${(1, 41)}$ and the maximum of ${f}$ is ${840}$ which occurs at ${x}$ = ${29}$ (Apply Stewart's to triangle ABC when knowing that BC = ${58}$ .) Adding these up, we get ${1 + 41 + 840 + 29}$ = ${911}$ or $\boxed{C}$ .

~ilikemath247365

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=aDajQGay0TQ

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

Page

Toolbox

Search