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2024 AMC 12B Problems/Problem 20: Difference between revisions

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==Solution 6 (AM-GM Inequality)==
==Solution 6 (AM-GM Inequality)==


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


~ilikemath247365
~ilikemath247365
==Solution 7 (Cheese)==


==Video Solution by SpreadTheMathLove==
==Video Solution by SpreadTheMathLove==

Revision as of 21:04, 10 November 2025

Problem

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 $m = \frac{a}2$, $n=\frac{a}2$, $a = m+n$, $d = x$, $b = 42$, $c = 40$.

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 (Median length formula)

Let the midpoint of $BC$ be $D$. Then, by the Median Length Formula: $2*AD^2 = AB^2 - BD^2 + AC^2 - CD^2$. If we let $BC = 2n$ and $AD = x$, then we get the relationship that: $x = \sqrt{1682-n^2}$. By the Triangle Inequality $2 < BC < 82$, so $1 < n < 41$. This means that the domain of x is $(\sqrt{1}, \sqrt{1681}) = (1, 41)$.

The rest follows Solution 1.

~mathwizard123123


Solution 6 (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

Solution 7 (Cheese)

Video Solution by SpreadTheMathLove

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

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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