Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 12B Problems/Problem 20"
(13 intermediate revisions by 3 users not shown)
Line 10: Line 10:
 
</math>
 
</math>
  
−
==Solution==
+
==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>.
Line 25: Line 25:
 
<cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath>
 
<cmath>\implies a = \sqrt{\frac{58^2}{2}-x^2}</cmath>
  
−
==Solution #1 ==
+
By triangle inequality, <math>2a+40>42 \implies a>1 </math> and <math>40+42>2a \implies a<41</math>
 +
 
 +
Let's tackle the first inequality: <cmath>\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1</cmath>
 +
 
 +
<cmath>\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2</cmath>
 +
 
 +
Here we use the property that <math>\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2</math>.
 +
 
 +
Therefore in this case, <math>x<41</math>.
 +
 
 +
For the second inequality, <cmath>\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2</cmath>
 +
 
 +
<cmath>\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2</cmath>
 +
 
 +
<cmath>\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1</cmath>
 +
 
 +
Therefore we have <math>1<x<41</math>, so the domain of <math>f(x)</math> is <math>(1,41)</math>.
 +
 
 +
 
 +
The area of this triangle is <math>\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)</math>. The maximum value of the area occurs when the triangle is right, i.e. <math>\theta = 90^{\circ}</math>. Then the area is <math>\frac{1}{2} \cdot 40 \cdot 42 = 840</math>. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is <math>40^2+42^2 = 58^2</math>. Thus the length of <math>x</math> is <math>29</math>.
 +
 
 +
Our final answer is <math>1+41+840+29 = \boxed{\textbf{911 } (C)}</math>
 +
 
 +
~KingRavi
 +
 
 +
==Solution 2 (Geometry) ==
 
[[Image:2024_amc_12B_P20.PNG|thumb|center|300px|]]
 
[[Image:2024_amc_12B_P20.PNG|thumb|center|300px|]]
  
Line 31: Line 56:
 
Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>,  
 
Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>,  
  
−
triangle <math>ACD</math>  has <math>3</math> sides <math>(40,42,2x)</math>
+
triangle <math>ACD</math>  has <math>3</math> sides <math>(40,42,2x)</math> , based on triangle inequality, 
−
as such,<cmath>2<  2x < 82</cmath>  
+
<cmath> 42 - 40 <  2x < 42 + 40 </cmath>  
−
<cmath>1\le x \le41</cmath>
+
<cmath>1 < x < 41</cmath>
 
so <cmath>p = 1, q=41</cmath>  
 
so <cmath>p = 1, q=41</cmath>  
  
Line 45: Line 70:
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 +
==Solution 3 (Trigonometry) ==
 +
Let A = (0, 0) , B =(b, 0) , C= (<math>c\cos\theta , c\sin\theta </math>)
 +
<cmath>
 +
M = \left(\frac{b + c\cos\theta}{2}, \frac{c\sin\theta}{2}\right).
 +
</cmath>
 +
 +
 +
<cmath>
 +
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}.
 +
</cmath>
 +
 +
When <math> \cos\theta = 1 </math>:
 +
<cmath>
 +
x = \frac{\sqrt{(c+b)^2}}{2} = \frac{c+b}{2} = \frac{42+40}{2} = 41.
 +
</cmath>
 +
 +
When <math> \cos\theta = -1 </math>:
 +
<cmath>
 +
x = \frac{\sqrt{(c-b)^2}}{2} = \frac{c-b}{2} = \frac{42-40}{2} = 1.
 +
</cmath>
 +
The domain of <math> f(x) </math> is the open interval:
 +
<cmath>
 +
\boxed{(1, 41)}.
 +
</cmath>
 +
 +
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, <math>{x}^2= \frac{1}{4}*(2{AB}^2+2{AC}^2-{BC}^2)</math>.
 +
So, <math>x^2=1682-\frac{1}{4}*{BC}^2.</math>
 +
 +
 +
We know that, by the Triangle Inequality, <math>2<BC<82</math>. Applying these to Apollonius, we have that the minimum value of <math>x</math> is <math>x=1</math> and the maximum value is <math>x=41</math> (both cannot be reached, however).
 +
 +
 +
The rest of the solution follows Solution 1.
 +
 +
 +
~xHypotenuse
 +
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2024|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:51, 22 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 2 (Geometry)

2024 amc 12B P20.PNG


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

triangle $ACD$ has $3$ sides $(40,42,2x)$ , 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


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_20&oldid=235274"