Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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</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>. | |
− | |||
− | |||
− | |||
− | 2 | + | 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>. |
− | so | + | |
− | which is achieved when A = 90 | + | Therefore we get the equation <math>2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2</math> |
− | < | + | |
− | x = 29 | + | <math>2a^2 + 2x^2 = 42^2 + 40^2</math>. |
− | s= 29 | + | |
− | p+q+s+r = 1 + 41 + 29 + 840 = | + | 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> | ||
+ | |||
+ | 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|]] | ||
+ | |||
+ | |||
+ | 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> , based on triangle inequality, | ||
+ | <cmath> 42 - 40 < 2x < 42 + 40 </cmath> | ||
+ | <cmath>1 < x < 41</cmath> | ||
+ | so <cmath>p = 1, q=41</cmath> | ||
+ | |||
+ | <cmath>2\cdot f(x) = 40 \cdot 42 \cdot \sin(A) \le 2\cdot840</cmath> | ||
+ | so <math>r = 840 </math> | ||
+ | which is achieved when <math>A = 90^\circ</math> , then <math>\angle ACD = 90^\circ</math> | ||
+ | <cmath>(2x)^2 = 40^2 + 42^2 </cmath> | ||
+ | <cmath>x = 29</cmath> | ||
+ | <cmath>s= 29 </cmath> | ||
+ | <cmath>p+q+s+r = 1 + 41 + 29 + 840 = \fbox{\textbf{(C) } 911}</cmath> | ||
~[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 | ||
+ | |||
+ | |||
+ | ==Solution 5 (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 | ||
+ | <math>\frac{40 + 42 + 2a}{2}</math> which is just <math>{a + 41}</math>. 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 | ||
+ | <math>\sqrt{(a^2 - 1)(41^2 - a^2)}</math>. | ||
+ | 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 | ||
+ | <math>a^{2} < </math><math>41^{2}</math>. | ||
+ | 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: | ||
+ | <math>\frac{((a^2 - 1) + (41^2 - a^2))}{2}</math> ≥ <math>\sqrt{(a^2 - 1)(41^2 - a^2)}</math>. 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 | ||
+ | <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>. | ||
+ | |||
+ | ~ilikemath247365 | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=aDajQGay0TQ | ||
==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}} |
Latest revision as of 01:53, 20 December 2024
Contents
[hide]Problem 20
Suppose , , and are points in the plane with and , and let be the length of the line segment from to the midpoint of . Define a function by letting be the area of . Then the domain of is an open interval , and the maximum value of occurs at . What is ?
Solution 1
Let the midpoint of be , and let the length . We know there are limits to the value of , and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length to and , and doesn't contain any information about the median. Therefore we're going to have to write the side in terms of and then use the triangle inequality to find bounds on .
We use Stewart's theorem to relate to the median : . In this case , , , , , .
Therefore we get the equation
.
Notice that since is a pythagorean triple, this means .
By triangle inequality, and
Let's tackle the first inequality:
Here we use the property that .
Therefore in this case, .
For the second inequality,
Therefore we have , so the domain of is .
The area of this triangle is . The maximum value of the area occurs when the triangle is right, i.e. . Then the area is . The length of the median of a right triangle is half the length of it's hypotenuse, which squared is . Thus the length of is .
Our final answer is
~KingRavi
Solution 2 (Geometry)
Let midpoint of as , extends to and ,
triangle has sides , based on triangle inequality, so
so which is achieved when , then
Solution 3 (Trigonometry)
Let A = (0, 0) , B =(b, 0) , C= ()
When :
When : The domain of is the open interval:
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, . So,
We know that, by the Triangle Inequality, . Applying these to Apollonius, we have that the minimum value of is and the maximum value is (both cannot be reached, however).
The rest of the solution follows Solution 1.
~xHypotenuse
Solution 5 (AM-GM Inequality)
By letting BC equal , we can use Heron's formula to calculate the area. Notice the semi-perimeter is just which is just . Next, by Heron's formula, the area of ABC is: which simplifies to the . We now know that the domain of is just the domain of . This domain is very easy to calculate. We see that 1 and . Because is always positive, we see that is in the open interval . Now, we find the maximum of . By the AM-GM inequality, we have: ≥ . Simplifying and letting = , we get that ≤ = . We know by AM-GM that = if and only if 1 = . Solving, = . Therefore, we have found the domain of is the open interval and the maximum of is which occurs at = (Apply Stewart's to triangle ABC when knowing that BC = .) Adding these up, we get = or .
~ilikemath247365
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.