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Difference between revisions of "2013 AMC 12A Problems/Problem 12"

(This is a rough draft of #12 answer explanation)
 
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Because the angles are in an arithmetic progression, and the angles add up to 180 degrees, the second largest angle in the triangle must be 60 degrees. Also, the side opposite of the 60 degree angle must be the second longest because of the angle-side relationship. Any of the three sides, 4, 5, or x, could be the second longest side of the triangle.
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== Problem==
  
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The law of cosines can be applied to solve for x in all three cases.
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The angles in a particular triangle are in arithmetic progression, and the side lengths are <math>4,5,x</math>. The sum of the possible values of x equals <math>a+\sqrt{b}+\sqrt{c}</math> where <math>a, b</math>, and <math>c</math> are positive integers. What is <math>a+b+c</math>?
  
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When the second longest side is five, we get that 5^2 = 4^2 + x^2 - 2(4)(x)cos 60, therefore x^2 - 4x - 9 = 0. By using the quadratic formula,
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<math> \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44</math>
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x = (4 +- root(16 + 36)) / 2). Because root(52) is greater than 4, it must be positive root(52), so x = 2 + root(13).
 
  
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When the second longest side is x, we get that x^2 = 5^2 + 4^2 - 40*cos 60, and x = root(21).
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==Solution==
  
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When the second longest side is 4, we get that 4^2 = 5^2 + x^2 - 2(5)(x)cos 60, therefore x^2 - 5x + 9 = 0. Using the quadratic formula
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Because the angles are in an arithmetic progression, and the angles add up to <math> 180^{\circ} </math>, the second largest angle in the triangle must be <math> 60^{\circ} </math>. Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, <math> 4 </math>, <math> 5 </math>, or <math> x </math>, could be the second longest side of the triangle.
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x = (5 +- root(25 - 36)) / 2. However, because root(-11) is imaginary, there is no possible way that the longest side can be 4.
 
  
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Adding the two other possibilities gets 2 + root(13) + root(21), with a = 2, b=13, and c=21. a + b + c = 36, answer choice A.
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The law of cosines can be applied to solve for <math> x </math> in all three cases.
 +
 
 +
When the second longest side is <math> 5 </math>, we get that <math> 5^2 = 4^2 + x^2 - 2(4)(x)\cos 60^{\circ} </math>, therefore <math> x^2 - 4x - 9 = 0 </math>. By using the quadratic formula,
 +
<math> x = \frac {4 + \sqrt{16 + 36}}{2} </math>, therefore <math> x = 2 + \sqrt{13} </math>.
 +
 
 +
When the second longest side is <math> x </math>, we get that <math> x^2 = 5^2 + 4^2 - 40\cos 60^{\circ} </math>, therefore <math> x = \sqrt{21} </math>.
 +
 
 +
When the second longest side is <math> 4 </math>, we get that <math> 4^2 = 5^2 + x^2 - 2(5)(x)\cos 60^{\circ} </math>, therefore <math> x^2 - 5x + 9 = 0 </math>. Using the quadratic formula,
 +
<math> x = \frac {5 + \sqrt{25 - 36}}{2} </math>. However, <math> \sqrt{-11} </math> is not real, therefore the second longest side cannot equal <math> 4 </math>.
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 +
Adding the two other possibilities gets <math> 2 + \sqrt{13} + \sqrt{21} </math>, with <math> a = 2, b=13 </math>, and <math> c=21 </math>. <math> a + b + c = 36 </math>, which is answer choice <math>\text{(A)}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=XQpQaomC2tA
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(NOTE: Problem 12 starts at 8:50 in the youtube video  ~jyang66)
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~sugar_rush
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=11|num-a=13}}
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[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:06, 25 December 2020

Problem

The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,5,x$. The sum of the possible values of x equals $a+\sqrt{b}+\sqrt{c}$ where $a, b$, and $c$ are positive integers. What is $a+b+c$?

$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44$

Solution

Because the angles are in an arithmetic progression, and the angles add up to $180^{\circ}$, the second largest angle in the triangle must be $60^{\circ}$. Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, $4$, $5$, or $x$, could be the second longest side of the triangle.

The law of cosines can be applied to solve for $x$ in all three cases.

When the second longest side is $5$, we get that $5^2 = 4^2 + x^2 - 2(4)(x)\cos 60^{\circ}$, therefore $x^2 - 4x - 9 = 0$. By using the quadratic formula, $x = \frac {4 + \sqrt{16 + 36}}{2}$, therefore $x = 2 + \sqrt{13}$.

When the second longest side is $x$, we get that $x^2 = 5^2 + 4^2 - 40\cos 60^{\circ}$, therefore $x = \sqrt{21}$.

When the second longest side is $4$, we get that $4^2 = 5^2 + x^2 - 2(5)(x)\cos 60^{\circ}$, therefore $x^2 - 5x + 9 = 0$. Using the quadratic formula, $x = \frac {5 + \sqrt{25 - 36}}{2}$. However, $\sqrt{-11}$ is not real, therefore the second longest side cannot equal $4$.

Adding the two other possibilities gets $2 + \sqrt{13} + \sqrt{21}$, with $a = 2, b=13$, and $c=21$. $a + b + c = 36$, which is answer choice $\text{(A)}$.

Video Solution

https://www.youtube.com/watch?v=XQpQaomC2tA

(NOTE: Problem 12 starts at 8:50 in the youtube video ~jyang66)

~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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