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

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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,  
 
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,  
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<math> x = \frac {5 + \sqrt{25 - 36}}{2} </math>. However, <math> \sqrt{-11} </math> is not real, therefore the second longest side cannot equal 4.
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<math> x = \frac {5 + \sqrt{25 - 36}}{2} </math>. However, <math> \sqrt{-11} </math> is <math> not real </math>, therefore the second longest side <math> cannot </math> equal <math> 4 </math>.
  
 
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> A </math>.
 
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> A </math>.

Revision as of 13:46, 7 February 2013

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 $x^2 = 5^2 + 4^2 - 40cos 60^{\circ}$, 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 $A$.

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