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 "2013 AMC 12A Problems/Problem 12"
m
Line 1: Line 1:
−
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, 4, 5, or x, could be the second longest side of the triangle.
+
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.
  
−
The law of cosines can be applied to solve for x in all three cases.
+
The law of cosines can be applied to solve for <math> x </math> in all three cases.
  
−
When the second longest side is five, 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,  
+
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>.
 
<math> x = \frac {4 + \sqrt{16 + 36}}{2} </math>, therefore <math> x = 2 + \sqrt{13} </math>.
  
−
When the second longest side is x, we get that <math> x^2 = 5^2 + 4^2 - 40cos 60^{\circ} </math>, therefore <math> x = \sqrt{21} </math>.
+
When the second longest side is <math> x </math>, we get that <math> x^2 = 5^2 + 4^2 - 40cos 60^{\circ} </math>, therefore <math> x = \sqrt{21} </math>.
  
−
When the second longest side is 4, 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,  
 
<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.
 
<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.
  
 
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 14:45, 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 $A$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_12&oldid=50814"