Difference between revisions of "2013 AMC 12A Problems/Problem 12"
m |
|||
Line 1: | Line 1: | ||
+ | == Problem== | ||
+ | |||
+ | 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>? | ||
+ | |||
+ | <math> \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
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. | 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. | ||
Line 12: | Line 20: | ||
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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=11|num-a=13}} |
Revision as of 18:39, 22 February 2013
Problem
The angles in a particular triangle are in arithmetic progression, and the side lengths are . The sum of the possible values of x equals where , and are positive integers. What is ?
Solution
Because the angles are in an arithmetic progression, and the angles add up to , the second largest angle in the triangle must be . Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, , , or , could be the second longest side of the triangle.
The law of cosines can be applied to solve for in all three cases.
When the second longest side is , we get that , therefore . By using the quadratic formula, , therefore .
When the second longest side is , we get that , therefore .
When the second longest side is , we get that , therefore . Using the quadratic formula, . However, is not real, therefore the second longest side cannot equal .
Adding the two other possibilities gets , with , and . , which is answer choice .
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 |