Difference between revisions of "2013 AMC 12A Problems/Problem 12"
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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>. | 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== | ||
+ | https://www.youtube.com/watch?v=XQpQaomC2tA | ||
+ | |||
+ | ~sugar_rush | ||
== See also == | == See also == |
Revision as of 19:12, 24 November 2020
Contents
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 .
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.