Art of Problem Solving
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

2013 AMC 12A Problems/Problem 12: Difference between revisions

← Older editNewer edit →
mNo edit summary
Aplus95 (talk | contribs)
265 edits
No edit summary
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 17:39, 22 February 2013

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

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
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_12&oldid=51487"