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 "2014 AMC 12B Problems/Problem 24"

m
(→‎Solution)
Line 20: Line 20:
  
 
According to the first equation, <math> e=\frac{d^2-9}{10} </math>. Plugging this into the third equation results in <math> d^3-109d-420=0 </math>. The only positive root of this cubic is <math> d=12 </math>. Substituting into the first and second equations gives <math> e=\frac{27}{2} </math> and <math> f=\frac{44}{3} </math> and thus the sum of all diagonals is <math> 3d+e+f=\frac{385}{6} </math>. Our answer is therefore <math> 385+6=\boxed{391} </math>.
 
According to the first equation, <math> e=\frac{d^2-9}{10} </math>. Plugging this into the third equation results in <math> d^3-109d-420=0 </math>. The only positive root of this cubic is <math> d=12 </math>. Substituting into the first and second equations gives <math> e=\frac{27}{2} </math> and <math> f=\frac{44}{3} </math> and thus the sum of all diagonals is <math> 3d+e+f=\frac{385}{6} </math>. Our answer is therefore <math> 385+6=\boxed{391} </math>.
 +
 +
*This solution requires solving a cubic - however, I thought that it was in the rules that the AMC 12 cannot ask for one to solve a cubic - perhaps I am mistaken?
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:20, 27 December 2014

Problem

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$

Solution

Note that $ABCD$ and $BCDE$ are isosceles trapezoids. They must be cyclic quadrilaterals, so we can apply Ptolemy's Theorem. Let $d=AC=BD=CE$, $e=AD$, and $f=EB$. Then we have:

$d^2=10e+9$

$d^2=3f+100$

$de=10d+42$

According to the first equation, $e=\frac{d^2-9}{10}$. Plugging this into the third equation results in $d^3-109d-420=0$. The only positive root of this cubic is $d=12$. Substituting into the first and second equations gives $e=\frac{27}{2}$ and $f=\frac{44}{3}$ and thus the sum of all diagonals is $3d+e+f=\frac{385}{6}$. Our answer is therefore $385+6=\boxed{391}$.

  • This solution requires solving a cubic - however, I thought that it was in the rules that the AMC 12 cannot ask for one to solve a cubic - perhaps I am mistaken?

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_24&oldid=66686"