Difference between revisions of "2014 AMC 12B Problems/Problem 24"

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==Solution==
 
==Solution==
  
Note that <math> ABCD </math> and <math> BCDE </math> are isosceles trapezoids. They must be [[Cyclic quadrilateral|cyclic quadrilaterals]], so we can apply [[Ptolemy's Theorem]].  
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Use Ptomley's theorem, get three equations, and play with them a bit.
Let <math> d=AC=BD=CE </math>, <math> e=AD </math>, and <math> f=EB </math>. Then we have:
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You find one diagonal is 12, and solve for the other 2.
  
<math> d^2=10e+9 </math>
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The sum of the numerator and denominator is then 391
 
 
<math> d^2=3f+100 </math>
 
 
 
<math> de=10d+42 </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 00:20, 27 January 2015

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

Use Ptomley's theorem, get three equations, and play with them a bit. You find one diagonal is 12, and solve for the other 2.

The sum of the numerator and denominator is then 391

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
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

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