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

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==Solution==
 
==Solution==
  
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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.
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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.
  
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<math> d^2=10e+9 </math>
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The sum of the numerator and denominator is then 391
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<math> d^2=3f+100 </math>
 
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<math> de=10d+42 </math>
 
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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>.
 
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*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 23:20, 26 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 (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

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