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

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(→‎Solution)
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</cmath>
 
</cmath>
  
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<math>c</math>, being a length, must be positive, implying that <math>c=12</math>. Plugging this back into equations <math>(1)</math> and <math>(2)</math> we find that <math>a = \frac{44}{3}</math> and <math>b= \frac{135}{10}=\frac{27}{2}</math>.
+
Or similarly (to check):
  
−
We desire <math>3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math>
+
<cmath>
 +
\begin{align*}
 +
\frac{c^2-9}{10}c &= 10c+42\\
 +
\frac{c^3-9c}{10} &= 100c + 420\\
 +
c^3-9c &= 100c + 420\\
 +
c^3-109c-420 &=0\\
 +
(c-12)(c+7)(c+5)&=0
 +
\end{align*}
 +
</cmath>
 +
 
 +
<math>c</math>, being a length, must be positive, implying that <math>c=12</math>. In fact, this is reasonable, since <math>10+3\approx 12</math> in the pentagon with apparently obtuse angles. Plugging this back into equations <math>(1)</math> and <math>(2)</math> we find that <math>a = \frac{44}{3}</math> and <math>b= \frac{135}{10}=\frac{27}{2}</math>.
 +
 
 +
We desire <math>3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math>
  
 
== 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 22:04, 14 February 2016

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

Let $a$ denote the length of a diagonal opposite adjacent sides of length $14$ and $3$, $b$ for sides $14$ and $10$, and $c$ for sides $3$ and $10$. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:

\begin{align} c^2 &= 3a+100 \\ c^2 &= 10b+9 \\ ab &= 30+14c \\ ac &= 3c+140\\ bc &= 10c+42 \end{align}

Using equations $(1)$ and $(2)$, we obtain:

\[a = \frac{c^2-100}{3}\]

and

\[b = \frac{c^2-9}{10}\]

Plugging into equation $(4)$, we find that:

\begin{align*} \frac{c^2-100}{3}c &= 3c + 140\\ \frac{c^3-100c}{3} &= 3c + 140\\ c^3-100c &= 9c + 420\\ c^3-109c-420 &=0\\ (c-12)(c+7)(c+5)&=0 \end{align*}

Or similarly (to check):

\begin{align*} \frac{c^2-9}{10}c &= 10c+42\\ \frac{c^3-9c}{10} &= 100c + 420\\ c^3-9c &= 100c + 420\\ c^3-109c-420 &=0\\ (c-12)(c+7)(c+5)&=0 \end{align*}

$c$, being a length, must be positive, implying that $c=12$. In fact, this is reasonable, since $10+3\approx 12$ in the pentagon with apparently obtuse angles. Plugging this back into equations $(1)$ and $(2)$ we find that $a = \frac{44}{3}$ and $b= \frac{135}{10}=\frac{27}{2}$.

We desire $3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}$, so it follows that the answer is $385 + 6 = \fbox{\textbf{(D) }391}$

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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