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

Revision as of 06:46, 31 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

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:

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

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:

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

$c$ , being a length, must be positive, implying that $c=12$ . 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 $3c+a+b = 3\dot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}$ , so it follows that the answer is $385 + 6 = 391$ .

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

Revision as of 06:46, 31 January 2015

Problem

Solution

See also

@@ Line 10: / Line 10: @@
 ==Solution==
-qwerqwer
+Let <math>a</math> denote the length of a diagonal opposite adjacent sides of length <math>14</math> and <math>3</math>, <math>b</math> for sides <math>14</math> and <math>10</math>, and <math>c</math> for sides <math>3</math> and <math>10</math>. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:
+<cmath>
+\begin{align}
+c^2 &= 3a+100 \\
+c^2 &= 10b+9 \\
+c &= 30+ab \\
+ac &= 3c+140\\
+bc &= 10c+42
+\end{align}
+</cmath>
+Using equations <math>(1)</math> and <math>(2)</math>, we obtain:
+<cmath>
+a = \frac{c^2-100}{3}
+</cmath>
+and
+<cmath>
+b = \frac{c^2-9}{10}
+</cmath>
+Plugging into equation <math>(4)</math>, we find that:
+<cmath>
+\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*}
+</cmath>
+<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>.
+We desire <math>3c+a+b = 3\dot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = 391</math>.
 == See also ==
 {{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}