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

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Let <math>BE = a</math>, <math>AC = CE = BD = b</math>
 
Let <math>BE = a</math>, <math>AC = CE = BD = b</math>
  
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By [[Ptolemy's theorem]] for quadrilateral <math>ABCE</math>, <math>3c + 140 = ab</math>, <math>a = 3 + \frac{140}{b}</math>
+
By [[Ptolemy's theorem]] for quadrilateral <math>ABCE</math>, <math>AB \cdot DE + BC \cdot AE = BC \cdot AC</math>, <math>3c + 140 = ab</math>, <math>a = 3 + \frac{140}{b}</math>
  
−
By [[Ptolemy's theorem]] for quadrilateral <math>ACDE</math>, <math>3a + 100 = b^2</math>
+
By [[Ptolemy's theorem]] for quadrilateral <math>BCDE</math>, <math>CD \cdot BE + BC \cdot DE = BD \cdot CE</math>, <math>3a + 100 = b^2</math>
  
 
<math>3(3 + \frac{140}{b}) + 100 = b^2</math>, <math>b^3 - 109 b -420 = 0</math>, <math>(b-12)(b+7)(b+5) = 0</math>, <math>b = 12</math>
 
<math>3(3 + \frac{140}{b}) + 100 = b^2</math>, <math>b^3 - 109 b -420 = 0</math>, <math>(b-12)(b+7)(b+5) = 0</math>, <math>b = 12</math>
Line 151: Line 151:
 
<math>a = 3 + \frac{140}{12} = \frac{44}{3}</math>
 
<math>a = 3 + \frac{140}{12} = \frac{44}{3}</math>
  
−
By [[Ptolemy's theorem]] for quadrilateral <math>ABDE</math>, <math>AD \cdot a = 14b + 30</math>, <math>AD = \frac{27}{2}</math>
+
By [[Ptolemy's theorem]] for quadrilateral <math>ABDE</math>, <math>AE \cdot BD + AB \cdot DE = AD \cdot BE</math>, <math>AD \cdot a = 14b + 30</math>, <math>AD = \frac{27}{2}</math>
  
 
<math>12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \boxed{\textbf{(D) }391}</math>
 
<math>12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \boxed{\textbf{(D) }391}</math>

Revision as of 07:00, 1 January 2023

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 $m+n$ ?

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

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=1-2vT_GIceA

Solution 1

Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF \perp AE$. [asy] size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair O,A,B,C,D,E0,F; O=origin; A= dir(198); path c = CR(O,1); real r = 0.13535; B = IP(c, CR(A,3*r)); C = IP(c, CR(B,10*r)); D = IP(c, CR(C,3*r)); E0 = OP(c, CR(D,10*r)); F = foot(C,A,E0); dot("$A$", A, A-O); dot("$B$", B, B-O); dot("$C$", C, C-O); dot("$D$", D, D-O); dot("$E$", E0, E0-O); dot("$F$", F, F-C); label("$c$",A--C,S); label("$c$",E0--C,W); label("$7$",F--E0,S); label("$7$",F--A,S); label("$3$",A--B,2*W); label("$10$",B--C,2*N); label("$3$",C--D,2*NE); label("$10$",D--E0,E); draw(A--B--C--D--E0--A, black+0.8); draw(CR(O,1), s); draw(A--C--E0, royalblue); draw(C--F, royalblue+dashed); draw(rightanglemark(E0,F,C,2)); MA("\theta",A,B,C,0.075); MA("\pi-\theta",C,E0,A,0.1); [/asy] In $\triangle CFE$ we have $\cos\theta = -\cos(\pi-\theta)=-7/c$. We use the Law of Cosines on $\triangle ABC$ to get $60\cos\theta = 109-c^2$. Eliminating $\cos\theta$ we get $c^3-109c-420=0$ which factorizes as \[(c+7)(c+5)(c-12)=0.\]Discarding the negative roots we have $c=12$. Thus $BD=AC=CE=12$. For $BE=a$, we use Ptolemy's theorem on cyclic quadrilateral $ABCE$ to get $a=44/3$. For $AD=b$, we use Ptolemy's theorem on cyclic quadrilateral $ACDE$ to get $b=27/2$.

The sum of the lengths of the diagonals is $12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}$ so the answer is $385 + 6 = \fbox{\textbf{(D) }391}$

Solution 2

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 into equation $(5)$ to check:

\begin{align*} \frac{c^2-9}{10}c &= 10c+42\\ \frac{c^3-9c}{10} &= 10c + 42\\ 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 $385 + 6 = \fbox{\textbf{(D) }391}$

Solution 3 (Ptolemy's but Quicker)

Let us set $x$ to be $AC=BD=CE$ and $y$ to be $BE$ and $z$ to be $AD$. It follow from applying Ptolemy's Theorem on $ABCD$ to get $x^2=9+10z$. Applying Ptolemy's on $ACDE$ gives $xz=42+10x$; and applying Ptolemy's on $BCDE$ gives $x^2=100+3y$. So, we have the have the following system of equations:

\begin{align} x^2 &= 9+10z \\ x^2 &= 100+3y \\ xz &= 42+10x \end{align}

From $(3)$, we have $42=(z-10)x$. Isolating the x gives $x=\dfrac{42}{z-10}$. By setting $(1)$ and $(2)$ equal, we have $x^2=9+10z=100+3y$. Manipulating it gives $3y=10z-91$. Finally, plugging back into $(2)$ gives $x^2=100+10z-91=10z+9$. Plugging in the $x=\dfrac{42}{z-10}$ as well gives

\begin{align*} \left(\frac{42}{z-10}\right)^2 &= 10z+9\\ 10z^3 - 191z^2 + 820z + 900 &= 1764\\ 10z^3 - 191z^2 + 820z - 864 &= 0\\ (5z-8)(2z-27)(z-4) &=0 \end{align*}

It is impossible for $z<10$ for $x<0$; that means $z=\frac{27}{2}$. That means $x = 12$ and $y = \frac{44}{3}$.

Thus, the sum of all diagonals is $3x+y+z = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = 385/6$, which implies our answer is $m+n = 385+6 = \fbox{391 \textbf{(D)}}$.

~sml1809

Solution 4

2014AMC12BProblem24Solution4.png

Let $BE = a$, $AC = CE = BD = b$

By Ptolemy's theorem for quadrilateral $ABCE$, $AB \cdot DE + BC \cdot AE = BC \cdot AC$, $3c + 140 = ab$, $a = 3 + \frac{140}{b}$

By Ptolemy's theorem for quadrilateral $BCDE$, $CD \cdot BE + BC \cdot DE = BD \cdot CE$, $3a + 100 = b^2$

$3(3 + \frac{140}{b}) + 100 = b^2$, $b^3 - 109 b -420 = 0$, $(b-12)(b+7)(b+5) = 0$, $b = 12$

$a = 3 + \frac{140}{12} = \frac{44}{3}$

By Ptolemy's theorem for quadrilateral $ABDE$, $AE \cdot BD + AB \cdot DE = AD \cdot BE$, $AD \cdot a = 14b + 30$, $AD = \frac{27}{2}$

$12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \boxed{\textbf{(D) }391}$

~isabelchen

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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