Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1: Changed two theorem names to links)
(11 intermediate revisions by 4 users not shown)
Line 8: Line 8:
 
\textbf{(E) }421\qquad</math>
 
\textbf{(E) }421\qquad</math>
  
−
==Solution==
+
== Solutions ==
 +
=== Solution 1 ===
 +
Let <math>BE=a</math>, <math>AD=b</math>, and <math>AC=CE=BD=c</math>. Let <math>F</math> be on <math>AE</math> such that <math>CF \perp AE</math>.
 +
<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 <math>\triangle CFE</math> we have <math>\cos\theta = -\cos(\pi-\theta)=-7/c</math>. We use the [[Law of Cosines]] on <math>\triangle ABC</math> to get <math>60\cos\theta = 109-c^2</math>.  Eliminating <math>\cos\theta</math> we get <math>c^3-109c-420=0</math> which factorizes as
 +
<cmath>(c+7)(c+5)(c-12)=0.</cmath>Discarding the negative roots we have <math>c=12</math>. Thus <math>BD=AC=CE=12</math>. For <math>BE=a</math>, we use Ptolemy's theorem on cyclic quadrilateral <math>ABCE</math> to get <math>a=44/3</math>. For <math>AD=b</math>, we use [[Ptolemy's theorem]] on cyclic quadrilateral <math>ACDE</math> to get <math>b=27/2</math>.
 +
 +
The sum of the lengths of the diagonals is <math>12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}</math> so the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math>
 +
 +
=== Solution 2 ===
 
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:
 
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:
  
Line 46: Line 92:
 
</cmath>
 
</cmath>
  
−
Or similarly (to check):
+
Or similarly into equation <math>(5)</math> to check:
  
 
<cmath>
 
<cmath>
Line 64: Line 110:
 
== 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}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:58, 25 September 2021

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 $m+n$ ?

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

Solutions

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 $385 + 6 = \fbox{\textbf{(D) }391}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_24&oldid=162730"