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Difference between revisions of "2024 AMC 12A Problems/Problem 19"

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The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
  
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==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}}
 
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Revision as of 19:18, 8 November 2024

Problem

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?

$\textbf{(A) }\frac{31}7 \qquad \textbf{(B) }\frac{33}7 \qquad \textbf{(C) }5 \qquad \textbf{(D) }\frac{39}7 \qquad \textbf{(E) }\frac{41}7 \qquad$

Solution 1

First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals. Let $AC=u$. We apply the Law of Cosines on $\triangle ACD$: \[u^2=3^2+5^2-2(3)(5)\cos120^\circ\] \[u=7\]


Let $AB=v$. Apply the Law of Cosines on $\triangle ABC$: \[7^2=3^2+v^2-2(3)(v)\cos60^\circ\] \[v=\frac{3\pm13}{2}\] \[v=8\]


By Ptolemy’s Theorem, \[AB \cdot CD+AD \cdot BC=AC \cdot BD\] \[8 \cdot 3+5 \cdot 3=7BD\] \[BD=\frac{39}{7}\] Since $\frac{39}{7}<5$, The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$.

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See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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