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

Revision as of 18:18, 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 $\boxed{\textbf{(D) }\frac{39}{7}}$ .

~lptoggled, formatting by eevee9406

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

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

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 The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
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 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}}
 {{MAA Notice}}

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

Revision as of 18:18, 8 November 2024

Problem

Solution 1

See also