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

2024 AMC 12A Problem 19 — Solution

2024 AMC 12A Problem 19

Below is the professionally curated solution for Problem 19 of the 2024 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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Concepts:cyclic quadrilateral law of cosines Ptolemy’s Theorem

Difficulty rating: 1930

19.

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 $ABCD?$

$\dfrac{31}{7}$

$\dfrac{33}{7}$

$5$

$\dfrac{39}{7}$

$\dfrac{41}{7}$

Solution:

In $\triangle ACD,$ the law of cosines gives $AC^2=9+25-2(15)\cos120^\circ=34+15=49,$ so $AC=7.$

Since $ABCD$ is cyclic, $\angle ABC=180^\circ-120^\circ=60^\circ.$ In $\triangle ABC$ with $BC=3$ and $AC=7,$ the law of cosines gives $49=AB^2+9-3AB,$ so $AB=8.$ By Ptolemy, $AC\cdot BD=AB\cdot CD+BC\cdot DA=8\cdot3+3\cdot5=39,$ hence $BD=\tfrac{39}{7}.$ This is shorter than $AC=7.$

Thus, the correct answer is D.

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