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

2024 AMC 12A Problem 19

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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}

\triangle ACD,

AC^2=3^2+5^2

-2\cdot3\cdot5\cos120^\circ

Find

AB

using

\angle ABC=60^\circ,

then apply Ptolemy:

AC\cdot BD=AB\cdot CD+BC\cdot DA

\triangle ACD,

the law of cosines gives

AC^2=9+25-2(15)\cos120^\circ

=34+15=49,

AC=7.

Since

ABCD

is cyclic,

\angle ABC=180^\circ-120^\circ=60^\circ.

\triangle ABC

with

BC=3

and

AC=7,

the law of cosines gives

49=AB^2+9-3AB,

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.