2024 AMC 10A Problem 22

Problem 22 of 25HarderGeometry

Let

\mathcal{K}

be the kite formed by joining two right triangles with legs

1

and

\sqrt3

along a common hypotenuse. Eight copies of

\mathcal{K}

are used to form the polygon shown below. What is the area of

\triangle ABC?

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Solution

Each half of a kite is a

30

60

90

triangle, so its edges have the shown lengths and directions. Take

A=(0,0)

and

AB

horizontal. The horizontal span in the figure is six unit lengths, so

B=(6,0).

Along the outer boundary from

A

C,

the three edges have vectors

\sqrt3(\cos30^\circ,\sin30^\circ),

\sqrt3(\cos90^\circ,\sin90^\circ),

and

(1,0).

Their sum is

(\tfrac52,\tfrac{3\sqrt3}{2}),

C=(\tfrac52,\tfrac{3\sqrt3}{2}).

Thus

AB=6

and the altitude from

C

\tfrac{3\sqrt3}{2},

giving area

\tfrac12\cdot6\cdot\tfrac{3\sqrt3}{2}=\tfrac{9\sqrt3}{2}.

Therefore, the answer is B.