2023 AMC 12A Problem 15

Question

Usain is walking for exercise by zigzagging across a

100

-meter by

30

-meter rectangular field, beginning at point

A

and ending on the segment

\overline{BC}.

He wants to increase the distance walked by zigzagging as shown in the figure below (

APQRS

). What angle

\theta=\angle PAB

=\angle QPC

=\angle RQB=\cdots

will produce a length that is

120

meters? (Do not assume the zigzag path has exactly four segments as shown; there could be more or fewer.)

\arccos\tfrac{5}{6}

\arccos\tfrac{4}{5}

\arccos\tfrac{3}{10}

\arcsin\tfrac{4}{5}

\arcsin\tfrac{5}{6}

The horizontal projection of each zigzag segment is its length multiplied by

\cos\theta

Add the horizontal projections of all the segments and compare the

100

-meter progress with the

120

-meter path

Every segment of the zigzag makes angle

\theta

with a horizontal side of the field. Therefore a segment of length

s

advances

s\cos\theta

meters horizontally. This remains true for the last segment even if it ends before crossing the full width of the field.

Adding the horizontal projections over the entire

120

-meter path gives

120\cos\theta=100.

Therefore

\cos\theta=\dfrac56,

so

\theta=\arccos\dfrac56.

Thus, the correct answer is A.

Problem 15 in other years