2021 AMC 12A Fall Problem 19

Attempt Problem 19 of the 2021 AMC 12A Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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19.

Let

x

be the least real number greater than

1

such that

\sin(x) = \sin(x^2),

where the arguments are in degrees. What is

x

rounded up to the closest integer?

10

13

14

19

20

\sin x = \sin x^2

requires

x^2 = x + 360k

x^2 = 180 - x + 360k

for an integer

k

The smallest

x \gt 1

comes from

x^2 = 180 - x

(with

k = 0

)

Equal sines require

x^2 = x + 360k

x^2 = 180 - x + 360k

for some integer

k.

The family

x^2 = x + 360k

first exceeds

1

k = 1,

giving

x \approx 19.5.

The family

x^2 = 180 - x + 360k

with

k = 0

gives

x^2 + x - 180 = 0,

x = \dfrac{-1 + \sqrt{721}}{2} \approx 12.93,

which is smaller.

Rounded up,

x = 13.

Thus, the correct answer is B.

Problem 19 in other years