2014 AMC 10A Problem 22

Attempt Problem 22 of the 2014 AMC 10A 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 2014 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

In rectangle

ABCD,

\overline{AB}=20

and

\overline{BC}=10.

Let

E

be a point on

\overline{CD}

such that

\angle CBE=15^\circ.

What is

\overline{AE}?

\dfrac{20\sqrt3}3

10\sqrt3

18

11\sqrt3

20

Answer: E

Concepts:special right triangle isosceles triangle angle chasing

Difficulty rating: 1950

Small hint:

Construct a helpful point making a

30

-

60

-

90

triangle

Big hint:

Show the constructed point is the same as

E

Solution:

Let

E'

be the point on

\overline{CD}

such that

AE'=AB=20

.

Since

AD=10

, triangle

ADE'

is a

30

-

60

-

90

triangle, so

\angle DAE'=60^\circ

and

\angle BAE'=30^\circ

.

Also

AE'=AB

, so triangle

ABE'

is isosceles. Its vertex angle at

A

is

30^\circ

, so each base angle is

75^\circ

.

Therefore

\angle CBE'=90^\circ-75^\circ=15^\circ

, so

E'=E

, and

AE=20

.

Thus, E is the correct answer.