2021 AMC 10B Spring Problem 20

Question

The figure below is constructed from

11

line segments, each of which has length

2.

The area of pentagon

ABCDE

can be written as

\sqrt{m} + \sqrt{n},

where

m

and

n

are positive integers. What is

m + n ?

20

21

22

23

24

Use the equal length segments to identify equilateral-triangle altitudes

Split the pentagon into two side pieces and the central isosceles triangle

Let

F

be the unlabeled point joined to

A,B,

and

C.

Because all the drawn segments have length

2,

triangles

ABF

and

CBF

are equilateral and lie on opposite sides of

BF.

Hence

\angle ABC=120^\circ,

so

[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3.

The same reasoning on the other side gives

[ADE]=\sqrt3.

Also, the Law of Cosines in

\triangle ABC

gives

AC^2=12,

and similarly

AD^2=12.

Thus the altitude of isosceles triangle

ACD

to its base

CD=2

is

\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.

Therefore

[ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}.

The pentagon’s total area is

\sqrt{12}+\sqrt{11},

so

m+n=12+11=23.

Thus, the answer is D .

Problem 20 in other years