2021 AMC 12B Spring Problem 17

Question

Let

ABCD

be an isosceles trapezoid having parallel bases

\overline{AB}

and

\overline{CD}

with

AB\gt CD.

Line segments from a point inside

ABCD

to the vertices divide the trapezoid into four triangles whose areas are

2,

3,

4,

and

5

starting with the triangle with base

\overline{CD}

and moving clockwise as shown in the diagram below. What is the ratio

\dfrac{AB}{CD}?

3

2+\sqrt2

1+\sqrt6

2\sqrt3

3\sqrt2

Answer: B

Concepts:trapezoid area ratio quadratic

Difficulty rating: 2010

Small hint:

The triangles on the bases have areas

4

(base

AB

) and

2

(base

CD

); write them as

\tfrac12 a h_a

and

\tfrac12 b h_b

Big hint:

With

a h_a=8,

b h_b=4,

and

(a+b)(h_a+h_b)=28,

the cross terms

a h_b

and

b h_a

satisfy a quadratic

Solution:

Let

AB=a,

CD=b,

and let the interior point be at heights

h_a

from

AB

and

h_b

from

CD.

The base triangles give

\tfrac12 a h_a=4

and

\tfrac12 b h_b=2,

so

a h_a=8

and

b h_b=4.

The total area is

2+3+4+5=14

=\tfrac12(a+b)(h_a+h_b),

so

(a+b)(h_a+h_b)=28.

Expanding,

a h_a+b h_b+a h_b+b h_a=28,

giving

a h_b+b h_a=16.

Let

u=a h_b

and

v=b h_a.

Then

u+v=16

and

uv=(a h_a)(b h_b)=32,

so

u,v=8\pm 4\sqrt2.

Finally

\dfrac{AB}{CD}

=\dfrac{a}{b}

=\dfrac{a h_b}{b h_b}

=\dfrac{u}{4}

=\dfrac{8+4\sqrt2}{4}

=2+\sqrt2.

Thus, the correct answer is B.