2020 AMC 12B Problem 18

Question

In square

ABCD,

points

E

and

H

lie on

\overline{AB}

and

\overline{DA},

respectively, so that

AE = AH.

Points

F

and

G

lie on

\overline{BC}

and

\overline{CD},

respectively, and points

I

and

J

lie on

\overline{EH}

so that

\overline{FI} \perp \overline{EH}

and

\overline{GJ} \perp \overline{EH}.

See the figure below. Triangle

AEH,

quadrilateral

BFIE,

quadrilateral

DHJG,

and pentagon

FCGJI

each has area

1.

What is

FI^2?

\dfrac73

8 - 4\sqrt{2}

1 + \sqrt{2}

\dfrac74 \sqrt{2}

2\sqrt{2}

Answer: B

Concepts:coordinate geometry area decomposition special right triangle

Difficulty rating: 1910

Small hint:

The four regions tile the square, so its area is

4

and its side is

2

Big hint:

\triangle AEH

is an isosceles right triangle of area

1,

so

AE = \sqrt2;

set coordinates and use the area of

BFIE

Solution:

The four regions have total area

4,

so the square has side

2.

Put

A = (0, 0),

B = (2, 0),

C = (2, 2),

D = (0, 2).

Since

\triangle AEH

is an isosceles right triangle with area

1,

we get

AE = AH = \sqrt2,

so

E = (\sqrt2, 0)

and

H = (0, \sqrt2).

Line

EH

is

x + y = \sqrt2.

Let

F = (2, t).

Its perpendicular distance to line

EH

is

FI = \tfrac{2 + t - \sqrt2}{\sqrt2}.

Write

s=\frac{FI}{\sqrt2}=\tfrac{2+t-\sqrt2}{2},

so the foot of the perpendicular is

I=(2-s,t-s).

The shoelace formula on

B=(2,0),

F,

I,

and

E=(\sqrt2,0)

gives

[BFIE]=s^2-(3-2\sqrt2).

Since this area is

1,

we get

s^2=4-2\sqrt2.

Then

FI^2 = 2s^2 = 8 - 4\sqrt2.

Thus, the correct answer is B.

2020 AMC 12B Problem 18

Answer: B

Difficulty rating: 1910

Small hint:

Big hint:

Solution:

Problem 18 in other years