2007 AMC 12B Problem 25

Attempt Problem 25 of the 2007 AMC 12B 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 2007 AMC 12B solutions, or check the answer key.

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25.

Points

A,

B,

C,

D,

and

E

are located in

3

-dimensional space with

AB=BC=CD

=DE=EA=2

and

\angle ABC=\angle CDE

=\angle DEA=90^\circ.

The plane of

\triangle ABC

is parallel to

\overline{DE}.

What is the area of

\triangle BDE?

\sqrt{2}

\sqrt{3}

2

\sqrt{5}

\sqrt{6}

Answer: C

Concepts:3D geometry coordinate geometry triangle area

Difficulty rating: 2400

Small hint:

Place

D=(-1,0,0),

E=(1,0,0),

and put

\triangle ABC

in a horizontal plane

z=k\gt0

Big hint:

The right angles at

D

and

E

put

A,C

on radius-

2

circles; use

AC=2\sqrt2

to find

k

Solution:

Set

D=(-1,0,0)

and

E=(1,0,0),

and let

\triangle ABC

lie in the plane

z=k\gt0.

Because

\angle CDE

and

\angle DEA

are right angles,

A

and

C

lie on radius-

2

circles centered at

E

and

D

in the planes

x=1

and

x=-1,

so

A=(1,y_1,k),\ C=(-1,y_2,k)

with

y_j=\pm\sqrt{4-k^2}.

Since

\angle ABC=90^\circ,

AC=2\sqrt2,

which forces

y_1=-y_2.

Taking

y_1=1,

y_2=-1,

gives

k=\sqrt3,

so

A=(1,1,\sqrt3),

C=(-1,-1,\sqrt3),

and

B

is

(1,-1,\sqrt3)

or

(-1,1,\sqrt3).

In the first case

BE=2

with

BE\perp DE;

in the second

BD=2

with

BD\perp DE.

Either way

\triangle BDE

has legs

2

and

2,

so its area is

\tfrac12(2)(2)=2.

Thus, the correct answer is C.

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