2007 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

Isabella’s house has

3

bedrooms. Each bedroom is

12

feet long,

10

feet wide, and

8

feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy

60

square feet in each bedroom. How many square feet of walls must be painted?

678

768

786

867

876

Concepts:perimeter area

Difficulty rating: 900

Small hint:

The four walls of a room form a band of height

8

wrapping around the floor

Big hint:

The painted area per room is (floor perimeter)

\times

(height)

-60

Solution:

The perimeter of each bedroom floor is

2(12+10)=44

feet.

So the wall area in one bedroom is

44\cdot8-60=352-60=292

square feet. Across all three bedrooms, Isabella paints

3\cdot292=876

square feet.

Thus, the correct answer is E.

2.

A college student drove his compact car

120

miles home for the weekend and averaged

30

miles per gallon. On the return trip the student drove his parents’ SUV and averaged only

20

miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

22

24

25

26

28

Concepts:distance rate and time harmonic mean

Difficulty rating: 990

Small hint:

Average mileage is total miles divided by total gallons, not the average of the two rates

Big hint:

The car uses

\frac{120}{30}

gallons and the SUV uses

\frac{120}{20}

gallons

Solution:

The trip home uses

\frac{120}{30}=4

gallons, and the return trip uses

\frac{120}{20}=6

gallons.

The round trip covers

240

miles on

10

gallons, so the average is

\dfrac{240}{10}=24.

Thus, the correct answer is B.

3.

The point

O

is the center of the circle circumscribed about

\triangle ABC,

with

\angle BOC=120^\circ

and

\angle AOB=140^\circ,

as shown. What is the degree measure of

\angle ABC?

35

40

45

50

60

Concepts:inscribed angle angle sum

Difficulty rating: 1190

Small hint:

The three central angles at

O

add up to

360^\circ

Big hint:

The inscribed angle

\angle ABC

is half of the central angle

\angle AOC

Solution:

The angles around

O

sum to

360^\circ,

so

\begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ. \end{aligned}

By the inscribed angle theorem,

\angle ABC

subtends the same arc

AC

as the central angle

\angle AOC,

so

\angle ABC=\tfrac12\angle AOC=50^\circ.

Thus, the correct answer is D.

4.

At Frank’s Fruit Market,

3

bananas cost as much as

2

apples, and

6

apples cost as much as

4

oranges. How many oranges cost as much as

18

bananas?

6

8

9

12

18

Concepts:ratio and proportion unit conversion

Difficulty rating: 1010

Small hint:

First convert

18

bananas into an equivalent number of apples

Big hint:

Then convert those apples into oranges using

6

apples

=4

oranges

Solution:

Since

3

bananas cost as much as

2

apples,

18

bananas cost as much as

12

apples.

Since

6

apples cost as much as

4

oranges,

12

apples cost as much as

8

oranges. Therefore

18

bananas cost as much as

8

oranges.

Thus, the correct answer is B.

5.

The

2007

AMC

12

contests will be scored by awarding

6

points for each correct response,

0

points for each incorrect response, and

1.5

points for each problem left unanswered. After looking over the

25

problems, Sarah has decided to attempt the first

22

and leave the last

3

unanswered. How many of the first

22

problems must she solve correctly in order to score at least

100

points?

13

14

15

16

17

Concepts:inequality

Difficulty rating: 1120

Small hint:

The

3

unanswered problems already contribute

3\cdot1.5

points

Big hint:

Solve

6x\ge100-4.5

and round the number correct

x

up to a whole number

Solution:

The three unanswered problems give

3\cdot1.5=4.5

points.

So Sarah needs at least

100-4.5=95.5

points from correct answers. Since

15\lt\dfrac{95.5}{6}\lt16,

she must solve at least

16

problems correctly, which would give her

16\cdot6+4.5=100.5

points.

Thus, the correct answer is D.

6.

Triangle

ABC

has side lengths

AB=5,

BC=6,

and

AC=7.

Two bugs start simultaneously from

A

and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point

D.

What is

BD?

1

2

3

4

5

Concepts:relative speed perimeter

Difficulty rating: 1080

Small hint:

Together the two bugs cover the whole perimeter, so each crawls half of it

Big hint:

One bug travels

A\to B\to D,

a distance of

AB+BD

Solution:

The perimeter is

5+6+7=18,

so each bug crawls

9

before they meet.

The bug going

A\to B\to C

reaches

D

on side

BC,

having traveled

AB+BD=9.

Since

AB=5,

we get

BD=4.

Thus, the correct answer is D.

7.

All sides of the convex pentagon

ABCDE

are of equal length, and

\angle A=\angle B=90^\circ.

What is the degree measure of

\angle E?

90

108

120

144

150

Concepts:equilateral triangle square (geometry)angle sum

Difficulty rating: 1330

Small hint:

The sides

AB,BC,EA

and the two right angles make

ABCE

a square

Big hint:

Then

\triangle CDE

is equilateral, and

\angle E=\angle AEC+\angle CED

Solution:

Because

AB=BC=EA

and

\angle A=\angle B=90^\circ,

quadrilateral

ABCE

is a square, so

\angle AEC=90^\circ

and

EC

equals the common side length.

Then

CD=DE=EC,

so

\triangle CDE

is equilateral and

\angle CED=60^\circ.

Therefore

\begin{aligned} \angle E&=\angle AEC+\angle CED \\ &=90^\circ+60^\circ \\ &=150^\circ. \end{aligned}

Thus, the correct answer is E.

8.

Tom’s age is

T

years, which is also the sum of the ages of his three children. His age

N

years ago was twice the sum of their ages then. What is

\frac{T}{N}?

2

3

4

5

6

Concepts:ages linear equation

Difficulty rating: 1290

Small hint:

N

years ago Tom was

T-N;

each child was

N

years younger

Big hint:

The children then totaled

T-3N;

set

T-N=2(T-3N)

Solution:

Tom’s age

N

years ago was

T-N.

His three children were each

N

years younger, so their ages then totaled

T-3N.

The condition gives

T-N=2(T-3N),

so

5N=T

and

\frac{T}{N}=5.

Thus, the correct answer is D.

9.

A function

f

has the property that

f(3x-1)=x^2+x+1

for all real numbers

x.

What is

f(5)?

7

13

31

111

211

Concepts:function substitution

Difficulty rating: 1200

Small hint:

Pick

x

so that

3x-1=5

Big hint:

With that

x,

compute

x^2+x+1

Solution:

Setting

3x-1=5

gives

x=2.

Then

f(5)=2^2+2+1=7.

Thus, the correct answer is A.

10.

Some boys and girls are having a car wash to raise money for a class trip to China. Initially

40\%

of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then

30\%

of the group are girls. How many girls were initially in the group?

4

6

8

10

12

Concepts:percentage invariant

Difficulty rating: 1330

Small hint:

Two leave and two arrive, so the group size stays the same

Big hint:

The two departing girls represent

40\%-30\%=10\%

of the group

Solution:

Since two girls leave while two boys arrive, the total group size is unchanged. The drop from

40\%

to

30\%

girls corresponds to the two girls who left.

So those two girls are

10\%

of the group, meaning the group has

20

people. The initial number of girls was

40\%

of

20,

or

8.

Thus, the correct answer is C.

11.

The angles of quadrilateral

ABCD

satisfy

\angle A=2\angle B=3\angle C=4\angle D.

What is the degree measure of

\angle A,

rounded to the nearest whole number?

125

144

153

173

180

Concepts:angle sum ratio and proportion

Difficulty rating: 1270

Small hint:

Let

x=\angle A;

then

\angle B=\tfrac x2,

\angle C=\tfrac x3,

\angle D=\tfrac x4

Big hint:

The four angles of a quadrilateral sum to

360^\circ

Solution:

Let

x=\angle A.

Then

\angle B=\tfrac x2,

\angle C=\tfrac x3,

and

\angle D=\tfrac x4.

The angle sum gives

x+\dfrac x2+\dfrac x3+\dfrac x4=\dfrac{25x}{12}=360,

so

x=\dfrac{12\cdot360}{25}=172.8\approx173.

Thus, the correct answer is D.

12.

A teacher gave a test to a class in which

10\%

of the students are juniors and

90\%

are seniors. The average score on the test was

84.

The juniors all received the same score, and the average score of the seniors was

83.

What score did each of the juniors receive on the test?

85

88

93

94

98

Concepts:mean percentage

Difficulty rating: 1330

Small hint:

Assume a convenient class of

10

students:

1

junior and

9

seniors

Big hint:

Total points must satisfy

10\cdot84=9\cdot83+s

Solution:

Take a class of

10

students, so there is

1

junior and

9

seniors.

The total of all scores is

10\cdot84=840,

and the seniors contribute

9\cdot83=747.

So the junior scored

840-747=93.

Thus, the correct answer is C.

13.

A traffic light runs repeatedly through the following cycle: green for

30

seconds, then yellow for

3

seconds, and then red for

30

seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?

\dfrac{1}{63}

\dfrac{1}{21}

\dfrac{1}{10}

\dfrac{1}{7}

\dfrac{1}{3}

Concepts:geometric probability

Difficulty rating: 1410

Small hint:

One full cycle lasts

30+3+30=63

seconds

Big hint:

A change is seen only if her interval begins within

3

seconds before one of the three color switches

Solution:

The cycle length is

30+3+30=63

seconds, with three color changes per cycle.

Leah sees a change exactly when her three-second interval starts within the

3

seconds before a switch. That gives

3\cdot3=9

favorable seconds out of

63,

a probability of

\dfrac{9}{63}=\dfrac17.

Thus, the correct answer is D.

14.

Point

P

is inside equilateral

\triangle ABC.

Points

Q,

R,

and

S

are the feet of the perpendiculars from

P

to

\overline{AB},

\overline{BC},

and

\overline{CA},

respectively. Given that

PQ=1,

PR=2,

and

PS=3,

what is

AB?

4

3\sqrt{3}

6

4\sqrt{3}

9

Concepts:area decomposition equilateral triangle triangle area

Difficulty rating: 1680

Small hint:

Split

\triangle ABC

into

\triangle PAB,

\triangle PBC,

\triangle PCA,

using the perpendiculars as heights

Big hint:

Their areas sum to

\tfrac12 s(1+2+3),

which also equals

\tfrac{\sqrt3}{4}s^2

Solution:

Let

s=AB.

Joining

P

to the vertices splits the triangle into

\triangle PAB,

\triangle PBC,

and

\triangle PCA,

with areas

\tfrac{s}{2},

s,

and

\tfrac{3s}{2}.

Their total is

3s,

which must equal the area

\tfrac{\sqrt3}{4}s^2

of the equilateral triangle. So

3s=\dfrac{\sqrt3}{4}s^2,

giving

s=\dfrac{12}{\sqrt3}=4\sqrt3.

Thus, the correct answer is D.

15.

The geometric series

a+ar+ar^2+\cdots

has a sum of

7,

and the terms involving odd powers of

r

have a sum of

3.

What is

a+r?

\dfrac{4}{3}

\dfrac{12}{7}

\dfrac{3}{2}

\dfrac{7}{3}

\dfrac{5}{2}

Concepts:geometric sequence

Difficulty rating: 1580

Small hint:

The even-power terms sum to

7-3=4

Big hint:

The odd-power series is

r

times the even-power series, so

3=r\cdot4

Solution:

The odd-power terms are

ar+ar^3+\cdots

=r(a+ar^2+\cdots),

that is,

r

times the even-power terms. The even-power terms sum to

7-3=4.

So

3=4r,

giving

r=\tfrac34.

Then

a=7(1-r)=\tfrac74,

and

a+r=\dfrac74+\dfrac34=\dfrac52.

Thus, the correct answer is E.

16.

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

15

18

27

54

81

Concepts:Burnside’s Lemma symmetry

Difficulty rating: 2000

Small hint:

The tetrahedron has

12

rotational symmetries; apply Burnside’s lemma

Big hint:

Count colorings fixed by the identity, the

8

vertex rotations, and the

3

edge rotations

Solution:

The rotation group of the tetrahedron has

12

elements: the identity,

8

rotations of order

3

about a vertex-face axis, and

3

rotations of order

2

about an edge-midpoint axis.

The identity fixes all

3^4=81

colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes

3^2=9

colorings; likewise each edge rotation swaps two pairs of faces and fixes

3^2=9.

By Burnside’s lemma the number of distinguishable colorings is

\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.

Thus, the correct answer is A.

17.

If

a

is a nonzero integer and

b

is a positive number such that

ab^2=\log_{10}b,

what is the median of the set

\{0,1,a,b,\frac{1}{b}\}?

0

1

a

b

\dfrac{1}{b}

Concepts:logarithm inequality median (data)

Difficulty rating: 2060

Small hint:

Since

b\lt10^b

for all

b\gt0,

we have

\log_{10}b\lt b

Big hint:

Rule out

b\ge1

by checking whether

a=\dfrac{\log_{10}b}{b^2}

can be a nonzero integer

Solution:

For

b\gt0,

we have

b\lt10^b,

hence

\log_{10}b\lt b.

If

b=1,

then

a=0.

If

b\gt1,

then

0\lt\dfrac{\log_{10}b}{b^2}\lt1,

so

a

could not be an integer.

Hence

0\lt b\lt1,

so

\log_{10}b\lt0

and

a=\dfrac{\log_{10}b}{b^2}\lt0.

Thus

a\lt0\lt b\lt1\lt\dfrac1b,

and the middle value of the sorted set is

b.

Thus, the correct answer is D.

18.

Let

a,

b,

and

c

be digits with

a\ne0.

The three-digit integer

\overline{abc}

lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer

\overline{acb}

lies two thirds of the way between the same two squares. What is

a+b+c?

10

13

16

18

21

Concepts:perfect square place value divisibility

Difficulty rating: 1930

Small hint:

Let the two squares be

N^2

and

(N+1)^2;

their difference is

2N+1

Big hint:

Subtracting the two positions gives

\overline{acb}-\overline{abc}=9(c-b)=\tfrac{2N+1}{3}

Solution:

Let the smaller square be

N^2,

so the larger is

(N+1)^2

and the gap is

2N+1.

Then

\overline{abc}=N^2+\dfrac{2N+1}{3},

\overline{acb}=N^2+\dfrac{2(2N+1)}{3}.

Subtracting,

\overline{acb}-\overline{abc}=9(c-b)

=\dfrac{2N+1}{3},

so

27(c-b)=2N+1.

Since

\overline{acb}

is farther along the interval,

c-b

is positive; and because the right side is odd,

c-b

is odd. If

c-b\ge3,

then

N\ge40

and

N^2

is not three digits.

So

c-b=1,

giving

N=13.

The points one third and two thirds of the way from

13^2=169

to

14^2=196

are

178

and

187,

so

a+b+c=1+7+8=16.

Thus, the correct answer is C.

19.

Rhombus

ABCD,

with side length

6,

is rolled to form a cylinder of volume

6

by taping

\overline{AB}

to

\overline{DC}.

What is

\sin(\angle ABC)?

\dfrac{\pi}{9}

\dfrac{1}{2}

\dfrac{\pi}{6}

\dfrac{\pi}{4}

\dfrac{\sqrt{3}}{2}

Concepts:cylinder rhombus trigonometry

Difficulty rating: 1830

Small hint:

Taping

\overline{AB}

to

\overline{DC}

makes a side of length

6

the base circumference, so the radius is

\frac{3}{\pi}

Big hint:

The cylinder’s height is the rhombus altitude

6\sin\theta;

set the volume equal to

6

Solution:

Let

\theta=\angle ABC.

The base circle has circumference

6,

so its radius is

\dfrac{6}{2\pi}=\dfrac3\pi.

The height of the cylinder is the rhombus altitude

6\sin\theta.

The volume is

\pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6,

so

\sin\theta=\dfrac{\pi}{9}.

Thus, the correct answer is A.

20.

The parallelogram bounded by the lines

y=ax+c,

y=ax+d,

y=bx+c,

and

y=bx+d

has area

18.

The parallelogram bounded by the lines

y=ax+c,

y=ax-d,

y=bx+c,

and

y=bx-d

has area

72.

Given that

a,

b,

c,

and

d

are positive integers, what is the smallest possible value of

a+b+c+d?

13

14

15

16

17

Concepts:coordinate geometry parallelogram optimization

Difficulty rating: 2040

Small hint:

The first parallelogram has area

\dfrac{(c-d)^2}{|b-a|};

the second has area

\dfrac{(c+d)^2}{|b-a|}

Big hint:

Subtracting gives

4cd=54|b-a|;

then minimize with small positive integers

Solution:

Two vertices of the first parallelogram lie at

(0,c)

and

(0,d),

and the other two have

x

-coordinates

\pm\dfrac{c-d}{b-a}.

Its area works out to

\dfrac{(c-d)^2}{|b-a|}=18.

The same computation for the second gives

\dfrac{(c+d)^2}{|b-a|}=72.

So

(c-d)^2=18|b-a|

and

(c+d)^2=72|b-a|.

Subtracting,

4cd=54|b-a|,

i.e.

2cd=27|b-a|.

Thus

|b-a|

is even, so

a+b

is smallest with

\{a,b\}=\{1,3\};

and

cd

is a multiple of

27,

so

c+d

is smallest with

\{c,d\}=\{3,9\}.

These satisfy all conditions, giving

a+b+c+d=1+3+3+9

=16.

Thus, the correct answer is D.

21.

The first

2007

positive integers are each written in base

3.

How many of these base-

3

representations are palindromes? (A palindrome is a number that reads the same forward and backward.)

100

101

102

103

104

Concepts:number base palindrome casework

Difficulty rating: 2100

Small hint:

Count base-

3

palindromes by their number of digits; note

3^6=729\lt2007\lt2187=3^7

Big hint:

Total all palindromes with up to

7

digits, then remove the

7

-digit ones exceeding

2007=2202100_3

Solution:

A palindrome is fixed by its first half. Counting base-

3

palindromes by length gives

2

of length

1

or

2,

6

of length

3

or

4,

18

of length

5

or

6,

and

54

of length

7.

That totals

2+2+6+6+18+18

+54=106

palindromes with at most

7

digits. Since

2007=2202100_3,

the

7

-digit palindromes larger than it are

2210122,

2211122,

2212122,

2220222,

2221222,

and

2222222,

which is

6

of them.

Therefore the count is

106-6=100.

Thus, the correct answer is A.

22.

Two particles move along the edges of equilateral

\triangle ABC

in the direction

A\to B\to C\to A,

starting simultaneously and moving at the same speed. One starts at

A,

and the other starts at the midpoint of

\overline{BC}.

The midpoint of the line segment joining the two particles traces out a path that encloses a region

R.

What is the ratio of the area of

R

to the area of

\triangle ABC?

\dfrac{1}{16}

\dfrac{1}{12}

\dfrac{1}{9}

\dfrac{1}{6}

\dfrac{1}{4}

Concepts:similarity area ratio centroid

Difficulty rating: 2220

Small hint:

Because both particles move linearly, the midpoint also moves linearly between key instants; call the traced triangle

XYZ

Big hint:

That triangle is centered at the centroid

O;

compare its circumradius

OZ

to

OC

Solution:

Let

D,E,F

be the midpoints of

BC,CA,AB,

respectively, and let

X,Y,Z

be the midpoints of

AD,BE,CF,

respectively. Track a third point halfway between the two particles. It is at

X

when the particles are at

A,D;

at

Z

when they are at

F,C;

and at

Y

when they are at

B,E.

Between these instants both particle positions vary linearly, so their midpoint traces the segments

XZ,ZY,YX.

Thus the enclosed path is the equilateral triangle

XYZ,

which by symmetry shares the center

O

of

\triangle ABC.

Because

Z

is the midpoint of the median

CF,

\begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned}

while

OC=\dfrac23 CF.

So the ratio of circumradii is

\dfrac{OZ}{OC}=\dfrac14,

and the area ratio is

\left(\dfrac14\right)^2=\dfrac{1}{16}.

Thus, the correct answer is A.

23.

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to

3

times their perimeters?

6

7

8

10

12

Concepts:right triangle Diophantine Equation Simon’s Favorite Factoring Trick

Difficulty rating: 2140

Small hint:

With legs

a\le b,

write

\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right)

and isolate the radical

Big hint:

Squaring and simplifying yields

(a-12)(b-12)=72;

check the factor-pair candidates in the original equation to remove extraneous ones

Solution:

Let the legs be

a\le b.

The condition is

\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right),

so

ab-6a-6b=6\sqrt{a^2+b^2}.

Squaring and simplifying gives

ab(ab-12a-12b+72)=0,

hence

(a-12)(b-12)=72.

The positive integer solutions are

(a,b)=(3,4),

(13,84),

(14,48),

(15,36),

(16,30),

(18,24),

(20,21).

The pair

(3,4)

is extraneous: its area is

6,

while its perimeter is

12

and three times that is

36.

So exactly

6

triangles work.

Thus, the correct answer is A.

24.

How many pairs of positive integers

(a,b)

are there such that

\gcd(a,b)=1

and

\dfrac{a}{b}+\dfrac{14b}{9a}

is an integer?

4

6

9

12

infinitely many

Concepts:divisibility greatest common divisor casework

Difficulty rating: 2340

Small hint:

Clearing

b

shows

\dfrac{14b^2}{9a}

is an integer, and since

\gcd(a,b)=1,

14

is divisible by

a

Big hint:

Similarly

9

is divisible by

b,

so

a\in\{1,2,7,14\}

and

b\in\{1,3,9\};

test the coprime pairs

Solution:

Let

k

be the integer value of the original expression. Multiplying by

b

and subtracting

a

gives

\dfrac{14b^2}{9a}=bk-a,

an integer. Since

\gcd(a,b)=1,

it follows that

14

is divisible by

a.

Multiplying instead by

9a

and subtracting

14b

gives

\dfrac{9a^2}{b}=9ak-14b,

so

9

is divisible by

b.

Thus

a\in\{1,2,7,14\}

and

b\in\{1,3,9\}.

Checking the coprime candidates, the expression is an integer only for

(a,b)=(1,3),

(2,3),

(7,3),

(14,3).

So there are

4

such pairs.

Thus, the correct answer is A.

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}

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.