2011 AIME II Problem 2

Attempt Problem 2 of the 2011 AIME II 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 2011 AIME II solutions, or check the answer key.

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2.

On square

ABCD,

point

E

lies on side

\overline{AD}

and point

F

lies on side

\overline{BC},

so that

BE = EF = FD = 30.

Find the area of square

ABCD.

Answer

Answer: 810

Concepts:square (geometry)coordinate geometry Pythagorean Theorem

Difficulty rating: 1970

Small hint:

Put

B = (0, 0),

C = (s, 0),

A = (0, s),

D = (s, s),

and write

E = (a, s),

F = (b, 0)

Big hint:

BE = FD

forces

b = s - a,

and then

EF = BE

forces

a = \frac{s}{3};

finish with

a^2 + s^2 = 900

Solution:

Let the side length be

s,

and place

B = (0, 0),

C = (s, 0),

A = (0, s),

D = (s, s).

Write

E = (a, s)

and

F = (b, 0).

Then

BE^2 = a^2 + s^2,

FD^2 = (s - b)^2 + s^2,

and

EF^2 = (a - b)^2 + s^2.

From

BE = FD

we get

a = s - b,

so

a - b = 2a - s.

Then

EF = BE

gives

(2a - s)^2 = a^2,

whose solutions are

a = \frac{s}{3}

and

a = s

(the latter identifies the separately named points

E

and

F

with corners

D

and

B,

making all three segments the same diagonal). So

a = \frac{s}{3}.

Now

900 = BE^2 = \frac{s^2}{9} + s^2 = \frac{10s^2}{9},

so the area is

s^2 = \frac{9}{10} \cdot 900 = 810.