2007 AIME II Problem 3

Question

Square

ABCD

has side length

13,

and points

E

and

F

are exterior to the square such that

BE = DF = 5

and

AE = CF = 12.

Find

EF^2.

Answer

Answer: 578

Concepts:Pythagorean Triple congruence (geometry)Pythagorean Theorem

Difficulty rating: 2300

Small hint:

Since

5^2 + 12^2 = 13^2,

angles

AEB

and

CFD

are right angles

Big hint:

Extend

EA

beyond

A

and

FD

beyond

D

to meet at

G,

and look for right triangles congruent to

AEB

Solution:

Since

5^2 + 12^2 = 13^2,

triangles

AEB

and

CFD

are right-angled at

E

and

F,

and they are congruent (sides

5,

12,

13

). Extend

EA

beyond

A

and

FD

beyond

D

until the two lines meet at

G.

Then

\begin{aligned} \angle GAD &= 90^\circ - \angle EAB \\ &= \angle ABE, \\ \angle GDA &= 90^\circ - \angle FDC \\ &= \angle DCF = \angle BAE. \end{aligned}

These two angles sum to

90^\circ,

so

\angle AGD = 90^\circ,

and triangle

AGD

is congruent to

BEA

(equal angles and hypotenuse

AD = BA = 13

). Hence

GA = EB = 5

and

GD = EA = 12.

Therefore

GE = GA + AE = 5 + 12 = 17

and

GF = GD + DF = 12 + 5 = 17,

with a right angle between them at

G,

so

EF^2 = 17^2 + 17^2 = 578.

2007 AIME II Problem 3

Answer: 578

Difficulty rating: 2300

Small hint:

Big hint:

Solution:

Problem 3 in other years