2015 AMC 12A Problem 19

Question

For some positive integers

p,

there is a quadrilateral

ABCD

with positive integer side lengths, perimeter

p,

right angles at

B

and

C,

AB = 2,

and

CD = AD.

How many different values of

p \lt 2015

are possible?

30

31

61

62

63

Answer: B

Concepts:Pythagorean Theorem counting integers in a range

Difficulty rating: 2010

Small hint:

Drop a perpendicular from

A

to

\overline{CD};

with right angles at

B,C,

that creates a rectangle

Big hint:

If

AE = x

and

DE = y

then

x^2 + y^2 = (2+y)^2,

which forces

x

to be even

Solution:

In every such quadrilateral

CD \ge AB.

Let

E

be the foot of the perpendicular from

A

to

\overline{CD};

then

CE = 2

and

AE = BC.

Let

x = AE

and

y = DE,

so

AD = 2 + y.

By the Pythagorean Theorem

x^2 + y^2 = (2+y)^2,

so

x^2 = 4 + 4y

and

x

is even. Writing

x = 2z

gives

y = z^2 - 1,

and the perimeter is

x + 2y + 6 = 2z^2 + 2z + 4.

Increasing values

z = 1, 2, 3, \dots

give the required quadrilaterals with increasing perimeter. For

z = 31

the perimeter is

1988,

and for

z = 32

it is

2116.

Therefore there are

31

possible values of

p \lt 2015.

Thus, the correct answer is B.