2015 AMC 10A Problem 24

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 < 2015

are possible?

30

31

61

62

63

Set

x=2k

from the right triangle in the diagram

Use

p=2k^2+2k+4

Let

BC=x

and

CD=AD=y

. Dropping the altitude from

A

to

CD

gives a right triangle with legs

x

and

y-2

, and hypotenuse

y

. Therefore

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

so

x^2=4(y-1)

.

Since

x

is an integer, write

x=2k

. Then

y=k^2+1

, and the perimeter is

\begin{aligned} &p=2+x+y+y \\ &=2k^2+2k+4. \end{aligned}

We need

2k^2+2k+4<2015

, or

k^2+k<1005.5

. This holds for

k=1,2,\ldots,31

, while

k=32

is too large. Thus there are

31

possible perimeters.

Thus, B is the correct answer.

Problem 24 in other years