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?

$30$

$31$

$61$

$62$

$63$

Solution:

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 $p=2+x+y+y=2k^2+2k+4.$

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.

2015 AMC 10A Problem 24

Solution:

Problem 24 in Other Years