# 2015 AMC 10A Problems/Problem 24

The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.

## Problem 24

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? $\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

## Solution 1

Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that $$x^2 + (y - 2)^2 = y^2.$$ Simplifying yields $x^2 - 4y + 4 = 0$, so $x^2 = 4(y - 1)$. Thus, $y$ is one more than a perfect square.

The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$), and so our answer is $\boxed{\textbf{(B) } 31}$

## Solution 2

Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$. Denote the intersection point of the perpendicular and $CD$ as $E$. $AE$'s length is $x$, as well. Call $ED$ $y$. By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$. And so: $x^2 = 4y + 4$, or $y = (x^2-4)/4$.

Writing this down and testing, it appears that this holds for all $x$. However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers. In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$, which is less than 2015. However, 64 gives us $2116 > 2015$, so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{\textbf{(B) } 31}$.

-jackshi2006

~ dolphin7

## Video Solution

~savannahsolver

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