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 $CD = AD = y$ 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

Solution 3

Let $AD=CD=a$. Construct point $E$ on line $CD$ so that $AE$ is perpendicular to $CD$, $AE=BC=b$. $CE=AB=2$, $DE=a-2$.

Because $\angle AED={90}^\circ$:

$(a-2)^2+b^2=a^2$
$a^2-4a+4+b^2=a^2$
$b^2+4=4a$
Note that from here we know that $b$ must be even.
$a=\frac{b^2+4}{4}$

We also know that $p < 2015$:

$p=AB+BC+CD+AD$
$p=2+b+a+a$
$p=2a+b+2$
$2a+b+2<2015$
$2a+b<2013$

Substituting $a$ in we get:

$\frac{b^2+4}{2}+b<2013$
$b^2+4+2b<4026$
$b^2+2b+1<4023$
$(b+1)^2<4023$
$b+1<\sqrt{4023}$, b is an integer
$b+1 \le 63$
$b \le 62$

From the triangle inequality:

$a-2+b>a$
$b>2$

But, $\triangle ADE$ does not have to exist. Quadrilateral $ABCD$ could be a square, in that case $b=2$.

So, $2 \le b \le 62$ and $b$ must be even. Count all the even numbers from $2$ to $62$, $\frac{62-2}{2}+1=\boxed{\textbf{(B) } 31}$.

~isabelchen

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc10a/398

~ dolphin7

Video Solution

https://youtu.be/9DSv4zn7MyE

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions


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