2015 AMC 10A Problem 24

Draw the shape, find a right triangle, and get an equation connecting the unknown sides.

1. The shape. The right angles at $B$ and $C$ mean $AB$ and $CD$ both stand straight up from $BC$, so they are parallel. $ABCD$ is a trapezoid with a short parallel side $AB = 2$, a long parallel side $CD$, and height $BC$.
2. Name the unknowns. Let $CD = AD = x$ and $BC = y$.
3. Make a right triangle. From $A$, drop a perpendicular to $CD$ and call its foot $E$. Then $ABCE$ is a rectangle, so $AE = BC = y$ and $CE = AB = 2$. That leaves $DE = x - 2$.
4. Pythagorean theorem in right triangle $AED$ (long side $AD = x$):
$$x^2 = y^2 + (x - 2)^2 = y^2 + x^2 - 4x + 4.$$
The $x^2$ cancels on both sides, leaving
$$y^2 = 4x - 4 = 4(x - 1).$$
5. Try small cases. $y^2$ must be $4$ times a whole number, so $y$ must be even. List the first few:

(The first one is the $2 \times 2$ square.)
6. General case. Write $y = 2k$. Then $x = k^2 + 1$ and the perimeter is
$$p = 2 + 2k + 2(k^2 + 1) = 2k^2 + 2k + 4.$$
This grows as $k$ grows, so each $k$ gives a different $p$.
7. The limit $p < 2015$. For $k = 31$: $p = 2 \cdot 961 + 62 + 4 = 1988$, which is less than $2015$. For $k = 32$: $p = 2 \cdot 1024 + 64 + 4 = 2116$, which is too big.
8. So $k$ can be $1, 2, \ldots, 31$: that is $31$ perimeters.

The answer is $\boxed{\textbf{(B)}\ 31}$.

Two right angles next to each other make a right trapezoid, and dropping a perpendicular from the top corner always gives a rectangle plus a right triangle. Because $AD = CD$, the $x^2$ terms cancel, leaving a simple link between $x$ and $y$. After that the geometry is gone, and the question is just counting whole numbers below a limit.

Estimate first to check your count. The perimeter is about $2k^2$, so $2k^2 < 2015$ means $k^2$ is about $1000$ and $k$ is about $31$. That points to $31$ and away from the choices near $62$, which come from counting $y$ values instead of $k$ values. Then check $k = 31$ and $k = 32$ exactly, as above.

Let $BC = y$ and $CD = AD = x$. The right angles at $B$ and $C$ make $AB$ and $CD$ both perpendicular to $BC$, so $AB \parallel CD$ and $ABCD$ is a right trapezoid with parallel sides $2$ and $x$ and height $y$.

Drop the perpendicular from $A$ to line $CD$, meeting it at $E$. Then $ABCE$ is a rectangle, so $AE = y$ and $CE = 2$, leaving $DE = x - 2$. In right triangle $AED$,
$$x^2 = y^2 + (x - 2)^2 \quad\Longrightarrow\quad y^2 = 4x - 4 = 4(x - 1).$$
So $y^2$ is a multiple of $4$, forcing $y$ even. Write $y = 2k$ with $k \ge 1$; then $x = k^2 + 1$.

The perimeter is
$$p = AB + BC + CD + DA = 2 + 2k + 2(k^2 + 1) = 2k^2 + 2k + 4.$$
This increases with $k$, so distinct $k$ give distinct $p$. The condition $p < 2015$ is $k^2 + k < 1005.5$: for $k = 31$, $k^2 + k = 992$ works, and for $k = 32$, $k^2 + k = 1056$ fails. Every $k$ from $1$ to $31$ is valid ($k = 1$ is the $2 \times 2$ square), so there are $31$ perimeters.

The answer is $\boxed{\textbf{(B)}\ 31}$.