2015 AMC 10A Problem 24

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 $\boxed{\textbf{(B)}\ 31}$.

Two right angles on the same side turn the quadrilateral into a right trapezoid, and dropping the perpendicular is the standard move that exposes a right triangle. Because $AD = CD$ appears on both sides of the Pythagorean equation, the squares cancel and leave a linear relation between $x$ and $y^2$; integrality then reduces the geometry to counting perfect squares below a bound. Watch for the parity constraint hidden in $y^2 = 4(x - 1)$.

Estimate first: $p \approx 2k^2$, so $k \approx \sqrt{1007} \approx 31.7$, which points to $31$ values and distinguishes (A)/(B) from the choices near $62$ that come from counting $y$ rather than $k$. Then confirm the boundary with $k = 31$ and $k = 32$ exactly.

• Counting every even $BC$ from $0$ to $62$ ($32$ values) or excluding the $2 \times 2$ square ($30$), or letting $BC$ be any integer up to $62$ and getting $62$.
• Forgetting that $AB = 2$ contributes to the perimeter, or writing $p = 2 + y + x$ with only one copy of the equal sides.
• Setting $DE = 2 - x$ with $x < 2$ as a separate case; the same equation results and it has no solution with $x < 2$.