2025 AMC 8 Problems/Problem 24

Problem

In trapezoid $ABCD$ , angles $B$ and $C$ measure $60^\circ$ and $AB = DC$ . The side lengths are all positive integers, and the perimeter of $ABCD$ is 30 units. How many non-congruent trapezoids satisfy all of these conditions?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

Let $a$ be the length of the shorter base, and let $b$ be the length of the longer one. Note that these two parameters, along with the angle measures and the fact that the trapezoid is isosceles, uniquely determine a trapezoid. We drop perpendiculars down from the endpoints of the top base. Then the length from the foot of this perpendicular to either vertex will be half the difference between the lengths of the two bases, or $\frac{b-a}{2}$ . Now, since we have a 30-60-90 triangle and this side length corresponds to the "30" part, the length of the hypotenuse (one of the legs) is $2 \cdot \frac{b-a}{2} = b-a$ . Then the perimeter of the trapezoid is $2(b-a)+a+b=3b-a=30$ . The only other stipulation for this trapezoid to be valid is that $b>a$ (which was our assumption). We can now easily count the valid pairs $(a,b)$ , yielding $(3,11),(6,12),(9,13),(12,14)$ . It is clear that proceeding further would cause $a \geq b$ , so we have $\boxed{\textbf{(E)}~4}$ valid trapezoids. ~cxsmi

Solution 2

Let $x$ be the length of $AB$ and $DC$ , and let $b$ be the length of the shorter base. Because $\angle B$ and $\angle C = 60^{\circ}$ , the length of the longer base is $b + \frac{x}{2} + \frac{x}{2} = b + x$ . Therefore, the perimeter is $3x + 2b = 30$ . The number of positive integer pairs $(x, b)$ is $(2,12), (4,9), (6,6), (8,3)$ , meaning the answer is $\boxed{\textbf{(E)}~4}$ .