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?

A. 0

B. 1

C. 2

D. 3

E. 4

1. 2025 AMC 8 官方解析

Answer (E): Let $a = AD$ and $s = AB = DC$. Draw a segment from $A$ to $\overline{BC}$ parallel to $\overline{DC}$, meeting $\overline{BC}$ at point $P$, forming parallelogram $ADCP$. It follows that $AP = DC = s$ and $PC = AD = a$. Isosceles $\triangle ABP$ has base angles measuring $60^\circ$, so it is equilateral and $BP = s$.

Then the perimeter of $ABCD$ is $L = 2a + 3s = 30$. To satisfy this equation with positive integer values, the number $a$ must be divisible by $3$ because $2a = 3(10 - s)$. The only possible values for $a$ are $3, 6, 9$, and $12$, each value corresponding to a different trapezoid. Greater multiples of $3$ will not lead to a positive integer value for $s$. Therefore there are $4$ different trapezoids that satisfy the conditions.

The table below lists the side lengths $a$, $s$, and $a + s$ for these 4 trapezoids, along with their perimeters.

OR

Let $a = AD$ and $s = AB = DC$. Drop perpendiculars from $A$ and $D$ to points $E$ and $F$ on base $\overline{BC}$, respectively, to form $30 - 60 - 90^\circ$ triangles $AEB$ and $DFC$.

In a $30 - 60 - 90^\circ$ triangle, the leg adjacent to the $60^\circ$ angle is half as long as the hypotenuse. Therefore $BE = FC = \frac{s}{2}$ and because length $EF$ equals $a$, base length $BC = \frac{s}{2} + a + \frac{s}{2} = a+s$. The solution proceeds as above.