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 $\boxed{\textbf{(E)}~4}$ .

~alwaysgonnagiveyouup

Solution 3

Let $x$ be the length of the legs of the trapezoid. Draw the angle bisectors of the 120 degree angles. Now you have 2 equilateral triangles, and another figure in between them. Let $x+a$ be the length of the shorter base of the trapezoid, and let $2x+a$ be the length of the longer base. Since the perimeter of the trapezoid is 30, $5x+2a = 30$ . Since $x>$ 0, and $a$ can be negative as long as $x+a>0$ , you get 4 solutions for $(x,a)$ , namely $(2,10), (4,5), (6,0), (8,-5)$ . Any solution with $x\ge10$ would lead to $x+a\le0$ . Hence, the answer is $\boxed{\textbf{(E)}~4}$

~adi2011

Solution 4

Drop altitudes from angle $A$ and angle $D$ . Then, two $30-60-90$ triangles and a rectangle are created. Let the hypotenuse of both of the triangles equal $x$ and let side $AD$ equal $y$ . Then, the distance from $B$ and $C$ to the feet of the altitudes that are closest to them is $\frac{x}{2}$ , and the distance between the feet of the two altitudes would be $y$ . The rest is similar to Solution 2 ~Soupboy0

Solution 5

Solution 6

Since the trapezoid perimeter is $30$ , we can name the top, congruent side lengths, and bottom variables. The top length is $A$ , the congruent side length is $C$ , and the bottom should be $B$ . (Since the top side is $A$ , and its parallel to the bottom, then the bottom side length should be $A + 2B,$ $B$ being one of the two lengths.) The perimeter of the trapezoid is $2A + 2B + 2C = 30,$ which can be divided by $2$ to become $A + B + C = 15$ Now, since the $B$ is the base of a 30-60-90 right triangle, and its angle opposite to the side is $30$ degrees. Therefore, $B$ is exactly half of side $C$ . We can test out cases of B. If $B = 1$ , then $C = 2$ , $A + 1 + 2 = 15$ which would give $A$ the value of $12$ . Keep repeating this process until $A$ is no longer a positive integer. Therefore, there are $\boxed{\textbf{(E)}~4}$ possible configurations of the trapezoid. ~Imhappy62789 ..

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
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 AJHSME/AMC 8 Problems and Solutions

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