AMC 8 2025 Problem 24
AMC8 2025 年第 24 题
四边形与多边形★★★★☆
In trapezoid , angles and measure and AB = DC . The side lengths are all positive integers, and the perimeter of is 30 units. How many non-congruent trapezoids satisfy all of these conditions?
在梯形ABCD中，∠B = ∠C = 60°，且AB = DC。该梯形的各边长均为正整数，其周长为30个单位。请问一共有多少个满足以上全部条件的非全等梯形？
- A.0
- B.1
- C.2
- D.3
- E.4
正确答案 · Correct Answer
E
解析 · Solution
解法1
常规解法·作两条高，再挑整数解
设上底 AD=a、下底 BC=b、两腰 AB=DC=L（，等腰梯形）。从 、 向下底作两条高，把梯形切成「直角三角形＋长方形＋直角三角形」。
每个直角三角形里，斜边是腰 ，底角 、顶角 。30-60-90 三角形三边比 ，最短边（对着 角的水平边）＝斜边的一半＝L/2。所以下底比上底长出两个 L/2：
周长：a+b+2L=a+(a+L)+2L=2a+3L=30。
找正整数解：，要 是正整数，30-3L 得是正偶数—— 必须是偶数且 L<10。逐个试：；；；。共 4 个，选 E。
易错提醒：L=10 给 a=0——上底缩没了，梯形塌成三角形，不算； 取奇数时 带小数，不合「各边均为正整数」。这两类都别混进来。
巧解·把梯形补成等边三角形
延长两腰 、，交于点 。 两个底角都是 ，第三个角也只能是 ——等边三角形，EB=EC=BC=b。又 ，小三角形 与它同角，也是等边的，EA=AD=a。
于是每条腰＝EB-EA=b-a，周长
约束：上底 逼出 ；腰 逼出 。所以 b=11,12,13,14，对应 a=3,6,9,12、腰 =8,6,4,2——与解法1 完全一致，共 4 个。
前置（等边三角形判定）：两个角是 ，第三个角 ；三个角相等的三角形三边相等。
Solution 1
*(Elementary — drop two heights, then pick integer solutions.)*
Let the top base be AD=a, the bottom base BC=b, and the legs AB=DC=L (an isosceles trapezoid with ). Drop heights from and , cutting the trapezoid into "right triangle + rectangle + right triangle."
In each right triangle the hypotenuse is the leg , with a base angle and a top angle. A 30-60-90 triangle has sides in ratio , and its shortest side (the horizontal one, opposite the angle) is half the hypotenuse, L/2. So the bottom base exceeds the top by two halves:
Perimeter: a+b+2L=a+(a+L)+2L=2a+3L=30.
Integer solutions: must be a positive integer, so 30-3L must be positive and even — even and L<10. Try them: ; ; ; . That is 4 trapezoids, choice E.
Watch out: L=10 forces a=0 — the top base vanishes and the trapezoid collapses into a triangle, so it doesn't count; odd makes a non-integer, violating "all side lengths are positive integers." Keep both out of your count.
*(Slick — complete the trapezoid into an equilateral triangle.)*
Extend the legs and to meet at . Triangle has two base angles, so its third angle is too — equilateral, with EB=EC=BC=b. Since , the small triangle has the same angles, so it is equilateral as well: EA=AD=a.
Each leg is then EB-EA=b-a, and the perimeter gives
Constraints: forces ; the leg forces . So b=11,12,13,14 with a=3,6,9,12 and legs 8, 6, 4, 2 — matching Solution 1 exactly: 4 trapezoids.
Prerequisite (spotting equilateral triangles): two angles force the third angle to be ; a triangle with three equal angles has three equal sides.
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