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2024 AMC 12A 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

等面四面体是指四个三角形面彼此全等的四面体。若一个等面四面体的各个面都是边长为整数的不等边三角形，则它的最小总表面积是多少？

A disphenoid is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?

3\sqrt3

3153\sqrt{15}

1515

15715\sqrt7

24624\sqrt6

答案：D
知识点：立体几何海伦公式
难度评级：2520
小提示：

等面四面体能由一个三角形构成，当且仅当该三角形是锐角三角形；总表面积为 44 个面的面积

A disphenoid can be built from a triangle exactly when that triangle is acute; its surface is 44 faces

大提示：

寻找面积最小的锐角不等边整数三角形；(3,4,5)(3,4,5) 是直角三角形，所以尝试 (4,5,6)(4,5,6)

Seek the smallest-area acute scalene integer triangle; (3,4,5)(3,4,5) is right, so try (4,5,6)(4,5,6)

解答：

等面四面体存在（可看作由一个长方体各面中心构成的四面体）当且仅当公共面三角形是锐角三角形，此时总表面积是单个面面积的 44 倍。把整数边长写成 u<v<wu\lt v\lt w。若 u≤3u\le3，则 w≥v+1w\ge v+1 且 u2+v2≤(v+1)2u^2+v^2\le(v+1)^2（只有 (u,v,w)=(3,4,5)(u,v,w)=(3,4,5) 时取等号），所以三角形不是锐角三角形。因此 u≥4u\ge4。

三角形 (4,5,6)(4,5,6) 是锐角三角形，因为 42+52>624^2+5^2\gt6^2。它的面积也是可能取到的最小值。任何锐角三角形的最大角至少为 60∘60^\circ，所以满足 (u,v)≠(4,5)(u,v)\ne(4,5) 的候选三角形面积至少为 12uvsin⁡60∘\tfrac12uv\sin60^\circ ≥12(4)(6)32=63\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3，这大于 (4,5,6)(4,5,6) 的面积 1574\tfrac{15\sqrt7}{4}。

由海伦公式，半周长 s=152s=\tfrac{15}2，面积为 152⋅72⋅52⋅32=1574\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}。总表面积为 4⋅1574=1574\cdot\tfrac{15\sqrt7}{4}=15\sqrt7。

因此正确答案是 D。

A disphenoid exists (as the tetrahedron formed by the face-plane midpoints of a box) exactly when the common face triangle is acute, and its total surface area is 44 times one face’s area. Write the integer side lengths as u<v<w.u\lt v\lt w. If u≤3,u\le3, then w≥v+1w\ge v+1 and u2+v2≤(v+1)2u^2+v^2\le(v+1)^2 (with equality only for (u,v,w)=(3,4,5)(u,v,w)=(3,4,5)), so the triangle is not acute. Thus u≥4.u\ge4.

The triangle (4,5,6)(4,5,6) is acute because 42+52>62.4^2+5^2\gt6^2. It also has the least possible area. The largest angle of any acute triangle is at least 60∘,60^\circ, so a candidate with (u,v)≠(4,5)(u,v)\ne(4,5) has area at least 12uvsin⁡60∘\tfrac12uv\sin60^\circ ≥12(4)(6)32=63,\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3, which is greater than 1574,\tfrac{15\sqrt7}{4}, the area of (4,5,6).(4,5,6).

By Heron’s formula with s=152,s=\tfrac{15}2, that area is 152⋅72⋅52⋅32=1574.\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}. The total surface area is 4⋅1574=157.4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7.

Thus, the correct answer is D.

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