2024 AMC 12A 第 24 题

先试着解答 2024 AMC 12A 第 24 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2024 AMC 12A 解答,或核对答案。

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
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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