AMC 8 2024 Problem 20
AMC8 2024 年第 20 题
立体几何★★★☆☆
Any three vertices of the cube PQRSTUVW, shown in the figure below, can be connected to form a triangle.
For example, vertices P, Q, and R can be connected to form an isosceles △PQR.
How many of these triangles are equilateral and contain P as a vertex?
如图所示，立方体PQRSTUVW上任意三个顶点相连可以构成一个三角形（例如，顶点P,Q，R相连可以构成一个等腰三角形△PQR）。请问在上述三角形中， 有多少个是以P为一个顶点的等边三角形？
- A.0
- B.1
- C.2
- D.3
- E.6
正确答案 · Correct Answer
D
解析 · Solution
解法1
唯一能够形成的等边三角形是通过正方形面的对角线。从P出发，你有3个可能的顶点，可以通过其中一个面对角线形成。因此，有3个可能的三角形。所以答案是3。
每个其他兼容点到P的边的数量必须是偶数，因此兼容点为R、V和T。由于P必须是三角形的一个顶点，我们必须从这三个点中选出两个。因此，答案为\(\binom{3}{2}=3\)。列举出来得到PRV、PRT和PVT。因此，答案是3。
观察立方体后，我们发现等边三角形只能由沿着立方体某个面的对角线（如PV）形成。由于题目要求三角形的一个顶点必须是P，可以形成的三条对角线是PT、PR和PV。现在，我们可以从列出的三条线中任选两条，并将它们与另一条线连接（例如，如果选择PT和PR，第三条对角线就是RT）。因此，选择三条对角线的方式有3种，所以答案为3，即3。
Solution 1
The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have 3 possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are 3 possible triangles. So the answer is 3, andliu766, e___
Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is List them out get , , and . Therefore, the answer is After looking at the cube, we realize that an equilateral triangle can only be formed lines that form a diagonal along a face of the cube (such as ). Because the problem has a condition that one of the triangle's vertex must be on , the three diagonals that can be formed are $PT, PR$, and . Now, we can choose any of 2 out of the 3 lines we have listed, and connect any of them with another line (for example, if we choose and , the third diagonal is ). Thus, there are 3 ways to choose the 3 diagonals, so the answer is 3, or 3
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