For example, the vertices 1-6-8 form an acute triangle while the vertices 1-2-3 form a right triangle.
Solution:
One approach is to look at each vertex and see how many acute and right triangles can be formed and then remove duplications at the end. This is not the most efficient way.
Another approach is as follows: To determine the number of acute triangles, we shall first find the probability that at a triangle formed will be acute and multiply that number by the total number of triangles that can be formed on the cube. Similarly, we can find the number of right triangles.
Let us start with an arbitrary vertex, say vertex 1. (Notice that we could have started with any vertex.) There are a total of 21 triangles that can be formed using this vertex.
Why are there 21? From the vertex 1, we can choose any of the other 7 vertices for the next vertex of the triangle. For the last vertex, we can choose any of the remaining 6 vertices, but now we have double-counted, since the triangle with vertices 1-2-3 is the same as the triangle with vertices 1-3-2. Thus, we have (7*6)/2 = 21 triangles.
There are three acute triangles, namely 1-6-8, 1-3-6, and 1-3-8.
There are eighteen right triangles, namely 1-2-3, 1-2-4, 1-2-5, 1-2-6, 1-2-7, 1-2-8, 1-3-4, 1-3-5, 1-3-7, 1-4-5, 1-4-6, 1-4-7, 1-4-8, 1-5-6, 1-5-7, 1-5-8, 1-6-7, and 1-7-8.
From that, we see that the probability of picking an acute triangle is 3/21 = 1/7 while the probability of picking a right triangle is 18/21 = 6/7.
In total, there are 56 different triangles that are possible. This comes from the formula for the binomial coefficient. In this case, we have 8 vertices and we are choosing 3 of them.
Thus, there are a total of (1/7)*56 = 8 acute triangles and (6/7)*56 = 48 right triangles that can be formed by connecting three vertices of a cube.
2 comments:
excellent method.
quite clear for me to understand.
thanks and keep up the good work.
@phynics Thanks
Happy Puzzling
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