Julia and Jason had a question about Problem 40 on Page 736 that I wanted to revisit since I don't think I answered the question to everyone's satisfaction. The problem, I think, could be clarified and improved as followed to make it clear what is being asked:

 

As currently written:

 

40. A red, blue, and green die are rolled. The number on each die represents a side length of a triangle. So 4, 2, 4 would represent an isosceles triangle.

a. How many combinations can be rolled?

b. What is the probability that you could build an equilateral triangle?

c. What is the probability that you could build a nonequilateral isosceles triangle.

 

Clarified and improved version:

 

40. A red, blue, and green die are rolled:

a. How many combinations can be rolled?

Let the number on each die represents a side length of a triangle. So 4, 2, 4 would represent an isosceles triangle.

b. What is the probability that you could build an equilateral triangle?

c. What is the probability that you could build a nonequilateral isosceles triangle.

 

Solution:

 

a. the number of combinations is (6 C 1) * (6 C 1) * (6 C 1) = 6 * 6 * 6 = 216.

 

b. to find the probability that you can build an equilateral triangle you need to know the number of possible “equilateral “ arrangements, it is easy to see there are only 6 i.e., (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), and (6,6,6). Therefore the probability of an equilateral arrangement of the three dice, P(equilateral) = 6/216 =1/36.

 

c. to find the probability of nonequilateral isosceles triangles you need to know the number of possible arrangements, lets start with the first die and let it have any of its 6 possible numbers, then one of the other two dice can only have 1 possible number (the matching one to make an isosceles triangle), and the remaining die can have only 5 possible numbers. By the Fundamental Counting Principle this leaves 6 * 1* 5 = 30 possibilities. But there are 3 dice so multiply 30 * 3 = 90. Therefore the probability of a nonequilateral isosceles triangle arrangement of the three dice, P(nonequilateral) = 90/216 = 5/12.

 

To further help you, if needed, to see what is going on in this problem I have listed the 216 possible “3 dice” combinations.

 

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36 combos

36 combos

36 combos

36 combos

36 combos

36 combos

6 x 36 = 216 total combinations

 

 

Further exploration: (looking only at the 90 nonequilateral isosceles triangle arrangements)

 

A question you may have at this point is: Are there some combinations of triangle sides that can be eliminated either because they are duplicates or because they do not make physical sense. You could eliminate the ones that not make physical sense and see what is left. A fundamental property of a triangle is:

 

“The sum of any two sides of a triangle must be longer than the remaining side.”

 

 Why is this true?

 

For example, 1,1,3 is not a valid possibility for a triangle since 1+1 < 3. Try using Geometer’s Sketchpad to prove this to yourself. Also, 1,1,2 is not a valid possibility for a triangle since 1+1 only = 2 and is not greater than 2 but the 1,1, 3 example is easier to visualize.

 

Below, the triangle arrangements that are not physically possible (NPP) are shown with the side that is too long indicated in red. A subset of these are the isosceles triangles that are not possible (NPPI, grey background). There are 23 of these.

 

You can further eliminate duplicate triangles if you assume, for example, that a 1,2,2 triangle is the same as a 2,1,2 triangle. This is a reasonable assumption if you are allowed to rotate the triangles.

 

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NPP=30

 

NPP=21

 

NPP=15

 

NPP=12

 

NPP=12

 

NPP=15

NPPI=10

 

NPPI=4

 

NPPI=2

 

NPPI=2

 

NPPI=2

 

NPPI=3