Dartboards

The Problem

The twins have been given a dartboard each for Christmas but they only have three rings on them so they are only simplified versions of the real thing. However, they do use three darts each just as you do in a real darts game.

1. Sue’s board looks like the drawing below. What possible totals can she score if all of her darts land on the board? (Assume that none of them land on the rings.)

1. Now Tim’s board looks the same as Sue’s but not all of the numbers on his are the same as Sue’s. The possible totals that Tim can make are

9, 11, 13, 15, 16, 18, 20, 23, 25, 30.

What numbers are on Tim’s board?

Teaching sequence

1. Introduce the problem with a discussion about the game of darts.
2. Display or draw the game board that is used in the problem.
3. Ask the students to identify the highest score that can be obtained from throwing three darts.
4. In pairs ask the students to find all the possible scores obtained from throwing three darts.
5. Share the possible scores.
6. Pose parts (b) of the problem for the students to work on. Check that they understand what is required.
7. As they work ask questions that focus on their manipulation of numbers using the number operations. Encourage them to justify their solutions.
   How did you begin part (b)? Why did you start there?
   How did you work out the numbers?
   What information was the most useful? Did you need it all?
   Are you convinced you have the solution? Might there be another one? Why or why not?
8. As the students work, encourage them to record their solutions so that they can be shared with others in the class.
9. Share solutions.

Extension to the problem

Suppose that you had a board something like Sue’s and Tim’s but with not necessarily the same numbers. How many possible totals could you get by using three darts?

Could any two of these totals be the same?

Solution

1. Let’s look at Sue’s board and let’s be systematic. Now, the darts might all fall into the same number ring; two might be the same and one different; and all the numbers might be different. So the totals would be

   2 + 2 + 2 = 6	5 + 5 + 5 = 15	9 + 9 + 9 = 27
   2 + 2 + 5 = 9	2 + 2 + 9 = 13
   5 + 5 + 2 = 12	5 + 5 + 9 = 19
   9 + 9 + 2 = 20	9 + 9 + 5 = 23
   2 + 5 + 9 = 16

   This gives ten totals: 6, 9, 12, 13, 15, 16, 19, 20, 23, 27.

2. Now for Tim’s board. If 3 darts ‘hit’ the same region then the totals will be a multiple of 3. Now we could look for the numbers among the list that were divisible by 3 but there are four of these and there are only three totals that can be made by using the same number twice. So let’s think a bit more.

Now the smallest number will be a multiple of 3 because that is the smallest number on the board hit with three darts. Since the smallest number is 9, then the smallest number on the board is 3.

But the same thing works in the opposite direction. The biggest number has to be a multiple of 3 too. Since the biggest total is 30, the biggest number is 10.

How can we get the other number from Tim’s board? What is the next smallest total? How could we get the next smallest total? Surely that’s formed by adding two of the smallest number on the board to the middle number on the board? So

11 = 2 x 3 + the middle number.

Actually in this problem we have far more information than we need.

Solution to the extension

Suppose that the three numbers were a, b and c. Then we could list the totals in the same way that we listed the totals on Sue’s board.

Three the same: 3a, 3b, 3c.
Two the same: 2a + b, 2a + c, 2b + a, 2b + c, 2c + a, 2c + b.
All different: a + b + c.

This gives 10 possible totals. This is at least consistent with the two examples that we’ve had so far.

Could any of these 10 totals be the same? First of all, let’s assume that a < b < c, otherwise there are going to be lots of totals that are the same. (We leave you to work out how many.) So 3a must be different from any other total since a is the smallest number. (If your class is not sure about this then they should do a little algebra to check it out.)

Similarly 3c must be the biggest number.

So let’s see what happens to 3b. Is it clear that 3b isn’t equal to any of 2a + b, 2b + a, 2b + c or 2c + b? So let’s try putting 3b = 2a + c. We’ll do this using an example. Try a = 5 and c = 11. Then b = 7. So what totals do we get with 5, 7 and 11 on the board?

Three the same: 15, 21, 33.
Two the same: 17, 21, 19, 25, 27, 29.
All different: 23.

So here there are only 9 totals. These are 15, 17, 19, 21, 23, 25, 27, 29, 33.

Must we always get at least 9 different totals?