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### Magic Triangle Puzzle

```
Date: 09/24/2001 at 14:18:00
From: Karen Froelich

My daughter (she's in 7th grade) was given a math assignment. I'll try
to explain it:

The top of the paper gives the following directions; the diagram is in
the shape of a triangle:

(  )
/    \
[14]  [20]
/        \
(  )--[22]--(  )

The numbers in the squares are the sums of the numbers in the circles
to which they are connected. Discover a method to help you find the
numbers to put in the circles.

(There are numbers given in the squares, for example: using the
diagram, put a 14 in the square on the left side, a 20 in the square
on the right side, and a 22 in the bottom square. Now find the missing
addends in the circles.

Is there a simple method for finding the missing addends? My daughter
did find the answers but says she can't explain the method, she just
does it in her head by elimination. But there must be a method.

Thank you so much for your time.
```

```
Date: 09/24/2001 at 16:27:38
From: Doctor Rick

Hi, Karen.

One trick that occurs to me right away is to find the sum of the
missing numbers. I know this trick because I've solved other similar
problems. The sum of the top and left circles is 14; the sum of the
top and right circles is 20; and the sum of the two bottom numbers is
22. If you add all three of these sums, you get 14+20+22 = 56. This
sum-of-sums counts each circle twice. Dividing by 2, we get the sum of
the three circles (counted once each), namely 28.

Another thing I notice is that all three numbers must be even. If one
is odd, then both the others must be odd in order to make the sums
with the first odd number even. But if all three numbers are odd, then
the sum of the three numbers must be odd. We know the sum is 28, which
is even.

Now we can consider differences between the numbers. Imagine any
number in the top circle, say it's 6. The number in the right circle
must then be 14, so the sum will be 20. The number in the left circle
must be 8, so the sum will be 14. The right number is 6 more than the
left number, because the sum of the top and right numbers is 6 more
than the sum of the top and left numbers.

Similarly, the left number is 2 more than the top number, because the
sum of the left and right numbers is 2 more than the sum of the top
and right numbers. Thus we know that the top number is smallest; the
left number is 2 more than the top number; and the right number is 6
more than the left number.

So far we know the sum of the three missing numbers, and how the
numbers compare to one another. You could go to a guess-and-check
approach: pick a number for the left circle, fill in the other two and
see if their sum is more or less than 28. Then adjust the left number
accordingly.

Or, you could notice this. Take away 2 from the left number and 8 from
the right number, and all three will be the same. What happens to the
total? We have taken away 10, so the total is 18. If all three numbers
are the same and their sum is 18, then each number must be 18/3 = 6.
From this you can find all the numbers.

I don't know what sort of method your daughter is expected to come up
with. I would try to help her put her method into words before showing
her my method. Did she notice anything about how the missing numbers
compare to one another? Can she say more clearly what she means by
"elimination"? She may have noticed a pattern without quite realizing
it. She doesn't need to solve the problem in as systematic a way as I
did, but it would be nice for her to get some insight into her own
process. Examining your method after you've solved a problem is an
important but often overlooked step in the problem-solving process.

Very soon (if not already) your daughter will be learning to do the
problem using algebra. This requires a lot less thinking once you have
mastered the methods of algebra. Here is an algebraic solution in
brief.

Let the number in the top circle be T. Let the number in the left
circle be L. Let the number in the right circle be R. Then we have
three equations:

T + L     = 14
T +     R = 20
L + R = 22

Add the three equations:

2T + 2L + 2R = 56
2(T + L + R) = 56
T + L + R = 28

Solve the first equation for L:

L = 14 - T

Solve the second equation for R:

R = 20 - T

Substitute both of these into the previous equation:

T + (14 - T) + (20 - T) = 28
34 - T = 28
34 - 28 = T
6 = T

Then substitute 6 for T in the equations for L and R to find the other
two numbers.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/24/2001 at 16:33:22
From: Karen Froelich

Dear Dr. Math:

Thank you so much for taking time to answer my question! We've got it
now! And yes, my daughter is in pre-algebra. This will help her a lot.

Sincerely,
Karen Froelich
```
Associated Topics:
Middle School Algebra
Middle School Puzzles

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