Magic Triangle Puzzle

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

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
Subject: Re: Addends

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
Subject: Re: Addends

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