Magic Triangle PuzzleDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/