Subject: magic triangles

Name: Sandy
Who is asking: Other
Level: Elementary

Question:
My tutoring student brought math homework today in the form of a "magic triangle". There are three spaces along each side for missing numbers. The sums of the numbers along each of the 3 sides should be the same. Use the numbers 4 through 9. Don't use any number more than once. The sum of the numbers on each side should be 20. What is the logic behind solving a problem of this kind?

You have the numbers 4, 5, 6, 7, 8, 9. Suppose that the sum on each side were to be 21 instead of 20. The starting clue would be that there are not many ways to use the 4 in a triplet of numbers summing to 21. The only choice is 4 + 8 + 9 = 21. The lower edge of the triangle should be

		8   4   9

You cannot put the 4 on a vertex because it cannot be used again in a triplet summing up to 21. Next, where should you put the 5? It cannot be paired with 8, because 8+5 = 13, so 8 more is missing to sum up to 21, and you are not allowed to use that number again. So you must put 5 on the side next to the 9:

		      5
		8   4   9

And now it is easy to complete the triangle:

		    7
		  6   5
		8   4   9

There is no recipe for such puzzles. Just use your imagination, trial and error, and it will work out. If you are tutoring, provide clues or encouragement when the student is about to give up, but let the student find the solution.

Footnote:
Claude is correct that there is no recipe for this puzzle, in fact I would have solved his problem another way. First notice that 4 + 5 + 6 ... + 9 = 39. If each side sum is 21 then the sum of the three side sums is 3 x 21 = 63. Summing in this way counts each of the numbers at a vertex twice and hence the sum of the numbers at the verticies is 63 - 29 = 24. But the only triplet of the original numbers that adds to 24 is 7, 8 and 9 so the verticies are 7, 8 and 9.