Problem 27. Six circles are arranged in the shape of a triangle as shown. John writes the digits from 1 to 6 inside the circles so that the sums of the numbers in the circles on all three sides of the triangle are the same. He then calculates the sum of the numbers in the three circles at the vertices of the triangle. How many possible values could he obtain for this sum? So we know that the three numbers along this side, the three numbers along this side, and the three numbers along this side all have the same sum, and we're interested in the sum of this circle, this circle, and this circle, the three of the vertices. So first, let's try to find some bounds on what the sum could be. So we know that the sum of all the numbers in the triangle is 21. We also know that we have three numbers inside of the vertices. The smallest they could be is 1, 2, and 3, and the largest they could be is 4, 5, and 6. So they're between 6 and 15. Let's figure out what else we know about the sum. So if we let S be the sum of each side, we know that all three sides have the same sum, so S is the sum of each side, and then if we sum each of the three sides, we get 3S. But 3S is not the sum of each of the six, because we're going to count the circles of the vertices twice. For example, this circle is part of this side and this side. This circle is part of this side and this side. So if we add this S, this S, and this S, we do get each of the circles once, but then we also get each of the ones of the vertices one more time. So we can write that 3S is 21, the sum of all the numbers, plus the sum of the vertices, which are counted twice. Now, 3S is clearly divisible by 3, and so is 21. So this tells us that the sum of the vertices is also divisible by 3. And since we know that the sum of the vertices is between 6 and 15 and is divisible by 3, the possibilities for the sum of the vertices are 6, 9, 12, and 15. Now we can try to see if it's indeed possible to get each of these sums, and we see that it is. This gives us a sum of 6, this gives us a sum of 9, this gives us a sum of 10, and this gives us a sum of 15. So all four of these are indeed possible, and the answer is D.

triangle

circle arrangement

number assignment

sum of vertices

divisible by 3