Question

The six whole numbers $10, 11, 12, 13, 14,$ and $15$ are placed in six circles — one at each of the three corners of a triangle and one at the midpoint of each side — so that the sum $S$ of the three numbers along each side of the triangle is the same. The largest possible value for $S$ is

$36$

$37$

$38$

$39$

$40$

Solution:

Adding the three side sums gives $3S = (10 + 11 + \cdots + 15) + (\text{sum of corners}) = 75 + (\text{sum of corners}).$ To maximize $S,$ place the three largest numbers $13, 14, 15$ at the corners, giving corner sum $42.$

Then $3S = 75 + 42 = 117,$ so $S = 39.$ This is achievable: with corners $13, 14, 15$ and midpoints $12, 10, 11,$ each side sums to $39.$

Thus, the correct answer is D .

1985 AMC 8 Problem 24

Solution:

Problem 24 in Other Years