1985 AMC 8 Problem 24
Attempt Problem 24 of the 1985 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
24.
In a magic triangle, each of the six whole numbers through is placed in one of the circles so that the sum of the three numbers on each side of the triangle is the same. The largest possible value for is
Answer: D
Small hint:
Adding the three side sums counts each corner number twice and each midpoint number once
Big hint:
Let be the sum of the corner numbers. Then put the three largest numbers at the corners
Solution:
Let be the sum of the corner numbers. Adding the three side sums gives because the six numbers sum to and every corner is counted one extra time. To maximize place the three largest numbers at the corners, giving
Then so This is achievable: put between corners and put between and and put between and
Thus, the correct answer is D .
Problem 24 in other years
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
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