1985 AJHSME Problems/Problem 24
Contents
Problem
In a magic triangle, each of the six whole numbers 
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
![[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]](http://latex.artofproblemsolving.com/8/3/4/83415da2658f8c65d5efcd1a4cc144e572e5a2a0.png)
Solution 1
To make the sum the greatest, put the three largest numbers 
and 
in the corners. Then, balance the sides by putting the least integer 
between the greatest sum 
and . Then put the next least integer 
between the next greatest sum (
). Fill in the last integer 
and you can see that the sum of any three numbers on a side is (for example) 
.
-by goldenn
Solution 2
Let the numbers, in clockwise order from the very top, be 
. Notice that 
. Manipulation will deal 
. Now we use the answer choices to our advantage. We first check for 
and find that 
, which is impossible due to the restrictions in the question. Having eliminated 40, we now move onto 
and find that this deals 
, which is possible (for example, let 
). Thus, the answer is 
.
-scthecool
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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