Difference between revisions of "1985 AJHSME Problems/Problem 24"
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| + | ==Problem== | ||
| + | In a magic triangle, each of the six [[whole number|whole numbers]] <math>10-15</math> is placed in one of the [[circle|circles]] so that the sum, <math>S</math>, of the three numbers on each side of the [[triangle]] is the same. The largest possible value for <math>S</math> 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> | ||
| + | |||
| + | <math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | To make the sum the greatest, put the three largest numbers <math>(13,14</math> and <math>15)</math> in the corners. Then, balance the sides by putting the least integer <math>(10)</math> between the greatest sum <math>(14</math> and <math>15)</math>. Then put the next least integer <math>(11)</math> between the next greatest sum (<math>13 +15</math>). Fill in the last integer <math>(12)</math> and you can see that the sum of any three numbers on a side is (for example) <math>14 +10 + 15 = 39</math> | ||
| + | <math>\boxed{\text{D}}</math>. | ||
| + | -by goldenn | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AJHSME box|year=1985|num-b=23|num-a=25}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 10:05, 10 June 2024
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 2
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
