1985 AJHSME Problems/Problem 24

Problem

In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is

$[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]$

$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution 2

To make the sum the greatest, put the three largest numbers $(13,14$ and $15)$ in the corners. Then, balance the sides by putting the least integer $(10)$ between the greatest sum $(14$ and $15)$ . Then put the next least integer $(11)$ between the next greatest sum ( $13 +15$ ). Fill in the last integer $(12)$ and you can see that the sum of any three numbers on a side is (for example) $14 +10 + 15 = 39$ $\boxed{\text{D}}$ . -by goldenn

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1985 AJHSME Problems/Problem 24

Problem

Solution 2

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