1985 AJHSME 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 1

A numeral can appear in a maximum of 2 sides in this triangle, so we can put the largest 3 numbers, 15, 14, and 13 at the corners, as shown in Figure 1.

Figure 1

Then, we can label the left empty circle as $a$ and the right empty circle as $a-1,$ because 14 is one more than 13, and we need to balance it out for the sum to be the same. The bottom empty circle can be named as $a+1$ because $13+14=15+13-1.$ We only can use the numbers $10, 11,$ and $12$ now, so $a=11,$ $a+1=12,$ and $a-1=10.$ Therefore, the largest value for $S$ is $15+11+13=10\cdot3+(5+3+1)=\boxed{39~\textbf{(D)}}.$

But, we have to make sure that all three sides sum to 39. The completed triangle is shown in Figure 2.

Figure 2

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1985 AJHSME Problem 24

Problem

Solution 1