# 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 1

Let the number in the top circle be $a$ and then $b$, $c$, $d$, $e$, and $f$, going in clockwise order. Then, we have $$S=a+b+c$$ $$S=c+d+e$$ $$S=e+f+a$$

Adding these equations together, we get \begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}

where the last step comes from the fact that since $a$, $b$, $c$, $d$, $e$, and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$

The left hand side is divisible by $3$ and $75$ is divisible by $3$, so $a+c+e$ must be divisible by $3$. The largest possible value of $a+c+e$ is then $15+14+13=42$, and the corresponding value of $S$ is $\frac{75+42}{3}=39$, which is choice $\boxed{\text{D}}$.

It turns out this sum is attainable if you let $$a=15$$ $$b=10$$ $$c=14$$ $$d=12$$ $$e=13$$ $$f=11$$

## 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

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 