In a magic triangle, each of the six [whole numbers](https://artofproblemsolving.com/wiki/index.php/Whole_number) $10-15$ is placed in one of the [circles](https://artofproblemsolving.com/wiki/index.php/Circle) so that the sum, $S$, of the three numbers on each side of the [triangle](https://artofproblemsolving.com/wiki/index.php/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]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$
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](https://artofproblemsolving.com/wiki/index.php/Equation) 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](https://artofproblemsolving.com/wiki/index.php/Sum) is $10+11+12+13+14+15=75$
The left hand side is [divisible](https://artofproblemsolving.com/wiki/index.php/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 \[a=15\] \[b=10\] \[c=14\] \[d=12\] \[e=13\] \[f=11\]