To each point in the sides and the interior of an acute-angle triangle, what is the maximum sum of the distance of these points to the 3 sides of the triangle?
a) the arithmetic mean of the 3 heights of the triangle;
b) the greatest side of the triangle;
c) the greatest height of the triangle;
d) 3 times the radius of the inscribed circumference;
e) the diameter of the circumscribed circumference.
It makes sense to me that the answer is c), but I'm having some difficulty showing that, because I don't know any results that deals with distances in that type of triangle (just in equilateral triangles). But I assume that, for the distance to be the maximum possible, the line defined by these points and the sides of the triangle should not be perpendicular.
The answer is c) the greatest height of the triangle. Here is a proof, based on this excellent answer by @CalvinLin to a similar question, as suggested by @dxiv in his comment to your question.
Let the triangle, not necessarily acute, have vertices $A,B,C$, sides $a,b,c$, altitudes (also called heights) $h_a,h_b,h_c$, and area $\Delta$. Consider point $P$ inside or on the triangle with distances $p_a,p_b,p_c$ to the triangle's sides. The triangles $PAB,PBC,PCA$ divide the main triangle into three smaller triangles, so their areas add to the area of the main triangle. We therefore have
$$\begin{align} 2\Delta &= ap_a+bp_b+cp_c \\ &\ge \min(a,b,c)p_a+\min(a,b,c)p_b+\min(a,b,c)p_c \\ &= \min(a,b,c)(p_a+p_b+p_c) \end{align}$$
We therefore get
$$p_a+p_b+p_c \le \frac{2\Delta}{\min(a,b,c)}$$
Since $\frac{2\Delta}a=h_a,\frac{2\Delta}b=h_b,\frac{2\Delta}c=h_c$ and a small $a$ goes with a large $h_a$ etc. we get
$$p_a+p_b+p_c \le \max(h_a,h_b,h_c)$$
We get equality when point $P$ is at the vertex with maximum height. Therefore the maximum sum of distances from the triangle's sides is indeed the greatest height of the triangle.
Here is another approach, my first one.
One way to solve multiple-choice questions is to try some examples and see which answers are incorrect by elimination. If all choices but one are eliminated, that remaining choice is the answer.
So let's try, as an example, the $45°$-$45°$-$90°$ triangle with legs of unit length, placed on the coordinate plane so the vertices are $(0,0), (1,0), (0,1)$. Then the hypotenuse is part of the line $x+y=1$. I know this is not an acute triangle, but we can reduce the right angle very slightly to get one, and the resulting values also change only very slightly. I choose this example because it is easy to analyze.
If the distances from point $P$ to the legs are $a$ and $b$, we see that $\triangle GPH$ in my diagram is also a $45°$-$45°$-$90°$ triangle with legs of length $1-a-b$, and the distance from $P$ to the hypotenuse is $\frac{\sqrt 2}{2}(1-a-b)$. The sum of the distances from point $P$ to the three sides is thus
$$\begin{align} S&=a+b+\frac{\sqrt 2}{2}(1-a-b) \\ &=\frac{\sqrt 2}{2}+\left(1-\frac{\sqrt 2}{2}\right)(a+b) \end{align}$$
This is clearly a maximum when $a+b$ is a maximum, and given that the point $P$ is below the hypotenuse $x+y=1$ we see that the maximum value of $a+b$ is $1$. Thus $S_{\text{max}}=1$. That sum of distances happens, for example, at vertex $B$ or $C$.
We may now eliminate choices. As for (a), the three triangle heights are $1,1,\sqrt 2$, so the arithmetic mean is not $1$. So (a) is out.
The greatest side of the triangle is $\sqrt 2$, so (b) is out.
The heights of the triangle are $1,1,\frac{\sqrt 2}2$, so the greatest is $1$, and (c) is a possibility.
The diameter of the circumscribed circle is the hypotenuse, $\sqrt 2$, so (e) is out.
We need to find the radius of the inscribed circle.
For the center of the inscribed circle, we need the distances to the sides to satisfy
$$a=b=\frac{\sqrt 2}{2}(1-a-b)$$
Solving this gives
$$a=b=1-\frac{\sqrt 2}2$$
Three times that clearly is not $1$, so choice (d) is out.