Jun 17, 2018

$D P + E P + P F = 5 \sqrt{3}$ regardless where P is in the triangle.
This is called Viviani's theorem: https://en.wikipedia.org/wiki/Viviani%27s_theorem

#### Explanation:

ABC is the triangle. It's height $h = 5 \sqrt{3}$ (use Pythagoras on for instance ADC)

Remember that the area of a triangle is $\frac{b a s e \cdot h e i g h t}{2}$.

Now all three triangles APC, BPC and APB have a base = 10 since all sides are equal and 10. Their total area is equal to the area of ABC

therefore $\frac{s i \mathrm{de} \cdot C M}{2} = \frac{s i \mathrm{de} \cdot h}{2}$
=$\frac{s i \mathrm{de} \cdot D P}{2} + \frac{s i \mathrm{de} \cdot E P}{2} \cdot \frac{s i \mathrm{de} \cdot F P}{2}$

Therefore $D P + E P + P F = M C = 5 \sqrt{3}$ regardless where P is in the triangle.

Jun 17, 2018

Another one from Rahul's Book!

I'd say that given we're given choices, the result doesn't depend on our choice of $P$.

Let's pick $P = G ,$ the centroid/orthocenter. We get an altitude ${5}^{2} + {h}^{2} = {10}^{2}$ or $h = 5 \sqrt{3}$. The medians meet $\frac{2}{3}$ from each vertex, and the remaining piece of $\frac{1}{3} h$ is perpendicular to the sides. We have three of these to add up, they add to $h$, choice (2).