<s>[INST] Shortest distance around a pyramid 
Transcript:

The diagram shows a square based pyramid with base PQRS and vertex O. All the edges are length 20 meters. Find the shortest distance, in meters, along the outer surface of the pyramid from P to the midpoint of OR.

The only way I have been able to solve this question is using a computer and the paper is non calculator, so there must be a faster, better solution.
My solution is creating a point X on OQ and labelling the midpoint of OR as M, and setting $\theta = OPQ$. I then calculated $PX + PM$ in terms of $\theta$ and found the minimum point of this function, in order to find the shortest possible distance. Finding the minimum point would be nowhere near possible under time constraint without a computer.
I have included the correct answer, so it is the working I am looking for.
Answer:

 D $10\sqrt7$

 [/INST] Draw the black and blue lines as shown below:
$\hspace{3cm}$
The length of blue line (using Cosine theorem):
$$c+d=\sqrt{20^2+y^2-2\cdot 20y\cos 60^\circ}+\sqrt{10^2+y^2-2\cdot 10y\cos 60^\circ}=\\
\sqrt{y^2-20y+400}+\sqrt{y^2-10y+100}$$
Set its derivative to zero:
$$(c+d)'=\frac{y-10}{\sqrt{y^2-20y+400}}+\frac{y-5}{\sqrt{y^2-10y+100}}=0 \Rightarrow \\
(y-10)^2(y^2-10y+100)=(5-y)^2(y^2-20y+400) \Rightarrow \\
3y^2-20y=0 \Rightarrow \\
y=\frac{20}{3}$$
Hence:
$$c+d=\sqrt{(\frac{20}{3})^2-20\cdot \frac{20}{3}+400}+\sqrt{(\frac{20}{3})^2-10\cdot \frac{20}{3}+100}=10\sqrt{7},$$
which is intuitively less since it passes through the two triangular faces while the black line passes through the quadratic base with larger area and the triangular face. 

Anyway, to verify it, consider the length of the black line (using the Cosine theorem):
$$a+b=\sqrt{(20\sqrt{2})^2+x^2-2\cdot 20\sqrt{2}y\cos 45^\circ}+\sqrt{10^2+x^2-2\cdot 10x\cos 60^\circ}=\\
\sqrt{x^2-40x+800}+\sqrt{x^2-10x+100}$$
Set its derivative to zero:
$$(a+b)'=\frac{x-20}{\sqrt{x^2-40x+800}}+\frac{x-5}{\sqrt{x^2-10x+100}}=0 \Rightarrow \\
(x-20)^2(x^2-10x+100)=(5-x)^2(x^2-40x+800) \Rightarrow \\
13x^2-40x-800=0 \Rightarrow \\
x=\frac{20}{13}(1+3\sqrt{3})$$
Hence:
$$a+b=\sqrt{x^2-40x+800}+\sqrt{x^2-10x+100}=10\sqrt{7+2\sqrt{3}},$$
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