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Question

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 $10\sqrt7$

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Hint:

The distance has to be a straight line along the net of the pyramid. There are two options:

enter image description here

The first line has lenght $10\sqrt{7+2\sqrt{3}}$, and the second one has $10\sqrt{7}$, which is the smaller one.

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  • 6
    $\begingroup$ I'd also suspect you can't dig under the pyramid $\endgroup$ – Zereges Jul 11 at 22:11
  • $\begingroup$ Minor nitpick: Is it allowed to talk about the net of the pyramid? $\endgroup$ – Eric Duminil Jul 12 at 7:57
  • $\begingroup$ Question: how do you reason about shapes that can not be flattened like such? $\endgroup$ – orlp Jul 12 at 11:01
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    $\begingroup$ @orlp A curve of shortest distance which connects two points on a curved surface is called a geodesic. Finding geodesics is in general nontrivial and involves solving systems of differential equations. $\endgroup$ – mechanodroid Jul 12 at 11:07
  • $\begingroup$ Supplementary question: is it true for all square-based pyramids, regardless of dimensions, that the shortest route will never involve "digging under" the base? $\endgroup$ – Michael Kay Jul 12 at 11:08
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Draw the net of the pyramid and use the cosine rule. enter image description here

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    $\begingroup$ How do you know this is shorter than going through the square face? $\endgroup$ – JiK Jul 12 at 10:34
  • $\begingroup$ @JiK, isn't it obvious? The hypotenuse of that right triangle is longer than 20 (longer than the length of the equilateral triangle) $\endgroup$ – Seyed Jul 12 at 11:10
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    $\begingroup$ Interestingly, if base length is $1$ and side length is $2$, the shortest route will be the one along the square face. $\endgroup$ – mechanodroid Jul 12 at 11:38
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    $\begingroup$ @JiK It isn't necessary to consider the square face as the problem statement talks about the outer surface of the pyramid. The way pyramids are usually displayed, a here, makes the outer surface not contain the base, which is the square. $\endgroup$ – Ingix Jul 13 at 10:24
  • $\begingroup$ @Ingix I don't understand. The outside of the square face is in the outer surface of the pyramid. How does the way it's displayed change it? Why can you use the PSO face, which is also hidden, but not the square face? $\endgroup$ – JiK Jul 13 at 13:30
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Hint: Draw the net diagram of the square pyramid and the you will find two points that are mid-point of $OR$. As you know that, minimum distance between any two points is the straight line joining them, just connect $P$ and midpoint of $OR$ by a straight line. Then just find the length of this line and exclude the length of this line that goes out of the net diagram. This length of line will be minimum length. Why?

Note: You can choose any one of the two midpoint because since the diagram is symmetric, the two lengths will be the same.

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Cut along OQ and OR with a scissors, fold out that face to put the faces OPQ and OQR in one plane. Then the shortest path is a straight line from vertex P to the midpoint M of an opposite side of a 60-120 rhombus, I think.

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Draw the black and blue lines as shown below:

$\hspace{3cm}$![enter image description here

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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