I don't think after playing 20 or so levels I was much smarter for it. It turned out to be a bunch of rules that were extremely limited. You do begin to appreciate trig much more.
If you are being given a geometry subject test and are prepared for it, this is fair game. If the question is just given out of the blue as a brain teaser, then no, the university is not mining useful signals at all.
Either they can't be gamed, or they select for people who can at least be bothered to do the work to game them.
Which is par for the course for an educational institution (and a business one could argue).
Or you can easily derive the radius if you remember that a meter was initially defined as 1/10,000,000 of the distance from the equator to the North Pole. So you can easily calculate the Earth's circumference, and from there the radius: 10,000,000 * 4 / 1000 / 3.1416 / 2 = 6366 km.
Wikipedia gives 6371 km as the mean radius.
Perhaps it's multiple choice.
A. Dreadfully sorry, but I'm afraid I haven't memorized the circumference of Earth.
B. What do you mean? Over the poles or 'round the equator?
C. Is that 10 km above at all points on the rope, or just one?
D. What's the rope made of? Jute? Sisal? Nylon? How far apart are the supports? Young's modulus...
E. I don't know about that, but I have inherited a rather large quantity of string in 3-inch lengths...2 x Pi x Cat would levitate the rope by the height of the cat along the entire circumference of the Earth, allowing a billion cats to go under it simultaneously.
You could even make do with no additional rope if you're allowed to take advantage of terrain.
Assuming a 0.3 meter high cat, you only need ~0.12mm of extra rope (about the width of two sheets of paper) to be able to pull it up at a point and let the cat through.
https://upload.wikimedia.org/wikipedia/commons/2/21/Geometri...
R ≈ 6.4E6
h = 0.3
d = √(2Rh + h²) ≈ 2.0E3
γ ≈ sin γ = d/(R + h) ≈ 3.1E-4
extra rope length = 2 × (d − γR) = 2R × (tan γ − γ) ≈ 2R × γ³/3 ≈ 0.13E-3
If kovek's question was like the first image there, it'd be a simple radius/circumference calculation (2 * pi * 5km). But it sounds like the second image to me - which makes it into more of a horizon line problem.
(I'm assuming all cats are shorter than all dogs.)
My interpretation is that the rope is pulled taut to a height of 10 km at a single point, so that it runs in a straight line to the horizon in either direction and lays on the ground the rest of the way; this is also relevant to the question of longest line of sight from a given height.
So yes, you could supply the answer in terms of r radius, but you do need to know the radius unless I'm missing something obvious.
I was not able to answer that question right away, and their hint to draw it helped a lot. I did not end up getting accepted.