Recent content by bob012345

This is a little exercise for fun. Determine the area of the darker petals to the total area. The four circles are identical. Have fun!

As @Filip Larsen suggests, 2 clockwise.

At the Earth’s poles, the answer is indeterminate. East and West do not exist.

It’s not homework other than the sense that I’m working from home!

I’m just interested in seeing the various responses to the question. What do people do with it.

TL;DR: How far is due East from due West on Earth? The question is simply how far is due East from due West on Earth? I hope this generates some fun discussion. Enjoy! Edit: This is meant to be taken as a puzzle.

I often wondered if there exists any vantage point where a Hubble image of a distant object for example, could be seen similarly by the human eye? For example this…

I call it a Moondoggle.

Expanding the original problem to be an ## nx1## rectangle, it turns out the angle B to add to angle A (defined by the ratio ##\large\frac{1}{n}##) to get ##\pi/4## is defined by the ratio ##\large\frac{n-1}{n+1}##. Here is an example by construction for ##n=4##. First, construct a circle of...

One can take the log of both side of the inequality and plot that. That plot looks identical but goes on exponentially further. The purple is blue overlapping red and boundary between blue and purple is where it’s failing due to large exponents. Also, it’s obvious but if you look at the...

The plot can also be generated easily on Desmos.

It can be shown that the yellow figure of the last graphic in post #8 can be completely contained in the blue figure. Take off an equivalent yellow area from the blue area yields only blue. That proves the original yellow shape was smaller than the blue thus less than a quarter circle.

Here is another way but it’s just a variation of @coquelicot above in post #7 but the lengths are not obvious so they must be calculated. Drop a perpendicular at point Q. Since point P is at the midpoint of the vertical line the distance PQ is $$\frac{1}{2\sqrt{2}}$$ Then the distance QM is...