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### Regular polygon of radius $1$ with diagonals: mysterious ring of radius $1/e$?

I have two graphs. They refer to white dots, which are small discs in the unit circle not crossed by any chord. The first graph shows the size of the largest white dots, for each polygon up to 200 ...
• 47.9k
1 vote

### How to show that $\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$

Substitute $x=\frac12(1+t)$ $$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\ dx =\frac14\int_{-1}^1\sqrt{1-t^2}\ dt=\frac{\pi}{8}$$
• 77k

### How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)?

$10 + 2 \sqrt 5$ has norm $100 - 5 \cdot 4 = 80.$ $5 + 2 \sqrt 5$ has norm $25 - 5 \cdot 4 = 5.$ The ratio of the norms is $\frac{80}{5} = 16,$ which is an integer and a square, so the ratio ...
• 133k
Accepted

### How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)?

Yes. There is a way to formalize this particular type of sum of square roots, similar to the way a determinant is developed for quadratic equations. We can write the generic form of the expression in ...
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• 184
1 vote
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Either the problem uses minimum instead of maximum, or the solution uses maximum instead of minimum. I've made a quick sketch of the problem. The area you care about is the intersection between $\... • 34.5k 0 votes ### Element of area in hyperbolic plane with polar coordinates on hyperboloid model Here's a calculation in change-of-coordinates language. (The use of$r$throughout may be simpler despite my comment; leaving that as an exercise. Also, it should be straightforward to use ... • 73.3k 3 votes ### What is the product of the areas of every regular polygon inscribed in a circle of area$1$? Not a closed form but some ideas.$$P=\prod\limits_{k=3}^{\infty}\frac{k}{2\pi}\sin{\left(\frac{2\pi}{k}\right)}=\frac{135 \sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)}}{16 \pi ^4}\,\,\prod\limits_{k=7}^{... • 228k 0 votes ### Finding the area of a implicit relation Being a topic of 8 years ago with already a very detailed answer, I certainly don't have the ambition to write innovative things, it's to be understood as a corollary that could be useful to someone. ... • 3,455 2 votes Accepted ### How to prove that the triangle areas between squares on a isosceles triangle are equal The sum of the angles around each vertex is$2\pi$. Since the sum of the angles formed by the squares is$\pi\$, so is the sum of the other two angles. The result then follows from the formula for the ...
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