The Proof is Trivial!

I've posted this problem before (not on this thread)

Problem 242 *

Problem 241 * / **

i) Find $\displaystyle \sin \left(\frac{\pi}{5}\right)$.

ii) Prove that $\displaystyle \pi \phi > 5$, where $\phi$ is the golden ratio.


I made this problem up and it is not straightforward for a single star.

Now, the arc of a polygon of circumradius 1 (which is enclosed by a surrounding circle of radius 1)...

Area of a regular n-gon with unit circumradius. The two-star method I intended for the question was to first find $\sin\left(\frac{\pi}{5}\right)$. It is a root of the equation $x(1 + 4(1 - x^2)(1 - 4x^2)) = 0$. Hence, get $\sin\left(\frac{\pi}{10}\right)$ and use it in

$\displaystyle \int_{0}^{\frac{1}{2}\phi^{-1}} 1 - \frac{1}{\sqrt{1 - x^2}}\ dx$

to obtain the inequality.

What specific ** technique is being used?

By parts is reverse product rule which is C3

Don't discourage people from posting questions. If you don't like a question, then don't do it.

...

Don't discourage people from posting questions. If you don't like a question, then don't do it.

Yes, I did, but still thought the answer was kind of irritating.

Yes that's right! Why did you think the answer is irritating?

It's just because I find it not-quite-so intuitive. Having the same probability of the three numbers (picked from all positive integers) being the sides of a triangle as they being not the sides of a triangle is puzzling. If we try to formalise such statements, then considering the probability space $(\Omega, \mathcal{F}, \text{P})$ with

i) $\Omega := \{\ 1, 2, 3, ...\ \}$, $\mathcal{F} = \mathcal{P}(\Omega)$,

ii) if $A \in \mathcal{F}$, then $\displaystyle \text{P}(A) = \sum_{a \in A} \text{P}(\{a\}) = \pi |A|$ where $\pi$ is the (equal) probability of any given positive integer.

makes no sense, since $\text{P}(\Omega) = \begin{cases} 0\quad\ \text{when}\ \pi = 0 \\ \infty\quad \text{when}\ \pi > 0 \end{cases}$.

Following the above, I also find other "intuitive" statements irritating. Like picking an even integer among all the positive integers, which has a 'probability' of exactly one half. It's maybe because I am broken; I don't know.

Nobody does my problems anymore .

Oh well, simple, boring things.. simply 'cause I'm interested in what proofs might emerge..

Problem 243*

Find and prove formulae for the sum of the first n odd, and first n even numbers.

Solution 243*

Odd: $\displaystyle \sum^n_{r=1}(2r-1) = 2 \sum^n_{r=1}r - n = n(n+1) - n = n^2$

Even: $\displaystyle \sum^n_{r=1} (2r) = 2\sum^n_{r=1} r = n(n+1)$

These problems are far too easy for this thread.

Okay, solve it without using Mathematics.

What universities are you considering (or have applied to, or have been accepted in)?

Nobody does my problems anymore .

Solve it using bananas. (Say this in Rowan Atkinsons' voice)

Most probably, I will take a gap year...