June 20th, 2014, 11:21 AM  #11 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 
I was looking forward for your question. This can be done by counting endpoints of diagonals, I think, which avoids counting diagonals and their meetpoints altogether. The number of triangles in the interior of diagonals with three endpoints is $\binom{6}{3}$. The number of triangles in the interior of diagonals with four endpoints is $4\binom{6}{4}$. Similarly, the number of triangles in the interior of diagonals with five endpoints is $5 \binom{6}{5}$ (note that the diagonals with 5 endpoints must meet themselves in a hexagon) Thus, the number of triangles is $$\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} = 110$$ 
June 20th, 2014, 03:38 PM  #13  
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Quote:
Code: prod(i=1,primepi(2014),g=2014;p=prime(i);(p^sum(j=0,log(g)\log(p),g\=p)+1))  
June 20th, 2014, 04:16 PM  #14 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I'm amused by how similar our programs are and yet how different they appear. The prod syntax is nicer (no need for a temporary variable to hold the sum) but generating the nth prime at each step is inefficient. Not that it matters when you're only looping up to 2014, of course! I think we're ready for a new question? 
June 21st, 2014, 03:31 AM  #15 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 
Here are the question, the one who gets one of the hard questions below gets the control. Easy : Let $f$ be a univariate complex function satisfying $f(x) = f(x + \omega_1) = f(x+\omega_2) = f(x + \omega_3)$ with $\omega_1, \omega_2, \omega_3$ being $\Bbb R$linearly independent. Can a function like that exist? If so, give example otherwise show proof. [Bonus point 2] Hard : Let $f$ be defined as above except $\omega_1, \omega_2, \omega_3$ being $\Bbb Q$linearly independent. Can such a function exist? If so, give example otherwise show proof. Hard (this one is for real analysis lovers) : Let $f : \Bbb R \to \Bbb R$ be a function such that for all $x, y \in \Bbb R$ one has $f(x + y^2)  f(x) \geq y$. Can such a function exist? If so, give example otherwise show proof. Balarka . Last edited by mathbalarka; June 21st, 2014 at 03:36 AM. 
June 21st, 2014, 04:39 AM  #16 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
What does it mean to say that the $\omega_i$ are $\mathbb{R}$ linearly independant? Aren't they constants?
Last edited by v8archie; June 21st, 2014 at 04:41 AM. 
June 21st, 2014, 06:22 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
\begin{align*} f(x + y^2)  f(x) &\ge y \implies f(x + y)  f(x) \ge \sqrt{y} \qquad \text{(the positive root since $\sqrt{y} \ge \sqrt{y}$)} \\[12pt] \text{For } a \in \mathbb{R}^+, n \in \mathbb{Z}^+ \qquad f(x + \frac{a}{n})  f(x) &\ge \sqrt{\frac{a}{n}} \\ f((x + \frac{a}{n}) + \frac{a}{n})  f(x + \frac{a}{n}) &\ge \sqrt{\frac{a}{n}} \\ f((x + \frac{2a}{n}) + \frac{a}{n})  f(x + \frac{2a}{n}) &\ge \sqrt{\frac{a}{n}} \\ &\cdots \\ f((x + \frac{(n1)a}{n}) + \frac{a}{n})  f(x + \frac{(n1)a}{n}) &\ge \sqrt{\frac{a}{n}} \\[6pt] \text{Summing the above equations gives} \qquad f(x+a)  f(x) &\ge \sqrt{an} \\ \end{align*} Then, letting $n \to \infty$ we see that $f(x+a)  f(x) \to \infty$. This is true for all $x$ and all $a \gt 0$ so no such function exists. Last edited by v8archie; June 21st, 2014 at 06:25 AM. 
June 21st, 2014, 06:47 AM  #18 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  
June 21st, 2014, 06:51 AM  #19 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  
June 21st, 2014, 07:31 AM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
I'm not familiar with the Basel problem. I first proved it to myself using l'Hopital's Rule on$$\lim_{y \to 0} \frac{f(x + y^2)  f(x)}{y} \ge 1$$ but I wasn't completely happy with the validity of the argument. 

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