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The Proof is Trivial!

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Original post by jack.hadamard
It's significantly more than that. Roughly, people get this wrong 95% of the time. :tongue:


Oh yeh... I missed all them. Sack this, I'm not counting triangles.
Original post by jack.hadamard
Problem 230

How many triangles are there?

triangle.jpg

I think more mathematicians should be interested in cognitive psychology. :tongue:


40

Spoiler

(edited 10 years ago)
Original post by jack.hadamard
Problem 230

How many triangles are there?

triangle.jpg

I think more mathematicians should be interested in cognitive psychology. :tongue:


Same answer as LotF, but I'll provide a method.

Spoiler

Original post by Lord of the Flies
40


Correct, (of course)! :biggrin:


Original post by ukdragon37
...


A nice way (maybe, I haven't actually tried it) would be to consider the triangle as a graph, rank the vertices and then try to count the 3-vertex cliques through subsequent edge contractions. Anyway, it's still too complicated.

Continuing your way...

Spoiler

Original post by jack.hadamard
A nice way (maybe, I haven't actually tried it) would be to consider the triangle as a graph, rank the vertices and then try to count the 3-vertex cliques through subsequent edge contractions. Anyway, it's still too complicated.


But it's exactly how you would program a computer to do it! :wink: I'm not a theoretical computer scientist in the American (combinatorial) sense, so I'm pleasantly surprised I came up with a method that got it right. :biggrin:
(edited 10 years ago)
Reply 1605
Avatar for Zakee
Zakee
Original post by jack.hadamard
Correct, (of course)! :biggrin:




A nice way (maybe, I haven't actually tried it) would be to consider the triangle as a graph, rank the vertices and then try to count the 3-vertex cliques through subsequent edge contractions. Anyway, it's still too complicated.

Continuing your way...

Spoiler




I'm not sure how you did it, but I realized that you could do something like this:

Each vertex going up contains a certain number of triangles. (the verticies being the points of intersection).

There is a recurrence relationship, where you have 1(1) triangle, then 2(2) triangles, 3(3) triangles, 4(4) triangles. Then as you go higher up the value starts to decrease so you have one region with 3(5) triangles, one region with 2(6) triangles and one region with 1(7) triangles.

Is it just a coincidence that the maximum number of repeated triangles within a region where two lines intersect (i.e, there are 4 regions with 4 triangles) is the maximum number of regions of a certain value to the power of the the shape which is being used? As 4 regions is my most of a repeated value, and I've a triangle, it's 4^3? = 64. Maybe I'm spewing poppycock. :tongue:
Original post by jack.hadamard
Problem 230

How many triangles are there?

triangle.jpg

I think more mathematicians should be interested in cognitive psychology. :tongue:


I got 60. :unsure:

Okay... apparently, I missed 4.
(edited 10 years ago)
Original post by jack.hadamard
Problem 230

How many triangles are there?

triangle.jpg

I think more mathematicians should be interested in cognitive psychology. :tongue:


25?
Let's stick something different into the thread..

Problem 231* (*** if you don't look at the spoiler)

Prove the annulus A={(x,y)∈R2:1≤x2+y2≤4} A = \{ (x,y) \in \mathbb{R}^2 : 1 \leq x^2+y^2 \leq 4 \} is homeomorphic to the cylinder C={(x,y,z)∈R3:x2+y2=1,0≤z≤1} C = \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2=1, 0 \leq z \leq 1 \}

Spoiler

Original post by Zakee

Is it just a coincidence that the maximum number of repeated triangles within a region where two lines intersect (i.e, there are 4 regions with 4 triangles) is the maximum number of regions of a certain value to the power of the the shape which is being used? As 4 regions is my most of a repeated value, and I've a triangle, it's 4^3? = 64.


Your question is not very clear to me (i.e. why expect such a relation), but I will guess no.

Original post by FireGarden

Problem 231* (*** if you don't look at the spoiler)


Even if you look at the spoiler, it is still ***. :smile:
Which method of proof do you identify with? :tongue:
Original post by ukdragon37
Which method of proof do you identify with? :tongue:


I like the Dirac and Thermodynamical one :biggrin:
Original post by ukdragon37
Which method of proof do you identify with? :tongue:


That is quite possibly the best thing I've ever read ever. And quite cutting edge given the time - in 1938, Quantum mechanics and Tauberian theorems were very much in their infancy in the 1930s. Love it.
Reply 1613
Avatar for Jkn
Jkn
11
Original post by ukdragon37
Which method of proof do you identify with? :tongue:

Awesome!

My favourites are "The Method of Inverse Geometry" and "The Schrödinger Method" :lol:

However, I can't help but feel as though it could be dramatically expanded upon by including methods from philosophy...
I can't believe it.... ten thousand words of category theory, done and submitted. I never thought I'd be able to :cry:

Now to get some sleep. :sleep:
Reply 1615
Avatar for Jkn
Jkn
11
Had a nice afternoon avoiding STEP and indulging in some IMO problems! These questions are from a paper sat by Timonthy Gowers and Imre Leader who got Gold (42/42) and Silver (37/42) respectively (where each question is work 7). Also note that, whilst most IMO problems nowadays are restricted to those who are comfortable with the techniques taught, these particular problems do not require such knowledge and so should be just as accessible to someone who has been preparing for a year as they would to someone who hasn't (though there are many intern ate solutions to each). Enjoy!

Problem 232*/**

Let 1≤r≤n1 \le r \le n and consider all subsets of rr elements of the set {1,2,⋯ ,n}\{1,2, \cdots , n \}. Each of these subsets has a smallest member.
Let F(n,r)F(n,r) denote the arithmetic mean of these smallest numbers.

Prove that F(n,r)=n+1r+1\displaystyle F(n,r)=\frac{n+1}{r+1}.

Problem 233*/**

Determine that maximum value of m3+n3m^3+n^3, where mm and nn are integers satisfying m,n∈{1,2,⋯ ,2013}m, n \in \{1,2, \cdots , 2013 \} and (n2−mn−m2)2=1(n^2-mn-m^2)^2=1.

Edit: Another fun problem (not from IMO):

Problem 234*

Find all possible n-tuples of reals x1,x2,⋯ ,xnx_1,x_2, \cdots ,x_n such that ∏i=1nxi=1\prod_{i=1}^{n} x_i = 1 and ∏i=1kxi−∏i=k+1nxi=1\prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i = 1 for all 1≤k≤n−11 \le k \le n-1

Problem 235*

Prove that 1+11!2!+122!3!3+⋯+1(n−1)(n−1)!n−1n!n>2(n2+n−1)n(n+1)\displaystyle 1+\frac{1}{1! \sqrt{2!}}+\frac{1}{2 \sqrt{2!} \sqrt[3]{3!}}+ \cdots + \frac{1}{(n-1) \sqrt[n-1]{(n-1)!} \sqrt[n]{n!}} > \frac{2(n^2+n-1)}{n(n+1)}
where n is a natural number greater than 1.
(edited 10 years ago)
Reply 1616
Avatar for Zakee
Zakee
THE BOLZANO-WEIERSTRASS METHOD AND SCHRODINGER METHOD FOR ME. :rolleyes:
Ok, I'll drop one of mine on here.

problem 238 *

Using methods encountered at A-level, find the mean value of f(x) over an interval [0,xa].
(edited 10 years ago)
Problem 236 *

Find

∫(sin(ln(x))+cos(ln(x))) dx\int (sin(ln(x))+cos(ln(x)))\ dx

Problem 237 */**

∫0∞x29(5x2+49)17 dx\int^{\infty}_{0} \frac{x^{29}}{(5x^2+49)^{17}}\ dx

For the last one, there is an easier way than partial fractions.
(edited 10 years ago)

Spoiler

Solution 236

Just by inspection really: xsin(ln(x))+Cxsin(ln(x))+C

Spoiler


Solution 238
I may be simplifying things too much here, but would it just be ∫0xaf(x)xadx\displaystyle \int_{0}^{x_a}\frac{f(x)}{x_a}dx?
(edited 10 years ago)

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