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    (Original post by jack.hadamard)
    It's significantly more than that. Roughly, people get this wrong 95% of the time.
    Oh yeh... I missed all them. Sack this, I'm not counting triangles.
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    (Original post by jack.hadamard)
    Problem 230

    How many triangles are there?

    Attachment 225644

    I think more mathematicians should be interested in cognitive psychology.
    40

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    ... base 16 :teehee:
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    (Original post by jack.hadamard)
    Problem 230

    How many triangles are there?

    Attachment 225644

    I think more mathematicians should be interested in cognitive psychology.
    Same answer as LotF, but I'll provide a method.

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    Lower right corner limited to one "slice" is 4 triangles, combining to give 10 total. Hence 4 slices give 40. Symmetrically for the lower left corner but any triangle which uses the bottom-most pieces have already been counted, so 3 triangles combining to give 6 then times 4 slices gives 24. 40 + 24 = 64.
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    (Original post by Lord of the Flies)
    40
    Correct, (of course)!


    (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...

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    You can obtain the recurrence f(n-1) + n + 1 = f(n) for the number of ways to combine regions in the triangle when you have edges connecting its down-left vertex to the edge on the right. Solving it, you get f(n) = \frac{1}{2}n(n+3) + 1. This can easily be proved by induction (once you know the formula). Then, by introducing edges from the down-right vertex on the left edge, you see that the total number of triangles is g(n,m) = (m+1)f(n) + f(m-1)(n+1) where n are vertices on the right and m on the left edge.

    Plugging n = m = 3 luckily gives 64, so the formula may be correct.

    By the way, I didn't count in this way first and got slightly off the true number.
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    (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! 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.
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    (Original post by jack.hadamard)
    Correct, (of course)!




    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:
    Show

    You can obtain the recurrence f(n-1) + n + 1 = f(n) for the number of ways to combine regions in the triangle when you have edges connecting its down-left vertex to the edge on the right. Solving it, you get f(n) = \frac{1}{2}n(n+3) + 1. This can easily be proved by induction (once you know the formula). Then, by introducing edges from the down-right vertex on the left edge, you see that the total number of triangles is g(n,m) = (m+1)f(n) + f(m-1)(n+1) where n are vertices on the right and m on the left edge.

    Plugging n = m = 3 luckily gives 64, so the formula may be correct.

    By the way, I didn't count in this way first and got slightly off the true number.

    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.
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    (Original post by jack.hadamard)
    Problem 230

    How many triangles are there?

    Attachment 225644

    I think more mathematicians should be interested in cognitive psychology.
    I got 60. :unsure:

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

    How many triangles are there?

    Attachment 225644

    I think more mathematicians should be interested in cognitive psychology.
    25?
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    Let's stick something different into the thread..

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

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

    Spoiler:
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    A glimpse into the geometric side of topology. A homeomorphism between two topological spaces says they are topologically identical, and consists of a pair of continuous functions  f:A\mapsto C, g: C\mapsto A , such that f and g are inverses of each other (more precisely, f(g(x)) = identity of C, g(f(x)) = identity of A). The question then is now simply to find such functions.
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    (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 ***.
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    Which method of proof do you identify with?
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    (Original post by ukdragon37)
    Which method of proof do you identify with?
    I like the Dirac and Thermodynamical one
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    (Original post by ukdragon37)
    Which method of proof do you identify with?
    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.
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    (Original post by ukdragon37)
    Which method of proof do you identify with?
    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...
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    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:
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    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 \le r \le n and consider all subsets of r elements of the set \{1,2, \cdots , n \}. Each of these subsets has a smallest member.
    Let F(n,r) denote the arithmetic mean of these smallest numbers.

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

    Problem 233*/**

    Determine that maximum value of m^3+n^3, where m and n are integers satisfying m, n \in \{1,2, \cdots , 2013 \} and (n^2-mn-m^2)^2=1.

    Edit: Another fun problem (not from IMO):

    Problem 234*

    Find all possible n-tuples of reals x_1,x_2, \cdots ,x_n such that \prod_{i=1}^{n} x_i = 1 and \prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i = 1 for all 1 \le k \le n-1

    Problem 235*

    Prove that \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.
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    THE BOLZANO-WEIERSTRASS METHOD AND SCHRODINGER METHOD FOR ME. :rolleyes:
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    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].
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    Problem 236 *

    Find

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

    Problem 237 */**

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

    For the last one, there is an easier way than partial fractions.
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    (Original post by james22)
    Problem 236 *
    Find\int (sin(ln(x))+cos(ln(x)))\ dx
    Solution 236

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

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    (Original post by hecandothatfromran)
    Ok, I'll drop one of mine on here.
    Using methods encountered at A-level, find the mean value of f(x) over an interval [0,xa].
    Can you edit your post to add: Problem 238*

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