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Carrotsticks

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Here are two not-so-difficult problems I thought of a few days ago. It should be do-able for a keen 3U student.

Problem #1:



Consider the following *poorly drawn* diagram. We start with a right angled triangle of hypotenuse length n. The 'base' of this triangle is the hypotenuse of another triangle, and so the process repeats until it can no longer continue. The length of the other side is kept as a constant 1.

Define the sum of k triangles to be S(k), such that



Show that:



Problem #2:



A man starts at point X. He walks 1m North. He then walks 2m East, then 3m South, then 4m West etc etc until he finally walks N metres either N/S/E/W, depending on N.

Find the distance the man is from the X after the 'lap' involving him walking N metres.

Interesting variation to Problem #2:

A man starts from some point X. He walks 1m North. He then walks 1/2m East, then 1/3m South, then 1/4m West etc and continues with this pattern indefinitely.

Will he reach a finite point? If so, how far is it from his original position X? <a< \sqrt="" n="" \frac{2n+3}{3}[="" tex]<="" html=""></a<>
 
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Nooblet94

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Here are two not-so-difficult problems I thought of a few days ago. It should be do-able for a keen 3U student.

Problem #1:



Consider the following *poorly drawn* diagram. We start with a right angled triangle of hypotenuse length n. The 'base' of this triangle is the hypotenuse of another triangle, and so the process repeats until it can no longer continue. The length of the other side is kept as a constant 1.

Let the sum of all the triangles be A

Show that:



Problem #2:



A man starts at point X. He walks 1m North. He then walks 2m East, then 3m South, then 4m West etc etc until he finally walks N metres either N/S/E/W, depending on N.

Find the distance the man is from the X after the 'lap' involving him walking N metres.

Interesting variation to Problem #2:

A man starts from some point X. He walks 1m North. He then walks 1/2m East, then 1/3m South, then 1/4m West etc and continues with this pattern indefinitely.

Will he reach a finite point? If so, how far is it from his original position X? <a< \sqrt="" n="" \frac{2n+3}{3}[="" tex]<="" html=""></a<>
"Let the sum of all triangles be A"
I'm assuming you mean the sum of the area of all the triangles?
 

lolcakes52

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Okay, I gave up on the second problem for now as im a bit rusty on my modular arithmetic.
 

kingkong123

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Oh and incase you didn't know about this (it will probably help but I hope it doesn't give too much away!):

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Alternating_harmonic_series

http://en.wikipedia.org/wiki/Leibniz_formula_for_π
for problem 1, i got

<a href="http://www.codecogs.com/eqnedit.php?latex=A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" title="A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" /></a>

is that any where near anything? am i on any right track?
 

Carrotsticks

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for problem 1, i got

<a href="http://www.codecogs.com/eqnedit.php?latex=A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" title="A = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{\sqrt{n^{2}-k}}" /></a>

is that any where near anything? am i on any right track?
Your expression is incorrect, but you are on the right track.

Since n is a finite value, we cannot sum the expression to infinity otherwise at some point (and onwards), the inside of the square root becomes negative.

Also how come you have a fraction for the sum?
 
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seanieg89

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I think Q1 is incorrect...if for example we set n=2, we get A = 2.073 and a lower bound of 2.89 approximately. (There are only three triangles in this case.)

It also seems a little difficult/fiddly to determine whether or not the sequence of triangles "wraps all the way around" the common point to all triangles. Is overlap allowed?

Second question is pretty cool, I will let more people attempt it before I post a solution :).
 

Carrotsticks

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Thanks sean. When I made the question, I forgot to properly define the number of triangles, and it got mixed up with the length N and even confused me!

Regarding the variation, I think it's absolutely fascinating how by simply walking up and down, left and right, we expose ourselves to the two *arguably* most important numbers in Mathematics.

And yeah overlap is allowed. It doesn't necessarily 'wrap all the way around'. The tail of all the triangles isn't necessarily the converging point.
 

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