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Another hard math problem

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I don't understand why my professor is giving us so many hard problems to do in such a short amount of time. Please I need some advice on this problem, the hw set is due tomorrow.

All the lines in the sketch have slope 0,1,or -1, or are vertical What do the Points Pn approach?
 

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JasonRox

Homework Helper
Gold Member
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The middle of the base of each new triangle?
 
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I have only tried to figure out what point the x coordinate approaches, which has been hard enough. What makes it hard is that I don't think you can come up with a formula for the general pattern of the x coordinates because of the way the pattern is going in the diagram (for example p1 and p2 have the same x coordinate, as well as P5 and P6). The x coordinates I found were 0,.5,.5 1/4,1/2,3/8,3/8 7/16, 3/8,13/32,13/32.................
 
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Please any hints anyone? My brain has become fried trying this problem.
 

Tide

Science Advisor
Homework Helper
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I haven't worked it out but I'd try something like this:

Start with the first vector from the origin extending 45 degrees to the hypotenuse of the large triangle. The second vector is rotated 45 clockwise with respect to the first and is reduced in magnitude - call it [itex]\alpha[/itex] which I think is [itex]\frac {1}{\sqrt 2}[/itex] in this case. And the process repeats.

The relationship between successive operations on the vector is
[tex]\left( x_{n+1}, y_{n+1} \right) = \alpha \left( y_n , -x_n \right)[/tex]
which represents a ("double") geometric series which I think you can solve. Notice how the x and y are swapped along with the sign change - if I did this correctly it represents the rotation from one vector to the next.

Once you've found [itex]x_n[/itex] and [itex]y_n[/itex] just add them up.
 
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I finally realized that if you keep going, P8 will create a triangle that has the same orientation as the original triangle. The coordinates for P8 are (3/8, 3/16). Then for P16, which should just created another smaller triangle that is similar to the big triangle, I got the coordinates (51/128, 51/256). So I am guessing that the points Pn--->(2/5, 1/5). The problem now for me is that I have been trying to come up with a formula for this sequence that has its limits =(2/5,1/5) in order to prove that Pn----->(2/5,1/5). Any hints now on the formula for the sequence?
 

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