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Dannyz
Posts: 5
Joined: 8/29/2013

Problem 276 
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The posted solution is only valid for x=a/sqrt2. I believe the assumption was made that triangle BDC was isoseles. The problem reads "In any given right triangle..." I believe the correct solution is y= a^2x/(a^2+2ax+2x^2). I will show the derivation of that if anyone is interested.



Thursday, August 29, 2013 at 6:45:44 PM 
Euclid
Posts: 3
Joined: 9/7/2013

Re: Problem 276 
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Dannyz is right! The problem has an infinite number of solutions for Y between 0 and X because BC and X are not fixed. As Danny mentioned, the solution given is true ONLY for the right triangle where BC = X*sqrt(2). The solution suggested that Y is maximized when H is the midpoint of BC, but you can easily see that as BC increases, H will move further and further right of center. A basic geometric interpretation quickly reveals that as BC approaches infinite, Y approaches X. This can be proved by taking the limit as BC approaches infinite of Danny's equation  using L'Hopital's rule, it simplifies to Y = X. The solution triangle used is interesting and unique, however, in that the line AD perfectly bisects the square.
Solving for the general equation of Y, substitution yielded this quadratic: Y^2(2x^22axa^2)+Y(2x^3+2xa^2+2ax^2)(a^2)(x^2) = 0. It looks even uglier when you plug this into the quadratic formula! But an amazing thing happens: b^24ac turns out to be a perfect square; terms cancel and the correct answer is the same as Danny's formula above: Y = (xa^2)/(x^2 + (x+a)^2). Interestingly, the other answer for Y this quadratic yields is X (which can be thrown out).
Danny, did you arrive at your formula using the same method?



Monday, September 09, 2013 at 8:05:09 AM 
Dannyz
Posts: 5
Joined: 8/29/2013

Re: Problem 276 
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Base on the figure given, I let y be the "vertical y"; so y=xxcos(theta). I let z be the "horizontal y"; so z=acos(theta)xsin(theta). I set y=z and found tan(theta)=(2ax+a^2)/(2x^2+2ax). From that I found sin(theta) and cos(theta) in terms of x and a and then found y.
I also solved the problem based on the approach in the published solution using the fact that angle(DBC) + angle(DCB) = 45. However that only establishes the "vertical y." You must then show the "horizontal y" is the same.
I have a pdf of my solutions, but do not know how to post them.



Monday, September 09, 2013 at 10:40:15 AM 
Dannyz
Posts: 5
Joined: 8/29/2013

Re: Problem 276 
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Sorryfinally got around to viewing my last reply. Should have written, "Based on the figure given,..." Also, I should have stated that theta is angle CAD.



Saturday, September 14, 2013 at 5:54:25 PM 
Euclid
Posts: 3
Joined: 9/7/2013

Re: Problem 276 
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Ahh interesting. I can see how you derived your trig formulas for y and z (though the aCos(theta) term required a substitution using similar triangles). But substituting y for z yielded: tan(theta) + sec(theta) = (a+x)/x. How did you solve for tan(theta)? If I substitute for tan & sec in terms of y and x, it turns this into a solvable quadratic, which yields the correct answer. But I am curious to see which steps you took.
If you'd care to show more work on this, that would be great. Otherwise, you can email your pdf solution to mathsputin@outlook.com and I'd be happy to look at it.



Friday, September 20, 2013 at 1:51:03 AM 
Dannyz
Posts: 5
Joined: 8/29/2013

Re: Problem 276 
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Since tan(theta) + sec(theta) = (a+x)/x, sec(theta) = (a+x)/xtan(theta). Square both sides and substitute tan^2(theta) + 1 for sec^2(theta); tan^2(theta) drops out and tan(theta) follows easily form there...I sent the pdf to the address given.



Saturday, September 21, 2013 at 9:44:07 AM 
Euclid
Posts: 3
Joined: 9/7/2013

Re: Problem 276 
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Great work! There is one rather difficult step in your solution, when solving for sin(theta): the term sqrt[(2ax+a^2)^2 + (2x^2+2ax)^2] becomes sqrt(a^4+4a^3*x+8a^2*x^2+8ax^3+4x^4)... Factoring this fourth degree, multivariable polynomial requires some craftiness  luckily it turns out to be a perfect square, negating the radical. I tried setting it equal to the product of 2 quadratic polynomials with middle terms bax and cax, then solving for b and c... and found that b = c = 2 which makes it a perfect square.
Which program did you use to write your solutions? The diagrams and work are very neat.



Monday, September 23, 2013 at 4:34:43 AM 
Dannyz
Posts: 5
Joined: 8/29/2013

Re: Problem 276 
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I really did not know sin(theta) simplified so nicely until I solved the problem using the second method. When I saw the clean solution for y, I knew (a^4+4a^3x+8a^2x^2+8ax^3+4x^4) had to be a perfect squarethen it was easy.
I drew the figures with AutoCAD, and plotted to CutePDF Writer. Then I used OpenOffice. I imported the PDF figures, used the equation editor to write the solution, and exported the file as a PDF.



Monday, September 23, 2013 at 6:14:15 PM 