Problem 276 


Name Message
Dannyz
Posts: 5

Joined: 8/29/2013
Topic  Problem 276       Flag »  Reply »
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
Reply  Re: Problem 276 Flag »  Reply »
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^2-2ax-a^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^2-4ac 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
Reply  Re: Problem 276 Flag »  Reply »
Base on the figure given, I let y be the "vertical y"; so y=x-xcos(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
Reply  Re: Problem 276 Flag »  Reply »
Sorry--finally 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
Reply  Re: Problem 276 Flag »  Reply »
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
Reply  Re: Problem 276 Flag »  Reply »
Since tan(theta) + sec(theta) = (a+x)/x, sec(theta) = (a+x)/x-tan(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
Reply  Re: Problem 276 Flag »  Reply »
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
Reply  Re: Problem 276 Flag »  Reply »
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 square-then 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


 
» Reply to Problem 276 Topic

In order to post a new message, you must first be logged into an account.

Website design and hosting provided by 270net Technologies in Frederick, Maryland.