anonymous
  • anonymous
So, I've got a figure, and step one on this is "On the diagonal AC find a point B_1 so that CB_1=AB." Now, my professor just let people in my class explain that you could do it with a piece of paper or string or ruler etc... but I wondered out loud if it could be done more rigorously, and the TA said she had done it algebraically. This has motivated me to attempt it. Below is figure to use, and what I have done so far. Could someone check my work thus far and let me know if I am on the right track?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Oh, also. We are given that AP = L, and AB = mL and AC= nL. (L capitalized for readability)
anonymous
  • anonymous
So, heres what I have. We have our 3 assumptions above. I apply the Pythagorean Thm, and says that \[(AB)^2+(CB)^2=(AC)^2\] and because the figure is a square, AB=AC. Giving is 2(AB)^2. If we replace the lengths with the given names we have now: \[2(ml)^2=(nl)^2\]
anonymous
  • anonymous
distributing the squares we've got: \[2*m^2*l^2=n^2l^2\] Now I just divide out the L^2 on each side, leaving is with \[2m^2=n^2\]

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anonymous
  • anonymous
My thought is that now I can solve for m (or n), and I can ignore the negative root because its a square and has to have positive sides lengths. So I get \[m = n(\sqrt2)/2\]
anonymous
  • anonymous
At this point I look at the figure And I think, we want CB_1 to equal AB. So, we take CA and subtract AB_1 from it, and the remainder equals AB right? So, \[CB_1=CA-B_1A\] and since we want CB_1 to equal AB we have \[CA-B_1A=AB\] Now this, given what we started with and found with the Pythagorean theorem allows us to solve with replacement right? So, we have \[nl-B_1A=ml\] and since we showed that m= n(sqrt(2)/2) we have reduced to one variable:\[nl - B_1A=n(\sqrt2/2)l\] which we clean up to be\[nl-n(\sqrt2)/2)l=B_1A\]
anonymous
  • anonymous
Can anyone take a look at all this and tell me if I am crazy or not? Is this all legal steps?
anonymous
  • anonymous
I feel like the last step here is just cleaning up the left side so\[nl(1-\sqrt2/2)=B_1A\] but I am stuck here. Any gurus that can take a look at my work please?
anonymous
  • anonymous
@satellite73 ? :)
anonymous
  • anonymous
@mathstudent55 pretty please hehe
anonymous
  • anonymous
In your second comment, you said AB=AC. Would you like to recant that? I think you meant \(AC=\sqrt2 AB\).
anonymous
  • anonymous
Oh, sorry, I think you meant \(AB=BC\)...
anonymous
  • anonymous
yes thats what I meant, just by the figure being declared as a square. Sorry!
anonymous
  • anonymous
Well your reasoning makes sense to me. I don't know how formal this proof is supposed to be but I'd accept it :)
anonymous
  • anonymous
Thanks, I'm glad it all makes sense so far. The proof should be sound, but doesn't have to be super rigorous. I am having a lot of trouble finishing this though. The problem says Find B_1 such that CB_1 = AB. I'm not sure I have done that , or am having trouble interpreting my final steps. Any suggestions on finishing/more clearly stating?
anonymous
  • anonymous
Maybe setting up the figures in the x-y plane might help. Assign an arbitrary origin that makes computation easy. I think your best bet might be with letting \(C=(0,0)\).

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