3-5-7-L Problem

[sallycats]

I have been stuck on this problem and would greatly appreciate any help!
I've played around with it, trying out the Law of Cosines but have not been able to find L. Am I on the right track? Or should I try something else to solve this?

 

[TchrWill]

A point within a square is 3 units from one corner, 5 units from the next corner, and 7 units from the next.

First our picture:
Draw yourself a square ABCD, A lower left, B upper left, C upper right, and D lower right. Label the sides of the square x (or L). Locate a point P 3 from A, 5 from B, and 7 from C. Draw the perpendicular from P to AB and label it y. Darw the perpendicular from P to BC and label it z. From our picture:

1--y^2 + z^2 = 25
2--y^2 + (x - z)^2 = 9
3--z^2 + (x - y)^2 = 49
4--Subtract (2) from (1)
5--Subtract (2) from (3)

Can you take it from here?

 

[sallycats]

Thank you so much for you help, TchrWill! I have been working with your suggestions, but still find myself stuck.
The picture you described was slightly different from the one I have. Here is what I have.

1 (L-y)[sup:ud6vonyk]2[/sup:ud6vonyk] + x[sup:ud6vonyk]2[/sup:ud6vonyk]= 25
2 (L-x)[sup:ud6vonyk]2[/sup:ud6vonyk] + y[sup:ud6vonyk]2[/sup:ud6vonyk]= 49
3 y[sup:ud6vonyk]2[/sup:ud6vonyk] + x[sup:ud6vonyk]2[/sup:ud6vonyk] = 9

From there, I attempted elimination by subtracting 1, 2, and 3 as you suggested.

I reached the following in doing so:
-2Lx + 2Ly = 24 (or Ly - Lx = 12 or y = 12/L + x or x = y - 12/L)
L[sup:ud6vonyk]2[/sup:ud6vonyk] - 2Lx = 40
L[sup:ud6vonyk]2[/sup:ud6vonyk] - 2Ly = 16

And I'm having difficulty taking it from there. I've tried substitution using various numbers, but have had no luck.
Again, any additional tips or suggestions you or anyone else may have would be a great help.
Thank you again!

 

[Denis]

Yikes...that's not an easy one, Sally ... even if it appears easy :wink:
Agree with what you got to:
L^2 - 2Lx = 40
L^2 - 2Ly = 16

So 2Lx + 40 = 2Ly + 16 : Lx + 20 = Ly + 8 : Ly - Lx = 12 : L = 12 / (y - x) ...phew! OK?

From your equation3, x^2 + y^2 = 9 : so y = sqrt(9 - x^2) ; substitute that in my equation above:
L = 12 / (sqrt(9 - x^2) - x) ; simplify to get: sqrt(9 - x^2) = (12 + Lx) / L
Square both sides and simplify: 2L^2 x^2 - 9L^2 + 24Lx + 144 = 0

OK; from your L^2 - 2Lx = 40, we get: x = (L^2 - 40) / (2L) ; substitute that in above, simplify:
L^4 - 74L^2 + 928 = 0 ; solve for L to get L = sqrt(58) or L = 4
L = 4 not possible, so L = sqrt(58) or ~7.62

Wanna "see" it? Draw a square, sides 7.62 inches (fits nice on 8by11 sheets!)
You'll be able to make the 3,5,7 lines "fit" nicely.

Btw, haven't looked for a simpler way; just carried on from where you were...

 

[TchrWill]

Try understanding this identical problem with different numbers after which you should be able to work out your version.

<< A point within a square is 3 units from one corner, 4 units from the next corner, and 5 units from the next. What is the area of the square? >>

First our picture:
Draw yourself a square ABCD, A lower left, B upper left, C upper right, and D lower right. Label the sides of the square x. Locate a point P 3 from A, 4 from B, and 5 from C. Draw the perpendicular from P to AB and label it y. Darw the perpendicular from P to BC and label it z. From our picture:

1--y^2 + z^2 = 16
2--y^2 + (x - z)^2 = 9
3--z^2 + (x - y)^2 = 25
4--Subtract (1) from (2) expanded yielding z = (x^2 + 7)/2x
5--Subtract (1) from (3) expanded yielding y = (x^2 - 9)/2x
6--Substituting (4) and (5) into (1) yields x^4 - 34x^2 + 65 = 0
7--Using the quadratic formula, x^2 = [34+/-sqrt(1156 - 260)]/2 = 31.966
8--Therefore, x = 5.654 and, of course, the area is x^2 = 31.966 sq. units.

 

[mmm4444bot]

Wanna "see" it? Draw a square, sides 7.62 inches Denis "the falcon" uses a very sharp pencil, indeed !

(fits nice on 8by11 sheets!) Is this what they sell in Canada?

[attachment=0:3uodofvm]BigRazz.JPG[/attachment:3uodofvm]

 

[sallycats]

This is terrific! Thank you both for your help.
I went through the example TchrWill provided, and my work produced the same results.
I then used the same process with my problem, with the same results as Denis.