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It's given distance between

  • $AB = 27$
  • $BC = 752$
  • $CD = 26.75$
  • $AD = 758$
  • $CE = 1$
  • $0 < FC < 752$

How do I find $FG = x$ for point $F$ on line $BC$? Is it even possible?

EDIT:

As André mentioned in comments, just by defining lengths doesn't make the structure "rigid" to calculate x.

What if we define $90 < \angle C < 120$, does this makes approximate result possible?

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None of your givens say where F is located. Or do the marks mean that BF=FC (it doesn't look even close to equal in the diagram)? –  coffeemath Nov 27 '12 at 22:13
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Also in the diagram, CD appears longer than AB. The way the diagram is actually labelled it looks like it should be ED which is of length 26.75, rather than the whole side CD as you write in your listed assummptions. –  coffeemath Nov 27 '12 at 22:17
    
F is some point between B and C. I want to find distances from BC to AE. 0 < FG < 752. AB is a bit longer than CD, so CD should appear longer than AB, drawing mistake. There are no other special marks on BC, ignore those. I'll try to redraw cleanly if it helps. –  marltu Nov 27 '12 at 22:23
    
Maybe you should define say a variable $y=BF$ where $0<y<752$, and say in your question that you seek a formula for $x$ in terms of $y$. At least that would have been clearer to me... –  coffeemath Nov 27 '12 at 22:27
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It is a good picture, not to scale but who cares? The issue I see is that the quadrilateral is not rigid. If you make the sides of the given lengths, out of thin rigid steel rods, but put some hinges at the corners, the structure will have some wiggle. Looks to me as if even if we know $FC$, the distance labelled $x$ is not fully determined. –  André Nicolas Nov 27 '12 at 22:28

1 Answer 1

Here is my thought on the question; If one has a straight line of length $758$, points $A(0,0)$ and $D(758,0)$ apart. If one constructs a circle of radius $27$ centered at point $A$ and a circle of radius $26.75$ centered at point $D$. Are there two unique points $B(a_1,b_1)$ and $C(a_2,b_2)$, $B$ on the first circle and $C$ on the second circle such that $BC=752$. We have the following in order, $B$ on the first circle, $C$ on the second circle, the length of $BC$, $$(a_1-0)^2+(b_1-0)^2=27^2 \\ (a_2-758)^2+(b_2-0)^2=26.75^2 \\ (a_1-a_2)^2+(b_1-b_2)^2=752^2 $$

Three equations are clearly insufficient for four unknowns, I don't know how WolframAlpha got real solutions.

Even if one does find unique solution to the above system, unless $G=A=E$, line $FG$ is not unique.

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