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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?

  • 0
    I agree with André, the thing is far from rigid. It has a whole continuum of possible shapes. You could compute $x$ from $FC$ and $\angle C$, e.g. using explicit coordinates based on these two variables.2012-11-28

1 Answers 1

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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.