# Thread: Geometrical construction...kinda impossible to me

1. ## Geometrical construction...kinda impossible to me

the problem:
construct a square given one of its points(A) and points P - on the BC and Q on the CD segment... the pic i've made is not the finished construction, i'm supposed to construct it from the three given points... it's just there so you see what i mean if my description aint good enough.

2. Originally Posted by noobatmath
the problem:
construct a square given one of its points(A) and points P - on the BC and Q on the CD segment... the pic i've made is not the finished construction, i'm supposed to construct it from the three given points... it's just there so you see what i mean if my description aint good enough.
Much to my surprise this construction must be possible - never mind how I placed the points A, P, Q.

The point B must be on a semi-circle over AP, the point C must be on a semi-circle over PQ and the point D must be on a semi-circle over AQ.

But unfortunately I haven't found yet a second condition to determine
- the length of the square's side OR
- the midpoint of the square OR
- the angle $\displaystyle \angle(BAP)$ (or the angle $\displaystyle \angle(QAD)$ OR
- ???

Sorry.

3. Orient so that A is at the Origin and B and C are as Quadrant I as possible. I didn't sit down to prove that this can be done. I'll just add it to my "Conjecture" list.

A: (0,0)
B: (a,b)
C: (c,d)

$\displaystyle c \leq a$ and $\displaystyle b \leq d$ I'm not certain these are necessary.

Family of Lines through B: $\displaystyle y - b = m(x - a) \implies mx - y + (b-ma) = 0$

Family of Lines through C: $\displaystyle y - d = n(x - c) \implies nx - y + (d-nc) = 0$

Lines through B that are perpendicular to lines through C.

$\displaystyle n*m = -1$

Distance from Origin to B-Family equals Distance from Origin to C-Family

$\displaystyle \frac{|b-ma|}{\sqrt{1+m^{2}}}=\frac{|d-nc|}{\sqrt{1+n^2}} \implies \frac{(b-ma)^2}{1+m^{2}}=\frac{(d-nc)^{2}}{1+n^{2}}$

After substituting m = -1/n and simplifying:

$\displaystyle (nb+a)^{2} = (d-nc)^{2}$ I have to admit I was a little stunned by the simplicity of this result.

This leads to two solutions: $\displaystyle n = \frac{d-a}{b+c}\;\;or\;\;n = \frac{a+d}{c-b}$

It's not perfectly clear to me how to select which of these is the right answer, but there's only two to choose from. It seems like we made it somewhere. Perhaps earboth's semi-circles can help make the choice.

Our four equation for the sides of the square, then...

mx - y + (b-ma) = 0

nx - y + (d-nc) = 0

y = mx

y = nx

A very challenging problem. Gear up your analytic geometry.

Note: Problems like these are very frustrating. I always think of them, "Ancient Greeks probably could solve this problem. I have WAY better algebra than they did. I should be able to solve it."

4. Well ok then...as i see nobody was able to solve this problem i thought i'd just post the solution...
rotate one of the points (P or Q) for 90 degrees around A so now if you rotated P, draw a line through P' and Q and there you go!