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Here's the set-up:

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Take three circles entered at $(-1,0),(0,0),(1,0)$ with radii $\sqrt 2$ and $1$.

Then pick $p$, any point on the right-hand circle. Reflect $p$ in the horizontal axis to get $-p$. Draw the line segment from $(0,0)$ to $-p$.

Then $q$ is the intersection of this segment with the left circle.

Prove that the distance between $p$ and $q$ is $2$.

I have proved this using complex analysis, but is there an intuitive way to do this using pure geometry?

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  • $\begingroup$ What’s the point of having the middle circle? $\endgroup$
    – amd
    Commented Mar 5, 2016 at 21:57
  • $\begingroup$ @amd to confuse both the enemies and allies, but mostly the student. $\endgroup$
    – CAGT
    Commented Mar 5, 2016 at 22:33
  • $\begingroup$ It could be a hint... $\endgroup$
    – Moti
    Commented Mar 5, 2016 at 22:55
  • $\begingroup$ I would just say it intuitively... since the centers of the circles are exactly 2 units away... all corresponding projections of an arc onto another arc, whose arc curvatures are identical, the distance will be the separation of the centers. $\endgroup$
    – CAGT
    Commented Mar 5, 2016 at 23:13

2 Answers 2

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A hint that might help. The following diagram will help with the hint:

enter image description here

ED=2 as well. Prove that AE is parallel to CD and that AB=EB (or BD=BC). Hope this helps to get the geometry based solution:)

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Might the complex analysis itself be a form of two-dimensional geometry? We can render complex numbers as vectors in the complex plane. Addition is obvious. Multiplication by a+bi corresponds to a linear transformation using the multiplier matrix

a b

-b a

Matrices having the above form are isomorphous with the complex numbers.

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