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We have a Point $A$ on the Plane $P$ and two circles $C1,C2$ on the same plane with radii $R1,R2$ on the same plane. ($R1$ and $R2$ are not necessarily equal).

We want to construct an isosceles right triangle ($45^\circ,45^\circ,90^\circ$) such that:

  1. One of the vertices is $A$

  2. The other vertices are on circles $C1$ and $C2$ (one per circle)

For this question, we can just use simple linear transformation like scaling, rotation, reflection and translation. How can we do this? Note that we know $C1 ,C2$ are placed such that at least one triangle with these conditions exists.

Also I want to know whether there is any online resources which have more of this type of questions? I mean constructing a shape with given conditions using simple linear transformations?.

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  • $\begingroup$ This does not necessarily exist, if one of the circle is too far away, there is no way for it to exist. $\endgroup$ – L KM Mar 23 '18 at 11:22
  • $\begingroup$ @LKM , I'll edit the question. We know $C1 ,C2$ are placed such that at least one triangle with this conditions exists. $\endgroup$ – titansarus Mar 23 '18 at 11:26
  • $\begingroup$ is the right angle at $A$ ? $\endgroup$ – Lozenges Mar 23 '18 at 11:30
  • $\begingroup$ @Lozenges, the question didn't mention it but I think it must be at $A$. $\endgroup$ – titansarus Mar 23 '18 at 11:31

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