# Construction an isosceles right triangle with transformation

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?.

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