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Claim : If the entire plane is rotated about any two points, say \(P_1, P_2\) one after the other by angles \(\theta_1, \theta_2\) respectively, then the total transformation of the plane is identical to rotation about some point \(P\) by an angle \(\theta_1+\theta_2\). Here we consider the position of \(P_2\) that do not change upon rotation about \(P_1\).
Proof : Fix the \(X-Y\) coordinate axes and let the coordinates of \(P_1\ ;\ P_2\) be \((x_1, y_1)\ ;\ (x_2,y_2)\) respectively. We represent a point by column matrix and we also know the rotation matrix as defined below. First we shift the axes parallely to \(P_1\) and then rotate by \(\theta_1\) to get the coordinates as,
Now, shift the axes parallely to \(P_2\) and then rotate by \(\theta_2\) to get,
and we will finally move the axes back to the actual origin \(O\) (to realize the transformation) which is located at \((-x_2,-y_2)\) with respect to \(P_2\) by subtracting \((-x_2,-y_2)\),
But now observe that this is just rotation by an angle \(\theta_1+\theta_2\) about the point \(P\) given by,
Hence proved the claim \(\blacksquare\).
Now, let us locate the position of point \(P_1=\tau_3\left(\tau_1\left(\tau_2\left(P\right)\right)\right)\). Notice that rotation about point \(A_2\) by \(\angle A_3A_2A_1\) and followed by rotation about point \(A_1\) by \(\angle A_2A_1A_3\) is equivalent to rotation about some point \(Q_1\) by an angle \(\angle A_3A_2A_1+\angle A_2A_1A_3\). Now this followed by a rotation about \(A_3\) by \(\angle A_1A_3A_2\) is totally equivalent to rotation about some point \(R_1\) by angle \(\angle A_3A_2A_1+\angle A_2A_1A_3+\angle A_1A_3A_2\). Note that here the angles are directed module \(360^{\circ}\) and since all these rotation angles have the same orientation (anti-clockwise), they all will have the same sign and add up to give \(180^{\circ}\). So, essentially \(P_1\) is a rotation of \(180^{\circ}\) about some point \(R_1\). Let us now find \(R_1\) by observing the final position of point \(A_2\) upon this transformation.
Let us consider the sides of triangle \(A_2A_3,\ A_3A_1,\ A_1A_2\) to be \(l_1,\ l_2,\ l_3\) respectively as shown. Let \(A_2''=\tau_3\left(\tau_1\left(\tau_2\left(A_2\right)\right)\right)\),
By \(\tau_2,\ A_2\) remains unaltered. Upon applying \(\tau_1,\ A_2\) goes to the point \(A_2'\) such that \(A_3A_2'=l_2-l_3\) (Here the lengths are directed). Now upon applying \(\tau_3,\ A_2'\) goes to \(A_2''\) such that \(A_2A_2''=l_1+l_3-l_2\). So, note that \(A_2''\) is also obtained by rotating about the point \(R_1\) by \(180^{\circ}\), where \(A_2R_1=\frac{l_1+l_3-l_2}{2}\), which is clearly the distance of point of tangency of incircle to the side \(A_2A_3\) from \(A_2\) and hence \(R_1=D\) as shown in the diagram. Also note that this point \(R_1\) is uniquely determined as rotation by \(180^{\circ}\) is nothing but reflection about it. Similarly, \(R_2=E,\ R_3=F\). Hence \(P_1, P_2, P_3\) are obtained by reflection of \(P\) about the points \(D,E,F\) respectively. But, these reflections can be realized as a homothety about \(P\) with scale factor \(2\). Hence, the radius of the circumcircle of \(\triangle P_1P_2P_3\) is \(2r\), where \(r\) denotes the inradius of the \(\triangle A_1A_2A_3\). We also know that \(OI^2=R(R-2r)\), where \(R, O, I\) denote the circumradius, circumcenter, Incenter of the \(\triangle A_1A_2A_3\) respectively. Hence, \[\boxed{Radius\ of\ \Gamma=R\geq 2r= Radius\ of\ circumcircle\ of\ \triangle P_1P_2P_3}\]
Comment out the condition on the \(\triangle A_1A_2A_3\) so that the above inequality turns to equality for any point \(P\) !
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