Locus of reflection:Special case

Question

In a triangle ABC, right angled at C, points A and B are fixed. C moves on a circle with AB as diameter.Suppose I is the incentre and the incircle touches the hypotenuse at F. What is the locus of the reflection of F across CI?

I proceeded with coordinates, taking $A(-1,0),B(1,0)$and $C(cos2\theta,sin2\theta)$.I obtained a parametric form of locus.But the coordinates of the reflection are quite complicated to determine a geometric form straightaway. Here is an image with the coordinates of the reflection.

How do I remove the parameter?Is there any other euclidean geometry approach?

Answer 1

The parameterization you give seems to not plot right so I tried to find another one. I start as you did

$C$: $(\cos(t),\sin(t))$

For the incenter plug the side lengths $a=2\sin(t/2) ,b= 2\cos(t/2),c= 2$ and $x(A)=-1,x(B)=1,x(C)=\cos(t),y(A)=0,y(B)=0,y(C)=\sin(t)$ in the formula $(\frac{ax(A)+bx(B)+cx(C)}{a+b+c},\frac{ay(A)+by(B)+cy(C)}{a+b+c}).$

$I$: $(\frac{(2\sin(t/2)\cdot (-1)+2\cos(t/2)\cdot 1+2\cos(t)}{2\sin(t/2)+2\cos(t/2)+2},\frac{2\sin(t)}{2\sin(t/2)+2\cos(t/2)+2})$

$CI$: $y-\sin(t)=m\cdot (x-\cos(t))$

$m=\frac{\sin(t)-\frac{2\sin(t)}{2\sin(t/2)+2\cos(t/2)+2}}{\cos(t)-\frac{(2\sin(t/2)\cdot (-1)+2\cos(t/2)\cdot 1+2\cos(t)}{2\sin(t/2)+2\cos(t/2)+2}}$

For $F,$ use that the incircle has equation $(x-\frac{(2\sin(t/2)\cdot (-1)+2\cos(t/2)\cdot 1+2\cos(t))}{(2\sin(t/2)+2\cos(t/2)+2)})^2+(y-\frac{(2\sin(t))}{(2\sin(t/2)+2\cos(t/2)+2)})^2=(\cos(t/2)+\sin(t/2)-1)^2$ and intersect with $y=0.$

$F$ $(\frac{\cos(t)-\sin(t/2)+\cos(t/2)}{\sin(t/2)+\cos(t/2)+1},0)$

We're interested in the reflection $F'$ of $F$ in the line $CI$ for $t\in[0,\frac{\pi}{2}]$ in the first quadrant (replacing $t=2\frac{t}{2}$ by double angle formulas): $${\tiny x(t)=-\frac{(((2\cos(t/2)\sin(t/2)+1)((\cos(t/2))^2-(\sin(t/2))^2)-4\cos(t/2)^2+2)(2\sin(t/2)\cos(t/2))^2 +((1-2\cos(t/2)^2)\sin(t/2)+2\cos(t/2)^3-\cos(t/2)-1)((\cos(t/2))^2-(\sin(t/2))^2)+\sin(t/2) +2\cos(t/2)^2-\cos(t/2)-1) }{((2\sin(t/2)^3+(2\cos(t/2)+2)\sin(t/2)^2-\sin(t/2)-\cos(t/2)-1)((\cos(t/2))^2-(\sin(t/2))^2) +\sin(t/2)+\cos(t/2)+1)}}$$ $$ {\tiny y(t)=-\frac{((((2\cos(t/2)^2+1)\sin(t/2)-2\cos(t/2)^3+3\cos(t/2))((\cos(t/2))^2-(\sin(t/2))^2) +(3-6\cos(t/2)^2)\sin(t/2)-6\cos(t/2)^3+3\cos(t/2)) (2\sin(t/2)\cos(t/2))^3 +(((4\cos(t/2)^2-4)\sin(t/2)-4\cos(t/2)^3)((\cos(t/2))^2-(\sin(t/2))^2) +(4\cos(t/2)^2-4)\sin(t/2)+4\cos(t/2)^3) (2\sin(t/2)\cos(t/2))) }{(((4\cos(t/2)+2)\sin(t/2)^3+(2\cos(t/2)+2)\sin(t/2)^2 +((-2\cos(t/2))-1)\sin(t/2)-\cos(t/2)-1) ((\cos(t/2))^2-(\sin(t/2))^2)^2 +(4\sin(t/2)^4+2\sin(t/2)^3+((-2\cos(t/2))-4)\sin(t/2)^2+2\cos(t/2)\sin(t/2) +2\cos(t/2)+2) ((\cos(t/2))^2-(\sin(t/2))^2)+2\sin(t/2)^2+\sin(t/2)-\cos(t/2)-1)}}$$

To answer the question in the comments, the area under the curve in the first quadrant is $I=\int_0^{\frac{\pi}{2}}yx'dt$ so that the area of the shape is $4I$ (in maxima CAS):

/* trigsimp(y*diff(x,t)) */
((24*cos(t/2)+16)*sin(t/2)^8+((-32*cos(t/2))-44)*sin(t/2)^7+(12-20*cos(t/2))*sin(t/2)^6+(52*cos(t/2)+50)*sin(t/2)^5+((-28*cos(t/2))-38)*sin(t/2)^4+((-2*cos(t/2))-2)*sin(t/2)^3+(8*cos(t/2)+8)*sin(t/2)^2+((-2*cos(t/2))-2)*sin(t/2))/(2*sin(t/2)^3-sin(t/2)^2+((-3*cos(t/2))-3)*sin(t/2)+2*cos(t/2)+2)
quad_qag(%,t,0,%pi/2,3,'epsrel=5d-8);

$\approx-0.07079632679489668$

making the area inside the whole locus $\approx 0.283184$.