Question

Let $P$ be a point on the ellipse $x^2 + 2y^2 = 2$ with foci $S$ & $S'$. The locus of incentre of $\triangle PSS'$ is a conic $C$.

Let $\alpha$ be the eccentricity of conic $C$. $[100\alpha^2]$ is equal to, where $[x]$ is greatest integer less than or equal to $x$.

Answer 1

82

Answer 2

1. Ellipse Properties:
   The given ellipse is

   $$x^2+2y^2=2\quad\Longleftrightarrow\quad\frac{x^2}{2}+\frac{y^2}{1}=1.$$

   Its foci $S$ and $S'$ are at $(1,0)$ and $(-1,0)$ because for an ellipse in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (with $a^2=2$, $b^2=1$), the focal distance is $c=\sqrt{a^2-b^2}=\sqrt{1}=1$.

2. Using the Definition of the Ellipse:
   For any point $P=(x,y)$ lying on the ellipse, the sum of distances from $P$ to the foci is constant:

   $$PS+PS' = 2\sqrt{2}.$$

   Denote

   $$b = PS'=\sqrt{(x+1)^2+y^2},\quad c = PS=\sqrt{(x-1)^2+y^2}.$$

   Hence, $b+c=2\sqrt{2}$ is constant.

3. Barycentric Formula for the Incentre:
   For triangle $PSS'$ with vertices $P=(x,y)$, $S=(1,0)$, and $S'=(-1,0)$, the incenter $I=(h,k)$ is given by

   $$h=\frac{a\cdot x+b\cdot 1+c\cdot (-1)}{a+b+c},\quad k=\frac{a\cdot y}{a+b+c},$$

   where $a$ is the side opposite $P$, i.e. $a=SS'=2$.
   Notice that by computing $b-c$ and using $b+c=2\sqrt{2}$, one obtains (after simplification)

   $$h=\frac{2x}{b+c}=\frac{x}{\sqrt{2}},\quad k=\frac{2y}{2+b+c}=\frac{2y}{2+2\sqrt{2}}=\frac{y}{1+\sqrt{2}}.$$

4. Finding the Locus of the Incentre:
   Express $x$ and $y$ in terms of $h$ and $k$:

   $$x = h\sqrt{2},\quad y = k(1+\sqrt{2}).$$

   Substitute into the ellipse equation $x^2+2y^2=2$:

   $$(h\sqrt{2})^2+2\Big[k(1+\sqrt{2})\Big]^2 = 2,$$

   which simplifies to

   $$2h^2+2k^2(1+\sqrt{2})^2=2.$$

   Dividing by 2 gives

   $$h^2+k^2(1+\sqrt{2})^2=1.$$

   This is an ellipse (let it be $C$) with horizontal semi-axis $a'=1$ and vertical semi-axis $b'=\dfrac{1}{1+\sqrt{2}}$.

5. Eccentricity of Conic $C$:
   Since $a'>b'$ (because $1>\frac{1}{1+\sqrt{2}}$), the eccentricity $\alpha$ is

   $$\alpha = \sqrt{1-\left(\frac{b'}{a'}\right)^2} = \sqrt{1-\frac{1}{(1+\sqrt{2})^2}}.$$

   Notice that

   $$(1+\sqrt{2})^2 = 3+2\sqrt{2}.$$

   Thus,

   $$\alpha^2 = 1-\frac{1}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}-1}{3+2\sqrt{2}} = \frac{2+2\sqrt{2}}{3+2\sqrt{2}}.$$

   Recognize that $3+2\sqrt{2} = (1+\sqrt{2})^2$; hence,

   $$\alpha^2 = \frac{2(1+\sqrt{2})}{(1+\sqrt{2})^2} = \frac{2}{1+\sqrt{2}}.$$

   Multiply numerator and denominator by $(\sqrt{2}-1)$ to rationalize:

   $$\alpha^2 = \frac{2(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = 2(\sqrt{2}-1).$$

6. Final Computation:
   We need $[100\alpha^2]$ where $[\,x\,]$ denotes the floor function:

   $$100\alpha^2 = 100 \cdot 2(\sqrt{2}-1)=200(\sqrt{2}-1).$$

   Approximating $\sqrt{2} \approx 1.414$,

   $$200(1.414-1)=200(0.414)\approx82.8.$$

   Taking the greatest integer $[82.8]=82$.