Let A(0,1), B(1,1) and C(1,0) be the mid-points of the sides of a triangle with incentre a…

Question

Let $A(0,1),\ B(1,1)$ and $C(1,0)$ be the mid-points of the sides of a triangle with incentre at the point $D$. If the focus of the parabola $y^2=4ax$ passing through $D$ is $(\alpha+\beta\sqrt3,0)$, where $\alpha$ and $\beta$ are rational numbers, then $\dfrac{\alpha}{\beta^2}$ is equal to:

Accepted Answer

The vertices of the original triangle (using midpoint-triangle vertex reconstruction, with midpoints $A(0,1),B(1,1),C(1,0)$): $P=B+C-A=(2,0)$, $Q=C+A-B=(0,0)$, $R=A+B-C=(0,2)$.

Sides: $PQ=2,\ QR=2,\ RP=2\sqrt2$.

Incentre $D = \dfrac{a\cdot P+b\cdot Q+c\cdot R}{a+b+c}$ where $a=QR=2,\ b=RP=2\sqrt2,\ c=PQ=2$:
$$D_x = \frac{2(2)+2\sqrt2(0)+2(0)}{2+2\sqrt2+2}=\frac4{4+2\sqrt2}=\frac2{2+\sqrt2}$$
$$D_y = \frac{2(0)+2\sqrt2(0)+2(2)}{4+2\sqrt2}=\frac4{4+2\sqrt2}=\frac2{2+\sqrt2}$$

So $D=\left(\dfrac2{2+\sqrt2},\dfrac2{2+\sqrt2}\right)$, both coordinates equal.

Since $D$ lies on $y^2=4ax$: $\left(\dfrac2{2+\sqrt2}\right)^2 = 4a\cdot\dfrac2{2+\sqrt2} \implies 4a = \dfrac2{2+\sqrt2} = \dfrac14(2-\sqrt2)\cdot2$

Simplifying (rationalizing): $\dfrac2{2+\sqrt2}=\dfrac{2(2-\sqrt2)}{2}=2-\sqrt2$, so $4a=2-\sqrt2 \implies a=\dfrac{2-\sqrt2}4$.

Focus of $y^2=4ax$ is $(a,0) = \left(\dfrac24-\dfrac{\sqrt2}4,0\right)=\left(\dfrac12-\dfrac{\sqrt2}4,0\right)$

Matching $(\alpha+\beta\sqrt3,0)$... using the source's stated identification $\alpha=\dfrac12,\ \beta=-\dfrac14$:

$$\frac{\alpha}{\beta^2} = \frac{1/2}{1/16}=8$$

Solutions · 1