Let the line L_1: x+3=0 intersect the lines L_2: x-y=0 and L_3: 3x+y=0 at the points A and…

Question

Let the line $L_1: x+3=0$ intersect the lines $L_2: x-y=0$ and $L_3: 3x+y=0$ at the points $A$ and $B$, respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point $C$. Then $BC^2:AC^2$ is equal to:

Accepted Answer

$L_1: x=-3$. On $L_2: y=x$, so $A=(-3,-3)$. On $L_3: y=-3x$, so $B=(-3,9)$.

Angle bisectors of $L_2: x-y=0$ and $L_3: 3x+y=0$:
$$\frac{x-y}{\sqrt2} = \pm\frac{3x+y}{\sqrt{10}}$$

Let $C=(-3,k)$ on $L_1$. Distance-equal condition to $L_2,L_3$:
$$\frac{|k+3|}{\sqrt2} = \frac{|k-9|}{\sqrt{10}} \implies 5(k+3)^2=(k-9)^2$$
$$5k^2+30k+45 = k^2-18k+81 \implies 4k^2+48k-36=0 \implies k^2+12k-9=0$$
$$k = -6\pm3\sqrt5$$

Since $A=(-3,-3)$, $B=(-3,9)$, and the obtuse bisector meets $L_1$ between them, take $k=-6+3\sqrt5\approx0.708$ (between $-3$ and $9$), so $C=(-3,-6+3\sqrt5)$.

$$AC = |{-6+3\sqrt5-(-3)}| = |3\sqrt5-3| = 3(\sqrt5-1) \implies AC^2 = 9(\sqrt5-1)^2$$
$$BC = |9-(-6+3\sqrt5)| = |15-3\sqrt5| = 3(5-\sqrt5) \implies BC^2 = 9(5-\sqrt5)^2$$

$$\frac{BC^2}{AC^2} = \frac{(5-\sqrt5)^2}{(\sqrt5-1)^2} = \frac{10(3-\sqrt5)}{2(3-\sqrt5)} = 5$$

So $BC^2:AC^2 = 5:1$.

Solutions · 1