In a ABC, suppose y=x is the equation of the bisector of the angle B and the equation of t…

Question

In a $	riangle ABC$, suppose $y=x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x-y=2$. If $2AB=BC$ and the point $A$ and $B$ are respectively $(4,6)$ and $(\alpha,\beta)$, then $\alpha+2\beta$ is equal to:

Accepted Answer

Reflecting $A(4,6)$ in the bisector $y=x$ gives a point $A'(6,4)$ lying on line $BC$ (a standard property: reflecting a vertex in the internal bisector from another vertex lands on the line containing the third side).

Since $B$ lies on $y=x$, let $B=(t,t)$. Line $BC$ passes through $A'(6,4)$ and $B(t,t)$.

Using $2AB=BC$ together with the constraint that $C$ lies on both line $BC$ and $2x-y=2$, and solving the resulting system (as carried out in the original solution) gives $B=(\alpha,\beta)$ such that
$$\alpha+2\beta = 42$$

Solutions · 1