Let the circumcentre of a triangle with vertices A(a,3), B(b,5) and C(a,b), ab0 be P(1,1)…

Question

Let the circumcentre of a triangle with vertices $A(a,3),\ B(b,5)$ and $C(a,b),\ ab>0$ be $P(1,1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_1,k_2)$, then $k_1+k_2$ is equal to:

Accepted Answer

Let $D$ = midpoint of $AC = \left(a,\dfrac{3+b}2\right)$; since $PD\perp AC$ and $AC$ is vertical (both $A,C$ have $x=a$), $PD$ is horizontal, so $\dfrac{3+b}2=1 \implies b=-1$.

Let $E$ = midpoint of $BC$; using $PE\perp BC$, and substituting $b=-1$, solving gives $a=5$ or $a=-3$; since $ab>0$ and $b=-1<0$, we need $a<0$, so $a=-3$.

$A(-3,3),\ B(-1,5),\ C(-3,-1),\ P(1,1)$.

Line $BC$: through $(-1,5),(-3,-1)$: slope$=3$: $y=3x+8$.

Line $AP$: through $(-3,3),(1,1)$: slope$=-\dfrac12$: $y=\dfrac{3-x}2$.

Intersection: $\dfrac{3-x}2=3x+8 \implies 3-x=6x+16 \implies -7x=13 \implies x=-\dfrac{13}7,\ y=\dfrac{3+\frac{13}7}2=\dfrac{17}7$.

$$k_1+k_2 = -\frac{13}7+\frac{17}7 = \frac47$$

Solutions · 1