Coordinate Geometry
Circumcentre via parallelogram construction
MJAT_TS3_P2
Grade 12
Question:
Two circles $C_1$ and $C_2$ intersect at point $A$. Tangents from $A$ to the two circles meet the circles at $B$ and $C$ respectively. Let $P$ be a point such that $PMAN$ is a parallelogram (where $M$ and $N$ are the centres of $C_1$ and $C_2$). If $AB+BC+CA = k(PA\sin A + PB\sin B + PC\sin C)$ for $k\in\mathbb{R}$, then find $k$.
Step-by-Step Solution
Key Concept: Since $PMAN$ is a parallelogram with $M,N$ as circle centres, $P$ is the reflection of $A$ through the midpoint of $MN$. This makes $PA\perp BC$ at $A$ (since $MN\perp$ to the common chord). In $\triangle ABC$, $P$ turns out to be the circumcentre, so $PA=PB=PC=R$.
$P$ is the circumcentre of $\triangle ABC$ with circumradius $R$. By comparison: $k=\mathbf{2}$.
Correct Answer: 2