Let $\mathrm{P}$ be a point inside a triangle $\mathrm{ABC}$ with $\angle \mathrm{ABC}=90^{\circ}$. Let $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ be the images of $\mathrm{P}$ under reflection in $\mathrm{AB}$ and $\mathrm{BC}$ respectively. The distance between the circumcenters of triangles $\mathrm{ABC}$ and $\mathrm{P}_{1} \mathrm{P} \mathrm{P}_{2}$ is

• A

  $\frac{\mathrm{AB}}{2}$
• B

  $\frac{\mathrm{AP}+\mathrm{BP}+\mathrm{CP}}{3}$
• Correct answer
  C

  $\frac{\mathrm{AC}}{2}$
• D

  $\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{2}$

Solution

$\mathrm{M}$ is circumcentre of $\triangle \mathrm{ABC}$ $\Rightarrow \mathrm{M}\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)$ $\& \mathrm{~N}$ is circumcentre of $\triangle \mathrm{ABC}$ $\mathrm{N}=(0,0)=\mathrm{B}\left(\right.$ Mid-point of $\left.\mathrm{P}_{1}, \& \mathrm{P}_{2}\right)$ So $\mathrm{MN}=\frac{\mathrm{AC}}{2}$

Hence, the correct answer is option: 'C'.