Question

Let $P$ be a point inside a triangle $ABC$ with $\angle ABC={90}^{\circ }$. Let $P_1$ and $P_2$ be the images of $P$ under reflection in $AB$ and $BC$ respectively. The distance between the circumcenters of triangles $ABC$ and $P_{1}P{P}_2$ is

• A. $\frac{AB}{2}$
• B. $\frac{AP+BP+CP}{3}$
• C. $\frac{AC}{2}$
• D. $\frac{AB+BC+AC}{2}$

Answer 1

The correct option is C $\frac{AC}{2}$


We know that, circumcenter of a right angle triangle lies exactly at the midpoint of the hypotenuse.
Let $R$ and $S$ be the midpoint of the hypotenuse $AC$ and $P_1{P}_2$ respectively.
$\therefore S=(\frac{c-c}{2},\frac{d-d}{2})=(0,0)\therefore R=(\frac{a+0}{2},\frac{0+b}{2})=(\frac{a}{2},\frac{b}{2})RS=\sqrt{{(\frac{a}{2}-0)}^2+{(\frac{b}{2}-0)}^2}=\frac{1}{2}\sqrt{a^2+b^2}=\frac{AC}{2}$