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(C) $	riangle ABC$ is a right-angled triangle with $\angle ABC = 90^{\circ}$.The circumcenter of $	riangle ABC$ is the midpoint of the hypotenuse $A

Vedclass · Accepted Answer

$\frac{AC}{2}$

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(C) $	riangle ABC$ is a right-angled triangle with $\angle ABC = 90^{\circ}$.The circumcenter of $	riangle ABC$ is the midpoint of the hypotenuse $AC$,denoted as $M$.By the properties of reflection,$AB$ is the perpendicular bisector of $PP_1$,and $BC$ is the perpendicular bisector of $PP_2$.Since $AB \perp BC$ at $B$,the point $B$ is equidistant from $P, P_1,$ and $P_2$ because $BP = BP_1$ and $BP = BP_2$.Thus,$B$ is the circumcenter of $	riangle P_1PP_2$.In $	riangle ABC$,$M$ is the midpoint of $AC$,so $BM = AM = MC = \frac{AC}{2}$ (property of the median to the hypotenuse in a right triangle).Therefore,the distance between the circumcenters $B$ and $M$ is $BM = \frac{AC}{2}$.

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 $\triangle ABC$ and $\triangle P_1PP_2$ is

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