Point in Octagon

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A point $P$ is located in a regular octagon $ABCDEFGH$ with side length $4$ such that $\angle GAP=60^{\circ}$ and $\angle FGP=\angle BAP$ . If the average of the shortest distance between $P$ to the extensions of each face can be expressed as $a\sqrt{b}+c$ , where $a,b,c$ are positive integers and $b$ is not divisible by any square, then what is $abc$ ?

The answer is 8.

1 solution

Shriram Lokhande
Jul 1, 2014

As in a regular octagon the opposite sides are parallel & shortest distance to a line is the distance of perpendicular from the point to the line . octagon octagon as in figure L is an point the shortest distance to opp. sides is FE and we have to find $\dfrac{4\text{EF}}{8}$ let EI be $x$ $16=2x^2$ $x=2\sqrt{2}$

EF = $4+4\sqrt{2}\}$ hence average shortest distance between p and extensions of each face is $2+2\sqrt{2}$

hence, $abc = \boxed{8}$

Point in Octagon

The answer is 8.

1 solution

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