Octagon inside a square!

I want to know a simple method for the following question: There is a square of side 1 unit.Mark the mid points of the sides,say X,Y,Z,W of sides AB,BC,CD,AD respectively.Now form a triangle with vertices X,C, and D.Similarly join Y,A and D forming a triangle.Do the same for other two.Now compute the area of the octagon formed at the center of the square.Hope u understand the problem!?

Note by Kishan K
4 years, 10 months ago

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Use coordinates. To avoid fractions, I'll scale everything by a factor of $$12$$.

WLOG, let $$A(0,0), B(12,0), C(12,12), D(0,12)$$. Then, $$X(6,0), Y(12,6), X(6,12), W(0,6)$$.

Let $$V_1, \ldots , V_8$$ be the vertices of the octagon going counterclockwise, with $$V_1$$ being closest to $$W$$.

Note that there is $$8$$-fold symmetry about the center $$M(6,6)$$.

Line $$AZ$$ is $$y = 2x$$. Line $$DX$$ is $$y = 12-2x$$. Their intersection is $$V_1(3,6)$$.

By symmetry about $$M(6,6)$$, the other odd numbered vertices are $$V_3(6,3), V_5(9,6), V_7(6,9)$$.

Line $$BW$$ is $$y = -\tfrac{1}{2}x+6$$. Line $$DX$$ is $$y = 12-2x$$. Their intersection is $$V_2(4,4)$$.

By symmetry about $$M(6,6)$$, the other even numbered vertices are $$V_4(8,4), V_6(8,8), V_8(4,8)$$.

From here, it is easy to calculate the area of one "slice" of the scaled octagon: $$[\Delta MV_1V_2] = \tfrac{1}{2} \cdot 3 \cdot 2 = 3$$. Therefore, the scaled octagon has area $$8 \cdot 3 = 24$$. Hence, the original octagon has area $$\dfrac{24}{12^2} = \boxed{\dfrac{1}{6}}$$.

- 4 years, 10 months ago

Coincidentally, this was a problem from the 1989 Dutch finals, it was put in a collection of the 100 best problems from Dutch olympiads over the years. I tried to avoid coordinates, as it's clear that that's going to work, but I tried to find a more elegant solution, e.g. by determining the area of the part of the square that's not part of the octagon. I didn't manage to do that though. If I encountered such a problem on an olympiad, I'd probably go for the 'safe' coordinate method.

- 4 years, 10 months ago