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!?

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TopNewestUse 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 \(\dfrac{24}{12^2} = \boxed{\dfrac{1}{6}}\). – Jimmy Kariznov · 4 years ago

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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. – Tim Vermeulen · 4 years ago

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