Pythagoras and Vecten Break Japan's Isolation
Here we have a third solution to a Sangaku problem: Five squares are arranged as in the applet. Show that the area of triangle KMN equals the area of the square BEKH. The other two can be found elsewhere. The solution has been suggested by Nathan Bowler; it does not invoke any algebraic means.
The solution is based on two key observations:
First, as in Solution #2, we observe that there are pairs of flank triangles in Vecten configuration. Thus we see that the areas of the four triangles ABE, DEK, BCH, KHI are equal, as are the areas of triangles KMN and DIK.
B is the midpoint of XY. Thus, when square BEKH is cut into 5 pieces by vertical lines through B and K and horizontal lines through E and H, in the manner of Proof #3 of the Pythagorean theorem, points U and V being directly above B, the quadrilateral DUIV is a parallelogram (with center O.) So that triangles DOU and IOV have equal areas, from which
Area(DIK) = Area(DUVIK).
Now, taking into account the first observation, we see that
Area(DEK) = Area(ABE) = Area(BEU) and
Area(IKH) = Area(BCH) = Area(BHV),
which implies the required identity
Area(KMN) = Area(DIK) = Area(BEKH).
The last step admits a very slight modification that invokes Proof #10 of the Pythagorean theorem. Indeed, for the same reasons as above,
Area(DUVIK) = Area(AEFGHC) = Area(BEKH).
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- J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #50
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