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Square in Square (Posted on 2012-11-17) Difficulty: 3 of 5
A small square is placed inside a big square. The vertices of the small square are joined to vertices of the large square so as to divide the region between the squares into four quadrilaterals, with areas, in order, a, b, c, d.
Prove that a+c=b+d.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution Another approach (spoiler) | Comment 2 of 3 |
Draw two perpendiculars from each of W,X,Y,Z to the two sides of the large square that meet at A,B,C,D respectively.

These cut off rectangles at each corner, with areas 2RA,2RB,2RC,2RD say,

and form trapeziums (which have WX,XY,YZ,ZW as their slant edges) with

areas Ta,Tb,Tc,Td say.

Thus:    [a + c] [b + d] =

[(RA + Ta + RB) + (RC + Tc + RD) ] [(RB + Tb + RC) + (RD + Td + RA)]

=   [Ta + Tc] [Tb + Td]                                                               (1)

Now, each pair of opposite trapeziums could be fitted together, with their slant edges coinciding, to form a rectangle with width p and length L p, where p is the projection of WX on AB (same for all other pairs) and L is the length of a side of the large square.

Thus     [Ta + Tc] = [Tb + Td],  and (1) now gives   [a + c] [b + d] = 0 and

therefore:          a + c = b + d

  Posted by Harry on 2012-11-18 12:58:43
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