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.
Let E be any point within the square ABCD. Then the opposite triangles ADE+BCE=ABE+CDE, by considering the rectangles whose diagonals are the vertices A,B,C,D, and E:
(1-a)*(1-c)+((1-(1-a)*(1-(1-c))=(1-a)*(1-(1-c))+(1-c)*(1-(1-a))=a+c-2ac, and then adding up the relevant diagonal triangular pieces into which each of these rectangles is divided.
It makes no difference if we add a square, say FGHI, with sides parallel to the sides of ABCD and centred on E, and subtract a half of it from each pair of areas, since obviously this does not affect the equivalence.
Last, we can construct lines:
(1) From F perpendicular to AB, and choose a point W at distance k from F somewhere inside ABCD,
(2) From G perpendicular to BC, and choose a point X at the same distance k from G somewhere inside ABCD,
(3) From H perpendicular to CD, and choose a point Y at the same distance k from H somewhere inside ABCD,
(4) From I perpendicular to AD, and choose a point Z at the same distance k from I somewhere inside ABCD;
producing the given diagram.
Now we are deducting from each pair of quadrilaterals: FGHI (which did not affect the equivalence) plus two of 4 identical triangles (FWX, GXY, HYZ, IWZ) which cannot affect the equivalence either, as was to be proved.
Edited on November 18, 2012, 3:54 am
Posted by broll
on 2012-11-18 03:52:58