All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Square in Square (Posted on 2012-11-17)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 possible solution | Comment 1 of 3

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

 Search: Search body:
Forums (0)