My Math Forum geometrical question about congruent rectangles

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 June 23rd, 2010, 12:42 AM #11 Newbie   Joined: Nov 2009 Posts: 12 Thanks: 0 Re: geometrical question about congruent rectangles I have created an image which I want to upload, but it doesn't work... is it due to my status or because of an error?
 June 23rd, 2010, 01:18 AM #12 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: geometrical question about congruent rectangles Hello MG85, I maybe have an image that represents your question. Now, in other words, you are to proof that the area where "2 books are on top of each other, so 2 books thick" is larger than the area where "the area is one layer thick, so no overlay" I hope, the english and the idea are correct. Hoempa
 June 24th, 2010, 02:56 AM #13 Newbie   Joined: Nov 2009 Posts: 12 Thanks: 0 Re: geometrical question about congruent rectangles To make it more understandable I created a geo file, as I didn't have webspace I am only able to set a link to a photoalbum... http://de.fotoalbum.eu/gezi999/a490791/00000001 it has to be proved that the green area is always bigger than half of the rectangle's area...
 June 24th, 2010, 03:32 AM #14 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: geometrical question about congruent rectangles Do you have an extra assumption that the centres of the 2 rectangles are on top of each other? The question, skipjack found, doesn't say so. Hoempa
 June 24th, 2010, 09:51 AM #15 Newbie   Joined: Nov 2009 Posts: 12 Thanks: 0 Re: geometrical question about congruent rectangles No, there is not an assumption like that, but I didn't know how to realize it with dynageo...
 June 24th, 2010, 11:45 AM #16 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: geometrical question about congruent rectangles Suggestion: Maybe you could reflect all the area's with "one rectangle" into the green zone and show there is some space left. Hoempa
 June 25th, 2010, 08:51 PM #17 Global Moderator   Joined: Dec 2006 Posts: 20,471 Thanks: 2038 Let p = (area of overlap)/(area of rectangle) and r = (length of rectangle)/(width of rectangle). It seems that p can be considerably larger than 1/2, but can be arbitrarily close to 1/2 if r is sufficiently large. If rectangle ABCD (where AB > BC) is placed first, one can place the second rectangle, A'B'C'D' (where A'B' > B'C'), so that D' is very close to D and A' almost lies on AB. In the limiting case, where D and D' coincide and A' lies on AB, I find that p = r² - r?(r² - 1), which tends to 1/2 as r tends to infinity. However, the above placement method isn't optimal if r is not large (since p tends to 1 as r tends to 1), and I haven't proved that it's optimal when r is large. Perhaps a better approach would be to consider the general case, using three variables to specify the position of the second rectangle in relation to the first. That would be a bit tedious, but shouldn't be unduly difficult.
 June 26th, 2010, 01:46 AM #18 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: geometrical question about congruent rectangles I have tried something yesterday, on the puzzle. I think it will be tedious, I don´t even know whether it´s possible to proof so. We could use the cosine rule, generally: $c^2=a^2+b^2+2ab\cos(\gamma)$ I don´t know the formula to calculate the surface of a triangle. I do know this: surface=0.5*side*height, but there must have been another. It used the sides of the triangle. Surface=$\sqrt{(something with sides)}$. Too bad I don´t know the formula. Maybe we could use that to define the surfaces of all areas without overlap. Referring to the image, MG85 posted, The area with overlap could be described by for example $\triangle IJK$, $\triangle KLE$, $\triangle EFG$, $\triangle GHI$ and then quadrangle IKEG, by splitting in for example $\triangle KEG$ and $\triangle KIG$ The formula for surface of triangles will be needed to do so, for we have the sides. Again, this method is tedious if possible to proof so Hoempa
 June 26th, 2010, 02:49 AM #19 Global Moderator   Joined: Dec 2006 Posts: 20,471 Thanks: 2038 The usual form of the cosine rule is c² = a² + b² - 2ab cos C. Heron's formula gives a triangle's area in terms of the lengths of the sides of the triangle, but is unlikely to be useful here.
June 26th, 2010, 03:12 AM   #20
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Re: geometrical question about congruent rectangles

Quote:
 Originally Posted by skipjack c² = a² + b² - 2ab cos C
I´ll edit that one, thanks

Anyway, it was worth trying, thanks for the reply!

### finding the area of a second rectangle that is larger than the first but congruent

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