Squares into square

R

Raoul

New member
Joined
Apr 13, 2015
Messages
6
Let [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math]D[/FONT] be a square, [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]a[/FONT] Is it possible to insert two disjoint squares, both of side [FONT=MathJax_Math]a[/FONT] into ABCD?
 
Last edited:
Raoul said:
Let [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math]D[/FONT] be a square, [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]a[/FONT] Is it possible to insert two disjoint squares, both of side [FONT=MathJax_Math]a[/FONT] into ABCD?
Click to expand...
i dont think so. they will atleast share a point.
 
Ishuda said:
To have two disjoint squares with sides of length a, how much room do you need horizontally? Vertically? How much do you have?
Click to expand...
Let's reformulate:
You have a 2a-sided square, and inside it you can draw 2 equal squares, in such a way they do not touch each other. Is it possible for these two squares to have sides of lenght a?
2 squares.jpg
 
I've got an idea.
Suppose the big square with side length[FONT=MathJax_Main]2[/FONT]2 is ABCD and the small square with side length [FONT=MathJax_Main]1[/FONT] is EFGH. If I prove that EFGH must contain I (the center of ABCD), the problem is solved.
 
Raoul said:
Let's reformulate:
You have a 2a-sided square, and inside it you can draw 2 equal squares, in such a way they do not touch each other. Is it possible for these two squares to have sides of lenght a?
View attachment 5215
Click to expand...

Now you have me thinking.....

How long is the diagonal of the big square? Could you position the two small squares so that each is INSIDE the large square, but there's still a little space between them (I'm thinking of placing the small squares so that two sides of each would be perpendicular to the diagonal of the big square)??? I'll keep on thinking about this, but at first glance, it seems possible.
 
Mrspi, the side of the big square is 2a. Consequently, the lenght of it's diagonal is 2a√2.
 
"a=1: longest rectangle with width 1 that can be inserted in
a 2by2 square is along square's diagonal."
Denis, I need a proof for that.
 
You must log in or register to reply here.
Top