11My take was different....I did this by taking 'n' rectangles....btw why are u guys so obsessed with 4 rectangles?
There's a square ABCD.....inscribed in a circle, this square is now partitioned into non-overlapping rectangles, and for each rectangle it's circum-circle is drawn. If the sum of areas of all these circles equals the area of the circle circumscribing ABCD, then prove that all these rectangles are squares.
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62WLOG assume that the origninal square had a side 1
a, 1-a and b, 1-b are the division of the square..
now there are 4 rectangles of sides...
a, b
a, 1-b
1-a, b
1-a, 1-b
Now can you try from here..
(I know some ppl could say that this is teh most stupid hint.. but I guess the question should not be very diffiuclt from here?)
1LET THE SIDE LENGTH BE p.
the four sides of the rectangles are a,b
p-a,p-b
p-a,b
a,p-b
then the respective radii of the circles circumscribing the rectangles are the half of corresponding diagonals
1/2(a2+b2)1/2
1/2(a2+b2-2pa-2pb+p2)1/2
1/2(a2+b2-2pa+p2)1/2
1/2(a2+b2-2pb+p2)1/2
thus by putting total area =area of circle it comes out to be
Î (a2+b2+p2-p(a+b))=Î /2(p)2
or, p2-2p(a+b)+2(a2+b2)=0
now this is a quadratic in p
hence discriminant D=-4(a-b)2
but D cannot be negative as p is real
hence D=0 implies a=b and equation becomes
p2-4pa+4a2=0
hence p=2a=2b
hence the four rectangles are squares
proooooved.
62That's correct.. i was hinting at was..
WLOG original side to be 1
IF sum of areas of circles is equal to the original circle,
then Sum of the squares of the diagonal will be equal to the square of the original diagonal..
2 = a2+b2 + (1-a)2+b2+ a2+(1-b)2+ (1-a)2+(1-b)2
2 = a2+b2 + 1+a2-2a+b2+ a2+1-2b+b2+ 1+a2-2a+1+b2-2b
2 = 4a2+4b2 +4-4a-4b
a2+b2 -a-b+1/2 = 0
(a-1/2)2+(b-1/2)2 = 0
so a=b=1/2
62Alternatively, after your step:
p2-2p(a+b)+2(a2+b2)=0
2p2-4p(a+b)+4(a2+b2)=0
now you could have written this as
(p-2a)2+(p-2b)2 =0
which is zero when p=2a=2b
hence the solution :)
1162oh.. ok.. din notice..
I guess even then the same logic will work..
and this time cauchy schwarz would probably come into picutre..
or even here the final expression can be expressed as the sum of perfect squares...
11What i did was to have (l_i,b_i) as the diamensions of the ith rectangle.....which gave me the area of the circle circumscribing it as \pi \frac{l_i^2+b_i^2}{4}
From which I get \pi \frac{r^2}{2}=\sum{\pi \frac{l_i^2+b_i^2}{4} \\ \implies \sum{l_ib_i=\frac{1}{2}\sum{(l_i^2+b_i^2)
Now A.M-G.M to prove equality when l=b.
\boxed{Done}
62Just for fun :D
\boxed{}
is enough ... you dont have to write \boxed{Done}
actually that sign says "end of proof"
11OOps....I forgot to mention that the side length of this square is r.
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