I am afraid, your solution does not work. The idea is not to divide the 3rd square, but the remaining white area. By doing what you suggest, we don’t get 4 equal parts of the white space.
From India, Delhi
1426From India, Delhi
Very good lateral thinking puzzle that helps u to think out of box. Keep posting. Regards, Shailesh
From India, Pune
From India, Pune
Hi, Nice One. As mentioned, because of the conditioning of the mind from 1st to 3rd and 3rd not being simple, you tend to flow in the water despite the title itself saying that thinking out of the box. Like others, looking at the answer only smile came to my face.Keep on sharing. Thanks.Regards
From India, Delhi
From India, Delhi
Excellent and very powerful. This also proves that presentations need not be very long. One can speak/present less and convey more. Thank you Gayatri Sivaram
From India, Delhi
From India, Delhi
Dear Gayatri, Your PPT was really a nice and simple one. The last two slides of the great Mahatma Gandhi’s ideology brought strength to your concept. Good Luck Regards, Denny Raj
From India, Thiruvananthapuram
From India, Thiruvananthapuram
Dear Gayatri, Your PPT was really a nice and simple one. The last two slides of the great Mahatma Gandhi’s ideology brought strength to your concept. Good Luck Regards, Denny Raj
From India, Thiruvananthapuram
From India, Thiruvananthapuram
hi gayatri , its very nice sometime we r not able 2 see d simple thing or action which can solve our difficult problems n we search out other difficult options
From India, Ghaziabad
From India, Ghaziabad
Hi Gayathri, Hey U have put Life’s Truest Funda in a nutshell. Well That’s something good which if understood can make people ever happy.
From India, Bangalore
From India, Bangalore
HI Rajkumar,
My apologies. There was a mistake.Read CG instead of AH in my quote " Now join AH". Do you now agree that the white area has two equal halves,CGFB & CGHD ? Right? When you join CF & CH, you now have four equal parts of the white area, viz.,CFB,CFG,CGH & CDH.
Try it & confirm if I am right !!
Thanks.
From India, Lucknow
My apologies. There was a mistake.Read CG instead of AH in my quote " Now join AH". Do you now agree that the white area has two equal halves,CGFB & CGHD ? Right? When you join CF & CH, you now have four equal parts of the white area, viz.,CFB,CFG,CGH & CDH.
Try it & confirm if I am right !!
Thanks.
From India, Lucknow
Dear Sreinivas
I tried your corrected solution too.
It simply does not work.
What it does is - divides the white space into two equal halves.
Anything beyond this produces 4 pieces which are asymetrical, un-identical and do not appear to have the same area.
Please illustrate with diagram your solution - if you think you have got it right.
Now, consider my solutions :
I have two solutions which produce four equal areas, but they are not symmetrical in shape.
(1) The first solution involves dividing the white space into three equal squares (as apparent prima facie); then divide each small square into 4 smaller squares, making total 12 smaller squares.
icking up any 3 adjacent squares (other than those used in the original solution) will give 4 shapes, each made out of 3 smallest squares .
(2) The second solution i similar, buthere you divide each small squares into 4 equal triangles -equilateral - by joining the diagonals of the smaller square.
You will get 12 small triangles, out of which you need to pick sets of 3; thus totaling 4 sets of equal triangles and hence 4 equal areas.
Remember, in both the solutions, the areas are equal but the shapes are not identical.
Regards.
From India, Delhi
I tried your corrected solution too.
It simply does not work.
What it does is - divides the white space into two equal halves.
Anything beyond this produces 4 pieces which are asymetrical, un-identical and do not appear to have the same area.
Please illustrate with diagram your solution - if you think you have got it right.
Now, consider my solutions :
I have two solutions which produce four equal areas, but they are not symmetrical in shape.
(1) The first solution involves dividing the white space into three equal squares (as apparent prima facie); then divide each small square into 4 smaller squares, making total 12 smaller squares.
icking up any 3 adjacent squares (other than those used in the original solution) will give 4 shapes, each made out of 3 smallest squares .
(2) The second solution i similar, buthere you divide each small squares into 4 equal triangles -equilateral - by joining the diagonals of the smaller square.
You will get 12 small triangles, out of which you need to pick sets of 3; thus totaling 4 sets of equal triangles and hence 4 equal areas.
Remember, in both the solutions, the areas are equal but the shapes are not identical.
Regards.
From India, Delhi
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Positive thinking.pdf
Lateral Thinking Puzzle ch.ppt