如果你认为 A=2B时,切割成3个三角形,然后拼成一个正方形,就算你小学几何60分吧.
现在
再问:
1. A和B在什么关系下,该长方形可以被直线切割成3块,切割线不可以平行与长方形的边,被切开的
3块面积不相等,
when A*B <= (A/2)^2 + B^2, it can be done; when A*B > (A/2)^2 + B^2, it can't be done. |
"A*B <= (A/2)^2 + B^2, it can be done" is equivalent to A <=2B from A*B <= (A/2)^2 + B^2 then 4A*B<= A^2 + 4B^2 0<= A^2 -4A*B+ 4B^2 0<= (A-2B)^2 gives roots 2B =>A ( take) and 2B <=A ( omit)  (I don't believe kids in elementary school have learnt this) |
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Given a retangle abcd, where ab=cd=A, and bc=da=B, the area of it is A*B. Cut it somehow and rearrange it to a square with an edge length sqroot(A*B) for the square should have the same area as that of the retangle. Here is the procedure to cut it. Pass through the corner abc, draw a line with a length of sqroot(A*B), and join the edge cd at e. (1st cut) Then daw a line perpendicular to line be, and cross the corner dab (2nd cut) Then move them around to make a square with edge length of sqroot(A*B). 本贴由[salmonfish]最后编辑于：2008-9-21 21:16:3 |
A and B are just signs not are exact numbers what you can have are a scissors and a compass only. i think that may be only a scissors or cutter we can use. |
I did not use anything, no scissors, no compass, no cutter, even no paper and pencils. Use brain only!Imagine a retangle with edge lengthes A and B, and A > B (there was no specific numbers in my mind, except for the two "signs", A and B). So the retangle has an area A*B. Somehow to make it into a square with the equal area, the square's edge SHOULD be sqroot(A*B) (who cares what number it would be here ).(For your convenience, I "print" the brain images here) Since parallel lines, to A and B, are out of the question, now IMAGINE a line passing (b) and joining the opposite long edge (A) at (e). Just LET length of this line, (be), be sqroot(A*B), for that such a line SHOULD exist, and MUST exist. This is beause for B Hence, Line (be) = sqroot(A*B) < length of line (bd) = sqroot(A^2 + B^2) Also, from A>B, we have A*B > B^2, and hence Line (be) = sqroot(A*B) > length of edge B If such a line is existed, I'll take it (So I have the line (be) with a length sqroot(A*B)). Still no scissors,...... Now imagine the second line which passes the point (a) and perpendicularly join line (be) at (g). Since the triangle (bce) is similar to the triangle (agb) (angle bce= angle agb = 90 deg, and angle bec= angle gba), we have line (ag)/A = B/sqroot(A*B) that is, line (ag) =A*B/sqroot(A*B) = sqroot(A*B) Ooooooooooops! If line (be) = sqroot(A*B)> sqroot((A/2)^2 + B^2) (that is (ce) longer than A/2), the line (ag) will be short than sqroot(A*B). So a restrictive condition has to be posted here. That is, (A*B) has to be < or = (A/2)^2 + B^2, or A <=2B. Now, imagine one can fold the retangle and tear it along line (be) first, and then along line (ag). Now you can shove the three pecies around until you gt the desired square. Told you that I did not use scissors, compass, cutter, even paper or pencils
本贴由[salmonfish]最后编辑于：2008-9-22 11:2:13 本贴由[salmonfish]最后编辑于：2008-9-22 13:42:16 |
For 1st question: If you can draw a line of a given length (number) with a certain accuracy, then I can daw a square root of that number with the same accuracy. Reversely, if you can't daw a sqroot(2)= 1.41423... on a paper, you also can't draw a line of length 2.00000... with the same accuracy. For 2nd question: As long as there EXISTS such a perpendicular line with a length of sqroot(A*B), then the asked question CAN BE DONE. Once it has been proved can be done, my job is over. Who is gonna do it in real life? Be my guest. |
Mentally, I can still cut it into the three pecies. |
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