# Thread: Interesting Area Challenge - HARD

1. ## Interesting Area Challenge - HARD

Below is an interesting problem which I struggle to solve. Does anyone have any ideas?

2. if x is the side of the square with area 1, now if we increase it with y and area 2, we have

(x + y)(x + y) = 2,

x^2 + 2xy + y^2 = 2

but x^2 = 1 and x = 1, we have

1 + 2(1)(y) + y^2 = 2, re-arranging

1 - 2 + 2y + y^2 = 0

y^2 + 2y - 1 = 0,

y1 = sqrt 2 - 1

y2 = - sqrt 2 - 1

Notice that y1 is the only positive, so y = sqrt 2 - 1

length of the the new square = x + y = 1 + (sqrt 2 - 1) = sqrt 2.

If your square with side square root of 2, that is irrational . . . . hard to measure, right?

-------------------------------------------------------------------------------------------------------

NOTE: In general, doubling the area of ANY square with a side of length x, multiply it with sqrt 2, the SQUARE that,

New Area = [(x)(sqrt 2)]^2

3. Originally Posted by pacman
if x is the side of the square with area 1, now if we increase it with y and area 2, we have

(x + y)(x + y) = 2,

x^2 + 2xy + y^2 = 2

but x^2 = 1 and x = 1, we have

1 + 2(1)(y) + y^2 = 2, re-arranging

1 - 2 + 2y + y^2 = 0

y^2 + 2y - 1 = 0,

y1 = sqrt 2 - 1

y2 = - sqrt 2 - 1

Notice that y1 is the only positive, so y = sqrt 2 - 1

length of the the new square = x + y = 1 + (sqrt 2 - 1) = sqrt 2.

If your square with side square root of 2, that is irrational . . . . hard to measure, right?

-------------------------------------------------------------------------------------------------------

NOTE: In general, doubling the area of ANY square with a side of length x, multiply it with sqrt 2, the SQUARE that,

New Area = [(x)(sqrt 2)]^2
Maybe 'hard', but not impossible. I know this challenge can be done, however I do not know how to do it! Any other attempts?

4. To start, if all you have is a straight-edge,
how will you draw the 1st square itself?

If so, do you have 2 choices: double it or get half of it?

Like, can a smaller square be drawn inside the original square?

5. Originally Posted by Wilmer
To start, if all you have is a straight-edge,
how will you draw the 1st square itself?

If so, do you have 2 choices: double it or get half of it?

Like, can a smaller square be drawn inside the original square?
The first square is already drawn.

Does anyone have any brainy ideas?

6. Originally Posted by yeah:)
I know this challenge can be done,....
If you haven't seen it done, HOW do you know?

Best I can do:

Draw the 2 corner to corner diagonals; they cross at square's center;
divides square in 4 identical triangles, each 1/4 of square's area.

Each "half-diagonal" is the required side length of a square equal to
half the original square.

Label square ABCD, A top left, going clockwise; diagonals cross at M.
Concentrate on right side of square, or side BC:
if you're able to draw a line from B perpendicular to DB,
and one from C perpendicular to AC, these 2 lines will cross, say at N.

Then you have the desired square: BMCN ; exactly half of ABCD.

But I know you'll say: "I have nothing I can use to make a 90degree angle!
Not even the end of my straight-edge..."

SO: go to corner A, tear off that corner! Since it's a 90degree corner,
use it at corner B and C to line up BN and CN...get it?

I know...I know: "I'm not allowed to use my fingers to tear off...."

Are you allowed a pencil with a square eraser?

7. this may help, make a square as much as you can, the hypotenuse will be the new side of the new square . . . . you may use a string to achieve it. the hypotenuse of the square if used as the side of the new square, doubles the AREA of the old square. i hope it helps you . . . .
or you may try this Methods of computing square roots - Wikipedia, the free encyclopedia

8. Originally Posted by yeah:)
The first square is already drawn.
The real difficulty is proving that a parallel line can be constructed with only a straight edge (or two). Generating a perpendicular is not difficult, but the parallel seems to be.

9. Originally Posted by pacman
this may help, make a square as much as you can, the hypotenuse will be the new side of the new square . . . . you may use a string to achieve it. the hypotenuse of the square if used as the side of the new square, doubles the AREA of the old square. i hope it helps you . . . .
or you may try this Methods of computing square roots - Wikipedia, the free encyclopedia

Pacman, wouldn't it be much easier if your SQRT(2) line was used as the
side of the original square? Then you need only to draw the 2 lines = 1.

If you're allowed a string, then I'm allowed a straight edge with a square corner!

10. Originally Posted by Wilmer
Pacman, wouldn't it be much easier if your SQRT(2) line was used as the
side of the original square? Then you need only to draw the 2 lines = 1.

If you're allowed a string, then I'm allowed a straight edge with a square corner!

Thank you all so much for all the wonderful ideas! However, I have still found no way of doing this without measuring something (for example, with the option where you have to draw a perpendicular, you need a protractor to do that). All you are allowed is a straight edge to draw straight lines with - you are not allowed to measure anything!

Any more ideas?

11. Originally Posted by yeah:)
Any more ideas?
NO. Give it up. Betya there's no way.

Or it'll be something like using the corner of the sheet of paper
on which the given square is printed...

12. Originally Posted by Wilmer
NO. Give it up. Betya there's no way.

Or it'll be something like using the corner of the sheet of paper
on which the given square is printed...
Do not give up! There is a way!!!

13. Originally Posted by Danny
Thank you very much for that, Danny.