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 April 5th, 2010, 01:51 PM #1 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 puzzler Almost everyone would agree that the point of intersection of two diagonals of a rectangle deserves to be called the "middle" of the rectangle. After all, this would be the "balance point" of the shape if the rectangle were cut from a uniform material and placed horizontally on the tip of a pencil. The three squares are adjoined to make an L-shape piece. Find the exact location of the balance point of this figure. http://t1.gstatic.com/images?q=tbn:Z_KJ ... +parts.gif Design an L-shaped piece whose balance point lies outside the figure!
April 6th, 2010, 10:03 AM   #2
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Re: puzzler

Quote:
 Originally Posted by alevis Almost everyone would agree that the point of intersection of two diagonals of a rectangle deserves to be called the "middle" of the rectangle. After all, this would be the "balance point" of the shape if the rectangle were cut from a uniform material and placed horizontally on the tip of a pencil. The three squares are adjoined to make an L-shape piece. Find the exact location of the balance point of this figure. http://t1.gstatic.com/images?q=tbn:Z_KJ ... +parts.gif Design an L-shaped piece whose balance point lies outside the figure!
Hi alevis,

I used the two stacked squares on the left as one rectangle and the square to the right and drew diagonals to locate the balance point for each. That's indicated by the red lines. I connected these points with a dark blue line.

Next, I used the top left square and the bottom two squares as a rectangle and drew diagonals to locate the balance points of each. That's indicated by the light blue lines. I connected these points with a dark blue line.

The intersection of the dark blue lines, I think, is the balance point for the figure.

I can't seem to upload a jpg or png file. Or, maybe I don't know how. I copied it to photobucket, though.

April 6th, 2010, 12:29 PM   #3
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Re: puzzler

Hello, alevis!

Quote:
 Almost everyone would agree that the point of intersection of two diagonals of a rectangle deserves to be called the "middle" of the rectangle. After all, this would be the "balance point" of the shape if the rectangle were made of a uniform material and placed horizontally on the tip of a pencil. The three squares are adjoined to make an L-shape piece. Find the exact location of the balance point of this figure.

$\text{The side of the unit squares are length }s.$

$\text{The area of the L-shaped piece is: }\:3s^2$

Code:
      |
2s * - - - *(a,2s)
|       |
- *       |
: | *     |    (2s,s)
: * - * - * - - - *
a |     o |       |
: |       *       |
: |       | *     |
- * - - - * - * - * -
| - - a - - :  2s

$\text{The balance point (centroid) lies on the line }\:y \:=\:x$

$\text{We want the line, }x\,+\,y \:=\:a,\,\text{ which bisects the area.}$

$\text{The area of the right triangle is: }\:A \:=\:\frac{1}{2}a^2$

$\text{It is to be equal to half the area of the L-shaped piece: }\:$

$\text{So we have: }\:\frac{1}{2}a^2 \:=\:\frac{1}{2}(3s^2) \;\;\;\Rightarrow\;\;\;a^2 \:=\:3s^2 \;\;\;\Rightarrow\;\;\;a \:=\:\sqrt{3}\,s$

$\text{The centroid lies on the lines: }\:x\,+\,y\:=\:\sqrt{3}\,s\:\text{ and }\:y \:=\:x$

[color=beige]. . [/color]$\text{Hence: }\:x\:=\:y\:=\:\frac{\sqrt{3}}{2}s$

$\text{The centroid is at: }\:G\,\left(\frac{\sqrt{3}}{2}s,\:\frac{\sqrt{3}}{ 2}s\right)$

Quote:
 Design an L-shaped piece whose balance point lies outside the figure.
Code:
    - * - - - *
: |       |
: |       |
: |       |
: * - - - *
: |       |   G
3s |       |  o
: |       |
: * - - - * - - - * - - - *
: |       |       |       |
: |       |       |       |
: |       |       |       |
- * - - - * - - - * - - - *
: - - - -  3s - - - - - :

$\text{The centroid is at: }\:\left(\sqrt{5}\,s,\:\sqrt{5}\,s\right)$

 April 6th, 2010, 07:24 PM #4 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Re: puzzler Thank you all
 April 7th, 2010, 05:19 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2207 The answers given by soroban were obtained using an incorrect method, whereas masters used a valid method, and some easy algebra based on that method finds that the centroid of the original L-shape is at ((5/6)s, (5/6)s). The L-shape soroban suggests does have its centroid just outside the figure, but at (1.1s, 1.1s).

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