interesting geometry challenge

[Don Shepherd]

Who will be the first to solve this?

[Han]

Looks like it should be 9/4.

Edit: I think they should have not specified the lengths \( ab \) and \(ac \).

[Don Shepherd]

Yes, congratulations. Exactly 1/4 of the area of the 3-inch cube.

[Thomas Radtke]

Yes, quite easy to see that the intersection area is a constant for reason of symmetry.

[Gerson W. Barbosa]

(09-15-2014 03:47 PM)Don Shepherd Wrote:  Yes, congratulations. Exactly 1/4 of the area of the 3-inch cube.

I did (3/2)^2 = 9/4, but I have to admit 3^2/4 is some milliseconds faster (only one number to square, or one memory space to access :-)

[Bill (Smithville NJ)]

(09-15-2014 03:42 PM)Han Wrote:  Looks like it should be 9/4.

Edit: I think they should have not specified the lengths \( ab \) and \(ac \).

All you really need to know is that the second square is at least as large as half of the size of the first square and the fact that its corner is in the center of the first square. Then it's just rotate the square so that it's sides bisects the first square's sides and it becomes obvious that it's 1/4 of the area of the first square.

This is a classic style of problem that gives more information than what is actually required to solve the problem. It's a good type of math problem where thinking about it and visualizing what is occurring is more important than working the math out. It's a good problem to give to a group of students to see which students attempt to set up unknowns and then solve for it and which students can quickly visualize the quick solution.

Bill

[Gerson W. Barbosa]

(09-15-2014 07:38 PM)Bill (Smithville NJ) Wrote:  

(09-15-2014 03:42 PM)Han Wrote:  Looks like it should be 9/4.

Edit: I think they should have not specified the lengths \( ab \) and \(ac \).

All you really need to know is that the second square is at least as large as half of the size of the first square and the fact that its corner is in the center of the first square. Then it's just rotate the square so that it's sides bisects the first square's sides and it becomes obvious that it's 1/4 of the area of the first square.

If one doesn't want to rotate the larger square, the given data can nevertheless be used to calculate the area:

\[\sqrt{\left (\frac{\sqrt{10}+3}{2}-\sqrt{\frac{5}{2}} \right )^{2}\left ( \frac{\sqrt{10}+3}{2}-2 \right )\left ( \frac{\sqrt{10}+3}{2}-1 \right )}=\frac{9}{4}\]


No one should try this during an examination, however :-)

Gerson.

P.S.: Well, when I can make the formula above show up right...

P.S. 2: Fixed, thanks to Thomas Klemm's assistance below.

[Thomas Klemm]

(09-16-2014 04:33 PM)Gerson W. Barbosa Wrote:  P.S.: Well, when I can make the formula above show up right...

\[\sqrt{\left (\frac{\sqrt{10}+3}{2}-\sqrt{\frac{5}{2}} \right )^{2}\left ( \frac{\sqrt{10}+3}{2}-2 \right )\left ( \frac{\sqrt{10}+3}{2}-1 \right )}=\frac{9}{4}\]

Is that close enough?

Cheers
Thomas

[Gerson W. Barbosa]

(09-17-2014 02:32 AM)Thomas Klemm Wrote:  

(09-16-2014 04:33 PM)Gerson W. Barbosa Wrote:  P.S.: Well, when I can make the formula above show up right...

\[\sqrt{\left (\frac{\sqrt{10}+3}{2}-\sqrt{\frac{5}{2}} \right )^{2}\left ( \frac{\sqrt{10}+3}{2}-2 \right )\left ( \frac{\sqrt{10}+3}{2}-1 \right )}=\frac{9}{4}\]

Is that close enough?

Cheers
Thomas

Thanks! I have yet to find out what is wrong in the script above. It was created on the equation editor that is linked in your article at the old articles forum. I've tested with a short equation and it worked perfectly. Perhaps I did something wrong.

Cheers,

Gerson.

[Thomas Klemm]

(09-17-2014 03:59 AM)Gerson W. Barbosa Wrote:  Thanks! I have yet to find out what is wrong in the script above. It was created on the equation editor that is linked in your article at the old articles forum. I've tested with a short equation and it worked perfectly. Perhaps I did something wrong.

\sqrt{\frac{5}{2} \right )

Here the closing parentheses } of \sqrt is missing. There's one too much at the end of the left hand side of the equation.

Cheers
Thomas

[Gerson W. Barbosa]

(09-17-2014 06:20 AM)Thomas Klemm Wrote:  

(09-17-2014 03:59 AM)Gerson W. Barbosa Wrote:  Thanks! I have yet to find out what is wrong in the script above. It was created on the equation editor that is linked in your article at the old articles forum. I've tested with a short equation and it worked perfectly. Perhaps I did something wrong.

\sqrt{\frac{5}{2} \right )

Here the closing parentheses } of \sqrt is missing. There's one too much at the end of the left hand side of the equation.

Thanks!

Gerson.

[Peter Murphy]

The surveyor's formula works here.

Simplify by doubling all linear dimensions (which quadruples all areas). Take the center of the small square as the origin of coordinates, and the coordinates of the vertices in question are then (0,0), (3,1), (3,-3), and (1,-3).

Then the shoelace version of the surveyor's formula yields 9 as the area of the figure.

See here:

<http://en.wikipedia.org/wiki/Shoelace_formula>

[Thomas Klemm]

(09-18-2014 02:34 PM)Peter Murphy Wrote:  Take the center of the small square as the origin of coordinates, and the coordinates of the vertices in question are then (0,0), (3,1), (3,-3), and (1,-3).

Then the shoelace version of the surveyor's formula yields 9 as the area of the figure.

There are programs in the General Software Library:
HP-48
HP-42S

Or then you can use Pick's theorem:


i = 6
b = 8
A = i + b/2 − 1 = 6 + 4 - 1 = 9

Cheers
Thomas

[Peter Murphy]

Thomas,

Re: Your Post #13

Thanks for adding Pick's theorem to the mix, and for the helpful links to shoelace-formula programs.

[CosmicTruth]

2.25
FIRST!
DID I WIN? DID I WIN?
:-))

[Don Shepherd]

(09-18-2014 11:08 PM)CosmicTruth Wrote:  2.25
FIRST!
DID I WIN? DID I WIN?
:-))

Well, no, Han won, post #2.