interesting geometry challenge
09-15-2014, 02:48 PM
Post: #1
 Don Shepherd Senior Member Posts: 749 Joined: Dec 2013
interesting geometry challenge
Who will be the first to solve this?
09-15-2014, 03:42 PM (This post was last modified: 09-15-2014 03:49 PM by Han.)
Post: #2
 Han Senior Member Posts: 1,882 Joined: Dec 2013
RE: interesting geometry challenge
Looks like it should be 9/4.

Edit: I think they should have not specified the lengths $$ab$$ and $$ac$$.

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09-15-2014, 03:47 PM
Post: #3
 Don Shepherd Senior Member Posts: 749 Joined: Dec 2013
RE: interesting geometry challenge
Yes, congratulations. Exactly 1/4 of the area of the 3-inch cube.
09-15-2014, 03:51 PM
Post: #4
 Thomas Radtke Senior Member Posts: 780 Joined: Dec 2013
RE: interesting geometry challenge
Yes, quite easy to see that the intersection area is a constant for reason of symmetry.
09-15-2014, 04:03 PM
Post: #5
 Gerson W. Barbosa Senior Member Posts: 1,487 Joined: Dec 2013
RE: interesting geometry challenge
(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 :-)
09-15-2014, 07:38 PM (This post was last modified: 09-15-2014 07:40 PM by Bill (Smithville NJ).)
Post: #6
 Bill (Smithville NJ) Senior Member Posts: 464 Joined: Dec 2013
RE: interesting geometry challenge
(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
09-16-2014, 04:33 PM (This post was last modified: 09-17-2014 03:47 PM by Gerson W. Barbosa.)
Post: #7
 Gerson W. Barbosa Senior Member Posts: 1,487 Joined: Dec 2013
RE: interesting geometry challenge
(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.
09-17-2014, 02:32 AM
Post: #8
 Thomas Klemm Senior Member Posts: 1,814 Joined: Dec 2013
RE: interesting geometry challenge
(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
09-17-2014, 03:59 AM
Post: #9
 Gerson W. Barbosa Senior Member Posts: 1,487 Joined: Dec 2013
RE: interesting geometry challenge
(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.
09-17-2014, 06:20 AM
Post: #10
 Thomas Klemm Senior Member Posts: 1,814 Joined: Dec 2013
RE: interesting geometry challenge
(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
09-17-2014, 03:52 PM
Post: #11
 Gerson W. Barbosa Senior Member Posts: 1,487 Joined: Dec 2013
RE: interesting geometry challenge
(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.
09-18-2014, 02:34 PM
Post: #12
 Peter Murphy Junior Member Posts: 44 Joined: Feb 2014
RE: interesting geometry challenge
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>
09-18-2014, 04:21 PM (This post was last modified: 09-18-2014 04:24 PM by Thomas Klemm.)
Post: #13
 Thomas Klemm Senior Member Posts: 1,814 Joined: Dec 2013
RE: interesting geometry challenge
(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
09-18-2014, 06:15 PM
Post: #14
 Peter Murphy Junior Member Posts: 44 Joined: Feb 2014
RE: interesting geometry challenge
Thomas,

09-18-2014, 11:08 PM
Post: #15
 CosmicTruth Member Posts: 164 Joined: May 2014
RE: interesting geometry challenge
2.25
FIRST!
DID I WIN? DID I WIN?
:-))

Thanks
~~~~8< Art >8~~~~

PS: Please post more 50G stuff :)
09-19-2014, 12:09 AM
Post: #16
 Don Shepherd Senior Member Posts: 749 Joined: Dec 2013
RE: interesting geometry challenge
(09-18-2014 11:08 PM)CosmicTruth Wrote:  2.25
FIRST!
DID I WIN? DID I WIN?
:-))

Well, no, Han won, post #2.
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