interesting geometry challenge

09152014, 02:48 PM
Post: #1




interesting geometry challenge
Who will be the first to solve this?


09152014, 03:42 PM
(This post was last modified: 09152014 03:49 PM by Han.)
Post: #2




RE: interesting geometry challenge
Looks like it should be 9/4.
Edit: I think they should have not specified the lengths \( ab \) and \(ac \). Graph 3D  QPI  SolveSys 

09152014, 03:47 PM
Post: #3




RE: interesting geometry challenge
Yes, congratulations. Exactly 1/4 of the area of the 3inch cube.


09152014, 03:51 PM
Post: #4




RE: interesting geometry challenge
Yes, quite easy to see that the intersection area is a constant for reason of symmetry.


09152014, 04:03 PM
Post: #5




RE: interesting geometry challenge  
09152014, 07:38 PM
(This post was last modified: 09152014 07:40 PM by Bill (Smithville NJ).)
Post: #6




RE: interesting geometry challenge
(09152014 03:42 PM)Han Wrote: Looks like it should be 9/4. 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 

09162014, 04:33 PM
(This post was last modified: 09172014 03:47 PM by Gerson W. Barbosa.)
Post: #7




RE: interesting geometry challenge
(09152014 07:38 PM)Bill (Smithville NJ) Wrote:(09152014 03:42 PM)Han Wrote: Looks like it should be 9/4. 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. 

09172014, 02:32 AM
Post: #8




RE: interesting geometry challenge
(09162014 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 

09172014, 03:59 AM
Post: #9




RE: interesting geometry challenge
(09172014 02:32 AM)Thomas Klemm Wrote:(09162014 04:33 PM)Gerson W. Barbosa Wrote: P.S.: Well, when I can make the formula above show up right... 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. 

09172014, 06:20 AM
Post: #10




RE: interesting geometry challenge
(09172014 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 

09172014, 03:52 PM
Post: #11




RE: interesting geometry challenge
(09172014 06:20 AM)Thomas Klemm Wrote:(09172014 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. Thanks! Gerson. 

09182014, 02:34 PM
Post: #12




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> 

09182014, 04:21 PM
(This post was last modified: 09182014 04:24 PM by Thomas Klemm.)
Post: #13




RE: interesting geometry challenge
(09182014 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). There are programs in the General Software Library: HP48 HP42S Or then you can use Pick's theorem: i = 6 b = 8 A = i + b/2 − 1 = 6 + 4  1 = 9 Cheers Thomas 

09182014, 06:15 PM
Post: #14




RE: interesting geometry challenge
Thomas,
Re: Your Post #13 Thanks for adding Pick's theorem to the mix, and for the helpful links to shoelaceformula programs. 

09182014, 11:08 PM
Post: #15




RE: interesting geometry challenge
2.25
FIRST! DID I WIN? DID I WIN? :)) Thanks ~~~~8< Art >8~~~~ PS: Please post more 50G stuff :) 

09192014, 12:09 AM
Post: #16




RE: interesting geometry challenge  
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