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Post by slippy0 on Mar 12, 2006 23:24:57 GMT -5
This was a thread made long long ago, but I felt we needed it again. sometime last year (8th grade) I made myself a geometry problem which I couldn't solve, nor could my math teacher, or one of the highest ranked math kids in our age, (they didn't spend much more than 10 minutes on it, I bet they would've gotten it if they spent more time on it). but anyways, heres the problem:  Circle inscribed in square with side of 10, fine "?" on a 3 hour plane trip home today I spent an hour on it and finally figured out what to do. I got it after teaching myself Trig. If you know trig, you merely have to draw a few nice lines, and do 2 trig functions and you'll have the answer, though its not directly obvious what. So first of all: can anyone figure it out? second: post anything else you're proud of |
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Post by Phil on Mar 13, 2006 14:10:24 GMT -5
Aw man,
I remember there's some kind of formula to figuring it out too. I'll be smacked if I can remember what it is too.
I've seen this type of thing before but I can't remember how it works. If it comes to me I'll post it.
I think you're going to be an engineer Slippy0
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Post by Kevin on Mar 13, 2006 16:25:26 GMT -5
I believe ?=8.
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Post by mattis on Mar 13, 2006 16:53:37 GMT -5
I can get the circumference and area of the circle but nothing else.
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Post by slippy0 on Mar 13, 2006 18:29:44 GMT -5
close, but its actually a really ugly number (assuming I did it right  ) |
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Post by Kevin on Mar 13, 2006 19:43:32 GMT -5
Well I did it the way any engineer would: I used a computer ;D.
I drew the picture on AutoCAD and then measured the length. Maybe it rounded too much, but it usually doesn't. It displayed the value as 8.00, but usually goes out 10 or so places if it's a long decimal.
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Post by GetZ on Mar 13, 2006 20:23:31 GMT -5
? = 2[5Cos(30°)] = 8.66 I think that's right. I started out with doing it a more complicated way by trying to put things in terms of where the circle intersects the line but then I just divided the line up into 2 small triangles instead  This is pretty similar to a problem that I came up with a year ago when trying to figure out what the effective diameter was for a telescope to 'see' a dim trio of galaxies in the constellation of Draco. I gradually masked more and more of the aperture of an 8 inch (in diameter) telescope with a piece paper until I was just able to detect the faintest of the three galaxies. I think, if I remember correctly, this occurred at about the point where I had covered 2/3 of the apertures diameter and so I then had to figure out a way to calculate the remaining area... At the time, it took a bit of thought and required setting up a remarkably similar problem as the one that you posted above. But now, armed with some heavy artillery, I can just use some calculus to do it in like... half a second  Congrats on not just tossing the problem away and eventually coming up with a solution  As for me... I recently failed my Calculus mid-term exam (like... A LOT!) and now my prof. thinks i'm "a VERY bright" person, and also some of the other math profs have been acknowledging my existence. University is great! ;D ;D ;D (I'm the most confusing person I know) (...Or, don't know. As the case may be..) (I'm confused again)... |
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Post by slippy0 on Mar 13, 2006 20:26:38 GMT -5
well now we have a problem, my CAD software agrees (sorta, I got 8.011). But my work makes sense, and I checked over using different software as well as a calc. all give me the same answer (which isn't close enough to 8 to be considered a case of rounding in our software) Heres my and answer: SPOILER ALERT!see if you can find any errors. (I didn't show any of the work cus that would be lots o writing, if you know basic trig and geometry, you should be able to figure it out ) |
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Post by Kevin on Mar 13, 2006 20:42:41 GMT -5
I just got done with my second calculus class today GetZ. My final exam was a little tough, but I think I did pretty well. I'm signing up for the third class in the calculus series forspring quarter... w00t to having my college freshman year of math done as a senior in high school.
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Post by GetZ on Mar 13, 2006 20:47:40 GMT -5
Oops I assumed that it was an equalateral triangle for some reason. My answer should then be, ? = 2[5Cos(26.5650512)] = 8.944271908 I'm not sure why either of the CADs are giving smaller answers though. Could you describe how you're using the programs to get the result? |
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Post by slippy0 on Mar 13, 2006 21:06:32 GMT -5
Oops I assumed that it was an equalateral triangle for some reason. My answer should then be, ? = 2[5Cos(26.5650512)] = 8.944271908 I'm not sure why either of the CADs are giving smaller answers though. Could you describe how you're using the programs to get the result? thats exactly what I got ;D So now lets yell at the CAD companies. |
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Post by Captain Awesome on Mar 13, 2006 21:08:24 GMT -5
see if you can find any errors. I see one right off the bat:  That little red angle in there should, by all logic, be 22.5 degrees. And the new right angles you drew in are most definitely not right angles. And yes, I know I'm a little late, but the stupid anti-aliasing on the important parts took forever to get rid of.  |
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Post by slippy0 on Mar 13, 2006 21:08:47 GMT -5
(I'm the most confusing person I know) (...Or, don't know. As the case may be..) (I'm confused again)... I know how that feels. |
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Post by slippy0 on Mar 13, 2006 21:12:31 GMT -5
see if you can find any errors. I see one right off the bat: That little red angle in there should, by all logic, be 22.5 degrees. And the new right angles you drew in are most definitely not right angles. And yes, I know I'm a little late, but the stupid anti-aliasing on the important parts took forever to get rid of. aaand youd be wrong breif description of why those are 90 degrees there are 4 centers of a triangle, one of these is the center of the circle which circumscribes it (which we have in this problem) This center is found by taking the perpendicular bysector of every side, and finding where they all intersect. so yah, basically, those lines are the perp. Bis. of those 3 sides, so they must be at a 90 degree angle. le picture:  Yah... I was kinda supposed to skip 9th and 10th grad math and be taking Pre-calc this year, but my school system was like "no", silly people and their non-exception making. |
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Post by Captain Awesome on Mar 13, 2006 21:18:54 GMT -5
But the triangle is not an equilateral triangle. The bisectors should line up like this:  EDIT: I'm talking about this one^^^^^^ ...yeah. |
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Post by slippy0 on Mar 13, 2006 21:24:29 GMT -5
...crap
its still 26.5650512, though, not 22.5
arctan 0.5=26.5650512
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Post by Captain Awesome on Mar 13, 2006 21:40:36 GMT -5
Wait... just kidding, it would be 26.565...
hold on...
dinner...
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Post by slippy0 on Mar 13, 2006 21:40:39 GMT -5
yaaaah... I think I'll let this sit for another year or two.
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