EPTEmpyrean's Math Problem of the Day - Page 2
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ghidorah
United States20 Posts
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PlayJunior
Armenia833 Posts
On December 15 2005 21:03 BigBalls wrote: Yes. Cut it in half and consider the semicircle with side 11, 7, 2. put the ends of the quadrilaterals on the semicircle. then use some properties about quadrilaterals and semicircles  And if the hexagon sides are 11, 11, 7, 7,2, 2? Then, you need to prove that a line connecting 2 vertices of hexagon is diameter of the circle...I don't think you can do that | ||
ManaBlue
Canada10458 Posts
 Too bad.Seems the math involved here is too esoteric for anyone but a university student with heavy math concentration to understand. | ||
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BigBalls
United States5354 Posts
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PlayJunior
Armenia833 Posts
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PlayJunior
Armenia833 Posts
On December 15 2005 22:17 BigBalls wrote: you can permute sides to 2 7 11 2 7 11 You cannot rely on any specific drawing. | ||
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Asta
Germany3491 Posts
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Jin
Canada439 Posts
I solved it to get the equation z^3 - 174z - 308 = 0 Factor that, you should get one of the roots as 14. ***Spoiler*** Draw two diagonals in the quadrilateral. The property you have to use is that the product of the two diagonals is equal to the sum of opposing sides multiplied together. Edit: and any inscribed angle made in a semicircle is a right angle *** End spoiler *** | ||
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BigBalls
United States5354 Posts
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FakeSteve[TPR]
Valhalla18444 Posts
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Jin
Canada439 Posts
On December 15 2005 20:59 wasted wrote: but we are looking for something like this? ![[image loading]](http://img206.imageshack.us/img206/1292/hexagon3kx.png) This diagram is wrong because the hexagon is inscribed. Certainly not a trivial question. Where is the next one? | ||
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nicetoknowyou15
Jamaica185 Posts
On December 16 2005 12:34 Jin wrote: The solution is 7 I believe. That's what Bigballs has as well. I solved it to get the equation z^3 - 174z - 308 = 0 Factor that, you should get one of the roots as 14. -308+z(z^2-174) z^2-174=0 z^2=174 square root of 174 = 13.2 ?? | ||
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Jin
Canada439 Posts
On December 16 2005 12:56 nicetoknowyou15 wrote: -308+z(z^2-174) z^2-174=0 z^2=174 square root of 174 = 13.2 ?? Um... z(z^2-174) = 308 z^2-174 != 0 -14 | 1 0 -174 -308 | 0 -14 -196 -308 -------------------- gives 1 14 22 0 so you get (z-14)(z^2 +14z +22) rest is trivial I realize the syntax might not make any sense but I can't draw it out in ascii. | ||
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-_-
United States7081 Posts
(until calculus 2 next semester and linear algebra the semester after that TT) | ||
Empyrean
17004 Posts
I'm always amazed at what this forum can do  | ||
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Empyrean
17004 Posts
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Bockit
Sydney2287 Posts
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Empyrean
17004 Posts
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Catyoul
France2377 Posts
Looks like I'm in time for day 2. Basically, we're looking for the highest power of 10 in 100000!, that is the minimum between the highest power of 2 and the highest power of 5. It's obvious that the highest power of 2 is higher, so we just have to find the highest power of 5 in 100000! To find that, we'll need to find how many multiples of 5, 5^2, 5^3,... there are between 1 and 100000, and that is simply : - 5 : 100000/5 rounded to the inferior = 20000 - 5^2 : 100000/25 = 4000 - 5^3 : 100000/125 = 800 - 5^4 : 100000/625 = 160 - 5^5 : 100000/3125 = 32 - 5^6 : 100000/5^6 = 6 - 5^7 : 100000/5^7 = 1 That makes 20000+4000+800+160+32+6+1=24999, which means we have 5^24999 in 100000!, and we have 24999 zeroes at the end of 100000! | ||
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Bockit
Sydney2287 Posts
On December 16 2005 19:22 Empyrean wrote: You don't know what a factorial is, do you I do, but what I said then was a guess  100000*99999*99998*99997*...* 1 | ||
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