First problem! I don't know how many I'll go to. Depends on when I run out of interesting problems
A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.
Find the radius of the circle.
Day two:
By the way, this one's easier.
How many zeros are at the end of 100000! ?
Day three:
We all know it's impossible to construct a circle with the same area as a square within the Archimedean constraints (used only straight-edge and compass, and must be done in a finite number of constructions).
Show how to construct a circle with the same area as a square if you remove one of those constraints.
Have fun.
And by the way, do you know about those nifty "Mira" things? They're basically red sheets of translucent reflective plastic. You can use them for constructions, and you can actually trisect an angle with one. They won't help you with this problem, though.
Day four:
This is from an old math competition. I don't remember which, exactly, but J. Riley wrote it.
If w, x, y, and z are finite, not all zero, and satisfy the equations
w = bx + cy + dz x = aw + cy + dz y = aw + bx + dz z = aw + bx + cy
Umm... not enough info? What are the interior angles of the hexagon? (or maybe it's not necessary, I dunno )
Edit: after further thought and some MS Paint diagrams, I've decided to demand a diagram for this question, so that we know exactly what this hexagon looks like. At the moment we don't even know which sides have which lengths, for one thing.
Edit: actually it depends on what you mean by inscribed. If we had a normal hexagon inscribed by the circle then every vertex would touch the circle, but if this particular hexagon was inside a circle only 2 vertices will touch, so find the largest radius you need.
actually he just giving us all a riddle.. he not asking for help, hence the title of the thread.. if he do neeed help, he could have just googled the question .. and the answer/explanation should be at first result.
On December 15 2005 19:59 Ender wrote: The hexagon is inside an eclipse, not a circle.
Edit: actually it depends on what you mean by inscribed. If we had a normal hexagon inscribed by the circle then every vertex would touch the circle, but if this particular hexagon was inside a circle only 2 vertices will touch, so find the largest radius you need.
First of all, an eclipse is a car, not a shape. Secondly, he said it's a circle and is asking for radius, something that an ellipse (I'm assuming that's what you ment) does not have.
That having been said, I have no idea how to do this problem. But I'd be interested to know now.
if that's true, then in this example the circle depends on the distance of the opposing sides with length 2. but this distance depends on the angle at the connections the 7 and the 11 side share. if you choose it to be 180°, you'd get a rectangle 2x18. from that point you could decrease the angles, resulting in "coffins" with different heights, thus hexagons with different "outcircles". or is the angle at the 7-11 connections somehow determined?
I also get 7, but I used MatLab's Fzero command on @(r)r-1/(cos(-asin(7/(2*r)) - asin(11/(2*r)))) because I've long forgotten how to simplify that kind of thing.
what about a hexagon with the side sequence 2-7-11-2-7-11 with the angles 180,180,360(or 0),180,180. that would result in a single line with the length 20, thus resulting in an outcircle with the radius 10. or is this construction forbidden by a property of hexagons?
It's hard to figure out the radius of the circle if you do not know the exact order of the lenght of each side and the angles of the hexagon. Or do you want us to figure out the the maximum or minimun radius?
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
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.
***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
Nice thread 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!
On December 16 2005 19:32 Catyoul wrote: Nice thread 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!
It's getting late (like I said before), but shouldn't you consider those things? - Overlapping, e.g. 5 and 25, you count them twice - You should multiply the corresponding power to the numbers (You need to subtract the numbers from each other, like I said before), so, assume they are right, they should be 20000*1 + 4000*2 +800*3 + 160*4 + 32*5 + 6*6 + 1*7 That's how many 5's you have in 100000!
Since there are obviously more 2's than 5's, number of 5's represent number of 0's, plus all the multiple of 10's, give you the correct solution (like you said).
On December 16 2005 19:32 Catyoul wrote: Nice thread 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!
It's getting late (like I said before), but shouldn't you consider those things? - Overlapping, e.g. 5 and 25, you count them twice - You should multiply the corresponding power to the numbers (You need to subtract the numbers from each other, like I said before), so, assume they are right, they should be 20000*1 + 4000*2 +800*3 + 160*4 + 32*5 + 6*6 + 1*7 That's how many 5's you have in 100000!
Since there are obviously more 2's than 5's, number of 5's represent number of 0's, plus all the multiple of 10's, give you the correct solution (like you said).
Every multiple of 25 is also a multiple of 5, every mult of 125 is already a mult. of 5, 25, 125 etc... so when he counted all multiples of 5, he counted all 25's once already, all 125's once as well... when he counted all 25's, he counted all 125's once again, all 625's etc... So he needs to add each successive power only once, leading to a pretty answer :D
I realize the syntax might not make any sense but I can't draw it out in ascii.
I forget why but when dealing with qudrilaterals, once you set it equal to zero and factor it, you can get solutions for anything that is left in the parenthasese by setting THAT equal to zero
So that's the answer I got ... I don't even know the meaning syntax ... like no clue what it is.
On December 16 2005 19:32 Catyoul wrote: Nice thread 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!
It's getting late (like I said before), but shouldn't you consider those things? - Overlapping, e.g. 5 and 25, you count them twice - You should multiply the corresponding power to the numbers (You need to subtract the numbers from each other, like I said before), so, assume they are right, they should be 20000*1 + 4000*2 +800*3 + 160*4 + 32*5 + 6*6 + 1*7 That's how many 5's you have in 100000!
Since there are obviously more 2's than 5's, number of 5's represent number of 0's, plus all the multiple of 10's, give you the correct solution (like you said).
Every multiple of 25 is also a multiple of 5, every mult of 125 is already a mult. of 5, 25, 125 etc... so when he counted all multiples of 5, he counted all 25's once already, all 125's once as well... when he counted all 25's, he counted all 125's once again, all 625's etc... So he needs to add each successive power only once, leading to a pretty answer :D
Well I've done probably something wrong because it nothing pretty so far. So every second multiple of 5 is a multiple of 10, but a multiple of 10 may have more than one 0, resulting in 11111 zeroes for all multiplies of 10 in 10000!. Then as already said every multiple of 25 is a multiple of 5 and so on... Making it 13021 zeroes for all numbers ending with 5. 11111+13021=24132
Damn, Catyoul's good I told you guys that one was easier.
Anyway, day three is up. I believe this is possible, but I actually don't know how to do it myself exactly. You can use higher order curves I believe, but I've yet to do it myself.
Oops, there's a hint.
Get on it
If no one gets it in a while maybe I'll just move on This one may take a few days, but in the meantime I'll post some more.
On December 17 2005 15:55 Empyrean wrote: Damn, Catyoul's good I told you guys that one was easier.
Anyway, day three is up. I believe this is possible, but I actually don't know how to do it myself exactly. You can use higher order curves I believe, but I've yet to do it myself.
Oops, there's a hint.
Get on it
If no one gets it in a while maybe I'll just move on This one may take a few days, but in the meantime I'll post some more.
it didn't bold correctly you forgot the slash on the second b nice problems and solutions =O
On December 15 2005 19:43 Empyrean wrote: Day three:
We all know it's impossible to construct a circle with the same area as a square within the Archimedean constraints (used only straight-edge and compass, and must be done in a finite number of constructions).
Show how to construct a circle with the same area as a square if you remove one of those constraints.
(edit: removed all the useless stuff that I wrote earlier )
I'm assuming that if we disregard the rule "only use a straightedge and a compass", then we are still restricted to using "realistic" tools of some kind? Otherwise, the problem trivializes when you have at your disposal, say, a ruler with infinite precision . Anyway I'm clearly no expert on constructions, but I think there should be some restriction on the set of tools you're allowed to use if you disregard that first rule.
The restriction that needs to be removed is the finite constructions one, because one needs to split the diameter into the desired irrational ratio with just rational cuts.
i did the square/circle thing back when I took an advanced geometry course, its kind of annoying, i wont bother doing it again i hate geometry with a passion.
problem 4 just looks annoying, ill give it a look later if i have some time
I'm pretty sure you can find day four on the internet somewhere. It was a past competition problem, so it should be posted somewhere. I don't even remember where I remember it from lol.
IMO, this problem is pretty elegent and relatively simple, so GL.
You have an object (of negligible size) that's being shot out with an initial horizontal velocity of 8 m/s (y velocity is 0 m/s). What is its acceleration at t = 0.25 s? What are its centripetal and tangential acceleration at this time?
y velocity is upward velocity? it would have no acceleration, except air resistance if it was just shot out of something.. centripetal would be 9.8 would it not? gravity..? i dont think i understand the question >.<
i have a little physics background -_- in a frictionless environment(i.e. no air resistance) the horizontal velocity will remain constant, therefore no acceleration(horizontally), and the only force acting on said body will be the force of gravity, meaning the object will have an acceleration downwards of -9.8 m/s^2 :| but i understand now what it is, exactly, you are asking. and i dont quite remember the right forumlas ;|
On January 07 2006 21:14 Gene wrote: i have a little physics background -_- in a frictionless environment(i.e. no air resistance) the horizontal velocity will remain constant, therefore no acceleration(horizontally), and the only force acting on said body will be the force of gravity, meaning the object will have an acceleration downwards of -9.8 m/s^2 :| but i understand now what it is, exactly, you are asking. and i dont quite remember the right forumlas ;|
Ok, it will only have a vertical acceleration, that is correct. So I guess you got the first part right.
However, in this case, there are two components that add up to g, and that is what the question is asking for - its centripetal acceleration (because the falling motion of the object will trace a parabola) and its tangential acceleration (the acceleration that is tangent to the curve at t=0.25s).
oh yeah, obviously, its the physics im lacking in, im taking physics right now, but its first period and i sleep every day. got 100 on the test but i totally forgot how to do this simple problem ;\ its going to annoy me until i remember too.
On December 17 2005 18:20 LastWish wrote: Ok I've got it! Catyouls solution is eventually right.
heh, I tried to solve it computationally and came sorta close at 24800 ;P
I'm only mentioning it because there were some interesting patterns that I saw to while trying to get that 24800 projection. One interesting thing is that for any N>0 number of zeroes, there are only five numbers that end in N zeroes. The other was that there was an interesting pattern in how the number of zeroes increased: it would increase by 1 every 5 numbers, but after 6 times of increasing this way, it would always skip a number the number of zeroes would increase by 2 instead of 1.
So the skipped numbers were 5, 11, 17, 23, 29.. (skip), 30, 36, 42, 48, 54, 60.. (skip), 61, 67, 73, 79, 85, 91... (skip) etc
Anyway, just thought it was a cool pattern. I don't mean to tarnish the the elegance of the pure math presented here; I wish I coulda figured it out that way myself. I was hoping that by analyzing this stuff that I'd get some insight, but alas it didn't. I did all this in Python if anyone is curious about it; support for long integers is pretty good so computing the larger numbers for testing wasn't so bad.
EPT
)
![[image loading]](http://img206.imageshack.us/img206/1292/hexagon3kx.png)
This one may take a few days, but in the meantime I'll post some more.
