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Martin

Sat Oct 09, 2004 12:45 pm


[contest] 10 bit math questions

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No, these aren't from my homework.

1.  In something of a whimsical mood, the great probabilistic geometer, Drew Diecaster, selected two distinct random points on a straight line segment L. What are the chances that he can construct a triangle from the three resulting lengths?


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mynameisbob

Sat Oct 09, 2004 5:51 pm



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100% chance....


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wtd

Sat Oct 09, 2004 5:54 pm



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I thought about answering, but I'm glad someone else was able to.  :)


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Cervantes

Sat Oct 09, 2004 5:56 pm



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You know that's not the answer for two reasons:

1) It's too easy
2) One example proves it wrong.  Assuming the line goes from 0 to 1, what if Drew picked his points at 0.1 and 0.9?  You can't make a triangle with that, because the three lines you'd have to work with are of lengths 0.1, 0.1, and 0.8.  To successfully make a triangle, the lengths of the two smaller lines must be greater than or equal to the length of the longest line.  (right?)

Very interesting problem Martin, I'm going to be thinking about this for a long time.  Longer than the 24 game with 5,5,5,1 8)
Did you think up that problem yourself?


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wtd

Sat Oct 09, 2004 6:04 pm



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D'oh.  This is what I get for being so sick.


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Mazer

Sat Oct 09, 2004 6:17 pm



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EDIT: this post never happened. HA!


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Cervantes

Sat Oct 09, 2004 6:26 pm



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Yes, you are wrong.  again, if it's a 0% chance, only one example will prove it wrong.  
Say Drew selects 0.4 and 0.8.  
Length one is 0.4, Length two is 0.4, and Length 3 is 0.2.
The two smaller lengths (in this case, its the smallest length and one of the larger lengths, since they are the same) total 0.6, the larger length totals 0.4.  Hence, you can make a triangle.

EDIT: That mistake never happened.  HA! 8)

wtd: I hope you get better soon.
I didn't think about it for long before I realized that I could write a program to give me an average.  I came up with an average of 24.964%.

Oh, and now that I think about it, I think the sum of the two shorter lines would have to be greater than that of the longer line, because if they were exactly equal you'd end of up with a straight line that consists of three line segments piled ontop of one another.


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Mazer

Sat Oct 09, 2004 6:47 pm



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Say Drew selects 0.4 and 0.8.  
Length one is 0.4, Length two is 0.4, and Length 3 is 0.3.
The two smaller lengths (in this case, its the smallest length and one of the larger lengths, since they are the same) total 0.7, the larger length totals 0.4.  Hence, you can make a triangle.
I see that you are right, not sure why I was thinking that... but in your example, I assume you chose the points like that so that it would represent a percentage of the actual length of the line (as we don't know the real length), but somehow your third length grew. It should be 0.2.


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wtd

Sat Oct 09, 2004 6:52 pm



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wtd: I hope you get better soon.

Thanks.  So do I.  :)


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Martin

Sat Oct 09, 2004 7:29 pm



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No, I didn't make up the problem myself. I can't tell you where I got it though, for obvious reasons ;)

By the way, all of your answers are wrong.

Here is my hint: http://mathworld.wolfram.com/TriangleInequality.html


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Cervantes

Sat Oct 09, 2004 9:15 pm



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By the way, all of your answers are wrong. 

even the answer I cheated to get?  It's small and in white text, in my second post.
I think this is more than a 10-bit problem :?


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Genesis

Sat Oct 09, 2004 9:43 pm



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25% chance...?

EDIT: And I see that Cervantes already got that number...pretty much. Didn't notice his small print when I posted this.


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Cervantes

Sat Oct 09, 2004 10:07 pm



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err, I put it small and white for a reason: so that people who didn't want to see the answer I had come up with wouldn't see it just by reading through the thread.  So that only people who were driven to insanity would look there for a clue. Oh well, I'm posting the answer (well, I think its the answer)

I'm reasonably certain:

again, using 0.0 to 1.0 line segment.

p1 = the first point placed
p2 = the second point placed


if p1 < 0.5 then
    if p2 > 0.5 and p2 < p1 + 0.5 then
        put "A triangle is made.  HuZaA!"
    end if
end if

similarly,

if p1 > 0.5 then
    if p2 < 0.5 and p2 > p1 - 0.5 then
        put "A triangle is made.  HuZaA!"
    end if
end if




Let's look at the first part, if p1 < 0.5.

-there is a 50% chance that p1 is less than 0.5 *see note
-there is a 50% chance that p2 is greater than 0.5
-there is a 50% chance that p2 is less than p1 + 0.5 (the range that p2 could be in for this to be true is between p1 and p1 + 0.5.  since p1 is less than 0.5, p1 + 0.5 is less than 1.0, so the full 0.5 is used, if you understand what I mean.  it's not as if p1 is 7.5 and p1 + 0.5 is 12.5, but since p2 can only be up to 1.0, there is a full 2.5 that isn't used, so in that case the chance of p2 falling in the range of p1 and p1 + 0.5 would be 25%.  but since p1 is less than 0.5, the chance of p2 being between p1 and p1 + 0.5 is always 50%
50% * 50% * 50% is 12.5%.
Then we consider the other side of the coin, if p1 were to be greater than 0.5.  That gives us another 12.5% to make a triangle.
12.5% + 12.5% = 25%.

As for how I came up with the if p1 > 0.5 and p2 < 0.5 and p2 > p1 - 0.5 etc., I just played around with different values, and began to see a threshold.
I don't think I can explain it.  I tried, several times, but always failed.  :?  just play around with the following numbers, making one of them 0.01 bigger or smaller, and then making the other 0.01 bigger or smaller.

_____|____1_______2______3___
-----|-----------------------
p1 = |  0.25    0.25    0.75
     |
p2 = |  0.50    0.50    0.50



*note: Coutsos said this before he deleted his post: we will use infinate precision, infinate decimal points, in this problem.  So there is a 50-50 chance of p1 being less than 0.5, as well as a 50-50 chance of p1 being less than or equal to 0.5


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Genesis

Sat Oct 09, 2004 10:50 pm



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Yeah that's kinda how I figured it out. Except I used pencil and paper. (And my trusty calculator.) I didn't even think to write a program, good idea!

I'm pretty sure it's right though.


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Martin

Sun Oct 10, 2004 12:44 am



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Good work guys. I'm on links right now (KDE takes a while to compile...) so I'll give you bits once I get this OS up and running. Yes, 0.25 is the right answer.


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Cervantes

Sun Oct 10, 2004 7:16 am



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Yeah that's kinda how I figured it out. Except I used pencil and paper. (And my trusty calculator.) I didn't even think to write a program, good idea!

I'm pretty sure it's right though.
I used pencil and paper too (but no calculator :P).  Writing a program is cheating, and I wanted to figure out why.


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bugzpodder

Sun Oct 10, 2004 9:55 am



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aww, i wanted the 10 bits :(  but i am too late


if it's a 0% chance, only one example will prove it wrong.  


That is mostly right.  But not quite.  I think the probability of you selecting a integer in the reals is 0%.  (yes, exactly 0)

i think it is same for rationals as well, but i am not 100% sure


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Martin

Mon Oct 11, 2004 1:29 pm



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2. Root 2 is Irrational 

Prove it. This one's for 20 bits.


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Andy

Mon Oct 11, 2004 2:50 pm



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if we were to square root 2, we'd get the good old integer 2... if we were to rewrite 2 as a rational number, we'd have 2x/x... since both 2x and x cannot both have an even number of factors, we know that at least one of them is not a perfect square. thus the square root of 2x/x  cannot be rational which means root 2 is irrational


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Martin

Sun Oct 17, 2004 3:59 pm



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Awesome.

3. Limits
This one might seem easy, but I want a proof, not just inspection. 40 bits.

Prove:
Lim (x^2 - 1) = 3
x -> -2


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Andy

Mon Oct 18, 2004 3:26 pm



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LS:
if we were to factor x^2-1, we could rewrite the limit as 

Lim ((x-1)(x+1))
x->-2

as x gets closer and closer to -2, x-1 gets closer and closer to -3, and x+1 gets closer and closer to -1, and -1*-3=3

RS:
3

LS=RS

Lim (x^2 - 1) = 3
x -> -2


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apomb

Mon Oct 18, 2004 4:51 pm



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I have a complaint about dodge ... or other mods replying to this thread, since its for bits, why should they be able to answer, mabe they can , if they know it, put hints or something, i just find it unfair... because bits are involved


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Andy

Mon Oct 18, 2004 4:56 pm



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more than one person can answer you know.. as long as the solutions are different


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apomb

Mon Oct 18, 2004 4:58 pm



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ya, but will Martin give bits to both people?


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Andy

Mon Oct 18, 2004 4:59 pm



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if ur solution's nice, i'll give u bits


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bugzpodder

Mon Oct 18, 2004 4:59 pm



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LS:
if we were to factor x^2-1, we could rewrite the limit as 

Lim ((x-1)(x+1))
x->-2

as x gets closer and closer to -2, x-1 gets closer and closer to -3, and x+1 gets closer and closer to -1, and -1*-3=3

RS:
3

LS=RS

Lim (x^2 - 1) = 3
x -> -2
actually you need a delta, epsilon proof.


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Andy

Mon Oct 18, 2004 5:00 pm



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sry bugz, not that far yet in calc wit Ing


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apomb

Mon Oct 18, 2004 5:08 pm



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Dodge: thanks for adressing my issue with such promptness
bugspodder :you are a fast poster!
-Compwiz


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Brightguy

Mon Oct 18, 2004 7:56 pm


Re: [contest] 10 bit math questions

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Prove:
Lim (x^2 - 1) = 3
x -> -2
Did you happen to get that question from your calculus assignment by any chance? :lol: That question was on mine at least.


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Andy

Tue Oct 19, 2004 3:13 pm



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martin's in uni...


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Brightguy

Tue Oct 19, 2004 5:12 pm


Re: [contest] 10 bit math questions

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Yes, in fact we are a part of the same faculty.  Not sure if the assignments are the same, though, as he might be getting "advanced" problems...  and working on special Putnam-level stuff on the side, of course. :o


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Andy

Tue Oct 19, 2004 5:23 pm



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just as an aside, hey martin, rumors say that ur smoking even john at math marks.. is this true?


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Martin

Tue Oct 19, 2004 5:52 pm



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I have no idea, but I really doubt it. My math mark is about 10% lower than john's I'd guess (at ~115%).

That question was out of the math 137 textbook, as I was trying to learn how to do Delta-Epsilon proofs.


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Andy

Tue Oct 19, 2004 7:08 pm



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y not just pay bugz a visit


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Brightguy

Tue Oct 19, 2004 7:42 pm


Re: [contest] 10 bit math questions

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So the advanced courses are actually a lot different, then?  I'm guessing there are a lot of proofs.  Anyway, here goes this one:

For all &#949; > 0 we must find a &#948; > 0 such that, for all x,
0 < |x - (-2)| < &#948; &#8658; |(x&#178; - 1) - 3| < &#949;

This simplifies to showing that there exits a &#948; so that:
If 0 < |x + 2| < &#948; then |x - 2||x + 2| < &#949;

Now, if |x + 2| < 1, then
-1 < x + 2 < 1
-5 < x - 2 < -3
|x - 2| < 5
and |x - 2||x + 2| < 5|x + 2|
but we want to show: |x - 2||x + 2| < &#949;
We can get this is true if |x + 2| < &#949;/5, because then
|x - 2||x + 2| < 5(&#949;/5)
|x - 2||x + 2| < &#949;

So, we have shown that if |x + 2| < 1 and |x + 2| < &#949;/5, then |x - 2||x + 2| < &#949;
So, take &#948; = 1 or &#949;/5, whichever is lower.  Then, the following must be true:
0 < |x + 2| < &#948; &#8658; |x - 2||x + 2| < &#949;

So, by definition of a limit, 
lim
x&#8594;-2     (x&#178; - 1) = 3

