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#1 Re: Exercises » Random question if you want a challenge » 2012-11-18 05:41:03

This is a problem from the extreme USAMO contest, which is 6 problems, 2 days, here's another one if you want another one:

Let ABC be a triangle with angle A=90degrees. Points D and E lie on AC and AB, respectively, such that angle ABD=angle DBC and angle ACE=angle ECB. Segments BD and CE meet I. Determine whether or not it is possible for segments AB, AC, BI, ID, CI, IE to all have interger side lengths.

#2 Exercises » Random question if you want a challenge » 2012-11-17 15:39:48

zehao1000
Replies: 2

Here's a random problem I found:

An integer is assigned to each vertex of a regular polygon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer m from each of the integers at two neighbouring vertices and adding 2m to the opposite vertex, which is not adjacent to either of the first two vertices.( The amount m and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integer, there is exactly one vertex which the game can be won.

(Note: it's not like I know how to do it, it's just for fun, because this would challenge most mathematicians.)

#3 Re: Introductions » Hello » 2012-11-17 15:27:36

Okay, I'll post a random question somewhere else.(Don't need to reply, in fact no one should or else the whole process will repeat again.)

#4 Re: Introductions » Hello » 2012-11-17 14:42:26

And maybe we should stop this chat since it's reaching the 40th posts? Maybe just some random math problems instead of asking what's your favourite topic, because this is becoming attractive, which it should'nt be.

#5 Re: Introductions » Hello » 2012-11-17 14:37:45

Very bad at it such as those type of problems that state " in how many ways can you get from point A to point B?" or those types that are like "how many numbers between 1-1000 do not have 1 as a digit?" too many formulaes and comfuse them.

#6 Re: Introductions » Hello » 2012-11-17 13:32:52

Geometry is easy, but logic problems are my favourite, for example, Einstein Riddles.

#7 Re: Introductions » Hello » 2012-11-17 13:15:44

It's pretty easy so yes, and I know a lot divisibility rules, but I'm better at geometry.

#8 Re: Introductions » Hello » 2012-11-17 12:41:43

Wikipedia; but I'm just assuming they're correct, since I heard that wikipedia isn't always accurate.

#9 Re: Introductions » Hello » 2012-11-17 12:06:01

Well, I read the story about it a few hours ago so I know about it now, so it is 3 primes for any even greater than 5 and 2 for any even greater than 2.

#10 Re: Introductions » Hello » 2012-11-17 09:22:40

No, I got it from my math teacher, only know that it is for 2 primes, and I made a mistake, 9 is not prime from 9+7, it should be 11+5=16, but interesting fact anyways.

#12 Re: Introductions » Hello » 2012-11-16 16:10:50

Do you know this unproven theory?
Any even number can be writen as the sum of 2 primes, for example: 8=5+3, 16=7+9, mathematicians have calculated over 1 trillion even numbers and all have this property, this is unsolved, and has fascinated mathematicians.

#13 Re: Introductions » Hello » 2012-11-16 15:32:30

I guess you're right, I must've forgot the statistics, another thing that interests me is paradoxes for example:" No one goes to that restaurant; it's too crowded", probably hear people say this before, but they never notice they're actually contradicting themselves.

#14 Re: Introductions » Hello » 2012-11-16 15:25:35

I mean 4% chance it will get shot down.

#15 Re: Introductions » Hello » 2012-11-16 15:24:22

Unfortunately, I have to wait another 362 days to do AMC 8 again, but by then, I'll do much better than this time, probably know all of trigonometry by then, and I find the birthday problem quite interesting, it takes only 23 people to be 50% sure that at least 2 people share the same birthday,and when there are 50 people, the probability meets 97% which is incredible, you can even do a survey on just 50 people and 97 out of 100 times, at least 2 people will same the same birthday, this is the type of math problems that interest me the most of all, another will be if a 4% is the probability that a fighter will not get shot down, then what is the probability that after 50 missions, you won't get shot down? The probability is a whooping 97%, almost to 100%, you can get this using Euler's Number.

#16 Re: Introductions » Hello » 2012-11-16 15:15:20

I'm 12, a seventh grader, so that's why AMC 8 is so hard for me;, more like an exam for high school kids.

#17 Re: Introductions » Hello » 2012-11-16 15:09:10

I find math quite a easy subject, and all my other classmates find math a very difficult subject, which is an advantage for me to win all the contests at my home school.

#18 Re: Help Me ! » Question on squares » 2012-11-16 15:05:44

Oh, thanks! Now I get the whole thing.

#19 Re: Introductions » Hello » 2012-11-16 15:03:59

I also like logical game theory problems, but I don't exactly even have an idea of the Law of Sines, Cosines, and Tangents, and I feel like learning calculus first, which is quite stupid, because trigonometry is way before calculus.

#20 Re: Help Me ! » Question on squares » 2012-11-16 14:58:49

Yes, that is correct, I'm just trying to understand the solution by myself, so if 26/81 is the probability, then m+n is equal to 26+81 which is 107, which is the final answer, but the solution is confusing, since it's all probability.

#21 Re: Introductions » Hello » 2012-11-16 14:54:41

I live in Canada, but can do American Mathematics Competitions still, because I go to a math school where they have lots of math contests.

#22 Re: Help Me ! » Question on squares » 2012-11-16 14:52:54

Because this problem was in a 2010 contest so though I still thought that it was 2012, but now the problem seems to make much more sense.

#23 Re: Introductions » Hello » 2012-11-16 14:50:53

My favourite thing to do is math contests, because most of the time, whenever the contest results are handed out, I always get the highest score, except the Fibonacci contest, because then I get more bragging rights to my friends, I'm gonna fail AMC for sure, 25 questions, only 40 minutes, time is too tight, and the problems are extra hard, only 96 seconds a problem.

#24 Re: Help Me ! » Question on squares » 2012-11-16 14:46:07

OH-NO, sorry! I got the whole problem wrong! It's suppose be 2010, not 2012 and 2010^2 has 81 factors, I can't believe that just a difference of 2 can make such a huge difference, now I hope that this problem is much more clear! I actually though that 5 was a factor of 2012!

#25 Re: Introductions » Hello » 2012-11-16 14:39:31

In fact, the first question I sent you is a question from the AIME contest because I preparing for my next contest, AIME is extremely hard 15 questions 3 hours exam, my dream is to somehow get in that contest by doing very well on the AMC 12, not going to be easy.

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