Interesting maths problems

[Coolydudey]

I am creating a thread for interesting maths problems. To post, please send them to me first so I can verify (PM). I have participated in the maths Olympiad, but I'm not going to go to quite that level here. I would like to keep discussion about the problems themselves to a bare minimum, so that we don't spam the thread with partially relevant comments.

These should all be solved with elementary math, including remainders (5=1 (mod 4)), and basic calculus (differentiation). You can stray a little in case I forgot something.

Problem 1: prove x=3 is the only positive solution of 3^x + 4^x = 5^x

Problem 2 (this is mine, and a fair bit harder): prove x=2 is the only positive integer value of x for which sqrt ( x^2 + (x+1)^2 ) is an integer.

Problem 3 (again mine, not so hard though): find all the integer values of x for which sqrt(1^2 + sqrt (2^2 + sqrt(3^2 + sqrt(...... +sqrt(x^2))...))) is an integer. Expansion of the problem is to say find the form of all the real values of x for which the given function is an integer (much harder).

I'll leave these for now.

 

[Coolydudey]

Problem 4: We are give a sequence A(n), n a positive integer, defined by: A(1)=1 and A(n+1)=A(n)+1/(n^3). Prove that for any n, A(n)<1.25 (or 5/4).

Problem 5: Given positive integers a,b, such that a/b<sqrt(5), show that sqrt(5)-a/b>1/(4ab)

Problem 6: A kid is in the centre of a square swimming pool. His teacher is at one corner. The kid wants to get out. The teacher can run three times as fast as the kid can swim. Prove that the kid can get out.

This problem set is harder than the previous one, except for problem 2.
Watch out, 2, 5 and 6 are real tough ones.

 

[Coolydudey]

Is anyone actually trying these? I'm more than happy to post them, but I don't want to be wasting my time...

Problem 7 (n!=n*(n-1)*....*3*2*1): find every positive integer m such that 1!*3!*....*(2m-1)!=((m(m+1))/2)!

Problem 8 (geometry): in a right angled triangle ABC (A=90 degrees), we take the height AD and the bisect or of the angle C CE (E is on AB). AD and CE intersect at Z. If ED and BZ intersect at H, prove that:
1)AB*AD=AB*AZ + AE*AD
2)area of AEHZ = area of BHD.

Problem 9 (geometry, quite a bit easier): in a parallelogram ABCD, we extend side AD towards D such that DE=AD. If AC intersects BE at Z, prove that DZ passes through the midpoint of BC.

I know that where I come from geometry is a lot more advanced than usual, so you may have trouble with 8, but you should be able to do 9.

 

[NinjaSurfer]

I think it should be reasoned first

why 1+1=2

before we proceed

1 + 1 does not necessarily equal 2

do we assume that there is no energy lost in the process of addition?

if energy is lost in the addition process, then 1+1 would < 2

 

[Coolydudey]

I think it should be reasoned first

why 1+1=2

before we proceed

1 + 1 does not necessarily equal 2

do we assume that there is no energy lost in the process of addition?

if energy is lost in the addition process, then 1+1 would < 2

Heehee.... 1+1=2 only if the base in which we are reasoning is >2.

 

[Beholder]

6 was easy,
The kid starts swimming towards the corner opposite from the teacher, halfway towards the corner, which is - sqrt(2)/4, he turns 45 degrees in the opposite direction from where the teacher is coming from, leaving him another 0.25 (this is all assuming the length of the edge of the pool is 1). So the kid did sqrt(2)/4 + 1/4 = 0.603. If the teacher doesn't change directions when the kid changes directions, he needs to pass 2.25 in order to get to where the kid is, so the time it would take him is 2.25/3 = 0.75. And 0.75>0.603.
If the teacher changes direction when the kid changes the teacher has to pass 1.06 + 1.06 + 1 + 3/4 = 3.871, and the time it would take him is 3.871/3 = 1.29, and 1.29 > 0.603.

 

[NinjaSurfer]

Heehee.... 1+1=2 only if the base in which we are reasoning is >2.

why can't 1+1=Blue ??? why not? It CAN!!!! OMG I'm so confused.

 

[BigApplePi]

I am creating a thread for interesting maths problems. To post, please send them to me first so I can verify (PM). I have participated in the maths Olympiad, but I'm not going to go to quite that level here. I would like to keep discussion about the problems themselves to a bare minimum, so that we don't spam the thread with partially relevant comments.

These should all be solved with elementary math, including remainders (5=1 (mod 4)), and basic calculus (differentiation). You can stray a little in case I forgot something.

Problem 1: prove x=3 is the only positive solution of 3^x + 4^x = 5^x

Problem 2 (this is mine, and a fair bit harder): prove x=2 is the only positive integer value of x for which sqrt ( x^2 + (x+1)^2 ) is an integer.

Problem 3 (again mine, not so hard though): find all the integer values of x for which sqrt(1^2 + sqrt (2^2 + sqrt(3^2 + sqrt(...... +sqrt(x^2))...))) is an integer. Expansion of the problem is to say find the form of all the real values of x for which the given function is an integer (much harder).

I'll leave these for now.

As a fellow math person these seem like good problems, but it would help to have more definition. More definition would make the challenge more user friendly.
1. Most people don't know what "mod" means. Which problems need this?
2. Define "^" just in case.
3. Are we using the standard base 10 here or is that part of the problem?
4. If no one answers after a while, give clues without giving away answers ... if that is possible.
5. Anything else you can think of that helps.

 

[Coolydudey]

why can't 1+1=Blue ??? why not? It CAN!!!! OMG I'm so confused.

If you define BLUE=1' then that is true.

As a fellow math person these seem like good problems, but it would help to have more definition. More definition would make the challenge more user friendly.
1. Most people don't know what "mod" means. Which problems need this?
2. Define "^" just in case.
3. Are we using the standard base 10 here or is that part of the problem?
4. If no one answers after a while, give clues without giving away answers ... if that is possible.
5. Anything else you can think of that helps.

Problem 2 may need the theorem that any perfect square is 0 or 1 (mod 4). Basically, y=z (mod x) means that y and z have the same remainder when divided (Euclidean division) by x.
^ is to the power of.
Base 10 is assumed unless otherwise stated

Clues (first 6):

Spoiler

1)I solved this using differentiation
2)couldn't solve this, but trying to get the square root of a quadratic and then proving that this was never an integer other than x=3 was a good path
3)think of what happens the moment you get an irrational from one of the roots
4)it is enough to prove 1/(2^3)+1/(3^3)+1/(4^3)+... (to infinity) has a limit below 0.25. Don't get involved with limits though. You can use either inequalities or proving that this is smaller than a certain integral.
5)your only path for this one is using some ingenious rearrangement, and perhaps inequalities if you know any.
6)solution for 6) given

I also wanted to address problem 6. I got slightly confused.
Problem 6: prove the kid can't get out (look at the original to see what I mean) if the teacher runs 6 times as fast.

 

[Vrecknidj]

Problem 1: prove x=3 is the only positive solution of 3^x + 4^x = 5^x

Did you mean "x=2" ?

How complex do you want the proof. There's something along the binomial expansion lines I could dream up, but...

Let A(x) = 3^x
Let B(x) = 4^x
Let C(x) = 5^x

At x = 2, A(x) + B(x) = C(x); 25 = 25
At x = 1.9, A(x) + B(x) > C(x); 21.99 > 21.28
At x = 1.8, A(x) + B(x) > C(x); 19.35 > 18.12
The gap continues to widen as x approaches about 1.3, and then closes, but there always will remain a gap. And, since we're only dealing with positive numbers, we only have to consider the range from 0 to 2 (non-inclusive), and so long as we can find, say, 10 values in that range that all work, something like the squeeze theorem will establish all the points in between those points.

At x = 2.1, A(x) + B(x) < C(x); 28.42 < 29.37
At x = 2.2, A(x) + B(x) < C(x); 32.32 < 34.39
The gap continues to widen (dramatically) as x gets larger. Thanks to exponential expansion, it becomes clear by observation that we don't have to consider many cases above x=2.5 to establish the case.

I assume you want something formal. I could start by subtracting 3^x from both sides of the equation and then dividing both sides of your proposed equation by (4^x) and have a 1 on one side of the equation and a nice difference on the other. In particular, 1 = (1.25)^x - (0.75)^x.

And that's even easier to plot to show intersections and demonstrate divergences of gaps between them. (For example: http://tinyurl.com/6povltp)

Were you thinking of something more formal?

Dave

 

[Vrecknidj]

Problem 1: prove x=2 is the only positive solution of 3^x + 4^x = 5^x

Problem 2 (this is mine, and a fair bit harder): prove x=3 is the only positive integer value of x for which sqrt ( x^2 + (x+1)^2 ) is an integer.

Problem 2 is just a kind of special case, inside-out version of problem 1.

y = [(x^2) + (x+1)^2)]^0.5

y^2 = (x^2) + (x+1)^2

But, if Problem 1 has been proven, then, since if x = 3 then (x+1) = 4, and problem 1 establishes the relationship between 3, 4, and 5 in this case.

But, again, I suppose you want something more formal. (And something which doesn't assume the solution to problem 1 as a proven and therefore useable claim.)

Dave

 

[Coolydudey]

In response to both your answers, three things.
First, try only to post solutions.

second, these solutions must be formal (I don't care what they're done with, but look at the constraints for what you're allowed to use at the beginning).

Third, problem 2 is similar yet ultimately unrelated to 1, because 1 has the unknown in the power and 2 has the unknown in the base. The knowledge of the solution to problem 1 doesn't solve problem 2.

 

[A22]

3^x + 4^x = 5^x
e^xln3 + e^xln4 = e^xln5
e^1.0986x + e^1.386x = e^1.61x
Divide it all by e^1.61x
e^(-0.5114x) + e^(-0.224x) = 1

Not sure if that could be of any help lol

 

[Coolydudey]

3^x + 4^x = 5^x
e^xln3 + e^xln4 = e^xln5
e^1.0986x + e^1.386x = e^1.61x
Divide it all by e^1.61x
e^(-0.5114x) + e^(-0.224x) = 1

Not sure if that could be of any help lol

No, not as I see it, but your second step can be useful.

 

[A22]

btw x=2 is the only solution

9+16=25

 

[sheepie]

I was going to suggest the use of the LaTex mod for this thread but it doesn't seem to work :'(. Would help with clarity as it makes pretty math. I was completely turned off by the formatting of the math of the thread. Have made an inquiry, if I can get it working quickly will give you an example. Its pretty easy especially f you're just using it to format formulae.

Also makes pretty reports and bibliographies a breeze. Seriously. Its a word processor that just lets you get down to the actual informing of people and it spits it out sexy & professional.

LaTex is just good.

A report I did in the course I did a bit of Latex in:
I'm going to find an appropriate place to plug Latex.
It's that good.