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uhm didnt work out
(why do sometimes this forum post when i click preview??)
My idea was...
How do you make four equilateral triangles with sides l, using 6 sticks with length l?
iirc this problem is considered one of the most difficult problems ever submitted in the IMO...:O
Good work My solution was slightly different (with x'=u for easier reading):
Let [x] be the greatest integer less than x, and let x' be the fraction part of x, ie x'=x-[x]. Let n be any integer ≥2. Evaluate the following integral in terms of n:
Is 0° considered an acute angle? If so, that is another solution.
Triangle PQR with area A is given in a plane parallell to the gravitational acceleration g, with the length QR perpendicular to g, and the point P is above QR. Now we let two blocks (considered as points) slide down PQ and PR. This gives the different times t1 respective t2 for each block. If we assume there is no friction, find the minimum of T=t1+t2 and determine the triangle when this occurs.
Okay let's try this one more time. I need 3 quadraric equation examples with a, b, and c (coeffients of x^2, x and the constants as.
1 integers
2. rational numbers
3. irrational numbers
Thank You
The question is just to find any quadratic where a,b and c are irrational. So you just need to chosse any three irrational numbers. Do you know what an irrational number is?
http://mathworld.wolfram.com/IrrationalNumber.html
For example you can choose a=√2, b=pi. I leave for you to find a third irrational number for c.
edit: about your siolution, instead of setting equality between the two functions to see where they meet, keep the inequality sign and consider the same two cases.
I just want to tip you about this upcoming event
http://wpc.puzzles.com/
It's an internet based competition in logic puzzles.
How many here are going to participate??
Its also possible to use the rearrangement inequality directly on the first inequality. However, I spotted an error on my solution to the second one:
Still dont get it, doesnt the puzzle clearly state that the bar must be cut into 11 pieces??
Or is it my english that is failing?
Just two inequalities:
1. let p>1 be a real number, and given a sequence {ak} of n positive numbers whose sum is n, prove that:
2. let α,β,γ,δ be the angles of a quadrilateral. Prove that:
Is this question learning on your own, or is it part of a class? If it's part of a class, then there should be some other way to do it. How much have you covered in terms of real analysis (sometimes called "Advanced Calculus", which is a horrible name)?
Yes, im doing it on my own, and I dont think I have been doing a lot real analysis. Just the basic high school(?) calculus, like basic integration (substitution/parts), basic differential equations etc. In class I have not done anything close to investigating the definition of a limit or definition of a convergent serie
In both questions, use that logx+logy=logxy, logx-logy=log(x/y)
Thanks for the reply Ricky!
You need to use the definition of a convergent sequence here.
Unfortunately I have never worked with definitions of convergent series before, but I can give it a try. Could you please give a short startup/link to get me going??
Note that since a > b, a/b > 1, so (a/b)^p > 1 which is really a^p / b^p > 1. Multiply both sides by b^p and you get a^p > b^p.
Now here is your challenge: why can't you use this?
I think that is the wrong bit, since wouldnt stating that "a/b>1 implies that (a/b)^p>1^p=1", use the theorem that we are trying to prove?
It struck me when reading about the axioms of inequalities, that this inequality wasnt that easy to straight ahead deduce from the axioms.
the inequality to be proven is given p>0, a>b>0, prove that a^p>b^p.
the axioms are:
1. for two real numbers x and y exactly one of the following cases i true:
a) x=y b) x<y c)x>y
2. if x<y, then for all z x+z<y+z
3. x<y, z>0, then xz<yz
4. if x<y and y<z, then x<z
proof:
lemma: for x,y,z,w>0, if x<y, w<z, then xw<yz.
By axiom 2 we have xw<yw, and again by axiom 2 wy<yz, and by axiom 4 xw<yz. Now assume p is natural, then for p=1 the inequaliy holds. Assume it holds for p=n, ie a^n>b^n. then by the lemma we can multiply left side with a, and right side with b and establishing a^(n+1)>b^(n+1), thus by the induction principle it holds for all p
Now my questions are:
Is this proof correct? Im mostly concerned about the last step when introducing a limit. Do I need to state anything about continuity of the function a^p? Or is it sufficient to use the proved lemma, thus I can continue to multiply the inequality with a^(pk)>b^(pk) and continue to infinity to establish the inequality.
Also, thus the introduction of a limit need extra proofs of the limit, if I want to deduce the inequality straight down from the axioms?
I know that I can just state the the function a^x is strictly increasing for a,x>0 and thus a>b -> a^x>b^x, but that does need extra proofs conecrning derivatives etc, right? (and its not so much fun )
Or have I missed a really obvious way to prove this kind of intuitive inequality?:
and last, a challenge for you, find a simplier proof.
Hint: start by proving this theorem:
A circle and a point A is given. Let a line through A intersect the circle in points B and C, and draw a line PD, such that PD is a tangent to the circle with tangent point D. Then |AB|*|AC|=|AD|².
Your problem will follow quite easily from this theorem.
#7
nice problem btw
Using the inequality between arithmetic and geometric metdium yields:
letting ab=1/c, bc=1/a, ca=1/b, yields
1/c=2/a=4/b -> b=2a and c=a/2
abc=1 ->a*2a*a/2=1 -> a=1, which gives that b=2, c=1/2
It is possible to solve it exactly too.
as mathsy said, the area is a²sin150/2=9 -> a=√(18/sin150)=6
the ramaining side is 2*sin75*a (drawing the height gives two right triangles), so the perimeter is p=2*sin75*a +a +a = 2a(sin75+1)
but