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2691= 2^4 + 3^4 + 3^4 + 5^4 + 5^4 + 5^4 + 5^4 + 13

the +b is just a translation, the hard part is the az. Try polar coordinates.

JaneFairfax wrote:

Vi har

och . Om så . .Det är ett skönt problem. Tack så mycket, Kurre.

Yes that is the trigonometric approach

Good job!

Have you been studying swedish? I think I remember that you wrote that you are interested in languages. Nicely done!

Anyway here is my method:

Reflect AB in the horizontal line and draw line BC as in 2. Now reflect the triangle ABC in AC to make a quadrilateral as in 3. Now, all angles in the quadrilateral ABCD are equal, and all sides are equal, thus it is a square, and since AC is a diagonal,

must be half of a right angle, ie1a2b3c2212 wrote:

HOW TO READ THAT LANGUAGE??? can translate into english pls? thx

I did translate the important parts of the text:

Kurre wrote:

Find

Altough I should maybe have written that the figure consists of three squares, but that is kind of obvious.

TheDude wrote:

Right?

No. You forget that there is a minus sign. We have 3c<-b. But:

there is a minus sign infront of the 3c/2, so we cant apply the inequality there.

**Kurre**- Replies: 4

http://g.imagehost.org/0615/3462.jpg

Find

This is an easy problem if you use the standard approach with trigonometry (the problem is for ~17 years old students). But I found a much nicer geometrical method, without using any trigonometry.

So my challenge for you is to find it.

(or solve it any other way without trigonometry is fine too )

http://www.mathisfunforum.com/viewtopic.php?pid=18619#p18619

This proof can be generalized so, if you know the formula for all sums with powers up to k, we can express the sum to the power k+1 in the other sums.

Im not following. You cant just substitute expressions for inequalities in other inequalities

TheDude wrote:

\\

Substitute these values into our previous inequality:

The Right hand side is ok, but not the left side. You have:

if im not missing anything.

We first use the cosine theorem on each of the squared sides:

Thus we get:

solving for and using the areaforumla ( etc) :

To minimize this sum we can for example use jensens inequality.

http://en.wikipedia.org/wiki/Jensen%27s_inequality

We first assume that the triangle is acute, so all angles are less than . Then is convex, since and . Thus Jensens inequality applies and we get:

so

What if one angle is obtuse?

Then we can argue as follows. WLOG we assume

Q.e.d

use the formula on http://en.wikipedia.org/wiki/Trig_identities#Linear_combinations

Ricky wrote:

I guess he means that, because we assumed that p,p+2 where the largest pair there does not exist any numbers with factors p+2k,p+2(k+1).

As I posted before, this statement is false. (p+2k)*(p+2(k+1)) has precisely those numbers as factors.

hm I meant prime factors, but maybe he does not use that they are prime factors, idk, im too lazy to read through the oproof atm.

Ricky wrote:

I'm having a lot of trouble understanding what it is you're trying to go for.

If we suppose that there exist a larger pair of the form p+2k and p+2(k+1) Than the set of numbers that have as factors p+2k and p+2(k+1) must be equal to 0 (since they do not exist).

We are supposing that there exists a larger pair where p+2k and p+2(k+1) are both prime? This doesn't make sense, we've already supposed that p and p+2 were the largest pair with this property.

And you say that the number of integers that have p+2k and p+2(k+1) as factors must be zero. This is false, just look at the number:

(p+2k)*(p+2(k+1))

This number has both those as a factor.

I guess he means that, because we assumed that p,p+2 where the largest pair there does not exist any numbers with factors p+2k,p+2(k+1). IM not sure though and I have not tried to understand the proof either.

Actually I made this thread to see what kinds of programs that existed, since I got the idea to create such a program myself. I have tried Geometers Sketchpad and Cabri now, and Cabri was really nice but was not exactly the type of program I had in mind, altough I guess I will not be able to make a program close as good. But Il maybe give it a try anyway

**Kurre**- Replies: 6

Are there any good programs for drawing geometrical pictures? I mean that has functions for lets say creating circumcircles, marking centroids/circumcentres/medians etc, basically a program that is designed for creating proofs/problems/solutions in euclidean geometry??

**#15**let k,n be positive integers, a a nonzero real, k<n+1 . Show that:

both with real analysis and by using residue calculus

*edit: i did a mistake so i dont know if its possible to do this using residues, but that does not mean it must be impossible*

what do you mean by block walking?

Then assume you have a set of 2n persons, and divide it into two sets A and B of n persons in each. Now you want to choose a comittee of n persons. Then for each choice there will be say k persons in set A, and n-k persons in set B, and summing will yield exactly the sum above.

eh how does one define length, width and height for a triangular prism?? :s

kean wrote:

sure！ it's right

\sum_{k=1}^n \zeta_k^m =0

it does not hold for all m and n