Math Is Fun Forum

no, a value for a polynomial can be a perfect square, without the polynomial being a perfect square of a polynomial...

I did not write that q must be prime

thanks

#14
let

Okey here is my proof:
Given triangle ABC, let D, M, N be midpoints of BC, AC, AB respectively (see picture). Insert a coordinate system along the median AD. We want to find the y component (perpendicular to AD) of the center of mass R for the triangle, which we call d. R will be the weighted sum (definition of center of mass) of the locations of the following three pieces: parallellogram ANDM, triangle BND, triangle DMC. The mass of ABC, BND, DMC and ANDM are m, m', m'' and m''' respectively.

But the center of mass of ANDM will lie on AD (can be seen by for example that ADN is congruent to DAM, so the y components will cancel), so we can skip that piece.

Furthermore, BND is congruent to DMC which both are similar to BAC, with a scaling factor of 1/2. Thus the distance to the center of masses R' and R'' for BND and DMC are a distance d/2 from their medians NE and MF (see picture). Also, since the areas for the triangles ADC and ABD are equal and they have the same base AD, the heights with base AD are equal, and equal h. Since the medians NE and EF are 1/2 of the height h away from AD, they have y components h/2 and -h/2. Also since BND and MDC are scaled by a factor 1/2, their areas (masses) are scaled by a factor 1/4, so m'=m''=m/4.

Now when im done explaining the picture we can come to the main part of the proof. Using the definition of center of mass one obtains (R''' being the location of the center of mass of ANDM):

cool eh?

#13 Let M be a (p-1)x(p-1) matrix where p is an odd prime number and each row has each element from {1,2,3,...,p-2,p-1} exactly once. Prove that p|det(M)

devanté

darn wrong thing over the e
(i typed that really slow)

#12
Let n be an integer with m prime factors (not necessarily distinct). suppose d|n and define

uhm okey that was much nicer than what I did

nice

Thanks! But I still cant find a field where to edit the title? i know what u mean, i have seen it before. I thought it had been too long since i created the topic..

Yea I also realized that. Its a really cool result

yes, if we can prove that it actually lie on one median. The masses of the two triangles that the median dividies the larger triangle obviously has the same mass, but that does not imply that the center of mass for the triangle is at the median. We need to prove that the distance from the median the the two center of masses for the smaller triangles are equal. That is relatively easy to do with the integral definition (its "obvious" in a sense that the mass after distance x m(x) are equal on both sides since the sides are both straight lines) but....I dont like it
Basically I made this thread because a friend told me that they proved this really easy by using vectors during a lecture, but today he showed the argument and it was in fact incorrect So maybe there is no simple method that I have missed...woho