New Discovery! (not really)

Since I have no life, I will share something that's pretty cool.

If y+mx=b where m>0, then the largest(maximum) value of xy (x times y) is b^2/4m.

I do have a proof of this but I don't know if it's useful or not. 

This theorem(?) really helps when you are finding the largest area of something. 

Proof-the Concurrency of medians of a triangle

Concurrency of Medians of a Triangle Theorem--The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.
 Proof:

Since CD bisects AB at point D and BE bisects AC at point F.
Point D and F are the midpoints
Connect D and F, giving us the midsegment of △ABC.
Using the Midsegment theorem
DF∥BC and DF=BC/2
DF/BC=1/2
Since DF∥BC and they're cut by transversal BF.
∠DFB≅∠FBC by the alternate interior angles theorem.
∠BOC≅∠DOF by the vertical angles theorem.
△DOF∼△COB By the AA Similarity Postulate.
DF/BC=FO/BO=DO/CO
Since BF/BC=1/2
FO/BO=1/2
Since BO+FO=BF
BF-BO=FO
(BF-BO)/BO=1/2
BF/BO-BO/BO=1/2
BF/BO-1=1/2
BF/BO=3/2
BO/BF=2/3 (Reciprocal Property)
BO=2BF/3 (Prove)
-------------------------------------------------------------------------

More interesting stuff

1. 1/6+1/12+1/20=?

=1/(2*3)+1/(3*4)+1/(4*5)
=1/2-1/3+1/3-1/4+1/4-1/5
=1/2-1/5
=3/10

You might ask: why 1/2*3=1/2-1/3?
let's solve 1/2-1/3 first
1/2-1/3=1/6  
you might notice that the solution is the difference of the two denominators over the product of the two denominators.
so let a=2 b=3
1/a-1/b=(b-a)/ab
a+1=b
so  [(a+1)-a]/a(a+1)
=1/a(a+1)=1/a-1/(a+1)
-------------------------------------------------------------
2. 1/(3*5)+1/(5*7)+1/(7*9)=?

(reminder: 1/2(x)=x/2)

=1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9)
=1/2(1/3-1/9)
=1/2(2/9)
=1/9

Why 1/3*5=1/2*(1/3-1/5)?
Using the same method
1/2(1/3-1/5)
=1/2(2/15)
=1/15
let a=3 a+2=5
1/2[(a+2)-a/a(a+2)]
=1/2[2/a(a+2)]
let k=2
1/k[k/a(a+k)]

1/3-1/5=2/15
1/a-1/(a+2)
=(a+2)/a(a+2)-a/(a+2)
=2/a(a+2)
let k=2
k/a(a+k)=1/a-1/(a+k)
so 1/k[k/a(a+k)]
=1/k[1/a-1/(a+k)]=1/a(a+k)

test:
1/5*7=1/5(5+2)
a=5 k=2
=1/2(1/5-1/7)