.
is rounded to the nearest integer and then divided by 
,
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Hey everyone! I want to start a proof marathon. Rules:
1. Nothing harder than middle school math.
2. Please provide an answer and a new problem so this thread doesn't die quickly.
3. Please number your proofs(S#)/problems(P#) so referencing is easy.
4. Use hide tags so there aren't any spoilers.
Here is the first problem (I'll start simple):
1. Nothing harder than middle school math.
2. Please provide an answer and a new problem so this thread doesn't die quickly.
3. Please number your proofs(S#)/problems(P#) so referencing is easy.
4. Use hide tags so there aren't any spoilers.
Here is the first problem (I'll start simple):
is true for all equations 
.
.
.
from both sides to get 
Simplifying this gives you the quadratic formula.
Where b is the number of lattice points on a polygon, and I is the number of lattice points in the interior of a polygon.
, but I'm sure a quick Google search can explain that! Any, the two steps that we need to prove are first that the theorem holds for rectangles, and then, for right triangles.
be the area, let 
be length, let 
be width, let 
be # of boundary points, let 
be the # of interior points.
;
;
,
.
cannot be expressed in the form 
where 
and 
are rational.
.
where 
.
is even. If 
where 
is an integer.
We use this to prove the pythagorean theorem. We know that the area of the big square is 
,
.
.
triangles have area 
.
.
![$(-\infty,2]\text{U}[3,\infty)$](http://latex.artofproblemsolving.com/f/f/f/fff94c5c79a7098e723d731abaa608ba56d3918e.png)
.
is even, where 
is a positive integer.
.
,
is true. Thus, we assume that 
,
is odd. Trying 
as 
Thus all natural 
points on a unit line, the distance between any given two adjacent points is 
.
?
unit" would be better, but I guess that's a mouthful...
of the points could be on one end and the last point could be on the other end: 
?
x 
such that Jill wins.
Show that if 
and 
then 
.
,
.
,
,
for 
.
.
,
,
,
,
and we see that the desired inequality is true.
.
is half the harmonic mean of 
.
.
for the last prime 
is prime, therefore, a contradiction.
