AoPS Academy
+3 w
, and 
. The spheres with centers , and touch the plane at 
, and 
, respectively. Prove that 
is called a 
if it satisfies the following conditions:
contains at least two elements.
For all 
with 
, we have 
, and among the two numbers 
and 
, exactly one is rational.
For all 
, 
is irrational. can have?
-sided die with sides 
, consider the following game:, you win immediately., in which case you win.
:
: Win
: Lose
: Lose
: Win.
. Prove that for every nonnegative integer 
, there is an odd number 
and a parabolic triangle with vertices at three distinct points with integer coordinates with area 
.
, take the triangle 
which as area 1. For 
, take 
. Assume 
from now on.
By Shoelace, the area is 
When 
, this is just 
WLOG 
, so that this is just 
all perfect squares, so we need a Pythagorean triple. Amusingly, we can do the following:
with 
Then, assign 
as 
, 
as 
, so 
is 
so the area would be 
so 
Therefore, we just need to show that for any positive integer there exists a Pythagorean triple whose product of side lengths has 
of .
as one of the sides, and making the other sides 
which will both be odd for , so the product has a of and we are done.
then
then 
then 
then 
? for no reason
as a vertex of the triangle. By Shoelace Theorem, the area of the triangle with vertices , 
, and 
is 
. Therefore, it suffices to specify a pair 
for each ., use 
. For , use 
. For 
, use 
. For 
, use the construction for 
but multiply both 
and by 
. 
, 
, 
, 
and 
, 
. The area of the triangle formed by these three points is absolute value of 
. Then, taking 
, we get 
, taking 
, we get 
and 
we get 
. Then, taking 
in the first case, 
in the second case, and 
in the last case, we are done. (Whether we are given 
, 
or 
.
, 
, and 
with 
, we claim that 
can be![\[(4^n, 4^{2n-1}+1, 4^{2n-1}-1).\]](http://latex.artofproblemsolving.com/2/e/e/2eeb37232b12e3b362324c6bbff9cf6a4e0f335d.png)
[for 
, if take , 
, 
as the vertices]
![\[a(b^2-a^2)=4^n\cdot m^2\]](http://latex.artofproblemsolving.com/2/5/0/2507fc28e3413202d65a73cfbb1f67ed5cb82a07.png)
by Shoelace (actually this isn't necessary at all lol just use normal base x height). Let 
for “convenience”. So 
, or 
.
. So, choosing 
and 
, 
, and 
.
holds, so we are done. 
has area 
. And triangles with vertices 
and 
have areas 
and 
, respectively. Note that 
is a complete residue class modulo 3, hence, for given , we can adjust 
such that 
is a perfect square and 
, where 
. Thus we're done.
, 
, 
.![$[XYZ]= \lvert 2a^3- 2ab^2\rvert$](http://latex.artofproblemsolving.com/d/c/4/dc411657bfd0dfcd22728b8615a4629afeb8342c.png)
“above” 
so that the area is ![$[XYZ]=2ab^2-2a^3$](http://latex.artofproblemsolving.com/d/6/b/d6b1daa3e571c7abea6b4af8177629f6c76b58d5.png)
Set 
and calculating, we get 
. This is a solution for 
. we may pick: 
, 
, 
.
into cases and construct a working and parabolic triangle.
, then 
and the parabolic triangle with vertices 
has area
, then 
and the parabolic triangle with vertices 
has area
, then 
and the parabolic triangle with vertices 
has area
![\[Area = \dfrac{1}{2} \left|(ab^2 + bc^2 + ca^2) - (ba^2 + cb^2 + ac^2) \right|\]](http://latex.artofproblemsolving.com/0/5/c/05c774fe2270ec3e4ed8da0ad522f9193590213f.png)
which is equal to :- 
and 
= 
, this will not matter, as then we can subtract the value a from b, c.
and 
= 
and so on is given by the determinant![\[\frac 12 \begin{vmatrix} x_1 & x_1^2 & 1 \\ x_2 & x_2^2 & 1 \\ x_3 & x_3^2 & 1 \end{vmatrix} = \frac 12(x_1-x_2)(x_2-x_3)(x_3-x_1).\]](http://latex.artofproblemsolving.com/d/e/f/def1dd5fd136c90809b6988b8f54b25a2e5be059.png)
Now take 
and 
, such that ![\[\frac 12 (x_1-x_2)(x_2-x_3)(x_3-x_1) = m^2\left(\frac{m^2-2}2\right)^2 \left(\frac{m^2+2}2\right)^2.\]](http://latex.artofproblemsolving.com/a/1/9/a194504dcf92097bd934b761986d6893303b602e.png)
By varying 
, we solve the problem for all 
. case, just use , , , lol.
and 
Prove that
Let 
Prove that
,
.
set the vertices at 
,
,
.
is odd for all integers 
,
,
.
,
and 
.
and 
.![$A = \frac{1}{2}(2^{2n+1})(2^{2n+1}+1)^{4} = [2^{n}(2^{2n+1}+1)^{2}]^{2}$](http://latex.artofproblemsolving.com/7/d/2/7d2fcd982bd4dd4ed41b4bd92c44bc7ca87d5f49.png)
.
,
,
.
.
instead of 
because it looks nicer 
Let the difference 
be equivalent to 
and let 
.
.
and 
for some odd integer 
.
.
,
.
a solution to 
It is well known that more solutions 
can be generated by raising 
to the power of a nonnegative integer, including the 
t
.
to the power of a power of 2. When we have a solution 
,
gives us the solution pair 
Note that this increases the 2-adic valuation of 
by one, and as 
,
.
,
![\[\left\{ \begin{array}{ll}
P=(2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
Q=(-2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
R=((2^{2n}+1)(2^{2n+1}+1), (2^{2n}+1)^2(2^{2n+1}+1)^2)
\end{array}\right. \]](http://latex.artofproblemsolving.com/1/6/7/16776fe8fee02bb87481fec488c2701d5ee6a57b.png)
to get ![$[PQR]=(2^n(2^{2n+1}+1))^2$](http://latex.artofproblemsolving.com/b/a/f/baf385b8652ca50fb4f8856209a16132dd0c982f.png)
,