[/quote]
,
with the following properties 
I
distinct red positive integers 
satisfying
or there are 
distinct blue positive integers 
satisfiying
be reals such that 
.Prove that
Where 
such that 
Prove that 
2. Let 
be sidelengths of a triangle such that 
Prove that 
Prove that
Where 
be positive real numbers such that 
.
knowing that the denominators are positive real numbers.
can be written in the form![\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]](http://latex.artofproblemsolving.com/9/e/9/9e96c07ea3b0750af8bed344df6e8a86fcf0d772.png)
where 
and 
are positive integers. Find the value of 
.
be three collinear points and 
three other collinear points. Let 
be the intersection of the lines 
and 
,
and 
.
be an odd prime and 
is a bijection and 
is an integer such that 
Show that 
is a multiple of 
.
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
of 
such that 
divides 
.
Prove that
Where 
satisfies
for some 
and for all 
,
of the functional equation 
for all 
for all 
.
be distinct polynomials of degree 
with non-zero coefficients. Suppose that they have 
common real roots counting multiplicity and 
common coefficients. Determine the maximum possible value of 
.
Y by bobjoe123, HamstPan38825, Adventure10, Rounak_iitr
Two incongruent triangles 
and 
are called a pair of pals if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let
and 
be the respective midpoints of sides 
and 
. The two sets of lengths 
and 
are identical 
-element sets of pairwise relatively prime integers.
Determine if there are infinitely many pairs of triangles that are pals of each other.
and 
are called a pair of pals if they satisfy the following conditions:(a) the two triangles have the same area;
(b) let
and 
be the respective midpoints of sides 
and 
. The two sets of lengths 
and 
are identical 
-element sets of pairwise relatively prime integers.Determine if there are infinitely many pairs of triangles that are pals of each other.
has twice the area of each triangle. So we are done if we can find and infinite number of relatively prime triples 
such that 
and triangles with side lengths 
have the same area.
and crunching away, we wind up with the formula 
.
satisfy 
(and there are an infinite number of such pairs), then it follows that we can let 
.
is approximately 
,
.
and 
be the median, the remaining side of the triangle has length 
So by the area formula
Without loss of generality, two triangles have 
and 
as sides respectively. Then 
which is equivalent to 
.
(there are infinitely many by recursion), and take 
.
are pairwise coprime.
and 
are equal - let 
without loss of generality. Then, by 2, we get that 
and 
.![\[2z=2AM=\sqrt{2(AB^2+AC^2)-BC^2}=\sqrt{2(x^2+y^2)-z^2}\implies BC=\sqrt{2x^2+2y^2-4z^2}\]](http://latex.artofproblemsolving.com/c/2/9/c290204793bbeb0c05867ee05cca0e36283c0b54.png)
By symmetry, we get that![\[YZ=\sqrt{2x^2+2z^2-4y^2}\]](http://latex.artofproblemsolving.com/4/2/8/42892cd75551a2cff0d86b2d152985fb5e7bfe05.png)
We have the following lemma:![\[\dfrac 12\sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}\]](http://latex.artofproblemsolving.com/8/d/e/8de9a03d2942c6946abd6133940aef30f122a60c.png)
Proof. We just need to show that this is equivalent to Heron's formula. We get that by the difference of squares formula
which is Heron's formula. 
![\[[ABC]=\dfrac 12\sqrt{x^2y^2-\left(\dfrac{x^2+y^2-(2x^2+2y^2-4z^2)}{2}\right)^2}=\dfrac 14\sqrt{4x^2y^2-\left(4z^2-x^2-y^2\right)^2}\]](http://latex.artofproblemsolving.com/1/b/0/1b0fe5089058822896e031412f2d4ab8f3d2b097.png)
By symmetry, we get![\[[XYZ]=\dfrac 14\sqrt{4x^2z^2-\left(4y^2-x^2-z^2\right)^2}\]](http://latex.artofproblemsolving.com/0/a/3/0a36c194fb10a62a1e413b0ca3c34fdf0bbdc090.png)
Equating, we get![\[4x^2(y^2-z^2)=4(x^2y^2-x^2z^2)=\left(4z^2-x^2-y^2\right)^2-\left(4y^2-x^2-z^2\right)^2=-5(3z^2+3y^2-2x^2)(y-z)(y+z)\]](http://latex.artofproblemsolving.com/0/c/a/0cacc5015fc835f6b49d8ace4ecfa05b892d06af.png)
Assuming 
,![\[-4x^2=15z^2+15y^2-10x^2\]](http://latex.artofproblemsolving.com/b/9/3/b93947d1ae42c569363f447507f10003c6459fc4.png)
which means that![\[2x^2=5y^2+5z^2\]](http://latex.artofproblemsolving.com/d/c/6/dc6e3f7bd7597bf315b5f3e00e5ef740520d5416.png)
Let 
,![\[y^2-10\left(\dfrac x5\right)^2=1\]](http://latex.artofproblemsolving.com/2/8/6/286890910e0c17bd1865a08e14432ceea9dcf509.png)
which has infinitely many solutions as a Pell equation. Thus, we can take any solution of![\[k^2-10\ell^2=1\]](http://latex.artofproblemsolving.com/6/6/2/6624eff66eedbedf01ec8e4bce0b42bbe08eff67.png)
and use 
.
and 
.
and 
are positive integers, the first is obviously satisfied. For the second, we note that it is equivalent to![\[25\ell^2+1\geq k^2=1+10\ell^2\]](http://latex.artofproblemsolving.com/d/a/3/da3b914ff38ed294fe5b8bf87ec2ccd0fbfa135e.png)
which is definitely true. Thus, taking the pair mentioned above indeed is a solution, and as there are an infinitely many 
,
equal to 
,
the triangle inequality is satisfied (which is not true because 
.
are the other sides then 
,
(that is, define 
)
,
and 
iff you can construct a triangle 
with three sides 
(both constructions are half of the parallelogram, and by the way, both triangles have the same area, so you can skip all the computations with medians.) If the median is 
,
and area 
or a triangle with sides 
and area 
,
,
and 
(otherwise the triangles are congruent). Then using the expanded form of Heron's formula 
for 
and 
you can find
,
,
.
,
or 
is either 
in the first one, and either 
in the second one. We need then 
,
,
,
.
or 
and 
,
somewhat close to each other is taking advantage of the fact that 
and set 
and 
(we'll take care of signs later). Then
,
and 
:
,
,
.
,
,
.
,
,
,
.
.
.
is odd and 
.
,
.
(which also takes care of the factor 
in 
,
,
,
.
and 
,
and 
.
and 
.
satisfies the triangle inequalities (so it's a necessary and sufficient condition). Analogously in 
.
and 
for the construction to work, but my original construction does not satisfy them. However, one could still use Pell equations (along the same lines as the first solution): consider any solution to 
,
,
,
,
.
,
,
.
without Heron's. Notice that since the areas of triangles 
and 
are the same, 
.
,
.
and 
to get ![\[\frac{a^2}{4}=c^2+m^2-2cm \cos \angle BAC\]](http://latex.artofproblemsolving.com/6/a/4/6a465e9dc8070dae4c13f4d8330fea5521a7f308.png)
and ![\[\frac{x^2}{4}=c^2+m^2-2cm \cos \angle YXW=c^2+m^2+2cm \cos \angle BAC.\]](http://latex.artofproblemsolving.com/3/c/9/3c9f31b258ff560033fd5d1c392bb98ae6143bda.png)
Add these equations to get 
.
to prove that.
has sides 
,
by the midpoint formula. Then, expanding Heron's formula out and substituting, we find that![$$[PQR]=\frac{1}{4}\sqrt{-16z^4-x^4-y^4+8z^2x^2+8z^2y^2+2x^2y^2}.$$](http://latex.artofproblemsolving.com/d/4/9/d49eb6f2aad5502d39f869274e03f9a8bbc0025c.png)
Now, WLOG let 
,
and 
.![$[ABC]=[XYZ]$](http://latex.artofproblemsolving.com/3/0/7/30756fadfdc0086da413812abc2934f05d6e594c.png)
becomes
which simplifies to 
.
,
.
has 
which we can verify satisfies the equation.
and likewise 
,
,
.
,
and 
,
,
,
which yields the same contradiction. Hence there are an infinite number of triples 
,
,
.
Now, equating both sides of the area of the two triangles yields 
This eventually simplifies to 
Now, we may fix 
It can be verified that these solutions also satisfy the triangle inequality.
and WLOG let 
Note that 
by Stewart's, so after a long computation with Heron we get that 
Now, let 
where 
which we have infinite solutions by Pell. Now, note that if we flip the triangle through 
and let 
we have that triangle ABD has sidelengths either 
or 
which both work by Triangle inequality as 
and finally the only way for 
which we can either discard or divide out from al
