[/quote]
,
has a unique solution in integers with 
and 
.
as 
fir 
.
for all 
.
and orthocenter 
.
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
,
are positive real numbers.Prove that
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such tha
(Adit Aggarwal, India.)
be positive reals such that 
.
be reals such that 
Prove that
.
of triangle 
meets 
at 
.
be a point on the (internal) angle bisector of 
such that 
.
meet line 
and 
.
is isosceles.
such that, for all positive integers 
and 
and 
Show that
with center 
.
the incenter of triangle 
the contact points of the incircle with sides 
of triangle 
the foot of the perpendicular line from 
.
passes from the antidiametric point 
of 
in the circle 
is a diametre of the circle 
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 
.
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 
,
,
.
,
divides both 
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
