Partitioning convex polygons into triangles of equal area and perimeter This post is based on https://math.stackexchange.com/questions/2822589/dissect-square-into-triangles-of-same-perimeter,  On a possible variant of Monsky's theorem and Cutting convex polygons into triangles of same diameter.
Question 1: Is this statement true: "For any convex polygon, there is some finite value of a positive integer n such that the polygon allows partition into n triangles all of which are of same area."?
Note: The above claim, if true, will automatically hold for infinitely many values of n - one only needs to subdivide the equal area triangular pieces further into equal area triangles. Further, replacing 'area' with 'perimeter' or 'diameter' above generates further questions.
Question 2: Is this claim true: "For any given convex polygon, there is at least one finite value of n such that the polygon allows partition into n triangles all of equal area and equal perimeter"?
Note: If question 1 has a negative answer, that would invalidate question 2. And there could be convexity-relaxed versions of both above questions.
An extension added on 15th August, 2022: If claim 1 is proved invalid (answer below makes progress in that direction), one could ask a further question: If a convex m-gon can be cut into some number n of triangles, what is the lower bound on the number of distinct areas/perimeters/diameters these n triangles can have - as a function of m and n?
 A: Conjecture: If $P=(0,1)$, $Q=(0,0)$, $R=(1,0)$, $S=(u,v)$, and $PQRS$ can be partitioned into triangles of equal area, then $u$ and $v$ are algebraically dependent.
Corollary of conjecture: There is only a zero-measure set of $S$ for which these $PQRS$ can be partitioned into triangles of equal area.
Corollary of conjecture: Among all quadrilaterals, only a zero-measure set can be partitioned into triangles of equal area, because all quadrilaterals are affinely equivalent to a quadrilateral of the above form.
The idea of this conjecture is that if we start with simple configurations of triangles, and gradually add more points, we never get enough degrees of freedom to solve for all of the areas being equal.

We get a first configuration of triangles by dividing the quadrilateral into two triangles. Then the equation of areas is of the form $A(u,v)=B(u,v)$, which relates $u$ and $v$.
If we divide one triangle into two, we get one degree of freedom to choose the dividing point, but add one constraint from the equality of one more area. If $x$ is the $x$-coordinate of that dividing point, the equality of areas is of the form $(\exists x)A(u,v,x)=B(u,v,x)=C(u,v,x)$; eliminating $x$ gives an algebraic relation between $u$ and $v$.
If we divide one triangle into three, we get two new degrees of freedom to choose the dividing point, but add two new constraints from the equality of two more areas. Here the equality of areas is of the form
$(\exists x,y)A(u,v,x,y)=B(u,v,x,y)=C(u,v,x,y)=D(u,v,x,y)$;  eliminating $x$ and $y$ gives an algebraic relation between $u$ and $v$.
No matter how we add points to the configuration, there are never enough degrees of freedom to solve all the equations for all values of $u$ and $v$; instead, we always get an algebraic relationship between $u$ and $v$.
Here are some examples of the equations that result from various configurations:
\begin{array}{}
PQR=RSP & u+v=2\\
QRS=SPQ & u=v\\
TPQ=TQR=TRS=TSP & u+v=2 \vee u=v \\
TRS=0,\ TPQ=TQR=TSP & u^2 - v^2 = 2u-v \\
& \\
PRU = PUT = QTV = QVS  &
3u+5=0 \vee 2u^2+6uv=u-2v+1 \\
= RSV = RVU = TQP = TUV & \phantom{3u+5=0}\vee\ 4u^2+6uv=3u-2v+1 \\
& \\
PQT = PTW = PWS &
u^2 - uv - 2v^2 = 3u- 7v\\
= QRU = QUT
& \ \vee\ 5u^2+3uv-2v^2=7u-3v\\
= RSV = RVU & \ \vee\ 
7u^2(u-4)+7v^2(8-3u-2v)=(u-v)(15u+v-50)
\\
= SWV = TUV = TVW & \ \vee\ 7u^2(u-3)+7v^2(2v-3)=uv(22u+15v-58)
\\
\end{array}
The conjecture above is equivalent to proving that eliminating variables from any configuration will always give a non-trivial relationship between $u$ and $v$, and I hope someone else will see how to prove that.
A: Question 1 indeed has a negative answer, as previous responders have speculated.  See e.g. the 2008 paper by Monsky and Jepsen:
Constructing Equidissections for Certain Classes of Trapezoids", Discrete Mathematics, 308 (23): 5672–5681, doi:10.1016/j.disc.2007.10.031, Zbl 1156.51304
(I copied this reference from the Wikipedia page on "equidissections".)
I think the simplest examples they prove have no equidissections are trapezoids $(0,0), (1,0), (0,1), (a,1)$ where $a$ is transcendental.
Interestingly there are still (algebraic) values of $a$ for which it is not known whether or not any equidissections exist.
