For a triangle $T = ABC$ we take the point $X$ on the side $(AB)$ such that $AX/AB=4/5$, the point $Y$ on the segment $(CX)$ such that $CY = 2YX$ and, if possible, the point $Z$ on the ray ($CA$ such that $\widehat{CXZ} = 180 - \widehat{ABC}$. We denote by $\Sigma$ the set of all triangles $T$ for which
$\widehat{XYZ} = 45$. Prove that all triangles from $\Sigma$ are similar and find the measure of their smallest angle.
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This problem is both terrible in itself and terribly worded (orl did copy it correctly from the original source): it's only true when the angles in the problem are directed (so with undirected angles, $\angle{XYZ}=135^\circ$) and $Z$ lies on the extension of $AC$ past $C$, not past $A$ as the problem states.
Indeed, we can prove that there are numerous solutions (distinct even up to similarity) if $Z$ lies on the extension of $CA$ past $A$.
Construct points $Y',C'$ on line $AB$ and $A'$ on line $CC'$ (so that $A,X,B,Y,C'$ are in that order) with figure $XYCAZ$ similar to $BY'C'A'C$. Then $\angle{CY'A}=45^\circ$ and $\triangle{A'BC'}\sim\triangle{AXC}$. Now define positive reals $b,u,v,t$ such that $AC=b$, $AX=4u$, $BX=u$, $CX=3v$, $BY'=vt$, $C'Y'=2vt$, $A'C'=bt$, and $A'B=4ut$. Also let $h$ denote the length of the altitude $CF$ from $C$ to $AXBY'C'$, so that $CY'=h\sqrt{2}$. Note that
\[\frac{h}{CC'}=\sin\angle{A'C'B}=\sin\angle{ACX}=\frac{2[ACX]}{AC\cdot CX}=\frac{AX\cdot h}{AC\cdot CX}=\frac{4uh}{3bv},\]so
\[CC'=\frac{3bv}{4u}.\]Since $FY'=h$, we find $AF=5u+vt-h$, $FX=h-u-vt$. By the Pythagorean theorem,
\begin{align}
h^2+(h+2vt)^2 &= 9b^2v^2/(16u^2) \\
h^2+(5u+vt-h)^2 &= b^2 \\
h^2+(h-u-vt)^2 &= 9v^2.
\end{align}Plugging (2) into (1) yields
\[h^2+(h+2vt)^2 = \frac{9b^2v^2}{16u^2}=\frac{9v^2}{16u^2}(h^2+(5u+vt-h)^2).\]Equating $h^2$ with (3), we find
\[h=\frac{4u^2-4tuv+9v^2}{8u},\]and putting this in (3) we now find
\[16u^4+32tu^3v+(80t^2-288)u^2v^2-144tuv^3+81v^4=0,\]or
\[(80u^2v^2)t^2+(32u^3v-144uv^3)t+(16u^4-288u^2v^2+81v^4)=0.\]The discriminant of this quadratic in $t$ is
\[\Delta_t=-64u^2v^2(64u^4-1296u^2v^2+81v^4),\]so for infinitely many pairs of fixed $(u,v)$, some $t$ are available. Note that for the problem to be true, $\triangle{XYZ}\sim\triangle{BY'C}$ must have sides in a constant ratio, i.e.
\[\frac{CY'}{BY'}=\frac{h\sqrt{2}}{vt}=\frac{4u^2-4tuv+9v^2}{4uvt\sqrt{2}}\]must be constant for all solutions $(u,v,t)$, a clear contradiction: for example, consider the solutions
\[(u,v,t)=\left(1,1,\frac{14+\sqrt{1151}}{20}\right)\]and
\[(u,v,t)=\left(2,1,\frac{2+\sqrt{4079}}{40}\right).\]
As a side note, the official solution claims that $\cot{A}=1$, $\cot{B}=1/2$, $\cot{C}=1/3$ is the unique solution and the official diagram has $Z$ on the extension of $CA$ past $A$, but it contradicts this diagram:
[geogebra]a644b0dd4b5149d22ad313c7131e8db8fc21782a[/geogebra]
I believe we need $\angle{XYZ}=135^\circ$ (undirected angle) for this to work, but it's ridiculous that the poser wrote $45^\circ$ (even if they meant directed angles, it's just completely silly and misleading).
The problem stated is true of instead of $AX/AB=4/5$, we have $AX/XB=4/5$. This is probably what is meant, as this version is assumed in the official solution of the problem.
It's false in its current formulation, there are many non-similar T which achieve the required angle condition.
