MHB Solve for $\overline{xyz}: \overline{xyz}\times \overline{zyx}=\overline{xzyyx}$

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The problem involves finding a three-digit number $\overline{xyz}$ such that when multiplied by its reverse $\overline{zyx}$, the result equals the five-digit number $\overline{xzyyx}$. Both $\overline{xyz}$ and $\overline{zyx}$ are defined as three-digit integers. The equation can be expressed as $\overline{xyz} \times \overline{zyx} = \overline{xzyyx}$. The challenge lies in determining the specific values of $\overline{xyz}$ that satisfy this equation. Solving this will require exploring the properties of three-digit numbers and their reversals.
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$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$
 
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Albert said:
$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$
$(100x+10y+z)(100z+10y+x)
=10000x+1000z+100y+10y+x-(1)$
for $10000xz<100000$
we have $xz<10$
compare both sides of(1):$xz=x\,\ ,or\,\, z=1$
(1) becomes :
$10000x+1000y(x+1)+100(x^2+y^2+1)+10y(x+1)+x=10000x+1000+100y+10y+x$
or $100y(x+1)+10(x^2+y^2+1)+y(x+1)=100+10y+y-(2)$
compare both sides of (2):
$\therefore y(x+1)<10,\,, and ,\, y(x+1)=y$
$for (x+1)>1 \,\,\therefore y=0 ,and \,\,x=3$
we get :$\overline {xyz}=301$
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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