There is a square $Q$ consisting of $(0,0), (2,0), (0,2), (2,2)$ 
There is a square $Q$ consisting of $(0,0), (2,0), (0,2), (2,2)$.
A point $P$ satisfies following condition:
The straight line passing through $P$ and dividing the area of square $Q$ in the ratio $1:3$ does not exist.

Can we know the locus of $P$ and the area of the locus?
 A: It is bounded by lines that divide the square exactly 1:3.
These are two forms:


*

*through $(0,a),(1,1/2),(2,1-a)$ or similar;  

*through $(0,a),(2/a,0)$ or similar.  


We need the envelope of the lines $(0,a),(2/a,0)$.  Find the intersection of the line through $(0,a),(2/a,0)$ and $(0,b),(2/b,0)$, then let $b\to a$.  (In the same way, you might find the tangent of a curve by taking chords from $a$ to  $b$, and then let $b\to a$.)
Once you have the equation of the envelope, find the area of the rounded triangle it covers from $(1,1/2)$ to $(1/2,1)$ to $(1,1)$.
A: We find the portion in the lower left quadrant of the square.  The boundary will be lines that cut off a triangle of area $1$ as shown in the figure below.  They have slope $m$ from $-2$ to $-1/2$, run from  $(0,\sqrt {-2m})$ to $(\sqrt{-\frac2m},0)$, so have equation $y=\sqrt {-2m}+mx$.  For a given $x$ in the range $[\frac 12, 1]$we want to find the $m$ that maximizes $y$. Taking the derivative and setting to zero gives $0=-\frac 12\sqrt{-\frac 2{m}}+x$ or $m=-\frac 1{2x^2}$  Plugging this in gives the envelope $y=\frac 1{2x}$.  To get the area, it is $4\int_{1/2}^1(1-\frac 1{2x})dx=2(1-\log (2)) \approx 0.613$

A: Here is a Geogebra approximation of the region in question (the light colored region in center):

