A Peculiar Locus Area

Let $A=(0,3)$ , $B=(0,0)$ , and $C=(4,0)$ . We can see that $AC$ can be expressed as the equation $y=-\dfrac{3}{4}x+3$ .

Note that the border of $\mathbf{L}$ must satisfy that any point $P$ on it has $[PEBF]=1$ . Suppose that $P=(x,y)$ . We must have $y=\dfrac{1}{x}$ in order to have $[PEBF]=1$ . Therefore, the equation of the border of the locus $\mathbf{L}$ is $y=\dfrac{1}{x}$ .

We have now reduced the problem to finding the area in between the two equations $y=-\dfrac{3}{4}x+3$ and $y=\dfrac{1}{x}$ . This calls for a definite integral! We must first find the upper and lower bound. This is when the two functions cross.

We set $-\dfrac{3}{4}x+3=\dfrac{1}{x}$ . Simplifying, we have that $3x^2-12x+4=0$ and using the quadratic equation, we obtain that $x=2\pm \dfrac{2\sqrt{6}}{3}$ .

Thus, we want to find the value of $\displaystyle\int_{2-\frac{2\sqrt{6}}{3}}^{2+\frac{2\sqrt{6}}{3}}\left(-\dfrac{3}{4}x+3\right)-\left(\dfrac{1}{x}\right)\text{d}x$

Taking the indefinite integral first, we see that $\displaystyle\int\left(-\dfrac{3}{4}x+3\right)-\left(\dfrac{1}{x}\right)\text{d}x=-\dfrac{3}{8}x^2+3x-\ln |x|+C$ . Now, we need to evaluate $\left.-\dfrac{3}{8}x^2+3x-\ln |x|\right|_{2-\frac{2\sqrt{6}}{3}}^{2+\frac{2\sqrt{6}}{3}}$

Plugging and chugging, the above turns out to be $2\sqrt{6}+\ln(5-2\sqrt{6})\approx 2.606547816$ .

Our desired answer is therefore $\left\lfloor 1000\dfrac{2.606547816}{6}\right\rfloor =\boxed{434}$ .

This really took me a lot of attempts to figure out. Here is my solution( Sorry for the bad image quality and clarity):

image

One can easily find the co-ordinates $P,Q,R,S$ by solving $xy=1$ and $\displaystyle\frac{x}{4}+\frac{y}{3}=1$ .

So now $[L]=[PQRS]-$ shaded area $=2\sqrt{6}-ln(5+2\sqrt{6})$ .

Thus we have $\displaystyle\lfloor1000\frac{[L]}{[ABC]}\rfloor=\boxed{434}$ .

if we put point B on the center of our cartesian coordinate, the equation of the hypotenuse is $3x+4y =12$

the key is the we want the find the area of loci $xy >= 1 \leftrightarrow y >= \frac{1}{x}$ bounded by $3x+4y =12$ and their point of intersection.

thus, we want to find the integral: $\int_{2-2/3\sqrt(6)}^{2+2/3\sqrt(6)} \frac{12-3x}{4} - \frac{1}{x} dx$

the rest is just some tedious algebra