Hyperbola halving a square

Each of the regions must have area $\tfrac12$ . Consider the bottom-left region. We find the area by integration: $A = \int_0^a 1\:dx + \int_a^1 \frac{a\:dx}x = \left[ x\right]_0^a + \left[ a \ln x\right]_a^1 = a - 0 + a \ln 1 - a \ln a = a - a \ln a.$ Set $A = \tfrac12$ and solve numerically: $a - a \ln a = \tfrac12.$ The solution is $a \approx \boxed{0.18668}$ .

We note that the pink region has an area of $\dfrac 12$ , and it is given by:

$\begin{aligned} \int_a^1 \left(1 - \frac ax \right) dx & = \frac 12 \\ x - a \ln x \ \bigg|_a^1 & = \frac 12 \\ 1 - a + a\ln a & = \frac 12 \\ a\ln a - a + \frac 12 & = 0 & \small \color{#3D99F6} \text{By numerical method} \\ \implies a & \approx \boxed{0.187} \end{aligned}$

We can create a simple rectangle where $\frac{1}{x}$ crosses $y=1$ , ( $a,1$ ), with area $a*1=a$ , and find the remaining area via integration.

$A = a + a\int_{a}^{1}\frac{1}{x}dx = a + aln|x|_a^1 = a + a(ln|1|-ln|a|) = a(1-ln|a|)$

The total area of the square is 1, meaning the equally divided segments must have area $\frac{1}{2}$ .

$a(1-ln|a|) = \frac{1}{2}$

I solved numerically to get $a \approx \boxed{0.187}$ .

Each region has an area of .5. Additionally, we can see that the intersection of the graph and the upper side of the rectangle must be 1=a/x and so x=a. Given this, our area must be the integral from a to 1 added to a. (a*1=a) The integral is alnx so our equation is a + aln(1) - aln(a) = .5... aln(1) = 0 so we can graph y=x and y=aln(a) + .5 for our answer: ~.187