Circle squared

Without loss of generality let the square be a unit square. Let the blue circle, aka circle $X$ , have radius $r$ and the red circle, aka circle $Y$ , have radius $R.$ Then the diagonal of the unit square can be divided into 4 sections: one each of the radii of the two circles and one each of the diagonals of squares with side lengths $r$ and $R.$ This results in the equation

$r(1 + \sqrt{2}) + R(1 + \sqrt{2}) = \sqrt{2}$

$\Longrightarrow r + R = \dfrac{\sqrt{2}}{1 + \sqrt{2}} = 2 - \sqrt{2} \Longrightarrow R = (2 - \sqrt{2}) - r.$

The combined area $A$ of the two circles is then

$A = \pi r^{2} + \pi R^{2} = \pi(r^{2} + ((2 - \sqrt{2}) - r)^{2}) =$

$2\pi(r^{2} - (2 - \sqrt{2})r + (3 - 2\sqrt{2})).$

Now $\dfrac{dA}{dr} = 2\pi(2r - (2 - \sqrt{2})) = 0$ when $r = \dfrac{2 - \sqrt{2}}{2},$ so the magnitude of the change in $A$ is least when circle $X$ has this radius. For this value of $r$ we see also that $R = r$ , i.e., the change in $A$ is least when the two circle have the same radius.

Thus, as this value for $r$ was obtained for a unit square, the desired fraction is $\dfrac{2 - \sqrt{2}}{2} = \boxed{0.29}$ to the nearest hundredth.

My approach is similar to Brian's.

Denote the radii by $r$ and $R$ and note that $r+R=2-\sqrt{2}$ , so that we can write $r$ and $R$ as $\frac{2-\sqrt{2}}{2}\pm d$ and $\frac{A}{\pi}=r^2+R^2=\frac{(2-\sqrt{2})^2}{2}+2d^2$ . The minimum is attained when $d=0$ and $r=R=\frac{2-\sqrt{2}}{2}\approx \boxed{0.29}$ . At that point, the rate of change of the $A$ is 0.

In my opinion Olawale Olayemi is right I took this approach considering :

1) The solution must be in a point where the function representing the sum of areas of circles is a maximum or a minimum since her derivative must be 0.

2) Assuming two solution exist, by symmetry we must have two points where derivative is 0 and second derivative must have same value and sign that should force to the existence of another points in between where derivative must equal 0, against hypothesis. Therefore equal radii circles is the only solution.

Take a unit circle for convenience. method. Draw a good diagram. Let r and R be the radii. Find one in terms of the other. Write the general expression for sum of areas ;differntiate to get minima.