Question

The area of the largest semicircle that can be inscribed in the unit square is A. Then [100A] is (where [ ] represents the Greatest Integer Function)

Answer 1

Since the figure is symmetrical and parallel to BD, $\angle QPB={45}^o$. Consider the point X on AD at which the semicircle is tangent to AD. A line extended from X that is perpendicular to the tangent will be parallel toAB, and will also pass through the middle of the semicircle diameter. Let the line meet BC at Y.
$OY=rcos{45}^o=\frac{r}{\sqrt{2}}$
Hence$1=AB=r+\frac{r}{\sqrt{2}}=r(1+1/\sqrt{2})$.
Thus $r=1/(1+1/\sqrt{2})$.
Rationalizing the denominator, we obtain $r=2-\sqrt{2}$ and area
$=\pi r^2/2=\pi (3-2\sqrt{2})$
Thus, the area of the largest semicircle that can be inscribed in the unit square is
$\pi (3-2\sqrt{2})\approx 0.539$.