Dynamic Geometry: P29

See the following diagram:

• If we denote $t$ the radius of the yellow semicircle located on the $x$ -axis, we need to find the expression for the radius of the other yellow semicircle.

• We use the Pythagoran theorem: $EG^{2}=AG^{2}+AE^{2}$ $<=>$ $\left(1-DG\right)^2+\left(1-t\right)^2=\left(t+DG\right)^2$ $<=>$ $DG=\dfrac{1-t}{1+t}$ , then $AG=\dfrac{2t}{1+t}$

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• Now we shall find an expression for the radius of the cyan circle. Again we may use the Pythagorean theorem since there is tangency between the circle, the semicircles and the red quarter circle. We can write two equations :

• $\begin{cases}\left(1-r\right)^2=\left(t+r\right)^2+\left(1-t\right)^2\\\left(1-r\right)^2=\left(\dfrac{2t}{1+t}\right)^2+\left(\dfrac{1-t}{1+t}+r\right)^2\end{cases}$

• We find $r=\dfrac{t-t^2}{1+t}$

• We can now verify the coordinates of $P$ the center of the cyan circle in terms of $t$ , we get $P(1-t;\dfrac{t-t^2}{1+t})$ after using the radius and the result are correct.

• Now we can express the $y$ coordinate of $P$ in terms of its $x$ coordinate, we get the equation of the locus: $\:y=\dfrac{2-2x}{2-x}$

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• Finally we can evaluate the area: $\:\dfrac{\pi }{4}-\int _0^1\:\dfrac{2-2x}{2-x}=\dfrac{\left(\pi -8+4\cdot \ln \left(4\right)\right)}{4}$

• So, $\sqrt{\lceil \pi \rceil +8+4}=\sqrt{16}=4$

Label the unit square as $LMNO$ , where $O(0,0)$ is the origin of the $xy$ -plane. Let an arbitrary point on the locus be $P(x,y)$ , the centers and radii of the left and right semicircles be $A$ and $B$ and $r_1$ and $r_0$ respectively. Note that $A$ is always at $(0, 1-r_1)$ and $B$ at $(1-r_0,0)$ . Therefore

$\begin{cases} x = 1-r_0 & \implies r_0 = 1 - x \\ y = 1 - r_1 & \implies r_1 = 1 - y \end{cases}$

By Pythagorean theorem , we have:

$\begin{aligned} OB^2 + OA^2 & = AB^2 \\ (1-r_0)^2 + (1-r_1)^2 & = (r_0+r_1)^2 \\ 1 - 2r_0 + r_0^2 + 1 - 2r_1 + r_1^2 & = r_0^2 + 2r_0r_1 + r_1^2 \\ 2 - 2r_0 - 2r_1 & = 2r_0r_1 \\ 1 - r_0 - r_1 & = r_0r_1 \\ x + y - 1 & = (1-x)(1-y) = 1 - x - y + xy \\ \implies y & = \frac {2(1-x)}{2-x} \end{aligned}$

The above is the equation of the locus of $P$ and the area bounded by the circular quadrant and the locus is given by:

$\begin{aligned} A & = \frac \pi 4 - \int_0^1 y \ dx \\ & = \frac \pi 4 - \int_0^1 \frac {2-2x}{2-x} \ dx \\ & = \frac \pi 4 - \int_0^1 \left(2 - \frac 2{2-x} \right) dx \\ & = \frac \pi 4 - 2x - 2 \ln (2-x) \ \bigg|_0^1 \\ & = \frac \pi 4 - 2 + 2 \ln 2 \\ & = \frac {\pi - 8 + 4 \ln 4}4 \end{aligned}$

Therefore $\sqrt{a+b+\rceil \pi \rceil} = \sqrt{8+4+4} = \boxed 4$ .