Dynamic Geometry: P16

Let $(x_{0}, \frac{1}{x_{0}})$ be an arbitrary point on the blue curve $y=\frac{1}{x}$ . The traveling rectangle has its center parametrically described as:

$x = \frac{x_{0}}{2},$

$y = \frac{1}{2x_{0}}$

which traces the yellow curve $y = \frac{1}{4x}$ . If we desire to compute the minimum sum of the rectangular perimeter and area, we can model this relationship as $f(x) = 2x + \frac{2}{x} + (x)(1/x) = 2x + \frac{2}{x} + 1$ whose first derivative (equal to zero) yields:

$f'(x) = 2 - \frac{2}{x^2} = 0 \Rightarrow x = 1$

and second derivative at $x=1$ yields:

$f''(x) = \frac{4}{x^3} \Rightarrow f''(1) = 4 > 0$ (a global minimum is attained when the rectangle is a unit square at the point $(1,1)$ on the blue curve).

The line $y=1$ intersects the yellow curve at $x = \frac{1}{4}$ , and the area bounded between the rectangle and the pink segment of $y = \frac{1}{4x}$ is computed per:

$\int_{1/4}^{1} 1 - \frac{1}{4x} dx \Rightarrow x - \frac{1}{4} \ln(x)|_{1/4}^{1} = \frac{3-\ln(4)}{4} = \boxed{\frac{3-2 \ln(2)}{4}}$

which yields $a=3, b =2, c=4 \Rightarrow \sqrt{a+b+c} = \boxed{3}.$

Consider a point $P \left(u, \dfrac 1u \right)$ on $y = \dfrac 1x$ . Then the rectangle formed has side lengths $u$ and $\dfrac 1u$ . The center of the rectangle is then $M = \left(\dfrac u2, \dfrac 1{2u}\right)$ . Replacing $\dfrac u2$ with $x$ , we get $M\left(x, \dfrac 1{4x}\right)$ . Therefore the equation of the locus of rectangle's center is $y = \dfrac 1{4x}$ .

The sum of area and perimeter of the rectangle is given by $x \cdot \dfrac 1x + 2x + \dfrac 2x$ . By AM-GM inequality :

$1 + 2\left(x + \frac 1x\right) \ge 1 + 2 \sqrt{x \cdot \frac 1x} = 3$

Equality or the area and perimeter is minimum when $x = \dfrac 1x = 1$ . Then the locus cuts the top side of the rectangle at $y=1 = \dfrac 1{4x} \implies x = \dfrac 14$ and the right side of the rectangle at $x=1$ and $y=\dfrac 14$ . Then the area bounded by the locus and the top-right corner of the rectangular is:

$\int_\frac 14^1 \left(1 - \frac 1{4x} \right) dx = x - \frac 14 \ln x \ \bigg|_\frac 14^1 = \frac 34 - \frac {\ln 4}4 = \frac {3-2\ln 2}4$

Therefore, $\sqrt{a+b+c} = \sqrt{3+2+4} = \boxed 3$ .

Here you can find a lot more sets of problems (including dynamic geometry) !