The Square-Circle Problem

Geometric solution

Let the side of the square = 2 units. Then AB = 2 and BP = 1, so AP = $\sqrt5$ . (pythagorean) $\\$ Let the radius of the circle = r units. Then OA = OP = 2. $\\$ Let M & N be the midpoints of AP & PQ respectively. $\\$ The centre O of the circle can be found by drawing the right bisectors of AP and PQ at M & N. $\\$ Note that since triangle PQC is a right isosceles triangle, PQ is perpendicular to diagonal AC and N is on the diagonal, so NO is on AC. Therefore AON forms a straight line. $\\$ Since $PQ = \sqrt2$ (pythagorean), $PN = \frac{\sqrt2}2 \\ AN =\sqrt{AP^2-PN^2}=\sqrt{(\sqrt5)^2-(\frac{\sqrt2}2)^2}=\frac{3\sqrt2}2$ . $\\$ $ON = AN - AO = \frac{3\sqrt2}2-r.\\ \text{By Pythagorean, } OP^2=PN^2+ON^2\\$ $\begin{aligned}\therefore r^2=&(\frac{\sqrt2}2)^2+(\frac{3\sqrt2}2-r)^2\\ r^2=&\frac12+\frac92-3\sqrt2r+r^2 \\ 0=&5-3\sqrt2r \\ r=&\frac5{3\sqrt2}=\frac{5\sqrt2}6 \end{aligned}$ So the perimeter of the circle $= 2\pi r=\frac{5\sqrt2\pi}3\approx 7.4\\$ Since the perimeter of the square is 8, the square's perimeter is greater than the circle's perimeter.

Suppose we draw a square ABCD centered on the cartesian plane where:

$A=(-1,1)$ ; $B=(-1,1)$ ; $C=(1,-1)$ ; $D=(1,1)$

Notice that each side of the square has distance 2. Then, we can say:

$P_{square} = 4 · 2 = 8$

We can calculate the middle points of the side BC and CD:

I will denote $M:=$ middle point of BC ; $N:=$ middle point of CD

$M=(0,-1)$ ; $N=(1,0)$

Also, we know a circumference goes trough points $A$ , $N$ and $M$

Therefore, grafically we have the following situation:

The equation of any circumference can be expressed as: $(x-a)^2+(y-b)^2=r^2$ , where the point $(a,b)$ is the center and $r$ is the radius.

Notice that circumference goes through points $A=(-1,1)$ , $N=(1,0)$ and $M=(0,-1)$ . Then, this 3 points satisfy the previous equation. Therefore, we have:

$(1)$ —> $(-1-a)^2+(1-b)^2=r^2$

$(2)$ —> $(1-a)^2+(-b)^2=r^2$

$(3)$ —> $(-a)^2+(-1-b)^2=r^2$

Using $(2)$ and $(3)$ :

$(2)$ —> $(1-a)^2+(-b)^2=r^2$ $\Longrightarrow$ $1 - 2a + a^2 + b^2 = r^2$

$(3)$ —> $(-a)^2+(-1-b)^2=r^2$ $\Longrightarrow$ $a^2 + 1 + 2b + b^2 = r^2$

Equating $(2)$ and $(3)$ :

$1 - 2a + a^2 + b^2 = a^2 + 1 + 2b + b^2$ $\Longrightarrow$ $2a + 2b = 0$ $\Longrightarrow$ $\boxed{a + b =0}$

Using $(1)$ and $(2)$ :

$(1)$ —> $(-1-a)^2+(1-b)^2=r^2$ $\Longrightarrow$ $1 + 2a + a^2 + 1 -2b + b^2 = r^2$

$(2)$ —> $(1-a)^2+(-b)^2=r^2$ $\Longrightarrow$ $1 - 2a + a^2 + b^2 = r^2$

Equating $(1)$ and $(2)$ :

$1 + 2a + a^2 + 1 - 2b + b^2 = 1 - 2a + a^2 + b^2$ $\Longrightarrow$ $4a - 2b + 1 =0$ $\Longrightarrow$ $4a - 2b = -1$ $\Longrightarrow$ $\boxed{2a -b =- 1/2}$

Now, using the two equations contained in boxes. We can calculate $a$ and $b$ using the reduction method:

$a = - 1/6$ ; $b = 1/6$

We know $(-1/6, 1/6)$ is the center of the circumference. Then, the circumference equation has the following form:

$(x + 1/6)^2 +(y-1/6)^2 = r^2$

Using the circumference goes through point $A=(1,1)$ , for example. We can calculate the radius:

$(-1+ 1/6)^2 +(1-1/6)^2 = r^2$ $\Longrightarrow$ $r^2=50/36$ $\Longrightarrow$ $r=(5\sqrt{2})/6$

Therefore, we can calculate the perimeter of the circumference:

$P_{circumference} = 2\pi r\ = 2 · 3.14 · (5\sqrt{2})/6 = 7.404...$

Finally, we can conclude $\boxed{P_{square} > P_{circumference}}$

Extra Problem

We have all the information calculated. We only need to calculate the areas:

$A_{square} = 2 · 2 = 4$

$A_{circumference} = \pi r^2 = 3.14 · (5\sqrt{2})/6 = 4.361...$

Finally, we can conclude $\boxed{A_{circumference} > A_{square}}$