Proof

I will explain why the area of circle is $\pi r^2$ ( $r$ is its radius ) . (使っている図が悪いので後日差し替えます。)

掛け算で三角形の面積を求めるものの、その他の積分計算や極限計算は使わないようにしました。掛け算の記号 ( $\times$ ) を省略することと、文字で数を表していることと、 $x ^2 = x \times x$ と、ルート ( $\sqrt{x}$ は $\sqrt{x} \times \sqrt{x} = x$ となる正の数 ) がわかれば中学校あるいは小学校の算数・数学で理解できると思います。

Here, we think about a circle whose radius length is 1, to make story simple. The circle is called unit circle.

Definition of the ratio of the circumferece of a circle to its diameter

Let us represent the ratio of the circumference of a circle to its diameter as $\pi$ . $\pi$ meets next equation.

$\textrm{(circumference)} = 2 \pi \times \textrm{(radius)}$

As for unit circle, circumference length is $2 \pi$ .

Practical value of $pi$ is known as 3.141592… .

Evaluate the area of circle with regular n-sided polygon

Let’s think inscribed regular n-sided polygon and circumscribed regular n-sided polygon. ( $n \geq 3$ )

Now, divide regular n-sided polygon to $2n$ right-angled triangles to calculate the area.

inscribed regular 14-sided polygon and circumscribed regular 14-sided polygon divided into 28 right-angled triangles

Then, pick up one right-angled triangle and define $x, y$ as the following.

$x$ : 内接正 $n$ 角形を分割した直角三角形の、直角と円の中心を結ぶ辺の長さ
$y$ : 内接正 $n$ 角形を分割した直角三角形の、直角を通り円弧に交わる辺の長さ

このとき外接正 $n$ 角形を分割した直角三角形の、直角を通り円に接する辺の長さは相似比から $\frac{x}{y}$ となります。上の図で赤色になっている円弧の長さは、円周 $2 \pi$ を $2n$ 分割した円弧なので $\frac{2 \pi}{2n} = \frac{\pi}{n}$ となります。

Now, calculate the area of one triangle, and those of inscribed and circumscribed regular n-sided polygons.

The area of a triangle parted from inscribed regular n-sided polygon is $\frac{1}{2} xy$ , one of a triangle parted from circumscribed regular n-sided polygon is $\frac{y}{2x}$ .

Multiple $2n$ , and we can get the area of inscribed and circumscribed regular n-sided polygon.

inscribed n-sided regular polygon : $nxy$
circumscribed n-sided regular polygon : $n \frac{y}{x}$

The area of circle is larger than inscribed n-sided polygon and smaller than circumscribed n-sided polygon, therefore the area of circle, $s$ , meets the following inequality.

$nxy \lt s \lt n \frac{y}{x}$

Evaluate $x , y$ with $n$

We got an inequality formula of $s$ , but we can’t calculate $s$ yet. Then evaluate $x$ , $y$ with inequality.

Evaluate $x$

円を分割したときの、三角形の辺と円周の交点から鉛直方向に線をひきます。すると円の直径は、鉛直方向の線により $n$ 個に分割されます。

分割された円の直径のうち、一番端の部分、図で言う紫の部分は、直径を単純に $n$ 等分した $\frac{2}{n}$ よりも小さいです。そして $x$ は半径 1 から紫の部分を引いたものに等しいですから、

$1 – \frac{2}{n} \lt 1 – \textrm{(purple)} = x .$

Evaluate $y$

$y$ is less than the arch. The circumference of unit circle is $2\pi$ , and we divided unit circle to $2n$ right-angled triangles, so the arch length is $\frac{2 \pi}{2n} = \frac{\pi}{n}$ . Therefore,

$y \lt \frac{\pi}{n} .$

And $y$ is larger than the arc of $x$ radius circle, $x \frac{\pi}{n}$ .

$x$ is larger than $1 – \frac{2}{n}$ , as we saw, then

$\left( 1 – \frac{2}{n} \right) \frac{\pi}{n} \lt x \frac{\pi}{n} \lt y .$

From the above,

$\left( 1 – \frac{2}{n} \right) \frac{\pi}{n} \lt y \lt \frac{\pi}{n} .$

Evaluate $s$ with $n$

Remove $x$ and $y$ from the inequality of $s$ and evaluate $s$ with $n$ .

$nxy \lt s \lt n \frac{y}{x}$

$x$ , $y$ の評価式から、

$\begin{eqnarray*} nxy & \gt & n \left(1 – \frac{2}{n} \right) ^2 \frac{\pi}{n} \\ & \gt & \left(1 – \frac{2}{n} \right) ^2 \pi , \\ n \frac{y}{x} & \lt & n \frac{\frac{\pi}{n}}{1 – \frac{2}{n}} \\ & \lt & \frac{n \pi}{n-2} . \end{eqnarray*}$

以上をまとめると

$\left(1 – \frac{2}{n} \right) ^2 \pi \lt s \lt \frac{n \pi}{n-2}$

n が大きくなると右辺と左辺が $\pi$ に近づいていくのがわかりますね。

Calculate Circle Area

上で得られた式から、円の面積が $\pi$ になりそうだということがわかります。 Confirm the area of circle is $\pi$ by proof of contradiction.

Evaluate Upper Side of Circle Area

Suppose the area of unit circle is larger than $\pi$ and define $s = \pi + \delta$ ( $\delta \gt 0$ ) .

As we saw, $s \lt \frac{n \pi}{n-2}$ . Simplify the inequality.

$\begin{eqnarray*} \pi + \delta & \lt & \frac{n \pi}{n – 2} \\ \delta & \lt & \frac{n \pi}{n – 2} – \pi \\ & \lt & \frac{n \pi – \left( n – 2 \right) \pi}{n – 2} \\ & \lt & \frac{2 \pi}{n – 2} \\ \left( n – 2 \right) \delta & \lt & 2 \pi \\ n – 2 & \lt & \frac{2 \pi}{\delta} \\ n & \lt & \frac{2 \pi}{\delta} + 2 \end{eqnarray*}$

The inequality should true for all $n$ , but $n$ that is not less than $\frac{2 \pi}{\delta} + 2$ doesn’t meet it (contradiction).

Therefore, the area of unit circle is not more than $\pi$ .

$s \leq \pi$

Evaluate Lower Side of Circle Area

Suppose the area of unit circle is smaller than $\pi$ and define $s = \pi – \delta$ ( $\delta \gt 0$ ) .

As we saw, $\left(1 – \frac{2}{n} \right) ^ 2 \pi \lt s$ . Simplify the inequality.

$\begin{eqnarray*} \left( 1 – \frac{2}{n} \right) ^2 \pi & \lt & s \\ \left( 1 – \frac{2}{n} \right) ^2 \pi & \lt & \pi – \delta \\ \left( 1 – \frac{2}{n} \right) ^2 & \lt & \frac{\pi – \delta}{\pi} . \end{eqnarray*}$

Now, $n \geq 3$ , so

$\begin{eqnarray*} 1 – \frac{2}{n} & \lt & \sqrt{\frac{\pi – \delta}{\pi}} \\ 1 & \lt & \sqrt{\frac{\pi – \delta}{\pi}} + \frac{2}{n} \\ 1 – \sqrt{\frac{\pi – \delta}{\pi}} & \lt & \frac{2}{n} \\ n \left( 1 – \sqrt{\frac{\pi – \delta}{\pi}} \right) & \lt & 2 \\ n & \lt & \frac{2}{1 – \sqrt{\frac{\pi – \delta}{\pi}}} . \end{eqnarray*}$

The inequality should true for all $n$ , but $n$ that is not less than $\frac{2}{1 – \sqrt{\frac{\pi – \delta}{\pi}}}$ doesn’t meet it (contradiction).

Therefore, the area of unit circle is not less than $\pi$ .

$\pi \leq s$

From the above, $\pi \leq s \leq \pi$ , normally $s = \pi$ .

The Life

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