# Proof: Triangle Inequality

Let’s prove $$|PR| \leq |PQ| + |QR|$$ for the destances among 3 points $$P, Q, R \in \mathbb{R}^n$$.

# (日本語) 互いに素なピタゴラス数が無限に存在することの証明

Sorry, this entry is only available in Japanese.

# (日本語) 数学: 球の体積を導出する

Sorry, this entry is only available in Japanese.

# Calculate Circle Area – Why is it $$\pi r^2$$ ?

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。)

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.

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 $$x$$

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

### 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*}

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

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

### 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 $$s = \pi$$ .