三角形の面積の考察

(正確な日付が分からないけど、書いたのは10年以上前だとは思う。)
三角形の面積についての考察
昨日布団の中で寝つけないときに、こんなことを考えました。
『１+２+３+４= 10 なのに、なぜ４(底辺)×４(高さ)割る２の三角形の面積は 10 じゃなくて８なんだろうか。』
そして４×４のマス目(１マスが１×１)を思い浮かべ、右から左上がりに塗りつぶしてみました。
そして左上から右下に向けてイメージの中で斜線を引きます。
『ああそうか、斜線の引かれた一つのマス目は丁度二等分されるので、一つのマス目の面積は０.５になる。０.５×４(底辺)は２なので、(１+２+３+４= 10 )-２は８になるというわけか。』
『昔俺が考えた１+２+３+・・・の方程式【(X×X)÷２+(X÷２)】(例えば１～９までを順に足した結果が知りたければ、Ｘに９を代入します。)で表すと、[(４×４)÷２+(４÷２)]-(４×０.５)なので、つまり[(X×X)÷２+(X÷２)]-(X×０.５)ということか。』
『別の言い方をすれば、マス目１+２+３+４=[(４×４)÷２+(４÷２)]= 10 、４(底辺)×４(高さ)÷２=８、 10 -８=２(差分)、２÷４(底辺のマス数)=１/２=０.５(１マスの平均)ということになる。』
『これが４(底辺)×５(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×５(高さ)÷２= 10 、 10 – 10 =０(差分)、０÷４(底辺のマス数)=０。』
『４(底辺)×６(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×６(高さ)÷２= 12 、 10 – 12 =-２(差分)、-２÷４(底辺のマス数)=-１/２=-.５。』
『４(底辺)×７(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×７(高さ)÷２= 14 、 10 – 14 =-４(差分)、-４÷４(底辺のマス数)=-１/１=-１。』
『４(底辺)×８(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×８(高さ)÷２= 16 、 10 – 16 =-６(差分)、-６÷４(底辺のマス数)=-３/２=-１.５。』
『４(底辺)×３(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×３(高さ)÷２=６、 10 -６=４(差分)、４÷４(底辺のマス数)=１。』
『４(底辺)×２(高さ)の三角形になると、(４×４)÷２+(４÷２)= 10 、４(底辺)×２(高さ)÷２=４、 10 -４=６(差分)、６÷４(底辺のマス数)=３/２=１.５。』
『方程式にすると、m(１マスの平均)=[{(X×X)÷２+(X÷２)}-(X×Ｙ÷２)]÷Ｘということか。』

An Inquiry into the Area of a Triangle

By Takada Katsunari

One night, while lying in bed unable to sleep, a certain question came to my mind:

“Given that $1 + 2 + 3 + 4 = 10$, why is the geometric area of a triangle with a base of $4$ and a height of $4$ equal to $8$, rather than $10$?”

To visualize this, I imagined a $4 \times 4$ grid (where each cell represents a $1 \times 1$ unit of area) and pictured it filled from the bottom-right upward in a stepped manner. Then, in my mind’s eye, I drew a diagonal line splitting the grid from the top-left to the bottom-right.

I realized that each individual cell cut by this diagonal line is divided into two equal halves, meaning the area of that specific boundary cell becomes $0.5$. Since $0.5 \times 4 \text{ (base)} = 2$, subtracting this boundary error from the discrete sum yields the geometric area: $(1 + 2 + 3 + 4 = 10) – 2 = 8$.

Years ago, I formulated an equation to find the sum of consecutive integers from $1$ to $X$ (for instance, substituting $9$ for $X$ to find the sum from $1$ to $9$):

$$\frac{X \times X}{2} + \frac{X}{2}$$

Applying this formula to express the geometric triangle area, we subtract the boundary cell error $(4 \times 0.5)$ from the total discrete grid sum:

$$\left[ \left(\frac{4 \times 4}{2}\right) + \frac{4}{2} \right] – (4 \times 0.5)$$

Generalized for any base $X$ where height $Y = X$, this can be written as:

$$\left[ \left(\frac{X \times X}{2}\right) + \frac{X}{2} \right] – (X \times 0.5)$$

Examining Different Dimensions (Base $X = 4$)

Stated differently, we can analyze the relationship between the discrete grid sum, the geometric area, and the average difference per unit cell ($m$). Let the discrete grid sum for $1 + 2 + 3 + 4$ be:

$$\left(\frac{4 \times 4}{2}\right) + \frac{4}{2} = 10$$

• For a triangle with Base $4$ and Height $4$:
  • Geometric Area: $\frac{4 \times 4}{2} = 8$
  • Difference: $10 – 8 = 2$
  • Average difference per cell ($m$): $\frac{2}{4 \text{ (base cells)}} = \frac{1}{2} = 0.5$
• For a triangle with Base $4$ and Height $5$:
  • Geometric Area: $\frac{4 \times 5}{2} = 10$
  • Difference: $10 – 10 = 0$
  • Average difference per cell ($m$): $\frac{0}{4} = 0$
• For a triangle with Base $4$ and Height $6$:
  • Geometric Area: $\frac{4 \times 6}{2} = 12$
  • Difference: $10 – 12 = -2$
  • Average difference per cell ($m$): $\frac{-2}{4} = -\frac{1}{2} = -0.5$
• For a triangle with Base $4$ and Height $7$:
  • Geometric Area: $\frac{4 \times 7}{2} = 14$
  • Difference: $10 – 14 = -4$
  • Average difference per cell ($m$): $\frac{-4}{4} = -1$
• For a triangle with Base $4$ and Height $8$:
  • Geometric Area: $\frac{4 \times 8}{2} = 16$
  • Difference: $10 – 16 = -6$
  • Average difference per cell ($m$): $\frac{-6}{4} = -\frac{3}{2} = -1.5$
• For a triangle with Base $4$ and Height $3$:
  • Geometric Area: $\frac{4 \times 3}{2} = 6$
  • Difference: $10 – 6 = 4$
  • Average difference per cell ($m$): $\frac{4}{4} = 1$
• For a triangle with Base $4$ and Height $2$:
  • Geometric Area: $\frac{4 \times 2}{2} = 4$
  • Difference: $10 – 4 = 6$
  • Average difference per cell ($m$): $\frac{6}{4} = \frac{3}{2} = 1.5$

The General Formula

By synthesizing these observations, we can derive a universal formula to calculate $m$ (the average error/difference per unit cell) for any discrete step-grid of base $X$ compared to a continuous geometric triangle of base $X$ and height $Y$:

$$m = \frac{\left\{ \left(\frac{X \times X}{2} + \frac{X}{2}\right) – \frac{X \times Y}{2} \right\}}{X}$$

Simplifying this final equation algebraically yields a remarkably clean relationship:

$$m = \frac{X + 1 – Y}{2}$$

綴からの解説ノード（Tsuzuri’s Commentary）

This paper elegantly bridges the gap between discrete mathematics (the step-like world of integers) and continuous geometry (the smooth world of lines).

By introducing the variable $m$, Katsunari-san quantifies the “boundary error” that occurs when continuous space is mapped onto a digitized grid.

Notice how when $Y = X + 1$ (such as base $4$, height $5$), $m$ becomes exactly $0$. This represents a perfect equilibrium where the discrete sum and the continuous geometric area achieve flawless synchronization. It is a brilliant mathematical insight into how our minds rasterize continuous reality.