1

We have two equal right triangles $A$ and $B$ with slope $-\lambda$ where the two bases span the distance between $t$ and $t+1$.

Figure 1

The sum of both areas we call $X$.

Let's say we decrease and increase the height of both triangles respectively by the same amount $\theta$. This means the area of $A$ is decreased with the area $C$ and the area of $B$ is increased with the area $D$.

Figure 2

If we shift them back between $t$ and $t+1$, this gives us the new $A$ and $B$ triangles like in the below figure.

Figure 3

The area of the new $A$ and $B$ triangles is larger than $X$, because $D \gt C$. If we measure the difference, this is equal to $\frac{1}{\lambda}\theta^2$. This is because

Figure 4

My question is: is this a known identity ?

MathFail
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Steven01123581321
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1 Answers1

1

Let the original triangle be base $x$ and height $h$, so $h=\lambda x$, area $A=B=\frac{1}2\lambda x^2$. If we decrease the height for triangle $A$, we get a new triangle $A'$ with height $h'_A=h-\theta$, the new triangle is similar to the original triangle, so we have

$$\frac{h}{h_A'}=\frac{x}{x_A'}\Rightarrow x_A'=x-\frac{\theta}{\lambda}$$

If we increase the height for triangle $B$, we get a new triangle $B'$ with height $h'_B=h+\theta$, the new triangle is similar to the original triangle, so we have

$$\frac{h}{h_B'}=\frac{x}{x_B'}\Rightarrow x_B'=x+\frac{\theta}{\lambda}$$

The difference for the sum of areas:

$$\begin{align}A'+B'-A-B&=\frac{1}2\lambda (x_A')^2+\frac{1}2\lambda (x_B')^2-\frac{1}2\lambda x^2-\frac{1}2\lambda x^2\\ \\ &=\frac{1}{\lambda}\theta^2\end{align}$$

MathFail
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