Third - order Intercept Point (IP3) Equation derivation

Could anyone elaborate on how to derive the following IP3 equation in the screenshot ?

Note: Extracted from https://www.keysight.com/upload/cmc_...ary.pdf#page=4

IP3out is a fictious output level where 3rd order products reach to the fundamental's level. 3rd order products rise 3 times faster than fundamentals(as the slops indicate) while you increase the input level.
D distance between two runners having velocities v and 3v closes in D/2v sec. In that time, the fundamental having velocity v takes D/2. As a result, output level(named as IP3out) becomes initial level plus D/2 as the formula states.

S. also this paper by Ken Kundert (p. 4).

Here's my graphical derivation for the example from this tutorial:

Hi,
You just have to use the equations of a straight line knowing one point and it's slope.

Find out the equations of the line that has slope one, and the equation of the line that has slope 3.

Use the fact that when the input is IIP3, the value of both lines is OIP3.

Why divide delta_P by (N-1) instead of just N ?

Just be cautious as an assumption of linearity is made. If the amplifier is not in the linear region actual results will be different.

Division factor is the difference of the slopes, s. my diagram above.