Students will learn how to solve equations with complex numbers by equating real and imaginary parts. Here's a lesson.

Conclude by giving your students these challenges:

- You Never Get a Six by NRICH
- 2020 Math Kangaroo Levels 5-6 Problem #24 by STEM4all
- 2018 AMC 8, Problem 12

Give your students the following three-part challenge:

Part 1: Suppose \(D\) occurs \(2/3\) along \(\overline{AB}.\)

How are the areas of \(\triangle ADC\) and \(\triangle ABC\) related? After you've found the solution, generalize from \(2/3\) to any rational.

Solution: \(\triangle ADC\) has \(2/3\) the area of \(\triangle ABC.\) That's because the base of \(\triangle ADC\) is \(2/3\) the base of \(\triangle ABC.\)

Part 2: Suppose \(D\) occurs \(2/3\) along \(\overline{AB},\) \(E\) occurs \(2/3\) along \(\overline{BC},\) and \(F\) occurs \(2/3\) along \(\overline{CA}.\)

How are the areas of \(\triangle DEF\) and \(\triangle ABC\) related? After you've found the solution, generalize from \(2/3\) to any rational.

Solution: Suppose \(\triangle ABC\) has area \(1.\) Then \(\triangle ADC\) has area \(2/3,\) as we've already established. Next, notice that \(\triangle CFD\) has \(2/3\) the area of triangle \(ADC,\) by the same reasoning. So \(\triangle CFD\) has area \((2/3) \cdot (2/3) = 4/9.\) If we remove the area of \(\triangle CFD\) from the area of \(\triangle ADC,\) we'll get the area of triangle \(ADF.\) So the area of triangle \(ADF\) must be \(2/3 - 4/9 = 6/9 - 4/9 = 2/9.\) But we could use the exact same reasoning for \(\triangle BEF\) and \(\triangle CFE.\) Since the area of \(\triangle DEF\) is the area of \(\triangle ABC,\) minus the areas of \(\triangle ADF,\) \(\triangle BED,\) and \(\triangle CFE,\) the area of \(\triangle DEF\) must be

$$1 - \dfrac{3 \cdot 2}{9} = \dfrac{9}{9} - \dfrac{6}{9} = \dfrac{3}{9} = \dfrac{1}{3}$$Thus, in summary, the area of \(\triangle DEF\) is \(1/3\) the area of \(\triangle ABC.\) For another elegant solution, go here and read the solution by Miranda and Sadaf from Greenacre Public School in Australia. It's very clever, and thus, it's unlikely your students will find this solution on their own. Even if they don't find it, you should still demonstrate it.

Part 3: Suppose we continue this pattern. That is, going counterclockwise around each triangle, we make a point \(2/3\) along each segment, and connect each point to form a new triangle. Suppose we do this \(n\) times, and label the innermost triangle \(\triangle DEF.\)

How are the areas of \(\triangle DEF\) and \(\triangle ABC\) related? After you've found the solution, generalize from \(2/3\) to any rational.

Solution: The first inner triangle will be \(1/3\) the area of \(\triangle ABC.\) By the same reasoning, the second inner triangle will be \(1/3\) the area of the first inner triangle, that is, \((1/3)^2.\) Thus, the \(n\)-th inner triangle will be \((1/3)^n\) the area of \(\triangle ABC.\)

Note: This challenge was inspired by the NRICH challenge Triangle in a Triangle.