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### Pappus' Theorem

Today we will learn about Pappus' theorem. This theorem states that if we take three points $1$, $3$, $5$ on a line, and another three points $2$, $4$, $6$ on another line, then the three intersection points of the following line pairs $$\{12, 45\}, ~\{23, 56\}, ~\{34, 61\}$$ are collinear.

Pappus' theorem looks very similar to  Pascal's hexagon theorem. In Pascal's theorem, we have a hexagon $123456$ inscribed in a circle and the intersection points of the three pairs of opposite sides $$\{12, 45\}, ~\{23, 56\}, ~\{34, 61\}$$ of the hexagon lie on a straight line.

So instead of being inscribed in a circle as in Pascal's theorem, in Pappus' theorem, we may say that the hexagon $123456$ "inscribed" in two lines and again, the intersection points of the three pairs of opposite sides $$\{12, 45\}, ~\{23, 56\}, ~\{34, 61\}$$ of the hexagon also lie on a straight line.

Different configurations of Pappus' theorem

Similar to Pascal's theorem, Pappus' theorem also has many configurations. The six vertices of the hexagon can be arranged on the two lines in any particular order. With each arrangement of the vertices, we have an instance of Pappus' theorem. Below is an example:

Now, why don't you try to draw for yourself your own picture of Pappus' theorem?! This theorem is so cool, isn't it?!

The picture below illustrates another interesting aspect of Pappus' theorem, that is the roles of the hexagon's vertices and the intersection points can be interchanged. As we can see, in the middle figure, the intersection points are interchanged with the vertices $1$, $3$, $5$; and in the right-hand-side figure, the intersection points are swapped with $2$, $4$, $6$.

Note that, in all three figures above, $u$ is always the intersection point of the pair $\{12, 45\}$, $v$ is always the intersection point of the pair $\{34, 61\}$ and $w$ is always the intersection point of $\{23, 56\}$.

Proving Pappus' theorem

We will use Menelaus' theorem to prove Pappus' theorem. Menelaus' theorem is a useful tool to prove the collinearity of points. The theorem is stated as follows:

Menelaus' theorem: Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the three lines $BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ are collinear if and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$

The proof that we are about to present is very similar to the proof of Pascal' theorem that we show in the previous post. Before going into detail, let us review that proof of Pascal' theorem. Let us consider the following figure.

On this figure, we see a Pascal's hexagon $P_1 P_2 P_3 P_4 P_5 P_6$ and the three intersection points $M_1$, $M_2$, $M_3$. To prove that $M_1$, $M_2$, $M_3$ lies on a straight line, we apply Menelaus' theorem for the triangle $ABC$. Since $M_1$ belongs to the line $BC$, $M_2$ belongs to the line $CA$ and $M_3$ belongs to the line $AB$, to prove that they are collinear, we show that $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{M_3 A}}{\vec{M_3 B}} = 1.$$

Now we do the same to prove Pappus' theorem. We will use Menelaus' theorem for the triangle $ABC$. The triangle $ABC$ is determined exactly in the same way as in the proof of Pascal's theorem: $A$ is the intersection of $P_4 P_5$ and $P_6 P_1$, $B$ is the intersection of $P_2 P_3$ and $P_4 P_5$, and $C$ is the intersection of $P_2 P_3$ and $P_6 P_1$.

Let us  write down the proof.

Applying Menelaus' theorem for the triangle $ABC$ with the following triples of collinear points $$\{M_1, P_6, P_5\}, ~~\{M_2, P_4, P_3\}, ~~\{M_3, P_2, P_1\}, ~~\{P_1, P_3, P_5\}, ~~\{P_2, P_4, P_6\},$$
we have $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{P_6 C}}{\vec{P_6 A}} \times \frac{\vec{P_5 A}}{\vec{P_5 B}} = \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{P_4 A}}{\vec{P_4 B}} \times \frac{\vec{P_3 B}}{\vec{P_3 C}} = \frac{\vec{M_3 A}}{\vec{M_3 B}} \times \frac{\vec{P_2 B}}{\vec{P_2 C}} \times \frac{\vec{P_1 C}}{\vec{P_1 A}}$$ $$=\frac{\vec{P_1 A}}{\vec{P_1 C}} \times \frac{\vec{P_3 C}}{\vec{P_3 B}} \times \frac{\vec{P_5 B}}{\vec{P_5 A}} = \frac{\vec{P_2 C}}{\vec{P_2 B}} \times \frac{\vec{P_4 B}}{\vec{P_4 A}} \times \frac{\vec{P_6 A}}{\vec{P_6 C}} = 1.$$ Multiply all the above identities, we derive $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{M_3 A}}{\vec{M_3 B}} = 1.$$ Thus, $M_1$, $M_2$, $M_3$ are collinear, and the theorem is proved.

So, once again today, by applying Menelaus' theorem in an effective way, we have proved Pappus' theorem. Pappus' theorem has a simple structure but it looks so elegant, this is why Pappus' theorem has been chosen as the logo for our Math Garden blog.

Hope to see you again in the next post.

Homework.

1. Let us call the purple points as Pappus' points, and the purple line connecting them as Pappus' line.
Now, let the six points stay fixed on the two lines. By different labeling $1,2,3,4,5,6$ to the six points, the questions are
• how many Pappus' lines can be generated?
• how many Pappus' points can be generated?
• each Pappus' point lies on how many Pappus' lines?
• each Pappus' line contains how many Pappus' points?

2. Instead of applying Menelaus' theorem for the triangle $ABC$ as in the above proof, choose $X$ as the intersection of $P_1P_2$ and $P_3P_4$, $Y$ the intersection of $P_1P_2$ and $P_5P_6$, and $Z$ the intersection of $P_3P_4$ and $P_5P_6$, prove Pappus' theorem by using Menelaus' theorem for the triangle $XYZ$.

3. Prove that if $P_2P_3$ is parallel to $P_5P_6$, and $P_3P_4$ is parallel to $P_1P_6$, then $P_1P_2$ is parallel to $P_4P_5$.