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Pappus's Hexagon Theorem

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Suppose that points $A,B$ and $C$ lie on the same straight line, and $D,E$ and $F$ lie on another one. Draw line segments $AE,BD,BF,CE,AF$ and $CD$.
Let $P$ be the intersection point of $AE$ and $BD$.
Let $Q$ be the intersection point of $AF$ and $CD$.
Let $R$ be the intersection point of $BF$ and $CE$.
Then $P,Q$ and $R$ are collinear. (i.e. The red dashed line and the red solid line are collinear.)

Discussion

To prove that three points are collinear, Menelaus’s Theorem is often useful. To apply the theorem to this problem, (TL;DR;C image below) one needs to find a triangle such that exactly one of points $P,Q$ and $R$ lies on the line formed by two vertices of the triangle, but not on the segment connecting those two vertices, while each of the remaining two points lies on a segment whose extension doesn’t contain any one of points $P,Q$ and $R$.


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It is obvious that points $Q$ and $R$ divide some segments internally. However, $P$ also has such property. How can one find a segment through $P$ such that $P$ externally divides the segment?

The answer can be find out by a careful observation of the two segments that pass through $P$. Since we want a segment so that $P$ externally divides the segment, we need at least two points on a line through $P$. Moreover, those two points should be on the same side of the line relative to $P$. This led us to define a new point. Let $S$ be the intersection of $AF$ and $BD$.


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Then the light blue triangle $\triangle B\color{blue}{S}F$ is what we need.

The proof

Since I have been tired of drawing figures using TikZ and typing equations with $\rm \LaTeX$, I don’t want to illustrate or prove any one of the four cases of Co-side Theorem, on which this proof heavily depends. I refer readers to Zhang J. Z.’s work.1

From the above figure, it’s clear that what we need to show is

\[ \frac{B\color{red}{P}}{\color{red}{P}\color{blue}{S}} \cdot \frac{\color{blue}{S}\color{red}{Q}}{\color{red}{Q}F} \cdot \frac{F\color{red}{R}}{\color{red}{R}B} = -1 \]

The main idea of this proof is simple: reduce everything to what is given. Thus, points $P,Q,R$ and $S$ (i.e. red and blue points) should be eventually eliminated. The tool that we’re going to use is Co-side Theorem.

\begin{equation} \begin{split} \frac{F\color{red}{R}}{\color{red}{R}B} &= \frac{S_{\triangle CEF}}{S_{\triangle CEB}} \\ \frac{\color{blue}{S}\color{red}{Q}}{\color{red}{Q}F} &= \frac{S_{\triangle CD\color{blue}{S}}} {S_{\triangle CDF}}\\ \frac{B\color{red}{P}}{\color{red}{P}\color{blue}{S}} &= -\frac{S_{\triangle AEB}}{S_{\triangle AE\color{blue}{S}}} \end{split} \end{equation}

On RHS of (1), one can see that the red points are gone, and we still need to remove the $S$. Moreover, dealing with length of segments is easier than their area because the former and the later are one-dimensional and two-dimensional quantities respectively. Therefore, we rearrange the area of the triangles to get

\begin{equation} \begin{split} \frac{B\color{red}{P}}{\color{red}{P}\color{blue}{S}} \cdot \frac{\color{blue}{S}\color{red}{Q}}{\color{red}{Q}F} \cdot \frac{F\color{red}{R}}{\color{red}{R}B} &= -\frac{S_{\triangle AEB}}{S_{\triangle AE\color{blue}{S}}} \cdot \frac{S_{\triangle CD\color{blue}{S}}}{S_{\triangle CDF}} \cdot \frac{S_{\triangle CEF}}{S_{\triangle CEB}} \\ &= -\frac{S_{\triangle AEB}}{S_{\triangle CEB}} \cdot \frac{S_{\triangle CEF}}{S_{\triangle CDF}} \cdot \frac{S_{\triangle CD\color{blue}{S}}} {S_{\triangle AE\color{blue}{S}}} \\ &= -\frac{AB}{CB} \cdot -\frac{EF}{DF} \cdot \frac{S_{\triangle CD\color{blue}{S}}} {S_{\triangle AE\color{blue}{S}}} \end{split} \end{equation}

However, those two triangles in (2) have no common side. Thus, we need to work out the area of the two coloured triangles in the figure below. (i.e. $\triangle CD\color{blue}{S}$ and $\triangle AE\color{blue}{S}$)


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From equation (2), observe that $AB/CB$ and $EF/DF$ are independent from each other. Then, we need to relate the area of those two colored triangles to the ratios of segment lengths in (2).

I still use Co-side Theorem since I can’t think of another theorem to eliminate S. In order to use that theorem, one has to find another triangle and compare it with one of those two coloured triangles. I’ve looked at the picture and the symbols to find out some clues. Since the letter $C$ is present in both $AB/CB$ and $\triangle CD\color{blue}{S}$, $C$ can’t be the “common side” of $\triangle CD\color{blue}{S}$ and the triangle that we want. Looking at $\triangle CD\color{blue}{S}$ again, it’s obvious that $D\color{blue}{S}$ is the “common side” because $D$, $S$ and $B$ are collinear.

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\begin{equation} \begin{split} \frac{S_{\triangle CD\color{blue}{S}}} {S_{\triangle AD\color{blue}{S}}} &= \frac{CB}{BA} \text{, or} \\ S_{\triangle CD\color{blue}{S}} &= \frac{CB}{BA} \cdot S_{\triangle AD\color{blue}{S}} \end{split} \end{equation}

The $CB/BA$ in (3) cancels the $AB/CB$ in (2). Therefore, we’re almost done!

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\begin{equation} \begin{split} \frac{S_{\triangle AE\color{blue}{S}}} {S_{\triangle AD\color{blue}{S}}} &= -\frac{EF}{FD} \text{, or}\\ S_{\triangle AE\color{blue}{S}} &= -\frac{EF}{FD} \cdot S_{\triangle AD\color{blue}{S}} \end{split} \end{equation}

Substituting (3) and (4) into (2), we get

\[ \begin{split} \frac{B\color{red}{P}}{\color{red}{P}\color{blue}{S}} \cdot \frac{\color{blue}{S}\color{red}{Q}}{\color{red}{Q}F} \cdot \frac{F\color{red}{R}}{\color{red}{R}B} &= -\frac{AB}{CB} \cdot -\frac{EF}{DF} \cdot \frac{\frac{CB}{BA} \cdot S_{\triangle AD\color{blue}{S}}} {-\frac{EF}{FD} \cdot S_{\triangle AD\color{blue}{S}}} \\ &= -1 \end{split} \]

Hence, by Menelaus’s Theorem, we can conclude that $P,Q$ and $R$ are collinear, and prove Pappus’s Hexagon Theorem


  1. Chou, S. C., Gao, X. S., & Zhang, J. Z. (1993, June). Automated production of traditional proofs for constructive geometry theorems. In Logic in Computer Science, 1993. LICS’93., Proceedings of Eighth Annual IEEE Symposium on (pp. 48-56). IEEE. (URL)

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