Results related to Half-Turns and Pencils of Lines

Definition 1: A half turn about a point $P$ is a rotation $R_{m}R_{l}$ where the axes of the reflections $l$ and $m$ are perpendicular and intersect at $P$.

Theorem 2: If $H_{P}$ is a half-turn, then $H_{P}=R_{v}R_{u}$ for any pair of perpendicular lines $u$ and $v$ that meet at $P$.

Suppose that $H_{P}=R_{l}R_{m}$ where $l$ and $m$ are a given pair of perpendicular lines meeting at $P$. Recall that MATH. Now, we have from the preceding section that there is a line $v^{\prime }$ passing through $P$ such that MATH. Observe that
MATH
and hence MATH. It follows that $u\bot v^{\prime }$ and hence, as there can be only one perpendicular to $u$ at $P$, $v=v^{\prime }$. Thus $H_{P}=R_{v}R_{u}$.

Theorem 3: Points and half-turns are in a one-to-one correspondence.

Theorem 4: Every line passing through $P$ is a fixed line under $H_{P}.$

Suppose that $s$ is a line passing through $P$. Let $t$ be a line perpendicular to $t$ passing through $P$. Note that, as $s\bot t$, MATH. It follows that MATH. As $s$ was arbitrary line passing through $P,$ $H_{P}$ fixes every line passing through $P$.

Theorem 5: $H_{P}=R_{l}$ if and only if $P$ is a pole of $l$.

First suppose that $P$ is a pole of $l$ and let $Z$ be a point on $l$. Let MATH and let $v$ be line perpendicular to $u$ passing through $P$. Note that $v$ is perpendicular to $l$ and hence $Z$ is a pole of $v$. It follows that MATH. Now MATH. As $Z$ was an arbitrary point of $l$, $l$ is pointwise fixed by $H_{P}$.

Also note that $H_{P}$ is an orthogonal collineation not equal to the identity map, thus $H_{P}=R_{l}$.

Now suppose that $H_{P}=R_{l}$ and let $u$ and $v$ be two perpendicular lines meeting at $P$. Note that $u$ and $v$ are both fixed by $H_{P}$ and hence by $R_{l}$. It follows that $u\bot l$ or $u=l$. If $u=l$, it follows that $R_{v}$ is the identity map (a contradiction), hence we must have that $u\bot l.$ A similar argument shows that $v\bot l$. As $u$ and $v$ are distinct perpendiculars that meet at $P$ it follows that $P$ is the pole of $l$.

The proof of the next theorem is omitted as it is fairly straightforward.


Theorem 6: The only fixed lines of $H_{P}$ are

  1. all lines passing through the point $P$ together with the polar line of $P$ if the plane is elliptic.

  2. all the lines passing through $P$ if the plane is nonelliptic.

Theorem 7:The only fixed points of $H_{P}$ are:

  1. $P$ and all the points of the polar line of $P$ if the plane is elliptic.

  2. $P$ alone if the plane is not elliptic.

If the plane is elliptic, the previous result yields that $H_{P}=R_{l}$ where $P$ is the polar line of $l.$ As the only fixed points of $R_{l}$ are the points on $l$ and the pole of $l$ ( $=P),$ the result follows.

Next we observe that if there is a point $P$ such that there is a point $Q$ with the properties that $P\neq Q$ and MATH then there is a line with a pole and hence the geometry is elliptic. (We are making no assumptions about parallel lines at this point in the argument.) Let MATH and $v$ be a line perpendicular to $u$ which intersects $v$ at $P.$ Note that, MATH - the last equality comes from noting that $Q$ on $v$ implies MATH - and hence $Q$ is either a pole of $u$ or on $u.$ If $Q$ were on $u,$then we would have $u=v$, which contradicts the assumption that $u$ and $v$ are perpendicular. Hence $Q$ is the pole of $u$ and, as there is a line with a pole, the plane is elliptic.

Proof

Thus, if the plane is non-elliptic, the only fixed point of $H_{P}$ is $P.$

Theorem 8: Let $P$ be a point and $l$ be a line. Then MATH if and only if MATH.

Suppose that MATH. We wish to show that MATH. Suppose that $u$ and $v$ are pair of perpendicular lines that meet at $P$ and let $u^{\prime }$ and $v^{\prime }$ denote $R_{l}(u)$ and MATH respectively. Observe that MATH and hence meet a point $Q.$ Also note that the construction yields that MATH. We will show that MATH and hence $P^{\prime }=Q$. First recall, from project 5, that MATH. Now:
MATH
where the last equality is the hypothesis of this direction of the proof. Thus $P^{\prime }=Q$.

Next we assume that MATH and show that MATH. Let $u$ be a line perpendicular to $l$ passing through the point $P$ and let $v$ be a line perpendicular to $u$ passing through $P$. Note that MATH and that, if $v^{\prime }$ denotes MATH then $u\bot v^{\prime }$. Observe that $P^{\prime }$ is at the intersection of $u$ and $v^{\prime }$. Now note that
MATH
(Note that this version of the proof does not require two separate cases (one for $P$ on $l$ and one for $P$ not on $l$).

Definition 9: Let $M$ be a motion. The motion $M$ is called an involution or said to be involutoric if $M\neq I$ and $M^{2}=I$.

The following theorem is an immediate consequence of the preceding theorem.

Theorem 10: Let $P$ be a point and $l$ be a line. Then $P$ is on $l$ if and only if $H_{P}R_{l}$ is involutoric.

First suppose that $H_{P}R_{l}$ is involutoric. Then MATH and hence MATH (as $H_{P}H_{P}=I)$. Thus it follows from the preceding theorem that MATH and hence either $P$ is on $l$ or $P$ is the pole of $l$. But if $P$ is the pole of $l$, then $H_{P}=R_{l}$ and thus $H_{P}R_{l}=I,$ which contradicts the definition of involutoric. Consequently, $P$ must be on $l$.

Now suppose that $P$ is on $l$. Then MATH and hence MATH. It follows that MATH. As $P$ is not a pole of $l$, $H_{P}\neq R_{l}$ and hence $H_{P}R_{l}\neq I$. The $H_{P}R_{l}$ is involutoric.

Theorem 11: Let $l$ and $m$ be distinct lines belonging to a pencil of lines through a point $P$. A line $x$ belongs to the pencil of lines through $P$ if and only if $R_{x}R_{m}R_{l}$ is a reflection.

That $P$ on $x$ implies that $R_{x}R_{m}R_{l}$ is a reflection was established in class (First Theorem on Three Reflections.).

We show that if $R_{x}R_{m}R_{l}$ is a reflection, then $P$ is on $x$. Suppose that $P$ is not on $x$ and drop a perpendicular $\ v$ from $P$ to $x$; let $P^{\prime }$ denote the intersection of $x$ and $v$. Since $v,m,$ and $l$ meet at $P$, there is a line $q$ such that MATH and $P$ is on $q$. Note that MATH and that
MATH
As $R_{x}R_{m}R_{l}$ is a reflection, MATH is involutoric and thus $P^{\prime }$ is on $q$. Now note that $P$ and $P^{\prime }$ are on both $q$ and $v$ and hence $q=v$. However, since MATH, this forces $l=m$. which is a contradiction. Hence $P$ is on $x$.

Definition 12: Let $l,m$ be lines. Then the set MATH is the pencil of lines determined by $l$ and $m.$

A pencil of the first kind is a set of lines that all have one point in common.

A pencil of the second kind is the set of all lines perpendicular to a given line.

A pencil of the third kind is a pencil which is not of the first or second kind.

Theorem 13:Let $l$ and $m$ belong to a pencil of lines perpendicular to a given line $t$. A line $x$ belongs to the pencil of lines perpendicular to $t$ if and only if $R_{x}R_{m}R_{l}$ is a reflection.

This is similar to the proof of the preceding theorem - just use the corresponding results for lines with a common perpendicular.

The proof of the following theorem is a good bit of work, hence we omit it.

Theorem 14: (Join Theorem) A pencil of the first kind and a pencil of the second or third kind have one and only one line in common.

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