15.8: Construction of a polar

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In this section we describe a powerful trick that can be used in the constructions with ruler.

$截屏2021-02-26 上午10.41.11.png$

Assume $\Gamma$ is a circle in the plane and $P\notin \Gamma$.Draw two lines $x$ and $y$ thru $P$ that intersect $\Gamma$ at two pairs of points $X$, $X'$ and $Y$, $Y'$. Let $Z=(XY) \cap (X'Y')$ and $Z'=(XY') \cap(X'Y)$. Consider the line $p=(ZZ')$.

Claim $\PageIndex{1}$

The constructed line $p=(ZZ')$ does not depend on the choice of the lines $x$ and $y$.

Moreover, $P \leftrightarrow p$ can be extended to a duality such that any point $P$ on the circle $\Gamma$ corresponds to a line $p$ tangent to $\Gamma$ at $P$.

We will use this claim without a proof, but the proof is not hard. If $P$ lies outside of $\Gamma$, it can be done by moving $P$ to infinity keeping $\Gamma$ fixed as a set. If $P$ lies inside of $\Gamma$, it can be done by moving $P$ to the center of $\Gamma$. The existence of corresponding projective transformations follow from the idea in Exercise 16.3.1.

The line $p$ is called the polar of the point $P$ with respect to $\Gamma$.

The point $P$ is called the pole of the line $p$ with respect to $\Gamma$.

Exercise $\PageIndex{1}$

Revert the described construction. That is, given a circle $\Gamma$ and a line $p$ that is not tangent to $\Gamma$, construct a point $P$ such that the described construction for $P$ and $\Gamma$ produces the line $p$.

Hint: Suppose $p = (QR)$; denote by $q$ and $r$ the dual lines produced by the construction. Then, by Claim $\PageIndex{1}$, $P$ is the point of intersection of $q$ and $r$.

Exercise $\PageIndex{2}$

Let $p$ be the polar line of point $P$ with respect to the circle $\Gamma$. Assume that $p$ intersects $\Gamma$ at points $V$ and $W$. Show that the lines $(PV)$ and $(PW)$ are tangent to $\Gamma$.

Come up with a ruler-only construction of the tangent lines to the given circle $\Gamma$ thru the given point $P\not\in \Gamma$.

Hint: The line $v$ polar to $V$ is tangent to $\Gamma$. Since $V \in p$, by Claim $\PageIndex{1}$, we get that $P \in v$; that is, $(PV) = v$. Hence the statement follows.

Exercise $\PageIndex{3}$

Assume two concentric circles $\Gamma$ and $\Gamma'$ are given. Construct the common center of $\Gamma$ and $\Gamma'$ with a ruler only.

Hint

$截屏2021-02-26 上午10.51.37.png$

Choose a point $P$ outside of the bigger circle. Construct the lines dual to $P$ for both circles. Note that these two lines are parallel.

Assume that the lines intersect the bigger circle at two pairs of points $X, X'$ and $Y, Y'$. Set $Z = (XY) \cap (X'Y')$. Note that the line $(PZ)$ passes thru the common center.

The center is the intersection of $(PZ)$ and another line constructed the same way.

Exercise $\PageIndex{4}$

Assume a line $\ell$ and a circle $\Gamma$ with its center $O$ are given. Suppose $O\notin \ell$. Construct a perpendicular from $O$ on $\ell$ with a ruler only.

Hint: Construct polar liens to two point on $\ell$. Denote by $L$ the intersection of these two liens. Note that $\ell$ is polar to $L$ and therefore $(OL) \perp \ell$.

Claim \(\PageIndex{1}\)

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

Exercise \(\PageIndex{4}\)