Given four points, $A$,$B$,$C$,$D$ ¿Is it posible to find $P$ and $Q$ such that they are harmonic conjugates to $A$, $B$ and $C$, $D$?

Question

This is the excercise 12 from section 4.10 in the book $\textbf{"An introduction to modern geometry"}$ from Levi S. Shively, it is also asked that one discuss different cases/posibilities. I have figured out (by testing in geogebra) that if the points are aligned in the manner $A$, $C$, $B$, $D$ it is not posible which discards the case when the initial points are pairs of harmonic conjugates between themselves.

In my continued testing I found that when the points are aligned in the manner $A$, $C$, $D$, $B$ or $A$, $B$, $C$, $D$ there exists such a pair $P$, $Q$ and it appears to be unique for every way of setting the points, what I mean is I can't simply place $P$ and build $Q$ from it, they are a unique pair but so far I have been unable to figure out how to find them given just my initial four points.

I belive finding the way of constructing this pair is the key to solving the problem but I've been thinking it for three days now and would apreciate it if anyone has an idea

Answer 1

Consider inversion with respect to an arbitrary circle with center outside the line $\ell$ on which $A, B, C, D$ lie. We denote the image of $x$ under this inversion by $x'$. Then $\ell'$ is a circle, and since inversion preserves cross-ratio, we have to find points $P, Q\in \ell$ such that the points $P', Q' \in \ell'$ are such that quadrilaterals $A'P'B'Q'$ and $C'P'D'Q'$ are harmonic. By properties of harmonic quadrilaterals, line $P'Q'$ has to pass through the common point $K'$ of tangents to $\ell'$ at $A'$, $B'$, and also pass through the common point $L'$ of tangents to $\ell'$ at $C'$, $D'$. Therefore, such points $P$ and $Q$ exist if and only if the line $K'L'$ intersects $\ell'$ at two points.

From this we can work out the construction of points $P$ and $Q$. 'Undoing' the inversion, we obtain the following construction of $P$ and $Q$:

• Choose an arbitrary point $X \notin \ell$
• Construct the circle $\omega_A$ passing through $X$ and $A$ which is tangent to $\ell$ at $A$
• Analogously, construct circles $\omega_B$, $\omega_C$, and $\omega_D$
• Find the common point $K\neq X$ of $\omega_A$ and $\omega_B$ (if you're unlucky and choose $X$ such that $\angle AXB=90^\circ$ then the circles $\omega_A$ and $\omega_B$ are tangent; if this happens you should choose another point $X$)
• Find the common point $L\neq X$ of $\omega_C$ and $\omega_D$ (there's similar problem as above, to avoid both problems just choose a point $X$ so that it does not lie on either the circle with diameter $AB$ or the circle with diameter $CD$)
• Draw the circle $\omega$ passing through $X$, $K$, and $L$. The points $P$ and $Q$ you want to construct are the common points of $\omega$ and $\ell$. If $\omega$ does not intersect $\ell$ or is tangent to $\ell$, there are no such points.

Answer 2

If you want to draw $P$ and $Q$, just base it on the construction in section 4.10. In this case, you need to draw a circle that is orthogonal to the circles $c_{AB},c_{CD}$ with diameters $AB,CD$ respectively. Find the intersection $R$ of the radical axis with the line $ABCD$. Then draw the circle $c$ with center $R$ that is orthogonal to either of $c_{AB},c_{CD}$. Then the intersections of $c$ with line $ABCD$ will be $P$ and $Q$.

• Construction of radical axis here: cut-the-knot.org.
• Construction of orthogonal circle here: mathisfun.com (draw circle centered at original point, and going through the tangent points).

Slightly more general construction at Constructing Orthogonal Circles Through 3 Points/Circles (look for ONE point and TWO circles section)