# Synthetic Geometry

the locus of points equidistant from two or more skew lines

Given two skew lines r and s, the condition on equality of distances of a point from them clearly gives a second degree relation so that the locus is a quadric surface. Let the line t perpendicular to r and s and meeting both meet them in R and S. Let M be their middle point. Let’s draw the plane $\pi$ through M parallel to both r and s. Let ${r}_{1}$ and ${s}_{1}$ be the lines in $\pi$ parallel respectively to r and s, ${b}_{1}$ and ${b}_{2}$ the two bisectors of ${r}_{1}$ and ${s}_{1}$. It is easy to see that this two lines are on the locus; their points have the same ”horyzontal” distance (along the direction of $\pi$) and the same ”vertical” distance from them (that is along the orthogonal direction t). So $\pi$ intersects  Q along these real lines and then it is its tangent plane at M and Q is hyperbolic. The line t=RS is perpemdicular to $\pi$, let ${T}_{\infty}$ its point at infinite. A point in the plane ${b}_{1}t$ has equal horizontal distance from r and s but unequal vertical distance. But moving the point in the plane ${b}_{1}t$ upwards in the vertical direction the ratio of the two vertical distances tends to 1. Then the line at infinite ${c}_{1}$ of the plane ${b}_{1}t$ is on Q, and similarly the line ${c}_{2}$ of intersection of the plane ${b}_{2}t$ wiyh the plane at infinite. Q is then a paraboloid and precisely, being ${c}_{1}$ and ${c}_{2}$ perpendicular, an equilateral paraboloid. M is the vertex of Q because its tangent plane $\pi$ is perpendicular to t that passes through ${T}_{\infty}$ point of intersection of the two lines at infinite of Q. The axes of Q is then t. We can build the locus Q in another manner. Let ${C(\alpha)}_{1}$ and ${C(\alpha)}_{2}$ be the two cylinders with axes r and s and ray  $\alpha$. Fixing $\alpha$, the quartic curve that constitutes their intersection is clearly formed of points equidistant from r and s (at distance $\alpha$) and so lies on Q. So we have a family of quartics parametrized by $\alpha$ on Q. They are all of type (2,2) because they lie on Q and , for example, on ${C(\alpha)}_{1}$ and being of degree four constitute their complete intersection. Obviously this family of quartics doesn’t exaust the eight dimensional system of quartics C(2,2). but nevertheless they enjoy the following properties: 1) for every point passes exactly one of them. Infact a point on Q has a precise equal distance from r and s and so determines $\alpha$ uniquely. 2) two different such curves are disgioint having different values $\alpha$ and hence different distance from r and s. If now we adjoin a third line l we can construct the quadric ${Q}_{1}$ between it and s, locus of points equidistant from them. If we intersect it with Q we obtain the locus of points equidistant from the three straight lines r, s, l: it is clearly a quartic C(2,2). Adjoining a fourth line m and constructing ${Q}_{2}$ between it and l and intersecting this one with the preceeding curve we obtain, in general, eight points locus of points equidistant from four general straight lines. Now Q is a equilateral iperbolic paraboloid and hence is ruled. To each straight line p on it we can associate a ruled hyperboloid as follows: from a point P on p draw the perpendicular PH to r and PK to s. As P moves on p, H and K are in biunivocal correspondence and are parametrized by the straight line p. So between r and s there’s a proiectivity that sends H to K. This proiectivity is a similitude since the two points at infinite correspond. Infact as P moves towards infinite the distances of HR and KS grow indefnitely. The lines joining H and K form, then, a ruled hyperboloid ${Q}_{p}$ that corresponds to p. For every p contained in Q the corresponding  ${Q}_{p}$ contains r and s. To every point $P\in p$ corresponds the line of a system of ${Q}_{p}$ that meets r and s in H and K in such a way that the triangle HKP is isoscele on the base HK. Every ${Q}_{p}$ contains r, s, and the line at infinite ${R}_{\infty}$${S}_{\infty}$, Any two ${Q}_{p\alpha}$ and ${Q}_{p\beta}$ intersect in a quadrilater constituted by r, s,${R}_{\infty}$${S}_{\infty}$ and a fourth variable line meeting r and s. In case $p={b}_{1}$ the corresponding hyperboloid ${Q}_{b1}$ contains the straight line t. In fact $M\in {b}_{1}$ and the foots of the perpendicular from M to r and s are R and S. So t=SR belongs to ${Q}_{b1}$ together with r, s, ${R}_{\infty}$${S}_{\infty}$.  But this works exactly for ${Q}_{b2}$ too. Then ${Q}_{b1}$ and ${Q}_{b2}$ intersect in the skew quadrilater t, r, s, ${R}_{\infty}$${S}_{\infty}$ and then touch each other at R, S, ${R}_{\infty}$, ${S}_{\infty}$. But two other important quadrics pass through this quadrilater. The first is the locus of Straight lines meeting r and s and perpendicular to r. They fill the ruled quadric ${Q}_{r}$ of lines meeting r, s, and the line in the plane at infinite perpendicular to r. Obviously it contains r and s and RS because this is ortogonal to both and meets both. Similarly for the locus ${Q}_{s}$ of lines meeting r and s and perpendicular to s. These two quadrics meet then in the lines r, s, t and consequently in a fourth line.  This line x must belong to the same system of t to compose a quadrilater. But it cannot meet both r and s in two points at finite X and Y distance elsewhere XY would be a second common perpendicular to r and s and the r and s would be parallel lines. x cannot meet r in X at finite distance and s at ${S}_{\infty}$ because ${S}_{\infty}$${R}_{\infty}$ is equal to the angle between r and s not right if r and s are in general position. If r and s would be orthogonal then X= R and consequently Y=S and we would find t again. Then this fourth line must be the line at infinite ${R}_{\infty}$${S}_{\infty}$. Then all ${Q}_{r}$,${Q}_{s}$,${Q}_{b1}$,${Q}_{b2}$ pass through this quadrilater and form a one dimensional linear system. Fixing the line p as above, also the lines PH fill a locus. It is obviously the quadric locus ${Q'}_{p}$ of lines meeting p and r and perpendicular to r. We have just seen that it contains r, p, and the line at infinite ${R}_{\infty}$ and ${P}_{\infty}$ and consequently a fourth line that is the common perpendicular to r and p. Similarly the lines PK fill a quadric locus ${Q''}_{p}$ containig p, s, ${S}_{\infty}$ and ${P}_{\infty}$ and the common perpendicular to p and s. ${Q'}_{p}$ and ${Q''}_{p}$ have the line p in common. But, if $p\neq b1,b2$, they can’t have any other lines in common. In fact if such a line would meet r and s at finite distance then it would be their common perpendicular RS. Then p would meet RS in its middle point M, and so could be only ${Q}_{b1}$ or ${Q}_{b2}$. Such a line can’t have H at finite distance end K at infinite because they are related to the distance of P on p. Then only the possibility of a line at infinity remains. But ${Q'}_{p}$ has at infinite the line ${r}_{\infty}$ perpendicular to r and ${P}_{\infty}{R}_{\infty}$, while ${Q''}_{p}$ has at infinite the line ${s}_{\infty}$ perpendicular to s and ${P}_{\infty}{S}_{\infty}$. But ${r}_{\infty}\neq{s}_{\infty}$ since r and s aren’t parallel. So the only possibility is that ${P}_{\infty}{S}_{\infty}={P}_{\infty}{S}_{\infty}$, hence ${P}_{\infty}\in {R}_{\infty}{S}_{\infty}$. Then ${P}_{\infty}\in {R}_{\infty}{S}_{\infty}\cap Q$. But also ${B}_{1\infty},{B}_{2\infty}\in {R}_{\infty}{S}_{\infty}\cap Q$ and then, if these three points are distinct, the line ${P}_{\infty}{S}_{\infty}$ lies on Q because it has three points of intersection qith it. But this is a contradiction because we know that $Q\cap {\pi}_{\infty}={b}_{1\infty}\cup {b}_{2\infty}$. If ${P}_{\infty}={B}_{1\infty}$ then p is the unique line ${b'}_{1}$ on Q passing through ${B}_{1\infty}$ other than ${b}_{1}$. And similarly for  ${B}_{2\infty}$. Then ${Q'}_{p}\cap {Q''}_{p}$ is given by the line p and a skew cubic curve unless p is ${b}_{1}$ or ${b}_{2}$ or ${b'}_{1}$ or ${b'}_{2}$.

Locus of points from which the proiections of two given skew lines on a given plane are perpendicular:

Let call a and b the two given skew lines and $\pi$ the given plane.$A=a\cap\pi$,$B=b\cap\pi$. From a point P let’s draw the planes <P,a> and <P,b> and we want that their intersections with $\pi$ be two perpendicular lines of $\pi$. The locus is a ruled surface: infact if P is on the locus every point Q on the line <P,a>$\cap$ <P,b> does work too (<Q,a>$\equiv$<P.a> and <Q.b>$\equiv$<P.b>…)!A plane <P,a> through a intersects $\pi$ in a’. This line is perpendicular to a unique line b’ through B. The correspondence <P,a> $\rightarrow$ <b,b’> is (1,1) between the two fasci of planes of axes a and b. The set of points of intersections of two corresponding planes (straight lines) is a 1+1 dimensional surface that is a quadric. We have again that it’s ruled. From the construction we see that the trace over $\pi$ of this quadric is given by points of intersections of perpendicular lines one through A and the other through B, that is a circle of diameter AB. The quadric is then a one falda hyperboloid.

Locus of points from which the two proiections of two given skew lines over a given plane are parallel:

The proof is the same as the preceeding one changing perpendicularity with parallelism. Again we obtain a ruled quadric those trace over $\pi$ is given by the line AB plus the line at infinite of $\pi$: two parallel lines always intersect there!!

Solve the following problem:

Given the triangle ABC, let’s draw from the vertex A, externally the triangle, two congruent segments AD and AE respectively perpendicular to AC and AE. From A let’draw the height AH and prolongate it to the point F such that FC = EB. Demonstrate that FB=EB.

Solution:

We must put ourselves in three dimensional space: let’s “fold” AD and AE to the same point D’ on the vertical to the plane of the triangle passing through A. This happens with a rotation of 90° havin axis AC and the other with axis AB. Observe that CD’=CD (and BD’=BE). According the three perpendiculars theorem D’H$\bot$ CB. But FH$\bot$CB too and then D’,H,F lie in the same plane perpendicular to BC in H. The right triangles CHD’ and CHF are equal because they have CD’=CD=CF and CH in common. Then D’H=HF. Then a rotation having for axix BC brings F to D’. But in its movement this rotation brings FB to D’B and so FB=D’B.

References

Luigi Campedelli : Lezioni di Geometria, Cedam vol I and II

Esercitazioni complementari du Geometria

M. Stoka, V. Pipitone: Esercizi e problemi di Geometria, Cedam

Conforto: Superfici razionali

M.C. Beltrametti, E. Carletti, D. Gallarati, G.M. Bragadin: Letture su curve, superficie e varietà proiettive speciali, Boringnieri, http://www.dima.unige.it/beltrame/libro.pdf
Lorenzo Roi     Approfondimenti di Geometria     http://www.lorenzoroi.net/geometria/Circonferenze.html