# 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 through M parallel to both r and s. Let and be the lines in parallel respectively to r and s, and the two bisectors of and . It is easy to see that this two lines are on the locus; their points have the same ”horyzontal” distance (along the direction of ) and the same ”vertical” distance from them (that is along the orthogonal direction t). So 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 , let its point at infinite. A point in the plane has equal horizontal distance from r and s but unequal vertical distance. But moving the point in the plane upwards in the vertical direction the ratio of the two vertical distances tends to 1. Then the line at infinite of the plane is on Q, and similarly the line of intersection of the plane wiyh the plane at infinite. Q is then a paraboloid and precisely, being and perpendicular, an equilateral paraboloid. M is the vertex of Q because its tangent plane is perpendicular to t that passes through 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 and be the two cylinders with axes r and s and ray . Fixing , the quartic curve that constitutes their intersection is clearly formed of points equidistant from r and s (at distance ) and so lies on Q. So we have a family of quartics parametrized by on Q. They are all of type (2,2) because they lie on Q and , for example, on 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 uniquely. 2) two different such curves are disgioint having different values and hence different distance from r and s. If now we adjoin a third line l we can construct the quadric 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 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 that corresponds to p. For every p contained in Q the corresponding contains r and s. To every point corresponds the line of a system of that meets r and s in H and K in such a way that the triangle HKP is isoscele on the base HK. Every contains r, s, and the line at infinite , Any two and intersect in a quadrilater constituted by r, s, and a fourth variable line meeting r and s. In case the corresponding hyperboloid contains the straight line t. In fact and the foots of the perpendicular from M to r and s are R and S. So t=SR belongs to together with r, s, . But this works exactly for too. Then and intersect in the skew quadrilater t, r, s, and then touch each other at R, S, , . 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 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 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 because X 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 . Then all ,,, 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 of lines meeting p and r and perpendicular to r. We have just seen that it contains r, p, and the line at infinite and and consequently a fourth line that is the common perpendicular to r and p. Similarly the lines PK fill a quadric locus containig p, s, and and the common perpendicular to p and s. and have the line p in common. But, if , 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 or . 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 has at infinite the line perpendicular to r and , while has at infinite the line perpendicular to s and . But since r and s aren’t parallel. So the only possibility is that , hence . Then . But also and then, if these three points are distinct, the line lies on Q because it has three points of intersection qith it. But this is a contradiction because we know that . If then p is the unique line on Q passing through other than . And similarly for . Then is given by the line p and a skew cubic curve unless p is or or or .

**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 the given plane.,. From a point P let’s draw the planes <P,a> and <P,b> and we want that their intersections with be two perpendicular lines of . The locus is a ruled surface: infact if P is on the locus every point Q on the line <P,a> <P,b> does work too (<Q,a><P.a> and <Q.b><P.b>…)!A plane <P,a> through a intersects in a’. This line is perpendicular to a unique line b’ through B. The correspondence <P,a> <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 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 is given by the line AB plus the line at infinite of : 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 CB. But FHCB 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

Il Prof. Vitali mi ha preparato per l’esame di Psicometria (Psicologia). Preparatissimo, gentile e disponibile ha fugato tutti i miei dubbi e mi ha presentato la statistica in un modo.. addirittura piacevole! Grazie Prof!