Section 3. Basic Constructions

Constr. 3.1. (Of the Polar.) Given a point P distinct from O. Construct its inverse, its polar, and in case P is outside K, the two tangents from P to K.
Soln. Case 1: P is inside or outside K: Dia. 3.1. Draw line OP cutting K at C,D. Draw another line thru P cutting K at A,B. Draw lines AC, BD, intersecting at F. Draw lines AD, BC, intersecting at E. Draw line EF cutting line CD at P'. Then line EF is the polar, P' the inverse of P. If EF cuts K at G,H, then PG, PH are tangents.

Case 2: P is on K. Pick another point Q on OP, outside K. Repeat case 1 to construct the polar of Q cutting K at G,H and OQ orthogonally at Q'. Since Q' is the midpoint of GH, we can use Constr. 1.2 to construct line PS || GH. Then PS is a tangent to K.

Proof. Since CD is a diameter, E is the orthocenter of CDF so that EF is orthogonal to CD. Also as in Constr 1.1, P' is the inverse of P. Therefore line EF is the polar of P. If line PH cut K again at another point H', then H,H',P,H should be a harm quad, which is algebraically impossible unless H = H'. Hence PH is a tangent. QED.
Constr. 3.2. (Of the Pole of a line.) Given a line L not thru O. Construct its pole, and the perpendicular from O to L.
Soln. Pick 2 points P,Q on L. then we make use of the fact that the pole L' of L is the intersection of the polars of P and Q.
Case 1: L does not cut K. (11 lines. Dia. 3.2;) With the same notation as Constr. 3.1, we pick any P on L. Then we repeat Constr. 3.1 to draw the points A,B,C,D,E,F, getting line EF as the polar of P. We could pick Q as any other point on L and draw 7 more lines to get the polar of Q. But we now try to save some of those lines, but choosing Q to be the point where L crosses line EF, the polar of P. Let the line EF cut K at G and H. Then we know the polar of Q must go thru P. Next, we choose one of the 4 points A,B,C,D. In the diagram we chose B. Let line QB cut K again at J. Then the lines HB and GJ cross at some point I which also lies on the polar of Q. Therefore the polar of Q is simply line PI. The pole L' of L is therefore the point where lines DF and PI cross. Case 2: L cuts K at 2 points A,B. (9 lines. Dia.3.3) Pick a point P on L outside or inside K. Let another line thru P cut K at C,D. Let lines AD cut BC at E, CA cut BD at F, so that line EF is the polar of P. Let line EF cut L at Q and cut K at G,H. Let lines BG and HA cut at I. Hence the polar of Q is line PI, which then cut line EF at the pole L' of L.

In both cases, OL' will then cut L orthogonally at some point N which is the inverse of L'. QED

Constr. 3.3. (Mid-point of a chord.) Given a chord AB of K. Construct the mid-point M of AB, as well as the rectangle ABCD circumscribed by K.
Soln. Elegant. No poles and polars. Only 9 lines. (Dia.3.4.) The problem is trivial if AB goes thru O. So assume otherwise. Draw diameters AC and BD. The idea is to draw ||gram ADBE, whose diagonals will intersect at M. Pick any point P on line DA outside K. Join PO, PB. Let PO and AB cross at Q. Draw line DQ crossing PB at R. Draw line AR, which will be || DB. Let lines RA, BC cross at E. Then ADBE is a ||gram, and so line DE crosses AB at mid-point M. QED
Constr. 3.4. Parallel line. Given a line L, and a point P not on L. Construct 3 equally spaced points on L. Construct a line L' thru P || L.
Soln. If L cut K, we can use Constr 3.3 to construct the 3 points. Otherwise use 3.2 to construct the pole N of L inside K, then let ON cut L at N'. Use 3.1 to construct the polar of N' which is a chord AB with N as midpoints. Then lines OA, ON, OB will cut L at 3 equally spaced points A', B', N'.

From these 3 equally spaced points on L we can draw a line L' thru P || L, using Constr 1.2. QED

Note: Admittedly, the above solution to Constr. 3.4 is perhaps achievable with fewer arcs. Readers please contribute ideas for that. Given a line L not through O, cutting K at 2 points A,B, the faster way of drawing line L' thru O || L is to draw diameter BD, then the mid-point M of AD, then line OM is L'.

Constr. 3.5. (Perpendicular from a point P to a line L.) Given a line L, and a point P on or not on L. Draw line through P perp. to L.
Soln. There are many cases:
Case 1: L goes thru O, P on L, P not on K or O. Draw inverse P' of P, then draw polar of P', which will go through P and perpendicular to L.
Case 2: L goes thru O, P on L, and on K or on O. Pick any point Q on line L outside K. Draw polar of Q crossing K and L at 3 equally spaced points. With that then use Constr. 1.2 to draw line thru P || the polar of Q.
Case 3: L goes thru O, P not on L, nor on K. Let L cut K at A,B. Let lines PA, PB will cut K again at C, D. Let lines AD, BC cross at E. Since P is the orthocenter of triangle ABE, line EP has to be perpendicular to L.
Case 4: L goes thru O, P on K but not on L. Let L cut K at A,B. Using Constr. 1.2, draw chord PQ || L. Draw diameters PR and QS. Then PS is perpendicular to L.
Case 5: O is not on L. A tedius way is to construct a line L' thru O || L, then draw line thru P || L' using cases 1 to 4 above. There are many special solutions for the cases when L cuts K and if P is on or not on L, on or not on K, or P = O. QED

Note: Up to now, we have been able to draw parallel lines and perpendiculars.

Constr. 3.6. (Shift a segment || O). Given a line segment AB. Draw vector OM || and equal to vector AB.
Soln. Case 1: O not on line AB. Join OA. Draw line OM || AB and line BM || OA, crossing at M.
Case 2: O is on line AB. Draw any line L' || L. Pick a point C on L', then draw ||gram ABCD. Next draw line DM || CO, cutting line AB at M. QED

Constr. 3.7. (Bisect any angle.) Draw angle bisector of any given angle on the plane.
Soln. Draw angle POQ with OP || BA and OQ || BC. Draw the mid-point N of BC, then line ON. Draw line BD || ON. Then BD bisects angle ABC.

Constr. 3.8. (Rotate any angle.) Given an angle AOB, with A,B on K, Given another point C on K. Draw D on K such that angle AOB = angle COD.
Soln. Draw chord BC. Draw line thru A || BC, cutting K at D. QED

Constr. 3.9. (Copy a triangle.) Given a triangle ABC and a point A' on a line L. Construct triangle A'B'C' congruent to ABC, with B on L. Dia. 3.9.
Soln. Shift triangle ABC parallelly to OPQ, by drawing parallelograms CAOQ, BAOP. Draw L' thru O || L, cutting K at X Let OP,OQ cut K at P',Q'. Rotate angle P'OX to Q'OY for some Y on K. Draw line PP'' || P'X cutting OX at P''. Similarly draw line QQ'' || Q'Y cutting OY at Q''. Then triangle POQ is congruent to P''OQ'', and OP'' is on L' parallel to L. Draw ||grams A'OP''B', A'OQ''C'. QED

Constr. 3.10. (4th proportional). Given 3 segments with lengths a, b, c. Construct a segment AB whose length is a b / c.
Soln. Shift the segments to the origin so that we have OP, OQ, OR with lengths a, b, c. Rotate, if necessary so that O,P,R are collinear, and OQ is a different line. Draw line RS || PQ, cutting OQ at S. Then from similar triangles, OS = a b/c. QED

Cor. 3.10.1. Given circle K having radius 1 unit, Given 2 segments with lengths a, b, we can construct segments with lengths a + b, a - b, a b, and a / b.
Soln. The cases for a b and a / b can be done with 3 segments one of which is a radius of K. For addition, shift the segments so they are parallel. Then if segments AB, CD haves lengths a, b, and are parallel, we can draw ||gram CDPO, then ||grams OPQB and OPBR, so that AQ will be a + b and AR will be b - a. QED

Constr. 3.11. (Triangles given SAS, AAS, RHS, and SSS.) Given 3 measurements of a triangle, either 2 sides and an included angle, or 2 angles and a side, or a right angle and 2 sides, or 3 sides, construct a triangle with such measurements.
Soln. Case SAS: This is easy. Shift the angle and the given segments to O. Rotate the segments about O to lie on the 2 arms of the given angle at O.

Case AAS: Draw a diameter AOB on K. Draw radii OC, OD on same sides of AOB so that angles AOC, BOD = the given angles. We have 3 subcases:
Subcase 1: If the given side is between the angles, then first draw a point B' on OB so that OB' has the given length. Next shift the side OD in parallel to become a line B'E cutting line OC at E. Then triangle AB'E is the final triangle.
Subcase 2: If the given side should face angle BOD, then first draw a point C' on OC so that OC' has the given length. Then shift the side OD in parallel to become a line B'C' cutting OB at B'. Then OB'C' is the final triangle.
Subcase 3: If the given side should face angle AOC, then it is similar.

Case RHS: Let h, a be the lengths of the hypotenuse and one side. Draw a point A in K so that OA = a / h. Draw chord BAC orthogonal to OA. Draw point A' on OA so that OA' = a. Draw line A'B' || AB, cutting OB at B'. Then OA'B' is the final triangle.

Case SSS: Dia. 3.10. This case is more difficult. Let a, b, c be given lengths of the triangle, with a < b < c. Draw segments OA, OB with lengths a' = a/c and b' = b/c. Let DOAC be a diameter. Our goal is to get a chord EAF so that AE = OB, then OAE is similar to the final triangle. Luckily, the lengths AF . AE = AC . AD, so that the length AF is known to be (1-a') (1+a') / b', so that length of EF is x = b' + (1-a') (1+a') / b'. Draw a point P in K so that OP = x / 2, then draw chord QPR orthogonal to OP. Then draw a chord HK || CD at distance QP from line CD. Then HK has length = EF. Draw the point E' on HK so that E'H = b'. Then OE' = OA. Rotate triangle OHE' thru angle E'OA to coincide with OEA. Finally, magnify OA to OA' where OA' = a, magnify OE to OE'' via the same ratio, and then OA'E'' is the final triangle. QED