Constructions by Ruler Alone, plus a Parabola

4. Constructions using the parabola.

Assume K is the parabola whose focus F is known and its axis is called the x-axis. Initially y-axis is not visible.

Constr. 3. (Polar). Given a point A not on K. Construct its polar. Soln. Draw 2 lines BAC, DAE thru A with B,C,D,E on K. Draw lines BD, CE to cross at H, and lines BE, CD to cross at G. Then line HG is the polar. Proof. Use the 3-D diagram on Thm. 1 to map the parabola to a circle, and the image of HG will be a polar of the image of A over the circle. QED Remark: this method is valid for any conic K.

Cor. 3.1. (The directrix, and lines // x-axis). Now we can construct the polar of focus F, which is the directrix, cutting x-axis at right angles at some point F'. Let K cross the x-axis at O, its vertex. Then O is the mid pt of FF', hence we can construct through any point a line parallel to the x-axis.

Cor. 3.2. Draw tangents from an exterior point A' and at a point A on K. Soln. A' is exterior of K, so its tangents touch K at the 2 points where its polar cuts K. To draw the tangent line at A, first draw line L1 thru A // x-axis. Then choose a point B west of A. Draw the polar of B, which is a chord CD with its mid pt E on L1. Now we can draw tangent line at A, which is parallel to line CED.

Cor. 3.3. (This is not a theoretical problem, but a practical problem). Given a line L that seems to be tangential to K. Draw its point of contact P accurately. Soln. Let L cut the y-axis at A and x-axis at B. Let O be the vertex of K. Draw //gms ABOC, then rectangle ACDO. Line CD then cut K at P. Proof. A has OP as polar, hence the y-coord of A is half that of P. Hence A is mid pt of BP. QED

Constr. 4. (Pole). Given a line L not // x-axis, draw its pole. Soln. Pick 2 points A,B on L. Draw the polars of A and of B. Then they will intersect at the pole of L.

Constr. 5. (Parallel lines). Given a line L and a point P not on L. Draw a line PP' // L. Soln. Draw pole Q of L. Then line QQ' // x-axis cutting L at Q'. If L already cuts K at C,D, then Q' is its midpoint, so we can draw line PP' // L. Otherwise draw polar L' of Q', which will be a chord with Q as mid pt, and we can draw PP' // L' similarly. Remark: Now we can translate any segment parallelly to anywhere in the plane, because we can draw //gms (parallelograms).

Constr. 6. (Perpendiculars). Given a line L and a point P, draw line PP' perp. to L. Soln. Draw the pole Q of L, then line L2 through Q // x-axis, cutting K at R. Draw focal chord RS, then the polar L3 of some point S' on ray RS exterior to K. L3 will be a chord of K. So we can finally draw line PP' through P // L3.

Constr. 7. (Bisecting and doubling a segment). Given 2 points A,B. Draw its mid pt C and the pt D so that B is the mid pt of AD. Soln. Draw any //gm ABB'A', then //gm A'B'DB, AB'BA". Then A"B' will bisect AB at C. Remark. Now we can draw perp. bisectors, and mirror images of a point or line in another line.

Constr. 8. (Bisecting an angle). Given an angle BAC. Draw its angle bisector AG. Soln. Case 1: AC // x-axis: Draw line FP // AB, cutting K at P, then line PP' // x-axis, then the polar L of any point Q on PP'. L will // the tangent at P'. Draw the line AG // L.
Case 2: AB, AC not // x-axis: Draw AB" // x-axis. Draw angle bisector AD' of angle BAB", then line BB' perp. to AD' cutting AD' at D and AB" at B'. Let AC cut BB' at E. Draw E' so that D is mid pt of EE'. Draw bisector AG' of angle B'AE' cutting BB' at G' Draw G so that D is mid pt of GG'. Then AG bisects angle BAC.

Constr. 9. (Copy a segment to a non parallel line). Given a triangle ABC. Draw point D on ray AC so that AD = AB. Soln. Draw angle bisector AE of angle BAC, then mirror image D of B across AE.

Constr. 10. (Copying a triangle). Given a triangle ABC and a line DE'. Draw a triangle DEG congruent to ABC so that E lies on DE'. Soln. Draw line AH // DE'. Draw angle bisector AI of angle BAH. Draw mirror images B',C' of B,C across AI. Draw mirror image C" of C' across AH. Finally, translate triangle AB'C" to DEG.

Constr. 11. (4th proportional). Given 3 segments of length x, y, z. Construct a segment with length x y / z. Soln. The usual high-school construction using 2 parallel lines.

Constr. 12. (Geometric mean). Given 2 segments of length x, y. Construct a segment whose length is their geometric mean. Soln. Make a segment of length 4 p, e.g. the latus rectum. Then use Constr. 11 to make a segment OX on the x-axis with length x y / 4 p. Then draw a line XY // y-axis cutting K at Y. Then XY has length sqrt ( x y ).

Constr. 13. (Triangle by RHS). Given 2 segments of lengths h, s, with h > s. Draw a right triangle with length of hypotenuse = h and that of one side = s. Soln. The third side has length t = sqrt(h^2 - s^2), which is the G.M. of (h + s) and (h - s). So draw segments of lengths h + s and h - s, then draw a segment with length t = their geometric mean. Draw a segment OY on y-axis with length t and a segment OX on x-axis with length s.

Constr. 14. (Triangle by SSS). Given 3 segments of lengths a, b, c. Draw a triangle with these as lengths of sides. Soln. Call the triangle ABC with A facing the side with length a, etc. The length of altitude = 2 * area / a = 2 sqrt( s (s - a) (s - b) (s - c)) / a, where s = (a + b + c)/2. So first draw a segment with length s, then those with lengths s - a, s - b, s - c. Next draw 2 segments of length x = G.M. of s and s - a, and y = G.M. of s - b and s - c. Finally, draw the 4th proportional x y / a, multiplied by 2.

5. Conclusion. It is now obvious that we can construct any point, and any line that can be constructed with straight edge and compass.

6. Exercises.
1. Find another way to do Constr. 9 without drawing right angles. Make use of the fact that for any point P on K, PF = PN where N is the intersection of the directrix with the line through P // x-axis.
2. Given a segment AB not // x-axis. Find a //gm ABCD with C,D being on K. (Try to use pole and polars.)