This article descibes some basic geometric constructions using only pencil, compasses and ruler. The ruler enables the drawing of straigth lines, the compasses are for drawing arcs and for the duplication of equal distances. Note: the scale of the ruler is not used. Request: English is not my native language. In case you discover mistakes in the following article, please inform me. Also suggestions for improvement are welcome. Contents Perpendicular bisector This is the the most important line in plane geometry.It holds all the points that have equal distance to two other (given) points. Construct a line perpendicular to AB that intersects AB in the center.
2. again draw arcs of same (other) radius with centers A en B. Arcs now intersect in Q 3. line PQ is the perpendicular bisector of AB
AS = BS AP = BP AQ = BQ LPAB = LPBA The circumscribed circle Given are 3 points: A,B and C.Construct the circle through A, B and C
2. construct the perpendicular bisector m of BC 3. center M of the circle is the intersection of l and m 4. draw circle with center M and radius MA = MB = MC
The perpendicular bisectors in a triangle intersect at 1 point, because this point has equal distance (radius) to all three points. Vertical perpendicular (point to line) Given are a line l and a point P not on l.Construct a line through P that is perpendicular to l.
2. draw arcs with equal radius with centers A en B, intersection of arcs is S 3. PS intersects l in Q 4. PQ is perpendicular to l
PQ is the distance of P to l Vertical perpendicular (line to point) Given are line lijn l with point P.Construct a line through P and perpendicular to l. method 1
2. draw arcs of equal radius with centers A and B, intersection of arcs is Q 3. PQ is perpendicular to l
method 2 This method applies when P is the end of a line.
2. Draw arcs of equal radius with centers A and P, intersection of arcs is S 3. Draw line through A and S 4. Extend AS, draw point Q such that SQ = AS 5. PQ is perpendicular to l
triangle PSQ has PS = SQ, so LQPS = LPQS = b in triangle APQ 2a + 2b = 180 degrees, so LAPQ = a + b = 90 degrees Bisector of an angle The bisector of an angle splits the angle in two equal parts.Given are lines l and m and their intersection P. Construct a line n that bisects the angle between l and m.
2. draw arcs with same radius with centers S and T, intersection is Q 3. PQ is the bisector of LP
Each point on n has equal distances to l and m (also because triangles are congruent) The inscribed circle Given is triangle ABC.Construct a circle that has AB, BC and AC as tangents.
2. construct bisector of LB...........(same) 3. M is intersection of bisectors and also the center of the inscribed circle. 4. construct line through M and perpendicular to AB. 5. MP is the radius of the inscribed circle.
So, M has equal distances to AB, BC and AC The bisectors of the angles of a triangle intersect in one point. Parallel line through point Given are line l and a point P, not on l.Construct a line through P that is parallel to l. method 1
2. call intersections A and B. 3. draw arc around P with radius AB 4. draw arc around B with radius AP 5. Q is the intersection of arcs ..3) and ..4) 6. PQ is the requested line
AB and PQ are therefore parallel lines. method 2 This is a fast way of parallel shifting.
2. place ruler along triangle 3. shift triangle along ruler until P is on edge. Division of line in equal parts We divide line AB in five equal parts.
2. using compasses, measure 5 equal parts on l, starting at A, ending in point S. 3. draw line BS 4. construct lines parallel to BS, see figure. 5. these parallel lines divide AB in five equal parts.
"parallel lines divide intersecting lines in parts having the same ratio" Because all parts on l are equal, parts of AB are also equal. Tangent of a circle Given is a circle with center M and a point P outside the circle.Construct a line through P that is tangent of the circle. method 1
2. construct point N in center of PM (perpendicular bisector construction) 3. draw circle with center N and radius MN 4. Q is intersection of circles 5. draw tangent PQ
method 2
2. adjust compasses to distance UV and draw circle with center M (circle has double length radius) 3. draw arc around P, radius is PM, intersecting big circle in point A. 4. draw line MA, Q is intersection with small circle 4. PQ is the tangent.
so PQ is perpendicular to MQ (the tangent of a circlre is perpendicular to the radius from center to tangent point) Duplicating an angle Given are lines l and m, their intersection point A and a point P on line p.Construct line s through P such, that the angle between s and p is equal to the angle at point A.
2. adjust compasses to AB and draw arc around P. (intersection with p is N, see figure) 3. adjust compasses to BC and draw arc around N. Intersection of arcs is Q. 4. draw line s through P and Q
Angles of 30, 45, 60, 72 degrees. We restrict ourselve to angles of integer degrees.Not every angle may be constructed by compasses and ruler. It is not possible to construct angles of 1, 2, 4, 5 , 10 ..........degrees. Only angles being a multiple of 3 are constructable. In all cases below, given is point A on line l. Construct a line through A with certain angle to l 30 degrees
2. draw circle with center M and radius MA 3. circle intersects l also in point B 4. draw arc around B with radius MA, intersection of circles is point P 5. LPAB = 30 degrees
LA = 30 degrees, because BP is chord of a 60 degree arc. (rule of Thales) 45 degrees
2. construct line through P' and perpendicular to AP' 3. adjust compasses to AP' and draw arc around P', intersection of arc and line is P 4. LPAP' = 45 degrees
60 degrees
2. adjust compasses to AB and draw arcs around A and around B 3. C is intersection of arcs 4. LCAB = 60 degrees
72 degrees This construction originates from the construction of a regular pentagon.
2. At B, construct line perpendicular to l 3. construct center of AB 4. make BP = AB/2 .... (BP = 1/2) see figure. 5. draw line AP 6. extend line by 1/2. ( AQ = AP + 1/2) 7. draw arcs around A and B with radius QA, intersection of arcs is D 8. LDAB = 72 degrees.
AS is bisector, AB = AS = SD (isosceles triangles) Let AB = 1 and AD = x. Application of bisector rule: x : 1 = 1 : (x-1) or: x^{2} - x - 1 = 0 x =
Trisection of an angle Earlier we agreed not to use the scale marks on a ruler.This constraint inhibits this construction. But if we allow ourselve to add 2 marks to the ruler, the construction becomes possible. We divide LBAC in three equal angles:
2. extend line CA 3. place ruler along CA and draw 2 pencil marks on it, one at point C and one at point A. 4. place ruler along point B and shift until marks are at S and the extension of CA. 5. LBQC = 1/3 * LBAC
According to the rule of Thales: x = 1/2 arc BC - 1/2 arc PS = 1/2 (3a - x) 2x = 3a - x x = a The square root of a*b Given are 2 line pieces with lengths a and b.Construct a linepiece with length root(ab).
2. regard AC as diameter of a circle, draw the circle. 3. construct line perpendicalar to AB, through B. 4. this line intersects circle in D. 5. BD is the right length: root(ab)
Pythagoras: a^{2} + x^{2} = p^{2} b^{2} + x^{2} = q^{2} p^{2} + q^{2} = (a+b)^{2} together: a^{2} + x^{2} + b^{2} + x^{2} = (a+b)^{2} a^{2} + 2x^{2} + b^{2} = a^{2} + 2ab + b^{2} x^{2} = ab Linepiece a * b Given are linepieces with lengths a and b and also a linepiece of length 1.(without the 1 we cannot know if multiplication makes a linepiece smaller or bigger) The construction is based on the rule that parallel lines divide intersecting lines in pieces with equal ratios.
2. on m, make SD = 1 and DC = b, see figure. 3. on l, make SA = a. 4. draw line AD 5. draw line BC parallel to AD 6. AB = a*b
AD is parallel to BC, so, SA : SD = AB : DC a : 1 = x : b x = ab Linepiece a / b
2. place distance 1 on m. 3. place distance b (SA) and distance a (AB) on l, see figure 4. draw line AD 5. draw line BC parallel to AD 6. DC = a/b
x = a/b Regular hexagon Starting with length AB of an edge, we construct a regular hexagon.
2. draw circle around M, radius MA 3. adjust compasses to length MA, draw arcs around A,B and intersections C, D, E, F 4. draw lines BC, CD, DE, EF, FA
So, the 6 angles in the center are 360 degrees together. Regular polygon with 9 edges. This construction is only possible by allowing the trisection tric, because angles of 20 or 40 degreesneed to be constructed. We start the construction from the length of edge AB.
2. extend AB 3. construct equilateral triangle NAB 4. draw arc around B, radius BN. 5. place 2 markers on ruler with distance AB 6. place ruler along point N, intersecting AB extension in point P, intersection with circle is Q. 7. shift ruler until PQ = AB (markers are at P and Q) 8. LNPA = 20 degrees. We reflect P to A. 9. construct perpendicular bisector of AP, intersection with NP is S 10. draw line through points A and S, LSAB = 20 degrees 11. extend AS, C is intersection with circle having center B and radius AB. 12. construct perpendicular bisectors of AB and BC, their intersection is point M. 13. draw circle around M with radius MB. 14. adjust compasses to AB (= BC) and draw arcs around A, B and intersections D, E, F, G, H, I with circle. 15. ABCDEFGHI is the polygon we wanted.
1...8 constructs LCAB = 20 degrees (holds arc of 40 degrees) point M is the intersection of 2 perpendicular bisectors of the edges. Another way to construct a 40^{0} angle This construction was invented by mr. Fedde Reeskamp. (Thanks for his nice work)From point B of equilateral triangle ABC with circumscribed circle a new circle is drawn such that points M,D,E are on a straight line. BE is the side of a regular 9 angle polygon. At first glance the ratio of the circle radius is 3:2. However this is not true. If this ratio is choosen the angle will be 38,93 degrees, not 40. |
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