Straightedge and compass constructions

[raman22feb1988]

Straightedge and then compass constructions

WIMS website: http://wims.unice.fr/wims/wims.cgi has many interesting mathematical tools.
Online calculators, interactive exercises, mathematical recreations.

Go to the ruler and compass section: It is equipped with many constructions. What lengths can you construct relative to a given length, with only a straightedge and compass? What regular polygons can you construct given the center and a vertex of the polygon? Straightedge is a ruler with no markings in it.

With straightedge and compass:
You can mark a random point on paper.
You can draw a line, line segment or ray between two points.
You can draw a circle with two points, center as first point, radius as distance between two points.
Mark the intersection of two lines, two circles or a line and circle.
Hide any unwanted points, lines or circles.

Marking up the middle of any two points, and then drawing perpendicular bisector of line segment are quite obvious from within these facilities. Bisection of angle is also possible.

Solution:
You can construct lengths of square roots of any integers. In a right angled triangle, the square of hypotenuse is the sum of squares of the other two sides.
You can divide a line segment into any number of equal parts. Suppose that you want N equal parts, take a parallel line with N+1 points that are equally spaced. Project the old line segment with this parallel line, with the end points of the line corresponding to the end points of the old line segment.

Regular polygons:
It is obvious to construct a regular polygon of 2N sides from a polygon of N sides, by bisecting the angles of the polygon.
You can construct equilateral triangle, in which the altitude is sqrt(3)/2 times its side. Note that altitude also bisects one of its sides.
A square is quite trivial to be constructed up, right?
A regular pentagon is possible, in which each angle is 108°. cos 72° = [sqrt(5)-1]/4.
Regular hexagon is possible from equilateral triangle, regular octagon is possible from square, and so on.
A regular polygon of 15 sides is possible from equilateral triangle and regular pentagon. Place their vertices over a circle, with one vertex of each coincident. Then, there exists atleast one vertex of triangle and one vertex of pentagon which are the adjacent vertices of the 15 sided regular polygon.
Though a regular polygon of 18 sides cannot be constructed from equilateral and regular hexagon because of the fact that 3 and 6 are not co-prime numbers.

In 1796, at the age of 19, Gauss realized that it is possible to construct a regular polygon of 17 sides by using only straightedge and compass.

A regular polygon of N sides can be constructed with straightedge and compass if and only if N is a power of 2 times 4 or product of one or more distinct prime Fermat Numbers.

The only known Fermat primes are 3, 5, 17, 257 and 65537. Then, there exists no more Fermat prime till 2^(2^32)+1 at all.
A length X relative to a given length can be constructed by using straightedge and compass, if and only if X is an element of quadratic extension field of rational numbers. This means that X is a root of an irreducible polynomial equation with rational coefficients, whose degree is a power of 2.

Note that cos (360/7)° is a root of the irreducible cubic 8x^3+4x^2-4x-1 = 0, and then that cos (360/9)° is a root of the irreducible cubic 8x^3-6x+1 = 0, so it is evident that these lengths cannot be constructed because the degree of the irreducible polynomial is 3, which is not a power of 2.
Note that there also exist some real numbers, which do not satisfy any polynomial equation with rational coefficients at all. These are called transcendental numbers. Pi and e are examples of transcendental numbers. Other irrational numbers are algebraic. Lengths of transcendental numbers also cannot be constructed with straightedge and compass only.

We can construct any rational length times a given length, because you can make any number of copies of the given length, and then you can divide up the given length into any number of equal parts.
A square root of any given length is possible, with the unit length. Take a line segment of given length plus the unit length. Draw a semicircle with this length as a diameter. Let P be a point, unit length away from the edge of this length. Then, the perpendicular distance between P and the semicircle arc gives up the square root of the given length. In this way, we can construct a length of fourth root of 2 from the square root of 2.
If you want to construct lengths which are roots of the quadratic equation x^2-ax+b = 0, then take up the coordinates of (a,b) and then mark up the coordinates of (0,1) too, as well. Draw up a circle with (a,b) and then (0,1) as the end points of the diameter. Then, the points where this circle cuts up the X-axis gives up the coordinates of the roots of this given quadratic equation.

If you take up a line segment with two parts of lengths 1 and b, and then project up the first part with length a, then the length of the projection of the second part will be ab (Multiplication of two lengths).
If you take up a line segment with two parts of lengths b and 1, and then project up the first part with length a, then the length of the projection of the second part will be a/b (Division of two lengths).

Some constructions that cannot be done up only with straightedge and then compass itself, at all:
Squaring of circle - Construct a square with equal area of that of the circle. Here we assume that circle of radius unit length. Involves construction of side of square of length sqrt(pi), that is transcendental, thus not at all possible with straightedge and then compass only.
Doubling a cube - Given a cube of unit length, construct a cube whose volume is double of that of the original cube. Here the new cube is of edge of length cube root of 2, which is a root of irreducible polynomial with rational coefficients of degree 3, x^3-2 = 0. So, it is not possible since.
Trisection of an arbitrary angle - Is not possible to construct up with a straightedge and then compass, as well.

Some constructions that are possible include up: Tangent to a circle from a point on the circle, Tangent to a circle from point outside the circle, Trisection of right angle, A line that is parallel to any given line, perpendicular from a point on the line, perpendicular from a point outside the line, angle bisectors, partitioning up a given line segment into any number of equal parts, centroid, orthocenter, circumcenter, incenter, circumcircle, incircle of a triangle, center of any given circle, diameter of circle, middle of two points, perpendicular bisector of any given line segment, parallelogram, square with any given side, square with any given diagonal.

Some constructions that I am not sure of, include up the following cases as well: Direct and then Inverse common tangents to two circles, Ten cases of the Appolonius problem, that is, circles that are tangential to any three given elements, where the given element may be either a point, line, or circle itself only, thus.

[raman22feb1988]

This is an excellent strategical puzzle to tune up your brains.

There is a box containing n holes. Two players put balls into these holes alternatively. (Assume that the box contains holes along a straight line, so that you cover up these holes in sequence.) Let player 1 start up with the game. Each player can put between min and max balls into the box in one turn. The person who puts the last ball will be the winner or loser, that is decided and fixed prior to the game, randomly.

The values of n, min, max are also decided and fixed prior to the game. For simplicity purposes, let us assume that 20 ≤ n ≤ 50 and then that 1 ≤ min < max ≤ 10.

If the person who puts the last ball is the winner, then for what values of n, depending upon min and max, will player 1 win the game, and then for what values of n will player 2 do so? Assume that both the players play so cleverly. Thus, what about the case if the person who puts the last ball is the loser? (If there is less than min slots remaining at the end, then the last player can put less than min balls during his last turn. Otherwise, the minimum number of balls that a player can put up within his turn is min only.)

The answer is as it follows up:

Here is the secret of the game. If the player who puts the last ball is the winner, then player 1 wins if N = 1 to max (mod min+max) player 2 wins if N = max+1 to min+max (mod min+max)

If the player who puts the last ball is the loser, then player 1 wins if N = min+1 to min+max (mod min+max) player 2 wins if N = 1 to min (mod min+max)

In either case, player 1 wins with probability (max)/(min+max) player 2 wins with probability (min)/(min+max).

Here is the explanation: If the player who puts the last ball is the winner, then that player who wins must reach N mod (min+max). Maintaining N mod (min+max) is the secret behind the game. If a player puts x balls, (between min and max), then the opponent will put the complement min+max-x, thus maintaining constant. If N mod (min+max) is between min & max, player 1 will acquire it within the first turn itself. If it is 0, then player 1 trivially loses up as player 2 will put complement in first turn itself. If it is max+1, then player 1 should play greater than or equal to max-min+2, otherwise player 2 will acquire max+1 (mod min+max). Player 2 will have to again keep up less than or equal to min (mod min+max) to avoid player 1 seeking max+1 (mod min+max). Going this way, player 2 wins. Also, similar case between max+1 and min+max-1, where player 1 plays higher, player 2 keeps up lower. If N (mod min+max) is less than or equal to min, then similarly player 1 keeps up lower, player 2 keeps up higher. Player 1 wins, within this case.

If the player who puts the last ball is the loser, then the winner must acquire between N-1 and N-min (mod min+max). For N (mod min+max) = min+1 and max+1, player 1 can acquire N-1 within the first turn itself. Between max+2 and max+min, player 1 can acquire something between N-1 and N-min during the first turn. For N = 1 (mod min+max), player 2 will acquire N-1 during the first turn itself, by putting up the complement of whatever player 1 puts. For N between 2 and min (mod min+max), player 2 always wins up, as putting complement will give up something between N-1 and then N-min only, thus.

Nice?