Roughly the idea is this- Imagine that you have a general angle \theta and can trisect it- that is, divide it into three angles each of measure \theta/3.
Now let \phi= \theta/3 so that the original angle is \theta= 3\phi
Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is cos(\theta)= cos(3\phi).
From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)
= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)
= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)
Since sin^2(\phi)= 1- cos^2(\phi) that is equal to
cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)
Now, as you take \theta going from 0 to \pi/2, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular \theta, we have 4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3} or 12 cos^3(\phi)- 9 cos(\phi)- 1= 0.
Now, drop a perendicular to from the end of the trisector, the line that makes angle \phi= \theta/3 with the base. The distance from the vertex to the foot of that perpendicular is X= cos(\phi).
That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying 12X^3- 9X- 1= 0.
Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.
And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of \sqrt{2}, an irrational number.
So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.
A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. x= \sqrt{2} is not rational but does satify x^2- 2= 0 and so is "algebraic of degree 2".
In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, x^2+ y^2= r^2, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.
The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order 2^0= 1, 2^2= 4, 2^3= 8, etc.
In the example I gave, 12x^3- 9x- 1= 0 is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".
All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.
Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!
(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is \pi and so if we could construct a square of that area, we would have constructed a side length of \sqrt{\pi} and \pi and \sqrt{\pi} are not "algebraic" of any order- they are "transcendental numbers".
The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length \sqrt[3]{2} which is algebraic of order 3, not a power of 2.
By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.
Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of \sqrt{2}. That number is, as I said before, algebraic of order 2 and so constructible. Roughly the idea is this- Imagine that you have a general angle \theta and can trisect it- that is, divide it into three angles each of measure \theta/3.
Now let \phi= \theta/3 so that the original angle is \theta= 3\phi
Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is cos(\theta)= cos(3\phi).
From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)
= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)
= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)
Since sin^2(\phi)= 1- cos^2(\phi) that is equal to
cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)
Now, as you take \theta going from 0 to \pi/2, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular \theta, we have 4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3} or 12 cos^3(\phi)- 9 cos(\phi)- 1= 0.
Now, drop a perendicular to from the end of the trisector, the line that makes angle \phi= \theta/3 with the base. The distance from the vertex to the foot of that perpendicular is X= cos(\phi).
That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying 12X^3- 9X- 1= 0.
Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.
And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of \sqrt{2}, an irrational number.
So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.
A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. x= \sqrt{2} is not rational but does satify x^2- 2= 0 and so is "algebraic of degree 2".
In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, x^2+ y^2= r^2, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.
The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order 2^0= 1, 2^2= 4, 2^3= 8, etc.
In the example I gave, 12x^3- 9x- 1= 0 is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".
All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.
Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!
(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is \pi and so if we could construct a square of that area, we would have constructed a side length of \sqrt{\pi} and \pi and \sqrt{\pi} are not "algebraic" of any order- they are "transcendental numbers".
The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length \sqrt[3]{2} which is algebraic of order 3, not a power of 2.
By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.
Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of \sqrt{2}. That number is, as I said before, algebraic of order 2 and so constructible. Roughly the idea is this- Imagine that you have a general angle \theta and can trisect it- that is, divide it into three angles each of measure \theta/3.
Now let \phi= \theta/3 so that the original angle is \theta= 3\phi
Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is cos(\theta)= cos(3\phi).
From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)
= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)
= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)
Since sin^2(\phi)= 1- cos^2(\phi) that is equal to
cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)
Now, as you take \theta going from 0 to \pi/2, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular \theta, we have 4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3} or 12 cos^3(\phi)- 9 cos(\phi)- 1= 0.
Now, drop a perendicular to from the end of the trisector, the line that makes angle \phi= \theta/3 with the base. The distance from the vertex to the foot of that perpendicular is X= cos(\phi).
That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying 12X^3- 9X- 1= 0.
Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.
And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of \sqrt{2}, an irrational number.
So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.
A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. x= \sqrt{2} is not rational but does satify x^2- 2= 0 and so is "algebraic of degree 2".
In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, x^2+ y^2= r^2, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.
The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order 2^0= 1, 2^2= 4, 2^3= 8, etc.
In the example I gave, 12x^3- 9x- 1= 0 is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".
All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.
Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!
(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is \pi and so if we could construct a square of that area, we would have constructed a side length of \sqrt{\pi} and \pi and \sqrt{\pi} are not "algebraic" of any order- they are "transcendental numbers".
The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length \sqrt[3]{2} which is algebraic of order 3, not a power of 2.
By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.
Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of \sqrt{2}. That number is, as I said before, algebraic of order 2 and so constructible.
)
