E-Storehow amazing mathematics really is.!!
We often wonder about the complexity and variety of the parts mathematics can be broken into...even i was left wondering ..when i saw that one such simple question can be proved by SO MANY WAYS ..
Neil Donaldson and Alex Godwin were students in S6 at Madras College, St Andrews, Scotland when they wrote this article giving eight different solutions to the problem Three by One.
Presented here are 8 distinct proofs that in the following diagram, a+b=c :
(Throughout these proofs the above naming conventions will be used.)
The distinct proofs are:
- Tan Angle Sum Formula Proof
- Sin Angle Sum Formula Proof
- Cosine Rule Proof
- Vectors Proof
- Matrices Proof
- Complex Numbers Proof
- Pure Geometry Proof
- Coordinate Geometry Proof
tan Angle Sum Formula Proof
From the diagram, a = tan -1 (1/3), b = tan -1 (1/2), and c = tan -1 1.
We have to prove:
| a + b | = | c | |
| or | tan -1 (1/3) + tan -1 (1/2) | = | tan -1 1 |
| or | tan[tan -1 (1/3) + tan -1 (1/2)] | = | 1. |
Using tan double angle formula, tan( x+y ) = [tan( x )+tan( y )]/[1-tan( x )tan( y )], we may rewrite the equation as:
| tan[tan -1 (1/3) + tan -1 (1/2)] | = |  |
| = | [1/3 + 1/2]/[1-(1/3)(1/2)] | |
| = | (5/6)/(5/6) | |
| = | 1 |
Hence the result is proved.
sin Angle Sum Formula Proof
From diagram:
sin( a+b ) = sin( c ), so a+b=c . Hence the result is proved.
Cosine Rule Proof
By altering the diagram:
It can be seen that x =3+2=5.
d can be found using the cosine rule ( a 2 + b 2 -2 ab cos C = c 2 ), so
| So | a + b | = | 180° - d |
| = | 180° - 135° | ||
| = | 45° | ||
| c = | tan -1 1 | = | 45° |
| So | a + b | = | c. |
Hence the result is proved.
Vectors Proof
 |  |
| c = | tan -1 1 | = | 45° |
| So | a + b | = | c. |
Hence the result is proved.
Matrices Proof
Let d = ( x,y ) be any point on the xy -plane.
Let d 1 be d rotated a degrees around the origin:
Let d 2 be d 1 rotated b degrees around the origin: (or another way: d rotated ( a+b ) degrees around the origin)
Let d 3 be d rotated c degrees around the origin:
So d 2 = d 3 .
Hence a rotation by a+b is the same as a rotation by c degrees. Hence a+b=c (as none of the angles are greater than 90°).
Complex Numbers Proof
z = x +iy = r exp( i tan -1 y/x )
Let z´ = z rotated 2( a + b ) degrees around origin.
| So | z´ | = r exp( i tan -1 y/x + 2 i ( a + b )) |
| = r [cos(tan -1 y/x + 2( a + b )) + i sin(tan -1 y/x + 2( a + b ))] | ||
| = r [cos(tan -1 y/x )cos(2( a + b )) - sin(tan -1 y/x )sin(2( a + b )) + i (sin(tan -1 y/x )cos(2( a + b )) + cos(tan -1 y/x )sin(2( a + b )))] |
Since cos(2( a + b )) = 2 cos 2 ( a + b ) - 1
| and | cos( a + b ) | = cos a cos b - sin a sin b |
 |
so we have cos(2( a + b )) = 0.
Similarly sin(2( a + b )) = 2sin( a + b )cos( a + b )
| and | sin( a + b ) | = sin a cos b + cos a sin b |
 |
so we have sin(2( a + b )) = 1.
| Therefore | z´ | = r [0 - sin(tan -1 y/x ) + i (0 + cos(tan -1 y/x ))] |
| = r [ i cos(tan -1 y/x ) - sin(tan -1 y/x )] |
 |
|
Let z´´ = z rotated by 90° around the origin. (Multiplying a complex number by i is equivalent to a 90° rotation.)
z´´ = i(x + iy) = -y + ix .
So a rotation of 2( a + b ) is equal to a rotation of 90°, hence
| 2( a + b ) | = 90° |
| a + b | = 45° |
Pure Geometry Proof
By looking at the leftmost unit square this diagram can be drawn:
 |
|
| a + b = c | < => | x + x + y = x + y + z , |
| < => | x = z |
Hence it must be proven that x = z :
 |
|
|  |
Hence ADF and ABC are similar triangles and x=z => a + b = c .
Coordinate Geometry Proof
Let B be the point ( x B ,y B ) on the line y = -(1/2) x , a distance of 1 from the origin.
The line perpendicular to y = -(1/2) x at B has equation:
This line intersects y = (1/3) x at A = ( x A ,y A ).
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