8 Methods for "Three By One"


by Neil Donaldson and Alex Godwin (S6) of Madras College, St Andrews

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:

  1. Tan Angle Sum Formula Proof
  2. Sin Angle Sum Formula Proof
  3. Cosine Rule Proof
  4. Vectors Proof
  5. Matrices Proof
  6. Complex Numbers Proof
  7. Pure Geometry Proof
  8. Coordinate Geometry Proof

tan Angle Sum Formula Proof

From the diagram, a = tan -1(1/3), b = tan -1(1/2), and c = tan -11.

We have to prove:

a + b = c
or tan -1(1/3) + tan -1(1/2) = tan -11
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.

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sin Angle Sum Formula Proof

From diagram:

sin( a+b) = sin( c), so a+b=c. Hence the result is proved.

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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 -11 = 45°
So a + b = c.

Hence the result is proved.

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Vectors Proof

c = tan -11 = 45°
So a + b = c.

Hence the result is proved.

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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°).

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Complex Numbers Proof

z = x +iy = r exp( i tan -1 y/x)

Let z´ = z rotated 2( a + b) degrees around origin.

So = 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 = 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)]
So cos(tan -1 y/x) = x/r
sin(tan -1 y/x) = y/r

= r 1/r (-y + ix)
= (-y + ix)

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°

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Pure Geometry Proof

By looking at the leftmost unit square this diagram can be drawn:

So a = x,
b = x + y,
c = x + y + z.
a + b = c <=> x + x + y = x + y + z,
<=> x = z

Hence it must be proven that x = z:

By splitting triangle ADE into two right angled triangles ADF and EDF it can be seen that
EF = FD =
AF = AE - FE =
AF : AB =
DF : BC =

Hence ADF and ABC are similar triangles and x=z => a + b = c.

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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).