8 Methods for Three by One

 
Proof


From the diagram, $a =\tan^{-1}(1/3), b =\tan^{-1}(1/2)$, and $c = \tan^{-1}(1)$.

We have to prove $a+b=c$, which is the same as proving that $\tan^{-1}(1/3)+\tan^{-1}(1/2)=\tan^{-1}(1)$.

Note that
$$
\begin{eqnarray}
\tan\left(\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{2}\right)\right)
&=&\frac{\tan\left(\tan^{-1}\left(\frac{1}{3}\right)\right)+\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right)}
{1-\tan\left(\tan^{-1}\left(\frac{1}{3}\right)\right)\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right)}\cr &=&\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}\cr &=&1 
\end{eqnarray}
$$
 

Proof
 
Using Pythagoras we can calculate the lengths of the diagonal lines:
 
 
 $$
\sin a = \frac{1}{\sqrt{10}}\quad \cos a = \frac{3}{\sqrt{10}}\quad \sin c = \frac{1}{\sqrt{2}}\quad \sin b = \frac{1}{\sqrt{5}}\quad \cos b = \frac{2}{\sqrt{5}}
$$
 
 
Using the identity $\sin(a+b) = \sin a \cos b + \sin b \cos a$ we see that
$$
\begin{eqnarray}
\sin(a+b) &=&   \frac{1}{\sqrt{10}}\cdot\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}\cdot \frac{3}{\sqrt{10}}\cr
&=& \frac{2+3}{\sqrt{10\cdot 5}}\cr
&=& \frac{5}{\sqrt{50}}\cr
&=&\frac{1}{\sqrt{2}}\cr
&=& \sin c
\end{eqnarray}
$$
 Hence the result is proved.
 

This method required us to extend the diagram as follows:
 
 
From the altered diagram it can be seen that $x=3+2=5$. 
 
Next, $d$ can be found using the cosine rule $c^2= a^2+b^2-2ab \cos C$.
 
Substitution of these values gives
$$
\cos d = \frac{-1}{\sqrt{2}}
$$
Thus $d=135^\circ$. Therefore $a+b = 180^\circ - d = 45^\circ = c$
 
 
Hence the result is proved.

  

 
 
 
 

Hence the result is proved.
 
 
 

Let $\bf{d} = (x, y)$ be any point in the $x-y$ plane. Let $\bf{d}_1$ be the point obtained by rotating $a^\circ$ about the origin.

  
 
 
 Let $\bf{d}_2$ be the point obtained by rotating $\bf{d}_1$ by $b^\circ$ around the origin. 
 
 
Finally, let $\bf{d}_2$ be the point obtained by rotating $\bf{d}$ by $c^\circ$ around the origin.
 
 
 
Hence a rotation by $a+b$ is the same as a rotation by $c$ degrees. Hence, $a+b=b$, as none of the angles is greater than $90^\circ$.  
 

 
 
By looking at the leftmost unit square this diagram can be drawn
 
 
 
 $a+b=c \Leftrightarrow x+x+y  = x+y+z \Leftrightarrow x=z$
Hence it must be proven than $x=z$:
 
 
 
 Split triangle ADE into two right angled triangles ADF and EDF
 
 
From this it can be seen that
$$
EF = DF = \frac{\sqrt{2}}{4}
$$
and
$$
AF = AE-FE = \sqrt{2} -\frac{\sqrt{2}}{4} = \frac{3\sqrt{2}}{4}
$$
  
Thus
$$
AF:AB = \frac{3\sqrt{2}}{4}:1
$$
and
$$
DF:BC = \frac{\sqrt{2}}{4}:\frac{1}{3} = \frac{3\sqrt{2}}{4}:1
$$

 
 
 Hence ADF and ABC are similar triangles and, therefore, $x=z$. Thus, $a+b=c$.
 

The coordinate geometry proof is based on the following diagram; the gradients are easy to read off the original image.

  
 Let $B$ be the point $( x_B ,y_B )$ on the line $y = -\frac{1}{2} x$  at a distance of $1$ from the origin. Since $\sqrt{x^2_B+y^2_B} = 1$ and $y_B = -\frac{1}{2}x_B$ we have
$$
\begin{eqnarray}
\sqrt{x^2_B+\frac{1}{4}x^2_B} &=& \sqrt{\frac{5}{4}x^2_B} = 1\cr
\Rightarrow x_B = \frac{2}{\sqrt{5}}\cr
\Leftrightarrow y_B = \frac{-1}{\sqrt{5}}
\end{eqnarray}
$$
Thus,
$$
B=\left(\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)
$$
We can now determine the equation of the line through $B$ perpendicular to the line $y = -\frac{1}{2} x$ to be
$$
y = 2x-\sqrt{5}.
$$
This line intersects $y=\frac{1}{3} x$ at $A$ which we can calculate to be
$$
A= \left(\frac{3}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)
$$
Now that we konw points $A$ and $B$ we can calculate the distance between them as
$$
|AB| = \sqrt{\left(\frac{3}{\sqrt{5}}-\frac{2}{\sqrt{5}}\right)^2 +\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)^2}= \sqrt{\frac{1}{5}+\frac{4}{5}}=1
$$
Referring back to the diagram shows that $a+b=45^\circ$ and the result is proved.
 

This method uses aspects of other proofs presented (and therefore could be shortened) but we include it because it gives a nice example of how complex numbers, matrices and all other methods fit nicely together.
 
 
 Let $r=\sqrt{x^2+y^2}$.
 
Now, the complex number corresponding to the point with coordinates $(x, y)$ in the argand diagram can be written as
$$
z= x+iy = r\exp\left(i\tan^{-1} \left(\frac{y}{x}\right)\right)
$$
Let $z'$ be the complex number obtained by rotating $z$ by $2(a+b)$ degrees. Then we can use the identities $\sin(a+b) = \sin a \cos b +\cos a\sin b$ and $\cos(a+b) = \cos a \cos b-\sin a \sin b$ to see that:
$$
\begin{eqnarray}
z' &=& r\exp\left(i\tan^{-1}\left(\frac{y}{x}\right)+2i(a+b)\right)\cr
&=& r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)+i\sin\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)\right]\cr
&=&r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)-\sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right)\right]\cr
&& +ir\left[ \sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)+\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right) \right]
\end{eqnarray}
$$
 
We  can simplify the arctangents by considering the following diagram:
 
 
 
 
From this we can see that $\cos\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{x}{r}$ and $\sin\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{y}{r}$
This simplifies the expression for $z'$ to
$$
z' = x\cos\left(2(a+b)\right)-y\sin\left(2(a+b)\right)+i\left(y\cos\left(2(a+b)\right)+x\sin\left(2(a+b)\right)\right)
$$
 
 Alex and Neil then used some double angle formulae and the fact that $\sin(a+b)=\cos(a+b)=\frac{1}{\sqrt{2}}$ to determine that 
$$
z' = -y+ix
$$
Since this is a complex number rotated through $90^\circ$ we can conclude that $a+b = 45^\circ$