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  <id>1335</id>
  <path>/www/nrich/html/content/98/06/art2/</path>
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  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;Consider this problem: for the following diagram, prove that $a+b=c$&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
 &lt;mdo:image alt=&quot;&quot; height=&quot;121&quot; src=&quot;problemStart.gif&quot; width=&quot;276&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
See how you would go about solving it. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Now, this problem has been shown on NRICH before. When first shown on NRICH it was solved in 8 different ways by a pair of students Alex and Neil. When we again showed the problem, Sigi sent us a lovely new geometric proof. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Sigi suggests that it would be interesting to look at the different proof methods and think about which are mathematically independent of each other in that they use genuinely distinct mathematical ideas rather than the same ideas dressed up in different ways. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
We agree with Sigi: Analyse each of the proof methods. How many genuninely distinct methods can be found? In what ways are they different from each other? Do you have a favourite proof? Do some proof methods seem to have potential for wider generalisation. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Sigi&amp;#39;s proof is &lt;a href=&quot;/content/98/06/art2/three-by-one.pdf&quot;&gt;found here&lt;/a&gt;.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
The distinct proofs from Alex and Neil are:&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow1&quot;&gt;Method 1: Tan Angle Sum Formula&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow2&quot;&gt;Method 2: Sin Angle Sum Formula&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow3&quot;&gt;Method 3: Cosine Rule&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow4&quot;&gt;Method 4: Vector&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow5&quot;&gt;Method 5: Matrices&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow6&quot;&gt;Method 6: Pure Geometry&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow7&quot;&gt;Method 7: Coordinate Geometry&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;#&quot; id=&quot;hideShow8&quot;&gt;Method 8: Complex Numbers&lt;/a&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div id=&quot;hidden8&quot; style=&quot;display: none;&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;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.&lt;/span&gt;&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 Let $r=\sqrt{x^2+y^2}$.&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
Now, the complex number corresponding to the point with coordinates $(x, y)$ in the argand diagram can be written as&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
z= x+iy = r\exp\left(i\tan^{-1} \left(\frac{y}{x}\right)\right)&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
Let $z&amp;#39;$ 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:&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
\begin{eqnarray}&lt;br&gt;&lt;/br&gt;
z&amp;#39; &amp;amp;=&amp;amp; r\exp\left(i\tan^{-1}\left(\frac{y}{x}\right)+2i(a+b)\right)\cr&lt;br&gt;&lt;/br&gt;
&amp;amp;=&amp;amp; r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)+i\sin\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)\right]\cr&lt;br&gt;&lt;/br&gt;
&amp;amp;=&amp;amp;r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)-\sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right)\right]\cr&lt;br&gt;&lt;/br&gt;
&amp;amp;&amp;amp; +ir\left[ \sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)+\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right) \right]&lt;br&gt;&lt;/br&gt;
\end{eqnarray}&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
We  can simplify the arctangents by considering the following diagram:&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;mdo:image alt=&quot;&quot; height=&quot;121&quot; src=&quot;complexnum.gif&quot; width=&quot;196&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
From this we can see that $\cos\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{x}{r}$ and $\sin\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{y}{r}$&lt;br&gt;&lt;/br&gt;
This simplifies the expression for $z&amp;#39;$ to&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
z&amp;#39; = x\cos\left(2(a+b)\right)-y\sin\left(2(a+b)\right)+i\left(y\cos\left(2(a+b)\right)+x\sin\left(2(a+b)\right)\right)&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt; Alex and Neil then used some double angle formulae and the fact that&lt;/span&gt; $\sin(a+b)=\cos(a+b)=\frac{1}{\sqrt{2}}$ &lt;span class=&quot;editorial&quot;&gt;to determine that&lt;/span&gt; &lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
z&amp;#39; = -y+ix&lt;br&gt;&lt;/br&gt;
$$&lt;br&gt;&lt;/br&gt;
Since this is a complex number rotated through $90^\circ$ we can conclude that $a+b = 45^\circ$&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
 &lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
     &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;We received another solution from Siri, which you can take a look at &lt;a href=&quot;/content/98/06/art2/three-by-one.pdf&quot;&gt;here&lt;/a&gt; .&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML/>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>4</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>1</keystage4>
  <keystage4plus>1</keystage4plus>
  <title>8 Methods for Three By One</title>
  <description>This problem in geometry has been solved in no less than EIGHT ways
by a pair of students. How would you solve it? How many of their
solutions can you follow? How are they the same or different? Which
do you like best?</description>
  <spec_group>Site Structural
    <specifier>TrendUpperSecondaryStudent</specifier>
  </spec_group>
  <spec_group>Trigonometry
    <specifier>Sine</specifier>
  </spec_group>
  <spec_group>Trigonometry
    <specifier>Tangent</specifier>
  </spec_group>
  <spec_group>Trigonometry
    <specifier>Sine rule</specifier>
  </spec_group>
  <spec_group>Vectors
    <specifier>Vectors</specifier>
  </spec_group>
  <spec_group>Advanced Algebra
    <specifier>Matrices</specifier>
  </spec_group>
  <spec_group>Advanced Algebra
    <specifier>Complex numbers</specifier>
  </spec_group>
  <spec_group>Admin
    <specifier>Not tagged</specifier>
  </spec_group>
  <spec_group>Secondary Mapping Document
    <specifier>MD Geometrical reasoning US</specifier>
  </spec_group>
</resource>