2911
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2011-02-01T00:00:01
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Mark four points on a flat surface so that there are only two different distances between them. One arrangement is as shown.<br></br>
<mdo:image alt="equilateral triangle " height="186" src="question%20triangle.JPG" width="222"></mdo:image><br></br>
<br></br>
How many more can you find?
<p>Are you sure that you have them all?</p>
<p> </p>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=6836&part=">Click here for a poster of this problem</a>.<br></br></mdoxml>
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An individual who is operating under the codename XYZ gave these
correct examples:<br></br>
<br></br>
Square - the $2$ different distances would be the SIDES (all same)
and $2$ DIAGONALS (equal)<br></br>
<br></br>
<mdo:image height="188" width="193" src="square.JPG" alt="square"></mdo:image><br></br>
<br></br>
Parallelogram with angles $60$, $120$, $60$ and $120$ and the
diagonal through the $120$ degrees angles is bisecting the angles.
i.e. it ends up being a Rhombus with one of the diagonals equal to
the sides.<br></br>
<mdo:image height="212" width="129" src="rhombus.JPG" alt="rhombus"></mdo:image><br></br>
<p class="editorial">Ellie offered the following as a solution:</p>
<div>A square, a diamond (kite) and a slanted square
(rhombus)</div>
<br></br>
<br></br>
<span class="editorial">The square and rhombus are given above. The
kite would look like this:</span> <br></br>
<br></br>
<br></br>
<br></br>
<mdo:image height="172" width="184" src="kite.JPG" alt="kite"></mdo:image><br></br>
<br></br>
<br></br>
<span class="editorial">Ray found one more:</span>
<p class="editorial"><mdo:image height="193" width="112" alt="isosceles triangle#" src="isosceles%20triangle.JPG"></mdo:image></p>
<p class="editorial"> </p>
<p class="editorial">And Ray and Kris found the last solution,
based on four vertices of a regular pentagon (the red lines are the
the length of the diagonals of a regular pentagon):</p>
<mdo:image height="137" width="250" src="trapezium.JPG" alt="trapezium"></mdo:image><br></br>
<br></br>
<p class="editorial">Well done to you all.</p>
<p class="editorial"></p>
<p class="editorial">You may have thought initially that:</p>
<p class="editorial"><mdo:image height="300" width="200" src="imposs2.png" alt="One points with all long lines from it, the rest of the lines short"></mdo:image>
</p>
<p class="editorial">is a solution. Unfortunately it is not. Since
if it was, then the blue triangle would be equilateral. Since all
the lengths are relative, we can fix the length of one side and use
this to find the lengths of the others:</p>
<p class="editorial"><mdo:image height="300" width="200" src="imposs1.png" alt="One points with all long lines from it, the rest of the lines short"></mdo:image>
</p>
<p class="editorial">We can then use Pythagoras' Theorem on the
bottom-right right-angled triangle to find that the missing length
in the triangle is the square root $3$. But then the length of the
unlabeled edge of the green right-angled triangle would be $x -
\sqrt{3}$. We can now use Pythagoras' Theorem on this triangle:</p>
<![CDATA[\begin{eqnarray} x^2 &=& (x-\sqrt{3})^2+1 x^2 \\
&=& x^2-2\sqrt{3}x+3+1 \sqrt{3}\\ x &=& 2 x \\
&=& \frac{2}{\sqrt{3}}\\ &=& \frac{2}{3}\sqrt{3}
\end{eqnarray} ]]>
<p class="editorial">But $x$ must be bigger than the square root of
$3$, and so this solution is impossible!</p>
<p class="editorial"></p>
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<mdoxml version="1.0">A problem that challenges students to use geometric reasoning -
easy to understand what is required, but not easy to find all the
arrangements...<br></br></mdoxml>
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<mdoxml version="1.0">There are more than $5$ different possible arrangements altogether
(including the one given).<br></br></mdoxml>
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<mdoxml version="1.0"><div></div><div></div><div><mdo:image width="193" height="188" src="square.JPG" alt="square"></mdo:image></div>
<div></div><div></div><div><mdo:image width="129" height="212" src="rhombus.JPG" alt="rhombus"></mdo:image></div><div></div><div></div><div></div><div><mdo:image width="184" height="172" src="kite.JPG" alt="kite"></mdo:image></div><div></div><div></div><div></div><div><mdo:image width="112" height="193" src="isosceles triangle.JPG" alt="isosceles"></mdo:image></div>
<div></div><div></div><div></div><div><mdo:image width="250" height="137" src="trapezium.JPG" alt="trapezium"></mdo:image></div>
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Keep Your Distance
Can you mark 4 points on a flat surface so that there are only two
different distances between them?
Using, Applying and Reasoning about Mathematics
Visualising