Xtra

1962 /www/nrich/html/content/98/06/15plus3/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> What is the value of $X$ in the equilateral triangle below? In how many different ways can you find this out? <mdo:image src="XtraDia1.gif" alt="Diagram for Xtra problem"></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Two solutions to this problem have been forthcoming from different students at the same school - Madras College. Thank you to Mike and Euan who used lots of trigonometry as well as to Thom who likewise resorted to double angles and the cosine rule and reduced the problem to solving a quadratic equation. Thom was also able to show the significance of the two roots. <div><mdo:image alt="Diagram of the problem" src="xtraprob.gif"></mdo:image></div> $$\eqalign{ \beta &=& \frac{\pi}{6} - \frac{\alpha}{2} \\ \cos\beta &=& \frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{3}}{2\sqrt{7}} + \frac{1}{2}\cdot\frac{1}{2\sqrt{7}} \\ \; &=& \frac{10}{4\sqrt{7}} = \frac{5}{2\sqrt{7}}}$$ <div><mdo:image alt="Diagram showing cos and sin of alpha/2" src="xtracos.gif"></mdo:image></div> Using the cosine rule on $\triangle ABP$ $$\eqalign{ 4 &=& x^2 + 7 - 2x\sqrt{7}\cos\beta \\ \; &=& x^2 + 7 - 5x}$$ Therefore $x^2 - 5x + 3 = 0$ $$x = \frac{5\pm\sqrt{13}}{2}$$ Both solutions satisfy the triangle inequality for $\triangle ABP$, namely $\sqrt{7} - 2 &lt; x &lt; \sqrt{7} + 2$. The diagram can be redrawn to show the trapezium $BPQC$ flipped down producing the much smaller equilateral triangle of side $x$ units. <div style="text-align: center;"><mdo:image alt="Diagram for the smaller value of x " src="xtrasmall.gif"></mdo:image></div> A solution which just needs Pythagoras's Theorem was sent in by Ewan from King Edward VII School, Sheffield. See if you can work it out from the diagram below, then reveal the hidden text to check your answer. <mdo:image src="Xtra.png"></mdo:image> <div class="toggle">The diagram shows the median AU at point A cutting the triangle in half. The length AU is in two parts, $y$ and $z$. Since the triangle is equilateral, $y+z=\frac{\sqrt 3}{2}x$. (You may like to prove this e.g. by trigonometry) To work out $y$ use Pythagoras's Theorem: \begin{align} y^2+\left(\frac{1}{2}\right)^2 &=\left(\sqrt 7\right)^2 \\ y^2 &= \frac{27}{4} \\ y &= \frac{3\sqrt 3}{2} \end{align} and use $y+z=\frac{\sqrt 3}{2}x$ to give $z = (x-3)\frac{\sqrt 3}{2}$. Then more Pythagoras's Theorem and our values found above give \begin{align} z^2+\left(\frac{x}{2} - \frac{1}{2}\right)^2 &=2^2 \\ \frac{3(x-3)^2}{4} + \frac{(x-1)^2}{4} = 4 \end{align} which simplifies to get the same equation as Thom and the others: $x^2 -5x + 3 = 0$. What a simple solution, Well done!</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> There are several different ways of solving <a href="http://nrich.maths.org/public/viewer.php?obj_id=1962">this problem</a>. It is a good example to bring home the point that, to be an expert problem solver, and to understand a piece of mathematics, it is not enough to find an answer. One should ask oneself "Have I used the best method?". This problem can be solved using trig formulae and the cosine rule but it can be solved more straightforwardly using only Pythagoras theorem. <h3>Possible approach</h3> Challenge the class to find different methods of solution. <h3>Key question</h3> How do we interpret the two solutions? Is the diagram correct for both solutions? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What are the ratios of the sides of a 30, 60, 90 degree triangle? </mdoxml> 2 3 0 0 0 1 1 Xtra Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations. Algebra Quadratic equations Admin Long problems Trigonometry Cosine rule 2D Geometry, Shape and Space Isosceles triangles 2D Geometry, Shape and Space Pythagoras' theorem