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.
Problem
What is the value of $X$ in the equilateral triangle below?
In how many different ways can you find this out?
Getting Started
What are the ratios of the sides of a 30, 60, 90 degree triangle?
Student Solutions
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.
$$\eqalign{ \beta &=& \frac{\pi}{6} - \frac{\alpha}{2} \\ \cos\beta &=& \frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{3}}{2\sqrt{7}} + \frac{1}{2}\cdot\frac{1}{2\sqrt{7}} \\ \; &=& \frac{10}{4\sqrt{7}} = \frac{5}{2\sqrt{7}}}$$
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 $\sqrt{7} - 2 < x < \sqrt{7} + 2$. The diagram can be redrawn to show the trapezium $BPQC$ flipped down producing the much smaller equilateral triangle of side $x$ units.
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.
\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!
Teachers' Resources
Why do this problem?
There are several different ways of solving this problem. 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.
Possible approach
Challenge the class to find different methods of solution.
Key question
How do we interpret the two solutions?
Is the diagram correct for both solutions?