Triangle Properties: STEP I 2004 Question 6 Solution

This is an interesting STEP question (as many of them are!) but it gets you to, unwillingly, prove some interesting properties of triangles. I also hope that this post can act as a train of thought needed throughout a STEP question.

At first glance there are probably a few viable ways of attacking the problem, some neater than others. The most obvious one is to go at it algebraically and try to achieve the desired result.

$\begin{align*}Midpoint\ of\ BC,\ M_{BC}\left(\frac{p_2 + p_3}{2}, \frac{q_2 + q_3}{2}\right)\end{align*}$

$\begin{align*}Midpoint\ of\ AC,\ M_{AC}\left(\frac{p_1 + p_3}{2}, \frac{q_1 + q_3}{2}\right)\end{align*}$

$\begin{align*}Gradient\ of\ line\ connecting\ M_{BC}\ and\ A,\ m_1 =  \frac{q_2 + q_3 - 2q_1}{p_2 + p_3 - 2p_1} \end{align*}$

$\begin{align*}Gradient\ of\ line\ connecting\ M_{AC}\ and\ B,\ m_1 =  \frac{q_1 + q_3 - 2q_2}{p_1 + p_3 - 2p_2} \end{align*}$

As you can see, this turns very messy, very quickly. I could persevere, find the equation of both the lines equate them and solve for x but that is going to be a humongous algebraic slog. There are then two options, persevere or go at it via a different method. Given that this is the first part of the question looking for a more concise method is undoubtedly the way to go.

Given that we are thinking about geometrical properties of lines and their midpoints it makes sense to use vectors to solve this problem. Most importantly it will likely cut down on the algebra a lot due to their concise notation.

$\begin{align*}\ \ Let\  \overrightarrow{OA} = \mathbf{a} = \begin{bmatrix} p_1 \\ q_1 \end{bmatrix},\ \overrightarrow{OB} = \mathbf{b} = \begin{bmatrix} p_2 \\ q_2 \end{bmatrix},\ \overrightarrow{OC} = \mathbf{c} = \begin{bmatrix} p_3 \\ q_3 \end{bmatrix} \end{align*}$

$\begin{align*}Let\ the\ midpoints\ of\ AB,\ AC,\ BC\ be\ denoted\ D,\ E,\ F \end{align*}$

$\begin{align*}\ \ \therefore \overrightarrow{OD} = \mathbf{\frac{1}{2}(a+b)},\ \overrightarrow{OE} = \mathbf{\frac{1}{2}(a+c)},\ \overrightarrow{OF} = \mathbf{\frac{1}{2}(b+c)}\end{align*}$

That has set up all of the required details to begin actually answering the question. The vector equations for the line joining A and the midpoint of BC and the line joining B and the midpoint AC, are the lines $AF$ and $BE$:

$\begin{align*}\mathbf{r_{AF}} = \mathbf{a} + \lambda (\mathbf{b + c - 2a}) \end{align*}$

$\begin{align*}\mathbf{r_{BE}} = \mathbf{b} + \mu (\mathbf{a+c-2b}) \end{align*}$

They intersect when $\mathbf{r_{AF}} = \mathbf{r_{BE}}$.

$\Rightarrow \mathbf{a} - \mathbf{b} = (\mu - \lambda)\mathbf{c} + (\mu + 2\lambda)\mathbf{a} - (\lambda + 2\mu)\mathbf{b}$

$\mathbf{c}$ is not parallel to $\mathbf{a}$ or $\mathbf{b} \Rightarrow \mu - \lambda = 0$

$\Rightarrow \mathbf{a} - \mathbf{b} = 3\lambda\mathbf{a} - 3\lambda\mathbf{b}$

This must then mean that $\begin{align*}\lambda = \frac{1}{3}\end{align*}$. Denote the point of intersection $Q$, it lies on $\begin{align*}\mathbf{r_{AF}},\ \lambda = \frac{1}{3}\end{align*}$.

$\begin{align*}\ \ \ \overrightarrow{OQ} = \mathbf{a} + \frac{1}{3}(\mathbf{b} + \mathbf{c} - 2\mathbf{a}) = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c}) = \frac{1}{3}\begin{bmatrix} p_1+p_2+p_3 \\ q_1+q_2+q_3 \end{bmatrix}\end{align*}$

$\begin{align*}\therefore Q\left(\frac{p_1+p_2+p_3}{3},\frac{q_1+q_2+q_3}{3}\right)\end{align*}$

We have found the point of intersection we now need to show that this lies on the line connecting C to the midpoint of AB, this is the vector equation $CD$.

$\begin{align*}\mathbf{r_{CD}} = \mathbf{c} + \omega (\mathbf{a + b - 2c})\end{align*}$

$\begin{align*}\ \ \overrightarrow{OQ} = \mathbf{r_{CD}} \Rightarrow \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} - \frac{2}{3}\mathbf{c} = \omega(\mathbf{a} + \mathbf{b} - 2\mathbf{c})\end{align*}$

This is satisfied by $\omega = \frac{1}{3}$ $\therefore\ Q\ lies\ on\ CD$.

Let us think about what we have actually shown here. We have shown that for any triangle $ABC$, if we connect the vertices to the midpoint of the opposite side they all meet at a common point (they are concurrent), this point is called the centroid. This has useful implications to finding the centre of mass of an object.

To answer the next part of the question we will also use vectors, this part seems almost tailor made to do so given that we are looking for when two lines are perpendicular.

$\begin{align*}\ \ \overrightarrow{OH} = \mathbf{h} = \begin{bmatrix} p_1 + p_2 + p_3 \\ q_1 + q_2 + q_3 \end{bmatrix} = \mathbf{a}+\mathbf{b}+\mathbf{c}\end{align*}$

$\begin{align*}\mathbf{r_{AH}} = \mathbf{a} + \lambda (\mathbf{h} - \mathbf{a}) = \mathbf{a} + \lambda (\mathbf{b} + \mathbf{c}) \end{align*}$

$\begin{align*}\mathbf{r_{BC}} = \mathbf{b} + \mu (\mathbf{c} - \mathbf{b}) \end{align*}$

If $AH \perp BC,\ (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{b}) = 0 \Rightarrow \mathbf{b} \cdot \mathbf{b} = \mathbf{c} \cdot \mathbf{c} \Rightarrow |\mathbf{b}| = |\mathbf{c}| \therefore p_2^2+q_2^2 = p_3^2+q_3^2 (1)$ as required.

A virtually identical method is taken for $BH \perp AC$.

$\begin{align*}\mathbf{r_{BH}} = \mathbf{b} + \lambda (\mathbf{a} + \mathbf{c}) \end{align*}$

$\begin{align*}\mathbf{r_{AC}} = \mathbf{a} + \mu (\mathbf{c} - \mathbf{a}) \end{align*}$

If $BH \perp AC,\ (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{a}) = 0 \Rightarrow \mathbf{a} \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{c} \Rightarrow |\mathbf{a}| = |\mathbf{c}| \therefore p_1^2+q_1^2 = p_3^2+q_3^2 (2)$.

The final part can be easily deduced if we subtract the two statements we have already shown to be true, eliminating $p_3,\ q_3$.

$\begin{align*}\ \ (1) - (2) \Rightarrow p_1^2 + q_1^2 = p_2^2 + q_2^2 \Rightarrow(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{b} - \mathbf{a}) = 0 \therefore AC \perp BH,\ BC \perp AH \Rightarrow AB \perp CH \end{align*}$

And that is the entirety of the question completed!

The line through a vertex which is perpendicular to the opposite side is called an altitude. We have shown that the three altitudes of a triangle meet at a common point (they are concurrent) if, this point is called the orthocentre.

This question is very nice for forcing you to prove interesting properties of triangles unwillingly, I highly recommend that you do this question yourself. If you do not understand what I have done at any point comment and I will explain.

Fibonacci Numbers and Continued Fractions

I feel I needed an updated post on the Fibonacci sequence with some of the more core ideas behind it, you can still access the old post.

The definition of the Fibonacci sequence is:

$\begin{align}F_{n+2}=F_{n+1}=F_n\ where\ F_0=1,\ F_1=1\end{align}$

You can see quickly that the first few values of the sequence are 1, 1, 2, 3, 5, 8, 13, ..., etc. this is a weakly increasing sequence, that is ${F}_{n+1}\ge F_n \forall\ n \ge 1$.

This clearly means that the sequence does not tend towards a limit, but does the ratio between terms tend towards a limit?

Consider that the $\lim_{n\to\infty}(\frac{F_{n+1}}{F_n})=x$. If we divide $(1)$ through by $F_{n+1}$ we get:

$\begin{align*}\frac{F_{n+2}}{F_{n+1}}=1 + \frac{F_n}{F_{n+1}}\end{align*}\tag{2}$

Clearly from our definition of the $\lim_{n\to\infty}(\frac{F_{n+1}}{F_n})=x$ we get that $(2)\equiv x=1+\frac{1}{x}$. This is easy enough to solve, we simply multiply through by $x$ and rearrange to get a quadratic we can solve.

$\begin{align*}x^2-x-1=0
\\\Rightarrow x = \frac{1\pm\sqrt{5}}{2}\end{align*}$

Only one of these two possible x values are logical, as the sequence is weakly increasing and all of the terms are greater than or equal to 0 clearly the ratio between terms is positive $\therefore\ x = \frac{1+\sqrt{5}}{2}$. This is the golden ratio, $\phi$.

Our definition that $x=1+\frac{1}{x}$ has interesting implications when we represent it as a recurrence relation, $x_{n+1}=1+\frac{1}{x_n}$. If we set $x_0 = 1$ then $x_1 = 1+\frac{1}{1},\ x_2 = 1 + \frac{1}{1+\frac{1}{1}}$, etc. This continues on such that $\lim_{n\to\infty}(x_{n+1})\equiv\phi = 1 + \frac{1}{1+\frac{1}{1+\frac{1}{\ddots}}}$, this is an example of an infinite continued fraction.

A continued fraction is essentially fractions within fractions (frac-cep-tions?), $a_1 + \frac{1}{a_2 + \frac{1}{\ddots \, + \frac{1}{a_m}}}$.

An infinite continued fraction is one that does not terminate after a finite sequence of iterations, $a_1 + \frac{1}{a_2 + \frac{1}{\ddots}}$. An irrational number can be continually, more accurately, approximated using infinite continued fractions.

If we want to represent any integers as an infinite continued fractions we can begin by thinking of the general continued fraction:

$\begin{align*}a=\frac{k}{1+\frac{k}{1+\frac{k}{\ddots}}} \Rightarrow a=\frac{k}{1+a} \Rightarrow a^2+a-k=0\end{align*}\tag{3}$

For a to be rational in this quadratic $1+4k$ must be a square number, clearly this square will be odd (odd + even = odd) from this you can quickly calculate the first few values of k to be $k = 2,\ 6,\ 12,\ 20,\ ...$, it can be quickly shown that the nth term $k_n=n^2+n$. Using the quadratic formula on $(3)$ we find that $a=\frac{-1+\sqrt{4(n^2+n)+1}}{2}\equiv\frac{-1\pm\sqrt{(2n+1)^2}}{2}\equiv\frac{2n}{2}=n$.

Therefore for all positive integers, a, we can represent them as an infinite continued fraction if and only if $k=a^2+a$. Or to put it mathematically:

$a=\frac{k}{1+\frac{k}{1+\frac{k}{\ddots}}},\ a\in\mathbb{N} \Leftrightarrow k=a^2+a$

This is one example of how useful (and fun!) continued fractions can be, they are also much more mathematically relevant than decimals as they provide a lot more useful information about the nature of the number itself.

See if you can express $\sqrt{2}$ as an infinite continued fraction. Leave any answers in the comment section.