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Triangle in a Triangle

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

This is a long Hints section - you may need to scroll down to find the help that best matches your interest

Vector Algebra


triangle labelled ready for vector reasoning
Vector algebra can take a little getting used to. You'll need to invest a bit of time, but if you do, you'll possess a really powerful thinking tool with many important applications.

So what do the symbols like $\mathbf{a}$, $\mathbf{b}$, $\mathbf{-a}$, $\mathbf{a + b}$, $\mathbf{a - b}$, $\mathbf{a + 2b}$, all stand for?

The letter $\mathbf{a}$ , typed in bold (or underlined when hand-written), means a quantity of shift or displacement - a specific distance in a specific direction.

In the problem we're doing here $\mathbf{a}$ will stand for the quantity of shift from $C$ to $A$ and $\mathbf{b}$ will stand for the quantity of shift from $C$ to $B$.

$\mathbf{a + b}$ means the combined effect, or total shift, when $\mathbf{a}$ and $\mathbf{b}$ are both applied.

$\mathbf{-a}$ means the opposite of $\mathbf{a}$, that is the shift that exactly undoes the effect of $\mathbf{a}$.

Important idea : if any two different vectors in the plane are selected every other shift in the plane can be expressed as a multiple (possibly a fraction and maybe negative) of the first "base" vector combined with a multiple of the second "base" vector.

Moreover, for any given shift there is only one multiples combination using the specified "base" vectors which is equivalent to it.

The illustration below shows a shift which is the combined effect of a single $\mathbf{a}$ taken with $2$ lots of $\mathbf{b}$.

diagram showing vector addition
The two "base" vectors must not be parallel of course, which would mean that one "base" vector was just a simple multiple of the other, and that's not really two different vectors in the sense required here.

OK, back to the triangle.

The vertices are $A$, $B$, and $C$, with points $X$, $Y$ and $Z$ on the sides.

$X$ opposite $A$, $Y$ opposite $B$ and $Z$ opposite $C$.

The "base" vectors could be any pair but we'll take the shift $C$ to $A$ as vector $\mathbf{a}$ and the shift $C$ to $B$ as vector $\mathbf{b}$.

The shift from $B$ to $A$ is the same as the joint effect of $\mathbf{a}$ and $\mathbf{-b}$ together.

Displacement $B$ to $A$ is usually written as $\mathbf{a-b}$ rather than $\mathbf{a+ -b}$

$C$ to $Y$ is $\frac{2}{3}$ $\mathbf{a}$

$B$ to $Z$ is $\frac{1}{3}$ $\mathbf{(a - b)}$

The next two are a little harder:

$C$ to $Z$ is $\mathbf{b}$ combined with the $\frac{1}{3}$ $\mathbf{(a - b)}$ above and together they make $\frac{1}{3} \mathbf{a} + \frac{2}{3}\mathbf{b}$ when simplified.

$B$ to $Y$ is $\frac{2}{3} \mathbf{a - b}$

Now let's put these vectors to work.

Remember the task : the internal lines are each in three parts because they intersect each other. We have to show that these intersections split each line in the ratio $3:3:1$

We will use vectors to describe the location of the point where $CZ$ and $BY$ intersect.

First we describe all the points that lie on $CZ$, next we describe all the points that lie on $BY$, then we see exactly what it would require for a point to fit both those descriptions and therefore be the point which lies on both lines.

The displacement vector from $C$ for any point along $CZ$ will have this form : $$\mu\left(\frac{1}{3}\mathbf{a}+\frac{2}{3}\mathbf{b}\right)$$ where $\mu$ is just some fraction value (multiple)

Careful with this next bit:

The displacement vector from $C$ to any point along $BY$ is $$\mathbf{b} + \lambda \left(\frac{2}{3}\mathbf{(a - b)}\right)$$ $\lambda$ is just a multiplier, called a scalar, and works like $\mu$

The single $\mathbf{b}$ describes the displacement from $C$ to $B$ then the extra shift moves the point up or down the line $BY$ from there.

We are after a point of intersection, so we're looking for a point which fits both descriptions $$\mu\left(\frac{1}{3}\mathbf{a}+\frac{2}{3}\mathbf{b}\right)$$ and $$\mathbf{b}+ \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{a}\right)$$
Using an idea from the beginning of this section : any displacement expressed as a combination of $\mathbf{a}$ and $\mathbf{b}$ will have its own unique pair of multipliers (scalars).

So for point $P$, which is on both lines, the amount of $\mathbf{a}$ will be the same in both descriptions, and likewise the amount of $\mathbf{b}$.

We are looking for $\mu$ and $\lambda$ values so that $$\frac{1}{3}\mu= \frac{2}{3} \lambda$$ (the quantity of $\mathbf{a}$ must be the same) and $$\frac{2}{3}\mu = 1 - \lambda$$ (the quantity of $\mathbf{b}$ must be the same)

The first equation tells us that $\mu = 2\lambda$. Substitute for $\mu$ in the second equation to get $$\frac{4}{3}\lambda = 1 - \lambda$$
leading to $$\frac{7}{3}\lambda =1$$ or $$\lambda =\frac{3}{7}$$ and therefore $$\mu =\frac{6}{7}$$
This result means that $P$ is $\frac{6}{7}$ of the way along $CZ$ and $\frac{3}{7}$ of the way along $BY$

The point $P$ splits $CZ$ in the ratio $6:1$ and splits $BY$ in the ratio $3:4$

We could now repeat the above process to find the intersection of $AX$ and $CZ$ by considering displacement vectors from $A$. And then continue to find the intersection of $BY$ and $AX$ by considering displacement vectors from $B$. But the outcome would be the same each time because it depends only on the ratio and that hasn't altered.

The general result is that the line from any vertex, $A$, $B$ or $C$, has two intersection points as it crosses the triangle. The nearer one is $\frac{3}{7}$ along (as $P$ was along $BY$) and the further one occurs $\frac{6}{7}$ along (as $P$ did along $CZ$)

So the partition ratio for any of these three lines is $3:3:1$

One last thought : new methods are often grasped a bit at a time.

Don't be surprised if you have to run through this reasoning several times, getting a little more from it each time.

Perhaps even have a rest and come back.

Working with someone else can often be helpful when you are trying to follow something new.

When you think you have it, why not take a clean piece of paper and try to write your own proof looking up as little as possible.

Dividing a line in a given Ratio (using only a straight edge and a pair of compasses)

Suppose the line $AB$ is to be divided into three equal parts.

line AB
Construct a line of any length from $B$ at any angle to $AB$.

AB with extra line
Open the compasses to any size setting and make three arcs along the new line, starting from $B$ and stepping on from the last arc to make the next.

These new intersections have been labelled $C, D$ and $E$.
AB with lined and arcs
$E$ is now joined to $A$ and then lines parallel to $EA$ are drawn from $C$ and from $D$, long enough to intersect the line $AB$.

line EA added to the diagram
all lines parallel to EA added
These lines cut $AB$ in the necessary positions for $AB$ to be divided into the required three equal parts.

Parallel lines are consructed by replicating the angle downwards fom $E$ to $A$, at point $C$ and point $D$.

If you need it, the next section will show you how to replicate an angle without using a protractor to measure it.

Replicating an Angle using a pair of Compasses but not a Protractor

Suppose we need a line downwards from $B$ at the same angle as the line downwards from $A$.
angle to replicate
Step 1. Make any convenient setting of the compasses and sweep an arc across both lines from $A$.

Step 2. With the same setting of the compasses sweep an arc below $B$.

diagram 1
Step 3. Use the $A$ figure to set the compasses so that they span between the two intersections of the arc with the lines from $A$.

Step 4. With this new setting, place the point of the compasses where the first arc (red) intersects the line from $B$ and mark off an arc (shown in blue) to intersect that first arc.

diagram 2
Step 5. Join $B$ to this point to complete the figure.

Why exactly does this replicate the angle?

Practise a few times, including cases where the given line from $B$ is not parallel to the corresponding line from $A$.

diagram 3