# Thread: equilateral triangles and the square root of three

1. ## equilateral triangles and the square root of three

Hi, I'm trying to understand this bit of geometry in a computer program. What it is is an equilateral triangle that's drawn from the bottom left corner. The first step in doing this is to give the position of the bottom left corner from the center of the triangle. It does it like this (-edge / 2, edge / (2 * Math.sqrt(3)). I'm sure that's quite obvious even if you don't understand java. edge if obviouslt the length of one of the 3 sides, the part before the comma is the x coordinate and after the y(which is at 0 at the top and increases as you work down the screen).

The first part is obvious -edge / 2, that's minus the edge length / 2, ok. But the next part is where I'm confused. what I need here if half of the alitude of the triangle, I know the ration is 1 , 0.5, sqrt(3) / 2, so I need a 1/4 of the sqrt of 3 but does dividing the length of the edge by 2 * the sqrt of three give the full length of the altitude, I just can't seem to understand how it works.

2. ## Altitude of a triangle

Hello AlyoshaKaz

Originally Posted by AlyoshaKaz
Hi, I'm trying to understand this bit of geometry in a computer program. What it is is an equilateral triangle that's drawn from the bottom left corner. The first step in doing this is to give the position of the bottom left corner from the center of the triangle. It does it like this (-edge / 2, edge / (2 * Math.sqrt(3)). I'm sure that's quite obvious even if you don't understand java. edge if obviouslt the length of one of the 3 sides, the part before the comma is the x coordinate and after the y(which is at 0 at the top and increases as you work down the screen).

The first part is obvious -edge / 2, that's minus the edge length / 2, ok. But the next part is where I'm confused. what I need here if half of the alitude of the triangle, I know the ration is 1 , 0.5, sqrt(3) / 2, so I need a 1/4 of the sqrt of 3 but does dividing the length of the edge by 2 * the sqrt of three give the full length of the altitude, I just can't seem to understand how it works.
I think what you need here is the fact that the altitudes of any triangle meet at a point that trisects each altitude.

So, if the sides of an equilateral triangle are each of length
$a$, the altitudes (as you say) are of length $\frac{a\sqrt{3}}{2}$ and the point of intersection of the three altitudes is at a distance

$\frac{1}{3}\times \frac{a\sqrt{3}}{2} = \frac{a}{2\sqrt{3}}$

from the base.

3. So the triangle is actually centered at the intersection of the three alitiudes rather than the center? I don't understand here how you get from one side of the equals to the other wouldn't it just be a * the square root of 3 / 6?

4. no one knows?

5. ## Altitudes and medians

Hello AlyoshaKaz
Originally Posted by AlyoshaKaz

So the triangle is actually centered at the intersection of the three alitiudes rather than the center? I don't understand here how you get from one side of the equals to the other wouldn't it just be a * the square root of 3 / 6?

You have asked two questions here:

1

Where is the centre of an equilateral triangle? By the centre of a triangle, we probably mean the centre of mass - in other words, the point at which a triangular shape will balance. In my first posting I said that this is where the altitudes meet - and this is correct for an equilateral triangle. However I also said that this also applies to any triangle. This isn't correct. Sorry! I should have said that, for
any triangle, the centre is where the medians meet.

Let me clarify this a little if I can.

• A median of a triangle is a line that joins a vertex to the mid-point of the opposite side. The medians of a triangle meet at the triangle's centre of mass - one-third of the way up a median from the mid-point of the side.
• An altitude of a triangle is a line drawn from a vertex to meet the opposite side at right-angles. If this opposite side is the base of the triangle, then the altitude is simply the height of the triangle - hence the use of the word 'altitude'.
• An equilateral triangle's altitudes and medians are the same line - they are the lines of symmetry of the triangle.

So, where is the centre of an equilateral triangle? It is at the point where the triangle's altitudes - its lines of symmetry - meet. You'll find that this is where a piece of card cut in the shape of an equilateral triangle will balance.

And whereabouts is this centre? The answer is that it's one-third of the way up an altitude from the base. There's no other point that can sensibly be called the centre of the triangle.

2

How do I get from $\frac{1}{3} \times \frac{a\sqrt{3}}{2}$
to $\frac{a}{2\sqrt{3}}$?

The answer lies in the fact that $\sqrt{3} \times \sqrt{3} = 3$
. Look:

$\frac{1}{3} \times \frac{a\sqrt{3}}{2}$

$=\frac{1}{\sqrt{3} \times \sqrt{3}} \times \frac{a\sqrt{3}}{2}$

$= \frac{1}{\sqrt{3}} \times \frac{a}{2}$
, by cancellation

$= \frac{a}{2\sqrt{3}}$

Is that OK now?

Hello AlyoshaKaz

You have asked two questions here:

1

Where is the centre of an equilateral triangle? By the centre of a triangle, we probably mean the centre of mass - in other words, the point at which a triangular shape will balance. In my first posting I said that this is where the altitudes meet - and this is correct for an equilateral triangle. However I also said that this also applies to any triangle. This isn't correct. Sorry! I should have said that, for
any triangle, the centre is where the medians meet.

Let me clarify this a little if I can.

• A median of a triangle is a line that joins a vertex to the mid-point of the opposite side. The medians of a triangle meet at the triangle's centre of mass - one-third of the way up a median from the mid-point of the side.
• An altitude of a triangle is a line drawn from a vertex to meet the opposite side at right-angles. If this opposite side is the base of the triangle, then the altitude is simply the height of the triangle - hence the use of the word 'altitude'.
• An equilateral triangle's altitudes and medians are the same line - they are the lines of symmetry of the triangle.

So, where is the centre of an equilateral triangle? It is at the point where the triangle's altitudes - its lines of symmetry - meet. You'll find that this is where a piece of card cut in the shape of an equilateral triangle will balance.

And whereabouts is this centre? The answer is that it's one-third of the way up an altitude from the base. There's no other point that can sensibly be called the centre of the triangle.

2

How do I get from $\frac{1}{3} \times \frac{a\sqrt{3}}{2}$
to $\frac{a}{2\sqrt{3}}$?

The answer lies in the fact that $\sqrt{3} \times \sqrt{3} = 3$
. Look:

$\frac{1}{3} \times \frac{a\sqrt{3}}{2}$

$=\frac{1}{\sqrt{3} \times \sqrt{3}} \times \frac{a\sqrt{3}}{2}$

$= \frac{1}{\sqrt{3}} \times \frac{a}{2}$
, by cancellation

$= \frac{a}{2\sqrt{3}}$

Is that OK now?