There are two triangles that were of particular interest to the Egyptians and they have some aspects that are curious and (I hope you'll agree) interesting to us as well. I call them the Knight Triangle and the Delta Triangle. Oddly, although mathematicians recognise a number of different triangles as 'special' I have not been able to find particular names for these two, in my view, quite special triangles and have thus taken the liberty of inventing names for them myself.
The Knight Triangle is a Right Triangle (one of its angles is a 90°, i.e. Right Angle) and its height is twice the length of its base or base is twice the length of its height. This resembles the movement of a knight in the game of chess: it can move in an "L" shape, one square horizontally and two vertically, or one vertically and two horizontally. Another way to define this triangle is to consider it to be a double square (doppelquadrat) divided by its diagonal.
The interesting thing about the Knight Triangle is that the length of its hypotenuse is its base times the square root of 5. This is easy to prove using Pythagorus' formula for calculating the lengths of the sides of a Right Triangle: a2 + b2 = c2
12 = 1;
22 = 4;
1 + 4 = 5;
square root of 5 = 2.2360679774997896964
What makes it particularly interesting is that the square root of 5 also turns up in the formula for calculating the Golden Ratio (phi or Φ or φ), which is equal to 1 plus the square root of 5 divided by 2.
The result of that calculation is 1.61803398874948482 (to the precision of my calculator).
The Delta Triangle is an isosceles triangle (two sides of equal length) which has a height the same length as its base. I call it a Delta Triangle because it resembles the upper case Delta character (Δ) in the greek alphabet. It can also be considered to be two Knight Triangles back to back.
There are two circles of interest related to a triangle: one that sits inside the triangle and touches each of the three sides, this is called the "inner circle"; and one that sits outside the triangle and touches each of the three points (or vertices), called the "circumcircle".
For a Delta Triangle you can calculate the radius of the inner circle (the "inner radius") with the following formula:
1/Φ is the reciprocal of Φ and, another point of interest, the reciprocal of Φ is the same as Φ - 1.
1/Φ or Φ -1 = 0.618033988749894
Thus, the inner radius of a Delta Triangle is the phi mean of half the base.
I discovered a way to find the inner radius, without using a calculator. All you need is a ruler and a pair of compasses. Draw your Delta Triangle (actually, this should work for any isosceles triangle) and draw a meridian line from the apex to the centre of the base. Using your compass, scribe an arc from the centre of the base up to the side, centred on the vertex at the end of the base. Draw a line from that same vertex through the centre of the arc and it will intersect the meridian line at the inner centre (the centre of the inner circle). Thus, the distance from that point to the base is the inner radius.
Incidentally, this is also an interesting demonstration of the relationship between Φ and π (the relationship of a circle's radius to its circumfrence.)
To calculate the circumradius (the radius of the circumcircle) simply multiply the height of the Delta Triangle by 0.625.
As proof for this assertion I will demonstrate how to find the circumradius without using a calculator.
Draw your Delta Triangle and a meridian line from the apex to the centre of the base. Using your compass find the centre of the side and draw a line through that point, perpendicular to the side, to the meridian line. Where it intersects the meridian is the circumcentre (the centre of the circumcircle) and the distance from there to the apex is the circumradius. (this method may work for all isosceles triangles.)
Note that I have marked the meridian line at 1/8ths. The circumcentre is at 5/8ths of the distance from the apex. Note also that a horizontal line half way up the triangle intersects the side at the same point as our circumradius and that point is directly above a point 1/4 the length of the base. That makes it very easy to use Pythagorus' formula to run the proof for these two triangles. I'll leave calculating the proof as an exercise for the reader.
5/8ths, by the way, is 0.625, QED.
The Egyptians wouldn't have expressed it as 5/8ths since (except in the case of 2/3rds) they expressed fractions as unit fractions added together, so 5/8ths would have been written 1/2 + 1/8.
There are a few simple methods to find phi:
Looking at the second method you will note that it is a Knight Triangle. The base of that triangle matches the meridian line of our Delta triangle and, magically, the phi mean of the meridian line falls at the circumcentre!
So we have discovered that for a Delta Triangle the inner radius is the phi mean of half the base and the circumradius is the phi mean of the height.
Another symmetry I noticed is that an arc with a radius equal to the circumradius, centred on the apex, where the arc meets the side of the Delta Triangle, a line perpendicular to the side from that point intersects the meridian at the inner centre.
I hope that you are beginning to see some of the fascination I have for these triangles and that you can perhaps understand my surprise that they are not included in the family of "special triangles" and more talked about in mathematics.
I'm sure the Delta Triangle has more interesting secrets hidden away that will reveal themselves to more study and contemplation. If you discover anything further, have spotted a mistake or have questions please feel free to drop me an email (deltaknight at declarepeace.org.uk)