Deep Secrets

The Square Root Algorithm

The basis of this technique lies in the observation that the area of a square is equal to the length of the square's side multiplied by itself. (For instance, a square that is 6 units long on a side will have an area of 6x6 = 36 square units.) Therefore, we can say that the length of a side is the "square root" of a square's area. Let us look at how this empirical understanding can be put to use.

Suppose that we want to find the square root of the number 5. The nearest 'perfect square' that is less than 5 is obviously 4; that is, 2 x 2 equals 4. (A perfect square is a number that is a whole number squared.)

The question, then, is whether there is a way to add the area of the remaining one unit to the area of the square containing four units so that we can construct a square containing an area of five units. Surely, if each of these five squares could be subdivided into sufficiently smaller segments, then those of the 'one unit' left over square to the right could be used to add layers around the perimeter of the larger 'four unit' square to the left.

In the diagram above, each original square has been equally subdivided into 8ths along each side, such that each original unit square now comprises 64 smaller squares, thereby causing the larger square to the left to comprise 4 x 64 = 256 smaller squares. However, even greater precision is possible if a finer subdivision is used. Because of the halving and doubling nature of Egyptian computation, I believe it likely that in most cases the Egyptian scribe would have imagined that each original unit square was subdivided into 64 x 64 equivalent smaller units. As a result, the original unit square on the right would then have 4,096 smaller squares, and the original four unit square on the left would have 4 x 4,096 = 16,384.

In addition, the perimeter of the original four unit square would now contain 128 + 128 + 126 +126 = 508 smaller squares, and not the perhaps expected 4 x 128 = 512. This is because as one counts the smaller squares on the perimeter, the corner squares do not get counted twice. The next step in the process is to take the smaller squares from the original unit square on the right and use them to layer around the larger square.

As one proceeds to do this, it will be noticed that even though there are 508 smaller squares on the perimeter of the square on the left, the corner squares each have 2 sides facing out. Therefore, there are in fact 512 sides facing out to which new squares must be matched up. When a layer of this one-on-one matching is done, however, there will still be gaps at each of the corners remaining to be filled. (See blue layer in Daigram 3).

This means that the new layer to be added will need yet an additional four squares to fill these corner gaps. Notice, then, that the new layer contains the same number of squares as the number of sides facing out in the previous layer (with this number of sides facing out being four more than the actual number of squares in this previous perimeter) plus the four squares needed to fill in the corner gaps. In other words, each and every new layer will require 8 more of the same size squares than were in the previous layer. To facilitate this overall procedure, an "8 units per layer" table can be tabulated and kept handy. (Such a table follows this discussion). Note that each new layer that is added will add the width of 2 smaller square units to the length of each side.

Returning to our example of the 5, the perimeter of the main square will contain 508 smaller squares, but have 512 sides facing out. As a result, it will be found that after taking units from the left-over square and using these to place 7 layers around the main square, 3,780 of the left-over square's 4,096 units will be used. This will leave 316 left-over smaller squares behind. Each side of the larger square will at this point then have a length equal to the beginning 128/64th units plus the 14/64th newly added units = 142/64th units. (The 7 newly added layers add 7 x 2 = 14 units to the length of each side).

A quick calculation will now reveal that if each of the 316 remaining smaller squares were imaginarily sliced into 9ths, there would be 9 x 316 = 2,844 of these slices (each being 1/9th of 1/64th of a unit thick, which equals 1/576th of a unit), and these 2,844 slices would be enough to layer almost exactly 5 times around the main square. (Since each side at this point will have 142 smaller squares, there are are 142 x 4 = 568 sides facing out. 2,844 divided by 568 equals 5.00704.) Therefore, the 5 can now be approximated as: 128/64 + 14/64 + 10/576. This can be further simplified to: 2 + 72/576 + 64/576, which can be reworked into a unit fraction format (as was used in Ancient Egypt), and be restated
as : 5 = 2 + 1/8 + 1/9.¹

The codified principles of this method are:

Principle 1) The perimeter of any square divided into S x S smaller units equals the number of these units per side multiplied by 4, take away 4. As a formula, this can be stated : (4 x S) - 4 = Perimeter. This principle holds true because the corner squares do not get counted twice.

Principle 2) The number of units in each new layer that is added equals the number of similarly sized units in the previous layer plus 8.

Principle 3) After N layers, the total number of same size squares that have accrued equals: N multiplied by the number of these size squares in the starting (base) perimeter plus the sum of the "8 units per layer" series for all the layers added.

For example, if the base perimeter has 508 squares, then the first layer added will have 508 + 8 squares. The second layer will have (508 + 8) + 8 squares, etc. At the Nth layer, the number of squares that will have accrued is (508 x N) plus the sum of the "8 units per layer" series after N layers.

Principle 4) Each layer that is added increases the length of a side of the larger square by 2 of the units in question. (Refer to Appendix Diagram 3).

Clearly, this table can be extended to whatever length may be deemed necessary. The Egyptian scribes' reliance on, and use of, various tables as an aid to computation is well documented.²

The Square Root of 3

The nearest perfect square that is less than 3 is 1, and so 1 becomes the beginning square. The perimeter of this square is 64 x 4 - 4 = 252/64ths. The 2 left-over squares will then have 4,096 x 2 = 8,192 smaller squares. The "8 units per layer" table will show that these 8,192 squares are enough to emplace 23 layers, leaving 188 smaller squares still remaining.

The beginning square will at this point have a side length of: 64/64ths + 46/64ths = 110/64ths. The number of sides facing out will be 110 x 4 = 440 sides. If the 188 remaining squares were each sliced into 40ths, then there would be enough of these slices to go around the new perimeter exactly 17 times (188 x 40 = 7520. And 7520 ÷ 440 = 17.)

Since each of these slices is 1/40th of 1/64 = 1/2560th of a unit thick, the side of the beginning square will now contain
64/64 + 46/64ths + 34/2560 units. An Egyptian scribe would likely have simplified this to: 1 + 1/2 +1/5 + 1/32 + 1/1280.³

The Square Root of 2

Following this same square root algorithm, and dividing by 64ths, will yield a square root approximation for the 2 of: 1 + 1/4 + 1/8 + 1/32 + 1/128, which is only .00015 too small.⁴

This approximation leads to an interesting coincidence of number. As was explained in an earlier section, the ancient Egyptian algorithm for finding the area of a circle implied an approximation for Pi that is, in essence, the equivalent of 256/81 (from the squaring of '8/9ths x 2'). Noting that Pi is the ratio of the circumference (perimeter) of any circle to that circle's diameter, once it was discovered that a square's diagonal was 2 times the length of the square's side, would not the scribes have sought a perimeter to diagonal relationship for a square? If they had, and if the above value for the 2 had been used, then that relationship would become 4 divided by 181/128 (note that the 2 approximation equals 1+1/4+1/8+1/32+1/128, which is the same as 181/128. Therefore, 4 ÷ 181/128 = 512/181). This means that the constant relationship of a square's perimeter to its diagonal can be understood to equal 512/181, which is the same as 2 x 256/181. It is interesting to wonder whether the coincidence of number between this value and that associated with Pi from the area algorithm may have had reinforcement significance for the ancient Egyptian choice of methods.

Square Root of Numbers Less Than 1

Since the algorithm method here described needs a beginning square, minor difficulties arise in the face of the necessity to take the square root of a number that is less than 1. Fractional quantities less than 1 will first need to be reconfigured into a common denominator form, such as 3/4 instead of the unit fraction form of 1/2 + 1/4. From here, the algorithm can be used to take the square root of numerator and denominator separately, followed by the consequent division to achieve the result. Either this, or the fraction could be multiplied by 4,096 (i.e., 64 x 64, or by 80 x 80, etc.) to see what fractional equivalent of a completed square it would be. For instance, in the case of 64 x 64, 3/4 could be rewritten 3,072 ÷ 4096. The square root of 3,072 could then be taken and the result divided by 4096 (= 64) to get the final answer.

The Inscribed Endecagon and Nonagon

The diagram of the inscribed hexagon can be used to find a computable length for the side of an inscribed equal-sided endecagon, an 11 sided polygon.

Appendix Diagram 4 shows this construction. The claim is that the line PG will do the job. 360 ÷ 11 yields an angle of 32 43' 38''. An inscribed chord of this angle will be .56338 the length of the circle's radius. PG works out to be only .0027 larger, an amount not likely to be discernible diagrammatically.

In this diagram, we already know the lengths of BC (=1), GC (=1/2), OG (=3/2), and of course, all the radii. The trick to finding PG is to realize that angle DAO is 15. Since triangle BOC is equilateral, we know angle BOG equals 30. Therefore, angle BOA must be 150 and DAO 15 (because triangle BOA is isosceles). Using the tangent of 15, we can find DO and hence YG. From this, YC can be found. DY ÷ YC will equal PG ÷ GC, and so PG is findable.

The inscribed pentagon can be used to find a computable length for the side of an inscribed equal-sided nonagon, a 9 sided polygon.

In the above diagram, lines have been drawn connecting each corner of the pentagon with the two corners farthest opposite, creating a five pointed star known as a pentagram. If a compass is set equal to the line segment EN, a chord of this length will prove a bit too large to fit exactly nine times around the circle. In contrast, a chord set equal to line segment AD will prove just a bit too small. However, if the compass is set to a length precisely mid way between the lengths of EN and AD, the resulting chord will create an apparently perfect equal-sided inscribed nonagon.

A study of the inscribed pentagon in Diagram 12 in the Pentagon section will show that line ON above is equivalent to line PR in that diagram, and hence to the sine of 18. Therefore, EN equals EO - ON = 1 minus .3090 = .6910. Similarly, as explained in footnote 5, AD is equal to 1 minus the tangent of 18, or 1 - .3249 = .6751.⁵ Mid way between these two findings is the length .68305, which is only about .001 short of ideal for the nonagon value.

If one is looking for a way to create equal-sided inscribed polygons via the diagrammatic method, then through trial and error, ways such as have just been described will eventually be found. To achieve the desired results, however, it is necessary to work with patience and precision using marking tools that are kept well sharpened.

Refining The Trigonometric Table to 1/2 Intervals

As stated in the text, a trigonometric table could have been worked out to 1 1/2 intervals via the sine subtraction formula (using 9- 7 1/2, for example), and then using an iteration of the process from there. How could it be refined further?

From the pentagon analysis, we saw that the ratios for the 4 1/2 angle could be found. The nonagon construction allows a similar finding for 5. The subtraction formula would then give a value for 1/2.

Another possibility is that with values for 4 1/2, 3 and 1 1/2 on hand, it was seen that at these small angles the length of the sine of the angle (opposite ÷ hypotenuse) increased by a constant value (equal to .02617). In other words, the sine of 3 degrees is twice that of 1 1/2 degrees, and the sine of 4 1/2 degrees is this same amount more than that of 3 degrees. If this was true (which it is), then it could be logically assumed that the sine of 1/2 degrees was 1/3rd of that for 1 1/2 degrees (which it also is). Why this is the case will hopefully be made clearer by the diagram below.

In the above, angle BOT is taken to be an angle less than 5 degrees. Line BG is the sine of angle BOT, and OB and OT are both radii. ON has been drawn perpendicular to BT, (BT being the chord of angle BOT), dividing both angle BOT and the line BT in half. As a result, line BN is half of the line BT, and is also the sine of angle BOX.

It is visually apparent that difference between the lengths of BG and BT is small, and also that when angle BOT is halved, the difference between the lengths of BN and BX is smaller still (BX being the chord of angle BOX). What we are seeing is that at very small angles, the sine of an angle and the chord of that angle are becoming very nearly equal in length. And so, at these very small angles the sine of a given angle is not only equal in length to 1/2 the chord of the angle which is twice as large, but it is also increasingly nearer to being equal to 1/2 the length of the sine of the angle which is twice as large. It is this latter occurrence which is being noticed in the computed results for the angles of 4 1/2, 3 and 1 1/2.

Euclid's Derivation of the 36 Angle Via the Golden Ratio ⁶

The following is a rewording into plain english of Euclid's presentation as it appears in his work, The Elements, written circa 300 B.C. As stated before, it is unlikely that Euclid was the originator of this derivation, but rather the codifier and disseminator. I include the following information because: 1) it is intrinsically interesting; and 2) it is interesting to compare this more studied approach with that shown in the Inscribed Pentagon section. It is as if at some point in antiquity it was felt that the empirical nature of the more ancient diagrammatic derivation of the inscribed pentagon (along the lines of that shown previously) was no longer deemed sufficiently rigorous. Lots of "little grey cells" were burned by whoever originally forged the following method.

In this first diagram, line JS has been drawn such that JM = MS. An arbitrary length, CS, is then added continuing JS to point C. A square is drawn on CS such that CT = CS = 'F'. Similarly, a square is drawn on MC.

The theorem now states that: JC x CS + MS² = MC²

Diagrammatically this means that the area of the rectangle bordered by the lines JC and CT (boxes 3, 4, and 1) plus the area of the square on MS (box 5) equals the area of the square on MC (boxes 3, 4, 5, and 1).

Inspection of the diagram will show that the proposed theorem is indeed true.

The next theorem in the chain is a special case of Theorem 6 above.

Theorem 11 shows how to find a point B on any line SK, such that:

SK x BK = SB², or written another way, so that: SK÷SB = SB÷BK.⁷

Line JS is drawn so that JM = MS. Square JSKD is drawn on JS. MK is swung upward to JS extended, creating MC = MK. Square CTBS is now drawn on CS, and as a result, CS = CT = SB = JP.

Employing the previous theorem, we have that : JC x CS + MS² = MC².

From the Pythagorean Theorem we know: MK² = SK² + MS².

Since MC = MK, then JC x CS + MS² = SK² + MS².

Subtracting MS² from both sides of the equation leaves: JC x CS = SK².

Since CS = CT, the previous equation means that the area of the rectangle bordered by JC and CT is equal to the area of the square on SK. Since both of these figures contain the area of rectangle JSBP, then the area of the square on CS (i.e., CS² ) must be equal to the area of the rectangle KD x BK. Therefore, CS² = KD x BK

Remembering that CS = SB and that KD = SK, we can restate this last equality to read: SB² = SK x BK, which is what we set out to prove.

This theorem states that if from a point C outside of a circle, a line is drawn tangent to the circle at point A, and another arbitrary line CJ is also drawn from C to intersect the circle at S and J, then: JC x CS = CA².

There are two possibilities within this arrangement:

1) where CJ passes through the center of the circle, and

2) where it doesn't.

The diagram above shows the first circumstance. In it, SM = MJ = MA, since all are radii of the circle. From Theorem 6 Book II we have:

JC x CS + MS² = MC². And since MS = MA, then:

JC x CS + MA² = MC².

From the Pythagorean Theorem we have: MC² = CA² + MA² (from right CAM).

Combining the last two equalities, and subtracting MA² from both sides of the equation, we find what we set out to prove:

JC x CS = CA²

The next diagram will explore the second possibility.

In this next diagram we see the situation where JC does not pass through the circle's center. OM is drawn perpendicular to JC from the circle's center making MS = MJ.

From Theorem 6, Book II we then have:

JC x CS + MS² = MC²

From the right triangles OMS and OMC we get the following:

MS² = OS² - OM² and MC² = OC² - OM²

These values for MS² and MC² can be substituted into the first formula:

JC x CS + OS² - OM² = OC² - OM², and so JC x CS + OS² = OC².

However, OS = OA (both are radii), and therefore JC x CS = OC² - OA².

Applying the Pythagorean Theorem to right triangle CAO we have:

CA² = OC² - OA². Substituting back into the previous equation, we find that the theorem is proven for this case as well:

JC x CS = CA².

Theorem 37, Book III is the converse of Theorem 36. It states that if from a point C outside a circle, two lines are drawn, such that one line intersects the circle at two points (S and J) and the other meets the circle at its perimeter (at a point A), and further if JC x CS = CA², then CA will be tangent to the circle at point A.

This theorem states that the angle between a line that is tangent to a circle and a chord that meets it, is equal to the angle that subtends that chord from its opposite side. In the above diagram, therefore, angle ZAC is to be proven equal to angle ABC. (Similarly, angle GAB will be equal to angle ACB).

We start by noting that OA = OB = OC as they are all radii, and from their respective isosceles triangles we can state:

OAB = OBA = 1

OBC = OCB = 2

OAC = OCA = 3 and so,

2( 1) + 2( 2) + 2( 3) = 180. Therefore,

1 + 2 + 3 = 90.

However, since ZG is tangent at A, angle OAZ equals 90 as well. We then have:

OAZ = 3 + ZAC = 90 = 1 + 2 + 3.

This means that: ZAC = 1 + 2 , which is what we wanted to prove.

Similarly, GAB + 1 = 90 = 1 + 2 + 3. And so,

GAB = 2 + 3.

This theorem is the final link in the chain, and it's a corker. It begins with Theorem 11, Book II (the second theorem in this progression) which states that a point B can be found on a line SK such that SK x BK = SB². Using SK as a radius, a circle is next drawn with point S as its center. Chord WK is drawn such that WK = SB and line segments SW and WB are then also added. The theorem now states that with these conditions met, angle WSK (i.e., 5) will equal 180 ÷ 5 = 36.

To enact the proof, a point O must be found such that OW = OB = OS, thus allowing a circumscribed circle to be drawn around triangle SBW.⁸ Once this is done, the preceding theorems in this section are then brought in to play their hand.

Line SBK was intentionally drawn to allow SK x BK = SB². Since SB = WK, this equality can be rewritten: SK x BK = WK².

From theorem 37, Book III, we see that WK is then tangent to triangle SBW's circumscribed circle at W. This allows us to prove, using Theorem 32, Book III, that angle KWB ( 3) is equal to angle WSB ( 5).

Since SK = SW, we know that triangle KSW is isosceles, and so: 1 = 3 + 4.

From the diagram we can see that both 2 and ( 5 + 4) need angle SBW to equal 180. Therefore: 2 = 5 + 4. But since 5 = 3, this means:

2 = 3 + 4. Therefore, 1 = 2.

This last statement means that triangle KWB must be isosceles, and so WB = WK = SB. Therefore, triangle SBW is also isosceles, and 4 = 5. A final tally can then be made of the angles in triangle SWK:

4 = 5

3 = 5

1 = 5 + 5

5 = 5

Triangle SWK will therefore contain 5 x ( 5) = 180, and 5 must therefore be 36. With this knowledge, an inscribed equal-sided decagon, and hence an inscribed equal-sided pentagon, can easily be constructed. This Euclid does in Theorem 11, Book IV.

1. This approximation equals 2.23611, which is only .0000432 too large. Note that 14/64 + 10/576 is the same as 136/576, and is simplified from there.
2. See Gillings, op. cit., p. 45 and pp. 168-9.
3. This equals 1.7320312, which is only .00002 too small. For computations needing a less exacting precision, the 1/1280 can be deleted, yielding an approximation which is only .0008 too small..
4. As a point of comparison, note that if the squares are divided into 80ths instead of 64ths, then 2 can be found to be: 1 + 1/5 + 1/7 + 1/14. This is also extremely accurate and at times may have been computationally preferable.
5. If a radius is drawn from O to L, EOL will equal 72. Since EOL would then be isosceles, OEL will be 54. ZEL is the same as ELK, and both equal 2x54 = 108. Triangles ZEL and ELK are both isosceles, and so KEL equals 36. Therefore, OEA equals 18 (i.e., 54 - 36). The tangent of OEA is the opposite÷adjacent = OA ÷ OE, or simply OA since OE is a radius and equals 1.
6. I am indebted to Asger Aaboe's treatment of this subject in his Episodes From the Early History of Mathematics, pp. 61 - 70.
7. The reader may recognize that this is an upside down version of Diagram 11 in the Pentagon section. If JS is made equal to 1, then MS = 1/2, MK = 5/2, and CS = SB = (5-1)÷2 = .618034.
8. Euclid's Theorem 5, Book IV presents a straightforward technique for achieving this that involves the drawing (until they intersect) of perpendiculars from the midpoints of two of a triangle's sides.