Friday, April 21, 2006

More on Equiangular Triangles

In today's blog, I review more proofs on equiangular triangles that are taken straight from Euclid's Elements. I use these properties in showing that the ratio of circumference to diameter for any circle is always the same (which is, of course, the definition of pi).

Lemma 1: If two triangles have one angle equal to one angle and the sides around the equal angle proportional, then the two triangles are equiangular.

Proof:

(1) Assume that ∠ BAC ≅ ∠ EDF and BA/AC = ED/DF

(2) There exists a point G such that ∠ FDG ≅ ∠ BAC and ∠ DFG ≅ ∠ ACB [See here for details on the construction.]

(3) ∠ B ≅ ∠ G [since the angles of a triangle add up to 180 degrees, see Lemma 4 here for details]

(4) triangle ABC is equiangular with triangle DGF [from step #2 and step #3]

(5) From the properties of equiangular triangles (see here), we know that:

BA/AC = GD/DF

(6) From our assumption in Step #1, we can conclude that:

GD/DF = ED/DF

(7) And from step #6, we can conclude that:

ED ≅ GD

(8) From Step #1 and Step #2, we can conclude that:

∠ EDF ≅ FDG

(9) We can now conclude that triangle DEF ≅ triangle DGF from Side-Angle-Side (see Postulate 1 here) since:

(a) DF ≅ DF

(b) ∠ EDF ≅ ∠ FDG [Step #8]

(c) ED ≅ GD [Step #7]

(10) ∠ ACB ≅ ∠ DFE since:

(a) ∠ ACB ≅ ∠ DFG [Step #2]

(b) ∠ DFG ≅ ∠ DFE [From Step #9, see here for properties of congruent triangles if needed]

(11) Finally, we can conclude that triangle ABC is equiangular with triangle DEF since:

(a) ∠ BAC ≅ ∠ EDF [By assumption in Step #1]

(b) ∠ ACB ≅ ∠ DFE [By Step #10]

(c) ∠ B ≅ ∠ E [Since angles of triangles add up to 180 degrees]

QED

Lemma 2: Those triangles which have one angle equal to one angle and which the sides about the equal angles are reciprocally proportional, are equal.

Proof:

(1) Let the sides of triangle ABC and triangle ADE be reciprocally proportional so that:

(2) Since triangle ABC and triangle ABD share the same height, we can conclude that (see here):

AC/AD = area triangle ABC/area triangle ABD

(3) Likewise, since triangle ADE and triangle ABD share the same height, we can conclude that:

AE/AB = area triangle ADE/triangle ABD

(4) Therefore, triangle ABC/triangle ABD = triangle ADE/triangle ABD.

(5) But then we have:

area triangle ABC * area triangle ABD = area triangle ADE * area triangle ABD

(6) And if we divide both sides by the area of triangle ABD, we get:

area triangle ABC = area triangle ADE.

QED

Lemma 3: Similar triangles are one to another in the square ratio of corresponding sides.

Proof:

(1) Let triangle ABC and triangle DEF be equiangular triangles such that:

∠ A ≅ ∠ D
∠ B ≅ ∠ E
∠ C ≅ ∠ F

(2) There exists a point G such that BG = [(EF)*(EF)]/BC

(3) From (#2) we have:

1/BG = BC/[(EF)*(EF)]

which implies that:

EF/BG = BC/EF

(4) From a Property of Equiangular Triangles (see here if needed), we know that:

AB/BC = DE/EF

so that:

AB/DE = BC/EF

(5) We can also conclude that triangle ABG has equal area to triangle DEF since:

(a) B ≅ ∠ E (step #1)

(b) AB/DE = EF/BG (from combining step #3 with step #4)

(c) Lemma 2 above

(6) We can also conclude that BC/BG = (CB)2/(EF)2 since:

BC*BG = EF*EF (step #2)

And if we multiply BC to both sides, we get:

(BC)2*BG = (EF)2*BC

If we divide BG from both sides and (EF)2 from both sides, we get:

(BC)2/(EF)2 = BC/BG

(7) Since triangle ABC and triangle ABG have the same height, we can conclude (see here):

CB/BG = area triangle ABC/area triangle ABG

(8) Applying setp #6, gives us:

area triangle ABC/area triangle ABG = (BC)2/(EF)2

(9) Finally, from step #5, we have:

area triangle ABC/area triangle DEF = (CB)2/(EF)2

QED

References

Wednesday, April 19, 2006

Euclid and pi

Pi, also know as Archimede's constant, is not mentioned in Euclid's Elements. The closest that Euclid comes is Proposition II in Book XII which states that two circles are to each other as the squares of their diameters.

Postulate 1: Law of Trichotomy

For any two values x,y, there are only three possible states:
(a) x = y
(b) x is less than y
(c) x is greater than y

This is one of the postulates of real numbers. See here for details on constructing real numbers.

Lemma 1: Similar polygons inscribed in circles are to one another as the squares on their diameters

Proof:

(1) By assumption, polygon ABCDE is similar to polygon FGHKL with BM and GN being the diameters of circles.

(2) From the property of similar polygon (see definition above), we have:

∠ BAE ≅ ∠ GFL

BA/AE = GF/GL

(3) triangle ABE is equiangular to triangle FGL [See Lemma 1 here for details]

(4) Since they are equiangular, we know that:

∠ AEB ≅ ∠ FLG

(5) Since both angles open on the same length of the circumference (see here):

∠ AEB ≅ ∠ AMB
∠ FLG ≅ ∠ FNG

(6) From (4) and (5), we can conclude that:

∠ AMB ≅ ∠ FNG

(7) Since BM and GN are both diameters of the circle (see here),
we can conclude that both ∠ BAM and ∠ GFN are right angles.

(8) Since the angles of a triangle add up to 180 degrees (see Lemma 4 here), we can conclude from step #6 and step #7 that triangle ABM is equiangular to triangle FGN.

(9) From the properties of equiangular triangles (see Lemma 3 here), we know that:
BM/GN = BA/GF

(10) From (9), we can conclude that:
(BM/GN)2 = (BA/GF)2 = BM2/GN2 = BA2/GF2

(11) From similar polygons (See Theorem here), we know that:
(Area of ABCDE)/(Area of FGHKL) = BA2/GF2

(12) Putting this all together, gives us:
BM2/GN2 = (Area of ABCDE)/(Area of FGHKL)

QED

Lemma 2: if A/B = C/D with A greater than C, then D is less than B.

Proof:

(1) Let A/B = C/D with A greater than C.

(2) So that AD = BC

(3) Now, D ≠ B since if D = B, then AD is greater than BC which contradicts step #2.

(4) Now, D cannot be greater than B since then AD is greater than BC which contradicts step #2.

(5) So, by the Law of Trichotomy (see Postulate above), we can conclude that D is less than B.

QED

Theorem: Two circles are to each other as the squares of their diameters.

Proof:

(1) Let C1 be the circle formed with diameter BD and area A1.

(2) Let C2 be the circle formed with diameter FH and area A2.

(3) Assume that A1/A2 ≠ (BD)2/(FH)2

(4) There exists an area S such that: (BD)2/(FH)2 = A1/S

(5) Assume that S is less than A2

(6) Then, there exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion, see Lemma 2.]

(7) We can inscribe a similar polygon into circle C1 [See here for details on this construction]

(8) From Lemma 1 above, we can conclude:

BD2/FH2 = (Area polygon AOBPCQDR)/(Area polygon EKFLGMHN)

(9) But then, from step #4:

BD2/FH2= A1/S

(10) So we can conclude that:

(Area polygon AOBPCQDR)/(Area polygon EKFLGMNH) = A1/S

(11) Since A1 is greater than Area polygon AOBPCQDR, we can conclude from step #15 that S is greater than Area polygon EKFLGMN from Lemma 3 above.

(12) But this is impossible since in step #6 we showed that S is less than this same regular polygon so we have a contradiction and we reject our assumption in step #5.

(13) Now, let's assume that S is greater than A2

(14) So this means that (FH)2/(BD)2 = S/A1

(15) Let T be the area such that S/A1 = A2/T

(16) We can see that T is less than A1 from Lemma 2 above.

(17) There exists a regular polygon that is greater in area than T but smaller than the area of the circle by the Method of Exhaustion (see Lemma 2)

(18) We can inscribe a similar polygon in A2.

(19) From Lemma 1 above, we can conclude:

FH2/BD2 = (Area polygon EKFLGMHN)/(Area polygon AOBPCQDR)

(20) Likewise from step #14, we have:
(FH)2/(BD)2 = S/A1

(21) And from step #15, this means that:

(Area polygon EKFLGMNH)/(Area polygon AOBPCQDR) = A2/T

(22) Now A2 is greater in area than polygon EKFLGMNH so that T must be greater than the area of polygon AOBPCQDR

(23) But this is impossible from step #17 so we have a contradiction and we reject step #13.

(29) We now apply the Law of Trichotomy (see Postulate above) and we are done.

QED

References

Monday, April 17, 2006

Derivative of sine and cosine

In today's blog, I bring together some of the results that I presented earlier to determine the derivative for sine and cosine.

Lemma 1: sin(A+B) - sin(A - B) = 2*cos(A)sin(B)

Proof:

(1) sin(A+B) = cos(B)*sin(A) + cos(A)*sin(B) [See here for proof]

(2) sin(A-B) = sin(A+(-B)) = cos(-B)*sin(A) + cos(A)*sin(-B) =

(3) since cos(-B) = cos(B) [See here] and sin(-B) = -sin(B) [See here], we get:

sin(A-B) = cos(B)*sin(A) - cos(A)*sin(B)

(4) sin(A+B) - sin(A-B) =

= cos(B)*sin(A) + cos(A)*sin(B) - cos(B)*sin(A) + cos(A)*sin(B) =

= 2*cos(A)*sin(B)

QED

Lemma 2: sin P - sin Q = 2*cos [(P+Q)/2 ] * sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A+B = (P+Q)/2 + (P-Q)/2 = (P+P+Q-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P-P+Q+Q)/2 = Q

(5) sin(P) - sin(Q) = sin(A+B) - sin(A-B) = 2*cos(A)*sin(B) [See Lemma 1 above]

(6) Putting it all together gives us:

sin(P) - sin(Q) = 2*cos(A)*sin(B) = 2*cos[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 1: d/dx(sin x) = cos x

Proof:

(1) Let y = sin(x)

(2) dy/dx = lim (Δx → 0) [sin(x + Δx) - sin(x)]/Δx

(3) From Lemma 2 above we know that:
sin(x+Δx) - sin(x) = 2*cos[(x+Δx+x)/2]*sin[(x+Δx-x)/2] =
= 2*cos(x + Δx/2)*sin(Δx/2)

(4) So, substituting (3) into (2) gives us:

dy/dx = lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ]

(5) Using the Product Law (see here), we know that:

lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ] =
lim(Δx → 0)[cos(x+Δx/2)] * lim(Δx → 0)[sin(Δx/2)/(Δx/2)]

(6) Now, if we set θ = Δx/2, we know that:
lim (θ → 0) [ sin(θ)/θ ] = 1 (See here for proof)

(7) We also know that
lim(Δx → 0)[cos(x + Δx/2)] = cos x

(8) This then gives us:
dy/dx = cos x * 1 = cos x.

QED

Lemma 3: cos(A+B) - cos(A-B) = -2*sin(A)sin(B)

Proof:

(1) cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) [See here for proof]

(2) cos(A-B) = cos(A+(-B)) = cos(A)*cos(-B)-sin(A)*sin(-B)

(3) Since cos(-B) = cos(B) and sin(-B) = -sin(B) [See here], we have:

cos(A-B) = cos(A)*cos(B)+sin(A)*sin(B)

(4) cos(A+B) - cos(A-B) =

=cos(A)*cos(B) - sin(A)*sin(B) - cos(A)*cos(B) - sin(A)*sin(B) =

= -2*sin(A)*sin(B)

QED

Lemma 4: cos P - cos Q = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A + B = (P+Q)/2 + (P-Q)/2 = (P+Q+P-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P + Q - P + Q)/2 = Q

(5) cos(P) - cos(Q) = cos(A+B) - cos(A-B) = -2*sin(A)*sin(B) [From Lemma 3 above]

(6) So,

cos(P) - cos(Q) = -2*sin(A)*sin(B) = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 2: d/dx(cos x) = -sin x

Proof:

(1) Let y = cos(x)

(2) dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx

(3) Now, from Lemma 4 above:

cos(x +Δx) - cos(x) = -2*sin[(x + Δx + x)/2]*sin[(x+Δx-x)/2] =
= -2*sin(x + Δx/2)*sin(Δx/2)

(4) Once again, applying the Product Rule for limits gives us:

dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx =
lim(Δx → 0)[ -sin(x + Δx/2) ] * lim(Δx → 0) [ sin (Δx/2)/(Δx/2)]

(5) Again, setting θ = Δx/2 gives us:
lim(Δx → 0)[ sin(θ)/θ ] = 1 [See here for proof]

(6) We can also see that:
lim(Δx → 0)[ -sin(x + Δ x/2) ] = -sin(x)

(7) Putting this all together gives us:
dy/dx = -sin(x)*1 = -sin(x)

QED

limit (θ → 0) sin θ/θ = 1

Today's proof is part of the review of basic properties that I use to determine the derivatives of sine x and cosine x. This is part of the larger story where I show how the Taylor Series can be used to define sin and cosine independently of Euclid.

To be clear, I am using Euclidean Geometry to determine the derivative for sin x and cosine x, then I am using the Taylor Series to show that these results are equivalent to an infinite series that makes no such assumption. These ideas form the foundation of Euler's Identity which is one of the most amazing results in all of mathematics.

Lemma 1: if (a/b) is greater than (c/d), then d/c is greater than b/a

Proof:

(1) Let n be a positive value such that 10n that is greater than (a/b).

(2) 10n / (c/d) is greater than 10n/(a/b) since any number can be divided more times by a smaller amount.

(3) But now, if we divide both sides of the equation by 10n, we get:

1/(c/d) is greater than 1/(a/b)

which means that:

d/c is greater than b/a

QED

Lemma 2: lim(θ → 0) sin θ/θ = 1

Proof:

(1) Let r be the length of the radius of circle O.

(2) The area for triangle OAB = (1/2)r*(r*sin θ) [See here for details if needed]

= (1/2)r2*sin(θ)

(3) The area of the sector OAB = (1/2)r2θ [Proof to be added later]

(4) The area of the triangle OAT = (1/2)r2tan(θ)

[Since the area of a triangle is (1/2)base*height, see here if needed, with base = r and height = r*tan(θ), see here if needed]

Now, since tan(θ) = sin(θ)/cos(θ) [see here if needed], then we have:

The area of triangle OAT = (1/2)r2(sin θ)/(cos θ)

(5) So, we can see that:

area of triangle OAB is less than area of the sector OAB which is less than area of the triangle OAT.

So that:

(1/2)r2(sin θ) is less than (1/2)r2θ which is less than (1/2)r2(sin θ)/(cos θ)

(6) Dividing all sides by (1/2)r2 gives us:

sin θ is less than θ which is less than sin(θ)/cos(θ)

(7) Now, if we divide (5) by sin θ (assuming sin θ ≠ 0), then we get:

1 is less than θ/(sin θ) which is less than 1/cos(θ)

Taking the reciprocal for each value gives us:

1 is greater than (sin θ/θ) which is greater than cos(θ).

(8) Now, we know that the limit (θ → 0) 1 = 1, by the Constant Law (see here).

(9) We know that lim(θ → 0) cos(θ) = 1 since:

(a) cos θ is a continuous function (see here if more details are needed)

(b) so this means lim (θ → 0) cos θ = cos 0 = 1 (see here for definition of continuous functions, see here for review of why cos 0 = 1)

(10) But now, we can apply the Squeeze Rule (see here), to get:

lim (θ → 0) (sin θ/θ) = 1.

QED