## Friday, May 05, 2006

### Euclid's Proof for the Pythagorean Theorem

In today's blog, I will go over Euclid's proof for the Pythagorean Theorem as presented in Euclid's elements. There is a very large number of different proofs for this result. The simplest involve tiles that show how squares of the two sides can be refitted to form a square on the hypotenuse. One proof was discovered by American president, James Garfield.

Theorem: In a right triangle, the hypotenuse squared is equal to the sum of the other sides squared.

Proof:

(1) Since BAC is a right angle, we can extend AC to make a square GFBA and we can extend AB to make a square ACKH. (See here for details on the construction)

(2) We can also build a square based on BC. (See here for details on the construction)

(3) Let AL be a line that is parallel to BD and CE. (See here for details on the construction)

(4) ∠ DBA ≅ ∠ FBC since ∠ FBA and ∠ DBC are right angles and we get this result if we add ∠ ABC to both.

(5) triangle DBA ≅ triangle FBC by Side-Angle-Side (see here) since:

(a) ∠ DBA ≅ ∠ FBC (#4)

(b) FB ≅ AB since they are sides of the same square.

(c) BC ≅ BD since they are sides of the same square.

(6) The parallelogram BL is double the triangle ABD since they have the same base (BD) and since they are in the same parallel. (see Corollary 3.1 here for details)

(7) Likewise, the parallelogram GFBA is double the triangle FBC for the same reason as step #6.

(8) From step #5, we can conclude that the parallelogram BL is congruent to the parallelogram GFBA.

(9) We can follow the same steps to show that parallelogram CL is congruent to parallelogram ACKH since:

(a) ∠ KCB ≅ ∠ ACE since ∠ BCE, ∠ ACK are right angles and we can add ∠ ACB to each.

(b) AC ≅ CK and BC ≅ CE since the sides of the same square are congruent.

(c) So that triangle KCB ≅ triangle ACE by Side-Angle-Side.

(d) parallelogram CL is double the area of triangle ACE

(e) parallelogram CA is double the aera of triangle KCB

(f) So therefore parallelogram CL ≅ parallelogram CA

(10) So that we have the square of BC is congruent to the square of AB added to the square of AC since:

(a) the square BDEC = parallelogram BL + parallelogram CL

(b) parallelogram BL ≅ square of AB

(c) parallelogram CL ≅ square of AC

QED

Corollary 1: In a right triangle, the hypotenuse is longest of the three sides of a triangle.

Proof:

(1) Assume that the hypotenuse h is equal or less than one side s1

(2) Then, h2 is less than s12 + s22

(3) But this contradicts the Pythagorean Theorem so we can reject our assumption.

QED

Corollary 2: sin2(θ) + cos2(θ) = 1.

Proof:

(1) From the main theorem, for any right triangle with base sides x and y and with hypotenuse z, we have:

x2 + y2 = z2

(2) If θ is the angle between z and x, then we have (see here):

sin θ = y/z

cos θ = x/z

So that:

sin2(θ) = y2/z2

cos2(θ) = x2/z2

(3) Now if we divide the equation in step #1, by z2 on both sides we get:

x2/z2 + y2/z2 = 1.

(4) Now inserting the values in step #2 gives us:

sin2(θ) + cos2(θ) = 1.

QED

## Tuesday, May 02, 2006

### Archimedes and the area of a circle

In today's blog, I will go over Archimede's proof that the area of a circle is (1/2)rc. When this proof is combined with Euclid's proof (most likely from Eudoxus), it is possible to show that for all circles, the ratio of C/D is constant.

Postulate 1: For a given chord of a circle, the segment of the circumference is longer than the chord.

In the example above, BDC is longer than BC.

Postulate 2: For a given segment of the circumference, lines connected above it combined are longer.

In the example above, AG + GE is greater than the segment of the circumference AE.

Lemma 1: The area of a regular polygon is (1/2)h*Q where Q is the sum of the perimeter.

Proof:

(1) A regular polygon can be divided up into a sum of triangles.

(2) The area of each triangle is equal to (1/2)*(GH)*(DE) [See here for details if needed]

(3) So, the area of the polygon = (# of sides)*(1/2)*(GH)*(DE)

(4) Let Q = the sum of the bases (for example, for the hexagon above, Q = DE + EF + FA + AB + BC + CD)

(5) Then, the area of the polygon = (1/2)*(GH)*Q

QED

Theorem: The area of a circle is (1/2)circumference * radius

Proof:

(1) Let K = (1/2)*C*R where C = circumference and R = radius.

(2) Let A = the area of the circle.

(3) Assume that A is greater than K

(4) We can inscribe a polygon inside A that is greater than K and less than A. [See Lemma 2 and the Method of Exhaustion here for details]

(5) So, the area of the polygon = (1/2)*Q*h where h is the distance from the center to the base and where Q is the perimeter of the polygon. [See Lemma 1 above]

(6) But Q is less than C (see Postulate 1 above) and h is less than R.

(7) So we have Area Polygon = (1/2)Q*h which is less than (1/2)*C*R

(8) But this contradicts step #4 so we reject step #3.

(9) Now, let's assume that K is greater than A.

(10) We can circumscribe a polygon P around A such that P is greater than A but less than K. [See Lemma 3 and Method of Exhaustion here for details.]

(11) From Lemma 1 again, we know that the area of this polygon is (1/2)*Q*h where Q is the perimeter of the polygon and h is the height.

(12) In the case of the circumscribed polygon (see diagram for Postulate 2), h = R.

(13) Using Postulate 2 above, we see that Q is greater than C.

(14) But then the area of the polygon is greater than K since (1/2)*Q*R is greater than (1/2)*C*R

(15) But this contradicts step #10 so we reject our assumption at step #9.

(16) Now, we apply the Law of Trichomoty (see here) and we are done.

QED

Corollary 1: For all circles, the ratio of Circumference to Diameter is constant

Proof:

(1) We know from Euclid (see here) that for any two circles C1 and C2 that:

Area1/Area2 = (Diameter1)2/(Diameter2)2

(2) We know from Theorem 1 above that:

Area of circle = (1/2)*radius*circumference.

(3) Combining step #1 with step #2 and using R=(1/2)D gives us:

[(1/2)*(1/2)D1*C1]/[(1/2)*(1/2)D2*C2] = [D1]2/[D2]2

(4) Canceling out (1/4)/(1/4) gives us:

[D1*C1]/[D2*C2] = [D1]2/[D2]2

(5) Multiplying both sides by (D2/D1) gives us:

C1/C2 = D1/D2

which means that:

C1*D2 = C2*D1

and finally that:

C1/D1 = C2/D2

QED

Definition 1: π

From the Corollary, we know that the ratio of the circumference to the diameter is constant. π is this ratio from circumference to diameter. In other words, π = C/D.

Corollary 2: The area of a circle is πr2.

Proof:

(1) From the theorem, the area of a circle is (1/2)(circumference)(radius)

(2) From the definition above, π = C/D

This means that C = D*π = 2*r*π

(3) Putting this all together gives us:

area = (1/2)(circumference)(radius) = (1/2)(2*r*π)(r) = πr2

QED

References

## Sunday, April 30, 2006

### Method of Exhaustion

In today's blog, I go over the proof for Eudoxus's Method of Exhaustion. This proof is used later in Euclid's proof that the areas of circles are in proportion to the squares of their diameters.

Theorem: Method of Exhaustion

Let x,y be any two positive real numbers where x is greater than y. If we continually remove a quantity v ≥ (1/2) from x, then eventually, we will be left with a value that is smaller than y.

Proof:

(1) We will only need to prove this for v = 1/2 since:

(a) This proof is done if we can show that there exists n such that (1/2)n*x is less than y.

(b) We only need to prove it for v=1/2 since we know that if v is less than (1/2), then:

vn*x is less than (1/2)*n*x.

(2) Let m be an integer such that m ≥ x/y + 1

(3) So that ym is greater than x

(a) Since ym is greater than y(x/y + 1) = x + y

(b) and x + y is greater than x since y is nonzero.

(4) There exists a value n such that 2n is greater than m.

(a) Let m = 1

(b) 21 = 2 is greater than 1.

(c) Assume this is true up to n.

(d) So that 2n is greater than n.

(e) Since 2n is greater than 1, we know that:
2n + 2n = 2*(2n) = 2n+1 is greater than n+1.

(f) So that this is true by induction.

(5) Thus, we see that y/x is greater than 1/2n

(a) ym is greater than x (step #2)

(b) so, m is greater than x/y

(c) and 1/m is less than y/x (or in other words, y/x is greater than 1/m)

(d) Now, 2n is greater than m so 1/2n is less than 1/m which also means that it is less than y/x.

(6) Thus (1/2)nx is less than y since y/x is greater than (1/2)n means that y is greater than (1/2)n*m.

QED

Now, let's look at two applications of the Method of Exhaustion.

Lemma 1: If a square is inscribed in a circle, its area is greater than half the area of the circle.

Proof:

(1) Let EFGH be a square that is inscribed in the circle above (see here for details on this construction).

(2) Let KLMN be a square that is circumscribed around the circle above (see here for details on this construction).

(3) The area of KLMN is greater than the area of the circle.

(4) The square EFGH is equal to (1/2) the area of KLMN since:

(a) Each square KFCE, FLGC, CGMH, ECHN are (1/4) the area of KLMN.

(b) Each triangle EFC, CFG, HCG, ECH is (1/4) the area of EFGH

(c) Each triangle EFC, CFG, HCG, ECH is (1/2) the area of the corresponding square KFCE, FLGC, CGMH, ECHN.

(d) Therefore, area EFGH = (1/2) * area KLMN

(5) Finally, area EFGH = (1/2)* area KLMN which is greater than the area of (1/2) the circle (from step #3).

QED

Lemma 2: For any area A that is smaller than the area of a circle C, there exists a regular polygon P that is smaller than C but greater than A.

Proof:

(1) Let us form a circle C on the diameter FH. [See here for details on construction]

(2) It is possible to find points E,G such EFGH forms a square that is inscribed in circle C [See here for details on this construction]

(3) The area of the square EFGH is greater than (1/2) the area of circle C [See Lemma 1 above]

(4) We can constructs points K, N, M, and L so that point K is midway between points E and F, N is midway between E and H, M is midway between H and G, and L is midway between G and F. [See here for details on this construction]

(5) Each triangle EKG, FLG, GMH, and HNE is greater in area than half the segment of the circle about it since:

(a) From each triangle, we can construct a rectangle that has an area greater than the circle segment. For example, around triangle EKG we can construct a rectangle as in the diagram above.

(b) Each half of the triangle is exactly 1/4 the size of the rectangle since each point bisects the side of the circumscribed square.

(c) So the triangle EKG is equal to (1/2) the area of the rectangle.

(d) Since the area of the rectangle is greater than the area of the segment, (c) gives us that the triangle EKG is greater than (1/2) the area of the segment.

(6) We can keep repeating this step so that each time, we remove more than (1/2) the area:

(a) We select the midpoints to each of the existing segments. [See step #4 for details]

(b) We form a triangle with these new midpoints from the points of the segment that this new point bisected.

(c) This set of triangles together forms a new regular polygon.

(d) Each triangle has more area than half the circle segment that had previously remained. [This is the same as the argument in step #5]

(11) So, then there eventually exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion above since each time we do step #6, we are subtracting greater than 1/2 of the remaining area from the circle.]

QED

Lemma 3: If a regular polygon P1 with area A1 is circumscribed around a circle C, then there exists a smaller polygon P2 with area A2 that can also be circumscribed around circle C such that:

A2 - C is less than (1/2)(A1 - C)

Proof:

(1) Let C be a circle formed from diameter BD with center E. [See here for details on construction]

(2) Circumscribe a square GHKF around circle C. [See here for details on construction]

(3) Find a point M that lies on circle C and is midway between A and B [See here for details on construction]

(4) Let ON be a line tangent to M such that N is on GA and O is on GB.

(5) ∠ GMN is a right angle [See here for details]

(6) GN is greater than MN [See the corollary here for details]

(7) MN ≅ AN since:

(a) ∠ EAN ≅ ∠ EMN since they are both right angles.

(b) AE,ME are both radii, so we have AE ≅ ME and therefore (see Theorem 1, here), ∠ EAP ≅ ∠ EMP

(c) Using subtraction of step #7b from step #7a gives us ∠ NAM ≅ ∠ NMA.

(d) And #7c gives us AN ≅ MN since congruent angles implies congruent sides (see Corollary to Theorem 1, here)

(8) So GN is greater than AN too

(9) triangle AMN and triangle GMN have the same height since they are both on the same base AG and since they share the same vertice at M.

(10) Now, we can conclude that area of GMN is greater than the area of AMN since they share the same height (step #9) but the base of GMN is greater than the base of AMN (step #8) since area = (1/2)base * height (See Lemma 2 here for details if needed)

(11) triangle GLA ≅ triangle GLB by Side-Angle-Side since:

(a) GB ≅ GA since they are the sides of the same square.

(b) ∠ BGL ≅ ∠ AGL since we can prove triangle GAE ≅ triangle GBE by Side-Side-Side.

(c) Both triangles share the same side at GL.

(12) So the area of OGN is greater than half the area of BGA.

(13) The argument in step #11 applies to each side and we can conclude that converting this polygon from a 4-sided to 8-sided regular polygon results in removing greater than (1/2) the difference between the area of the 4-sided polygon and the area of the circle.

(14) We can use the same method to divide an 8-sided polygon to a 16-sided polygon and so on.

QED

References