Welcome to the **Geometric Proofs of Pi **section of our **Measuring Pi Squaring Phi **web site.

This section is divided into two parts: (1) A Geometric picture of each Proof including an accompanying step by step summary of proof, and (2) A video of my stepping the reader through each step of each proof. The best way to view this section is to first do Part (1) View the Geometric Proof Photo, and then walk through Part (2), its video explanation. These geometric proofs are shown in chronological order of development as I learned more and more about how to discover the true value of Pi.

Note: For those of you who are familiar with the proof of Archimedes’ Circles of Arbelos (“Arbelos” is Italian for the “shoemaker’s knife” due to its geometric shape), you can easily see in my Proofs 1 and 2 that the sum of the arcs of the top half of the 4 blue circles tangent across the diameter of the yellow circle equals the arc of the top half of the yellow circle. These arcs are the semi-circles of the blue and yellow circles. In fact, I have drawn multiple levels of Archimedes’ Arbelos Circles in Proof 1. In Proof 1 Fig. 1, the sum of the arcs of the top half of two blue circles equals the top half of Circle C2, and then the sum of the arcs of the top half of two Circle C2’s equal the top half of the yellow circle (which is the yellow circle’s semi-circle). As an aside, the area under each Arbelos blade equals the area above each Arbelos blade. You may want to review the Arblelos Circles on the Internet to understand why the diameters and circumferences of the 4 blue circles respectively equal the diameter and circumference of the yellow circle. With this information, we can easily calculate the true value of Pi as I have shown in each Proof.

Geometric Proof 1 for True Value of Pi:

Walk-through for Geometric Proof 1 for the True Value of Pi

Note: A few readers have asked, “How do you know that the diameters of 4 blue circles fit exactly tangent across the diameter of the Big Yellow Circle in Proof 1 Fig 1?”

Answer:

Blue circle diameter = 2 / Pi, given where Circumf = 2, and diam = C / Pi, Proof 1 Fig 1,

Big Yellow circle diameter = 2 sqrt Phi, from Kepler’s Triangle Proof 1 Fig 1.

In Proof 1 Fig 1, we do not know yet what Pi is and so we ask, What is the value of Pi when the diameter of the Big Yellow Circle = 4 times the diameter of the Blue circle? The answer is:

2 sqrt Phi = 4 (2 / Pi), Solve for Pi

Pi = 4 / sqrt Phi,

Pi = 4 / 1.272019650… = 3.144605511…

Therefore, the diameter of 4 Blue circles fit exactly tangent across the diameter of the Big Yellow Circle in Proof 1 Fig 1 when Pi = 3.144605511… . And since Pi is a universal constant, not a variable, there is no need to look for another value of Pi.

Also, we can square the circumference of the Big Yellow Circle to the perimeter of Square YHWT, equate the two equations for C = P, and then solve for the value of Pi:

P, Perimeter of Square YHWT = 8, given in Proof 1 Fig 1,

C, Circumference of Big Yellow Circle = 2 Pi (sqrt Phi), from Kepler’s Triangle, Proof 1 Fig 1.

When C = P, what is the value of Pi?

Answer:

C = 2 Pi (sqrt Phi) = P = 8, Solve for Pi

Pi = 8 / (2 sqrt Phi),

Pi = 4 / sqrt Phi = 4 / 1.272019650… = 3.144605511…

Therefore, we have squared the circumference, C, to the perimeter, P, and equated C = P to solve for the only variable left, which is Pi = 4 / sqrt Phi. All of the above is clearly shown in Proof 1 Fig 1: (a) the Big Yellow Circle’s diameter is 4 times the diameter of the Blue circle, and (b) we have squared the Big Yellow Circle’s circumference to the perimeter of Square YHWT using the construction and sides of Kepler’s Golden Ratio Right Triangle.

Thanks for your questions. Time for you to step through the video of Proof 1 Fig 1:

Geometric Proof 2 for True Value of Pi:

Walk-through for Geometric Proof 2 for the True Value of Pi

Note: A few readers have asked, “How do you know that the diameters of 4 blue circles fit exactly tangent across the diameter of the Big Yellow Circle in Proof 2 Fig 2?”

Answer:

Blue circle diameter = Pi, given where Circumf = C = diam x Pi = Pi^2, Proof 2 Fig 2,

Big Yellow Circle diameter = 2 sqrt Phi (Pi^2 / 2), from Kepler’s Triangle Proof 2 Fig 2.

In Proof 2 Fig 2, we do not know yet what Pi is – except that we really do know from Proof 1 Fig 1, but let’s continue with Proof 2 Fig 2 as if we didn’t know — and so we ask, What is the value of Pi when the diameter of the Big Yellow Circle = 4 times the diameter of the Blue circle? The answer is:

2 sqrt Phi (Pi^2 / 2) = 4 Pi, Solve for Pi

Sqrt Phi (Pi) = 4 ,

Pi = 4 / 1.272019650… = 3.144605511…

Therefore, the diameter of 4 Blue circles fit exactly tangent across the diameter of the Big Yellow Circle in Proof 2 Fig 2 when Pi = 3.144605511… . And since Pi is a universal constant, not a variable, there is no need to look for another value of Pi.

Also, we can square the circumference of the Big Yellow Circle to the perimeter of Square YHWT, equate the two equations for C = P, and then solve for the value of Pi:

P, Perimeter of Square YHWT = 4 Pi^2, given in Proof 2 Fig 2,

C, Circumference of Big Yellow Circle = 2 Pi (sqrt Phi) (Pi^2 / 2), from Kepler’s Triangle, Proof 2 Fig 2.

When C = P, what is the value of Pi?

Answer:

C = 2 Pi (sqrt Phi) (Pi^2 / 2) = P = 4 Pi^2, Solve for Pi

Pi (sqrt Phi) = 4,

Pi = 4 / sqrt Phi = 4 / 1.272019650… = 3.144605511…

Therefore, we have squared the circumference, C, to the perimeter, P, and equated C = P to solve for the only variable left, which is Pi = 4 / sqrt Phi. All of the above is clearly shown in Proof 2 Fig 2: (a) the Big Yellow Circle’s diameter is 4 times the diameter of the Blue circle, and (b) we have squared the Big Yellow Circle’s circumference to the perimeter of Square YHWT using the construction and sides of Kepler’s Golden Ratio Right Triangle.

Proof 2 Addendum 1 Jan 04, 2018

Notice that one of the key attributes of Proof 2 is that I have defined BOTH the given perimeter of the Square YHWT and the Kepler-derived ordinal value circumference of the Yellow Circle A in terms of Pi: the side of YHWT is a given Pi^2, and the Kepler Triangle side of sqrt (Phi) x Pi^2 / 2 is the given radius of Yellow Circle A. Both equations — the perimeter, P, of Square YHWT and the circumference, C, of Yellow Circle A — are true. We merely want to find the true value of Pi that satisfies the equality of P = C.

Therefore,

Perimeter, P, of Square YHWT = 4 x (Pi ^2),

Circumference, C, of Yellow Circle A = radius x 2 Pi = sqrt (Phi) x (Pi^2 / 2) x 2 x Pi, or

reducing, Circumference, C, = sqrt (Phi) x (Pi^3).

Therefore, when “squaring” C to equal P, the value of Pi must be the same for both sides of the equation 4 x (Pi ^2) = sqrt (Phi) x (Pi^3), so that the Perimeter, P, equals (is “squared” to) the circumference, C.

Solving for Pi, when P = C,

4 x (Pi^2) = sqrt (Phi) x (Pi^3), Perimeter = Circumference

4 = sqrt (Phi) x Pi, divide both sides by Pi^2

4 / sqrt (Phi) = Pi, divide both sides by sqrt (Phi)

4 / 1.272019650… = Pi, sqrt (Phi) = 1.272019650…

3.144605511… = Pi conclusion

Substituting the value of Pi = **3.144605511…** in both equations for P = C, we get:

4 (3.144605511)^2 = **39.55417528…** for perimeter of Square YHWT, and

1.272019650 (3.144605511)^3 = **39.55417528…** for circumference of Yellow Circle A.

Everything balances on both sides of the equation P = C when Pi = **3.144605511…** .

If, however, we substitute Pi = **3.141592654…** in both equations for P = C, we get:

4 (3.141592654)^2 = **39.4784176…** for perimeter of Square YHWT, and

1.272019650 (3.141592654)^3 = **39.44059321…** for circumference of Yellow Circle A.

The same equations for P and C do not balance if we use Old Pi = **3.141592654…** as Pi. P does not equal C, even though we are using the same true given equations for both P and C. **That means Old Pi = 3.141592654… is WRONG!** It also means that any other value for Pi other than **Pi = 3.144605511…** would also be wrong because there are not 2 values for Pi. Pi is a universal constant, not a variable. And Proof 2 shows why true **Pi = 4 / sqrt (Phi) = 3.144605511… and no other value.**

QED. H. Lear

Thanks for your questions. Time for you to step through the video of Proof 2 Fig 2:

Geometric Proof 4 for True Value of Pi:

Walk-through for Geometric Proof 4 for the True Value of Pi

Geometric Proof 6 for True Value of Pi:

Walk-through for Geometric Proof 6 for the True Value of Pi