Relation with centroid G. Cut with a line going through G.

First of all let's calculate the centroid location.

It is not too hard to calculate directly with integral,
but the Guldin-Pappus relation gives this instantly.

Rotate the half disc through its diameter, the generated volume
(a full sphere) is V = 4/3 π R³

Area of half disc is S = πR²/2
and the Guldin relation gives : 2πOG × S = V

That is 2πOG × πR²/2 = 4/3 πR³, and after reducing :

OG = 4/(3π) R

Let's try now to split the half disc in two equal area parts through a segment MN.

Of course cutting in two quarter discs goes through G (by symmetry),
but in other cases, there is no reason for MN to go through G, and it is generally not the case.
We can be then interested in finding the envelope of MN.

We have two cases to consider :

- The easier, the two points M and N lie on the half circle.

Then the area of circle sector is 1/2 πR²/2 = constant, hence MN is constant and line MN envelopes an arc of circle.

The contact point is of course the midpoint of MN. - The 2nd case is M on the half circle and N on the diameter.

Let t = angle (OM,Oy). Area of triangle OMN should be equal to area of sector MOy, that is t.R²/2 = 1/2 R.ON.sin(t + π/2), hence :ON/R = - t/cos(t)

This allows to construct (calculate) point N for every location of M, and find the envelope.

The limit position of M is when |ON| = R, resulting into t_{0}= ±0.7390851332151607... that is 42.35°

Equation t = cos(t) gives this value with a 10^{-15}accuracy in just a few iterates of Newton method :

We could also depress repeatedly the "cos" key of any pocket calculator in radian mode, but this requires much more iterates !

Arc BC is an arc of circle with radius R.sin(π/4 - t

When N is in x', OT = R.sin(ONT) = R.sin(xOM/2).

G is strictly inside this curved triangle : above arc BC and below A, OA=R/2.

We get then the three cuts MN which go through G :

The trivial case Oy from case 2 (touching point with envelope in A)

Two other cases (from case 1) :

We get then cos(TOG) = OT/OG = R sin(π/4 - t_{0}/2) / (4R/(3π)) = 3 π sin(π/4 - t_{0}/2) / 4

Line MN going through G has an angle of φ = 17.854464...° with diameter xx'

Detailed calculations of envelope let as exercise.

Prove that T is midpoint of MN.Solution

A direct proof, without unnecessary calculations :

Consider two near lines MN and M'N' intersecting in T.

Area of 'triangles' TMM' and TNN' are equal, hence : TM.TM' = TN.TN'

When M'N' → MN, T becomes the touching point of MN and its envelope.

TM' → TM and TN' → TN.

At the limit, we get TM² = TN², QED.

Coordinates of T result immediately, hence the parametric equation of envelope.

Consider two near lines MN and M'N' intersecting in T.

Area of 'triangles' TMM' and TNN' are equal, hence : TM.TM' = TN.TN'

When M'N' → MN, T becomes the touching point of MN and its envelope.

TM' → TM and TN' → TN.

At the limit, we get TM² = TN², QED.

Coordinates of T result immediately, hence the parametric equation of envelope.

In a more traditional way, M has coordinats {R.sin(t), R.cos(t)} and N = {-R.t/cos(t), 0}

Equation of line MN is then :

y(sin(t)cos(t) + t) = x.cos²(t) + R.t.cos(t)

Differentiating wrt t results into :

2y.cos²(t) = -2x.cos(t).sin(t) + R(cos(t) - t.sin(t))

Solving this system for x and y gives the parametric equation of envelope :

x = R.(t + sin(t)cos(t))/(2cos(t)) y = R.cos(t)/2 |

We find again that touching point is midpoint of MN.