hi doc Proposition 10

Proposition 10 Sphere and the Cylinder I

 

Heres a version of proposition 10 that fills in a few details.

Proposition 10: If we take two tangents to the circular base of a cone and and form two triangles by connecting the points of tangency and the intersection point to the apex of the cone, then the area of the two triangles will in sum exceed the area of the cone segment cut off.

 

Proof: Let ACB be the circular base of the cone with C the midpoint of arc AB. Let O be the apex and let the two tangents at A and B meet at D. let the lines AO, BO, and CO be drawn.

 

We need to show that △ ADO + △ BDO > cone-segment-ACBO.

 

Let EF be tangent to the circle at C (and hence parallel to AB) with E and F on AD and BD respectively. ED + DF > EF. Adding AE+FB to both sides we get: AE+FB+ED+DF > AE+FB+EF. Then AD+BD > AE+FB+EF. But OA, OC, and OB are perpendicular to AD, EF, and BD,respectively.

Thus △ ADO + △ BDO > △ OAE + △ OEF +△OFB

Let G be the difference,

i.e. Let G=△ ADO + △ BDO - (△OAE + △ OEF +△OFB ) (call this *)

Now let L be the area between the circle and the tangents AE, EF, and FB.

There are two cases to consider:

I.                   G≥L

II.                 G<L

First case I :

Consider these two surfaces:

a. the pyramid with apex O and base AEFB excluding the face △OAB and

b. the cone segment OACB together with the circle segment ACB

These have the same extremities and thus by assumption 4, the area of a. > the area of b.

Now subtract segment ACB and we have:

△ OAE + △ OEF +△OFB + L > cone-segment-OACB

But G≥L. Therefore:

△ OAE + △ OEF +△OFB + G ≥ △ OAE + △ OEF +△OFB + L

> cone-segment-OACB

But (from *) the left hand side equals △ ADO + △ BDO proving case I.

 

Now for case II G < L. By repeated bisection we can find a circumscribed polygon such that the sum of the pieces inside the polygon but outside the circle is less than G. Call the sum of the areas of these pieces M where M<G.

 

△ OAE + △ OEF +△OFB > △ OAP + △ OPQ + +△OSB

adding M to both sides:

△ OAE + △ OEF +△OFB + M > △ OAP + △ OPQ + +△OSB + M (**)

 

But also pyramid OAPQRSB > cone-segment-OACB + segment ACB (assumption 4 again).

Subtracting the segment ACB we get:

△ OAP + △ OPQ ++△OSB + M > cone-segment-OACB

Combining this last with ** we have:

△ OAE + △ OEF +△OFB + M > cone-segment-OACB (***)

But now recall that:

△ ADO + △ BDO = G + △ OAE + △ OEF +△OFB

> M +△ OAE + △ OEF +△OFB

Combining these last lines with *** gives the desired result.

 

QED