If you look sign the guessed book

Bigger, and it involves pi somehow.

Mmmm . . .
Pi!

Let XP and XQ be normals to AC and AB. Since the three angle bisectors are concurrent, AX bisects angle A also and therefore XP = XQ.

Let's assume XD > XE.
Then ang(PDX) < ang(QEX)
Now considering triangles BXD and CXE,
the last condition requires that ang(DBX) > ang(ECX)

OR ang(XBC) > ang(XCB)
OR XC > XB Thus our assumption leads to : XC + XD > XE + XB
OR CD > BE which is a contradiction.

Similarly, one can show that XD < XE leads to a contradiction too. Hence XD = XE => CX = BX

From which it is so easy to prove..... the cycle polynomials for the five platonic solids (and I threw in the tesseract for good measure).

tetrahedron: (x1^4+3x2^2+8x1*x3)/12
cube: (x1^6+6x2^3+3x1^2*x2^2+8x3^2+6x1^2*x4)/24
octahedron: (x1^8+9x2^4+8x1^2*x3^2+6x4^2)/24
dodecahedron: (x1^12+15x2^6+20x3^4+24x1^2*x5^2)/60
icosahedron: (x1^20+15x2^10+20x1^2*x3^6+24x5^4)/60
tesseract: (32x6^4+x2^12+48x8^3+x1^24+24x1^2*x2^11+12x2^2*x4^5+32x3^8+12x4^6 +18x1^4*x2^10+12x1^4*x4^5)/192

Good to see ya brown eyes.

The unused space "between" the circles is 1/4 the size of the total unused space in the farm.

Is 3/4 "much bigger" than 1/4, I don't know, but it is bigger.

Then why are there crop circles?

Oh, and for the question asked, is the result the same no matter what size each circle is as long as they touch the edges?

Joe(ponders)Nation

Are you sure it was Farmerman who did this and not Visitors from another star? Maybe we should get the Mystery Van involved in this?

A: Farmerman has an ET infestation

in order of appearance:

jespah - deserving of a small piece of pi

George - she's apples

Tryagain - craft award for cutting and pasting blue sky

rosborne - incidently correct 4 the answer was in the question, lovely logic lad


Joe Nation - expanding the edges of (parenthetical research)

Seed - arrives on the starship that is sheridans 1750 play ' the rivals'

Region Philbus - explores the philbus-flung region




touching