Halloween Challenge: How to Carve the Pumpkin?

Favorite Answer

q1
(2a - R - r)^2 + (a - r)^2 = (R - r)^2
some more lines and we get
4R(a - r) = 5a^2 - 6ar + r^2
4R = 5a - r

q2
r = 5a - 4R
so a and r must have the same parity.
if we let a = 4k + r and R = 5k + r
5a - 4R
= 5(4k + r) - 4(5k + r)
= (20k - 20k) + (5r - 4r)
= r
a > 2r
4k + r > 2r
k > r/4

(a , r , R) = (4k + r , r , 5k + r)

q3
5a = 4R + r = 20k + 5r
= 20(k - 1) + 5(r + 4)
since r > 0, we fix r = 1
4(k - 1) > (r + 4)
k - 1 > 5/4
k > 9/4

k = 2, r = 5
k = 3, r = 1
a = 13
(13, 1, 16) and (13, 5, 15)

q4
R > a > r
R^2 = a^2 + r^2
(5k + r)^2 = (4k + r)^2 + r^2
r^2 + 10kr + 25k^2 = 2r^2 + 8kr + 16k^2
r^2 - 2kr - 9k^2 = 0
(r - k)^2 = 10k^2
r = k +- k√10 = k(1 +- √10)
k, r > 0 and k, r = integer
so there are no Pythagorean triple.

q5
can't say i'm a good judge of anything aesthetic. on the pumpkin, all i know is i like the taste of the juice. i just love the numbers in (15, 7, 17) and (27, 11, 31). the former because they are almost PT and the latter is a birthday of someone i know. Dsquared recommendation is best....Show more

pandect 244 has pretty much nailed parts 1)-4), but here are a few slight variations + an attempt at part 5).

(1) Compare sides of the right-angle triangle with vertices at:

(i) centre of larger circle;
(ii) midpoint of horizontal line passing through centres of small circles;
(iii) centre of smaller circle on right (or left):

(a-r)^2 = (R-r)^2 - (2a-R-r)^2
= 4(R-a)(a-r)
==> a-r = 4(R-a), since a ≠ r.
==> r + 4R = 5a.

(2) Start with: 4(R-r) = 5(a-r)
5 | LHS and gcd(4,5) = 1 ==> 5 | (R-r)
==> R - r = 5k, with k an integer
==> a - r = 4k.
Consider restrictions:
a > 2r ==> k >= [r/4]+1.
With k a positive integer, a = r+4k < r+5k = R
is automatically satisfied.
--> a = r + 4k, R = r + 5k, with k integer and k >= [r/4] + 1.

(3) Not difficult to prove that for fixed r and k, other solutions yielding the same value for "a" have the form r+4m, k-m, resp.,
i.e. a = r + 4k = (r+4m) + 4(k-m), with m a non-zero integer.
Also, for any value of a, we can select r such that r <= 4 and hence, m > 0.
With the constraints in place, we require:
k-m >= [r/4] + m + 1,
i.e. k >= [r/4] + 2m + 1,
and therefore
a >= (r+4[r/4]) + 8m + 4.
This is clearly minimised by choosing r = 1 and m = 1, so that a_min = 13.
(r'=1, R'=16; r''=5, R''=15.)

(4) From r + 4R = 5a
it follows that:
r ≡ a (mod 2),
so that r and a are both odd or both even.
If (r,a,R) forms a Pythagorean triple, then r and a must be both even (otherwise 2 ≡ 0 (mod 4), a contradiction).
==> a=2*a1, r = 2*r1.
Then, from r^2 + a^2 = R^2, it follows that R=2*R1.
==> r1 + 4*R1 = 5*a1.
The above argument can be repeated ad infinitum, which is impossible since a,r and R are finite, for example.


(5) Will try using the "Golden Ratio" φ a couple of times:
Will imagine the centre of the large circle to be the tip of the nose.

According to beauty studies, two measures for φ are:
(i) distance between inner points of eyes/eye width
(ii) distance between top of eyes and tip of nose/distance between tip of nose to top of mouth.

This translates to:
(2a-4r)/2r = φ
and
(2a-R)/sqrt(R^2-a^2) = φ

which yield the solutions:
r ≈ 0.276a, R ≈ 1.134a....Show more

certainly, the best pumpkin carving I ever did replaced into no longer a pumpkin in any respect. DH and that i lived in Key West for a collectively as, and pumpkins must be shipped in form the north, so there are not very many, and that they pass speedy. Went 3 days in the past Halloween to purchase a pumpkin and each place on the island replaced into bought out. while to a close-by industry and have been given a extensive cantaloupe. It replaced into so cool becasue as quickly as I placed the candle interior, it glowed orange form the interior. i could do it back right here, yet i've got yet to locate a cantaloupe sufficiently huge....Show more