Halloween Challenge: How to Carve the Pumpkin?

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.

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.

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