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“<strong>La</strong> <strong>matematica</strong> <strong>degli</strong> <strong>indovinelli</strong>” di Matteo Puzzle – <strong>matematica</strong>re@hotmail.com<br />

Considerazioni geometriche per il calcolo dei raggi:<br />

⎛π<br />

⎞ 3<br />

r1 = C1H = ⋅ tan ( C1BH)<br />

= ⋅ tan ⎜ ⎟=<br />

2 2 ⎝ 6 ⎠ 6<br />

2 3<br />

<br />

( ) ( ⎛π<br />

⎞ ⎛π<br />

⎞ ⋅<br />

3<br />

r2 = C1S⋅tan K1C1S ⋅ tan C2K1S)<br />

= r1⋅ tan ⎜ ⎟ = ⋅ tan ⎜ ⎟ =<br />

⎝ 6 ⎠ 2 ⎝ 6 ⎠ 18<br />

r<br />

3<br />

⋅<br />

5<br />

π ⎞ ⋅<br />

3<br />

⎟<br />

⎠<br />

7<br />

π ⎞ ⋅<br />

3<br />

⎟<br />

⎠<br />

2 n 1<br />

π ⋅ +<br />

⎛<br />

= ⋅ tan ⎜ =<br />

2 ⎝ 6 54<br />

⎛<br />

r4<br />

= ⋅ tan ⎜ =<br />

2 ⎝ 6 162<br />

............<br />

⎛ ⎞<br />

rn<br />

= ⋅tan<br />

⎜ ⎟ con n∈N<br />

2 ⎝ 6 ⎠<br />

calcolo dell’area <strong>degli</strong> infiniti cerchi:<br />

2 1 2<br />

∞ ∞ 2 2 ∞<br />

2 ∞ 2 ⋅n<br />

⋅ n+<br />

−( 2⋅ n+<br />

1<br />

⎛ ⎞ ⎛<br />

)<br />

⎞<br />

⎛π ⎞ ⋅3 π ⋅ ⎛ 1 ⎞ π ⋅<br />

⎛1⎞<br />

Λ<br />

C<br />

= π ⋅∑ ⋅ tan<br />

= π ⋅ = ⋅ = ⋅<br />

⎜ ⎜ ⎟ 2⋅ n+<br />

1<br />

n= 0 2 6 ⎟ ∑ ⎜ ⎟ ⎜ ⎟<br />

⎝ ⎠<br />

⎜<br />

n= 0 4 ⎟ ∑ ∑<br />

⎝<br />

⎠ ⎝ ⎠ 4 n=<br />

0⎝3 ⎠ 12 n=<br />

0⎝3⎠<br />

poiché (essendo una serie geometrica):<br />

∞<br />

ik<br />

kn a<br />

kn a r ⋅<br />

⋅<br />

∞<br />

⋅ ⋅<br />

∑a⋅ r = o più in generale: a r<br />

k<br />

∑ ⋅ =<br />

k<br />

n= 0 1−r<br />

n=<br />

i 1−r<br />

⎧a<br />

∈ <br />

⎪<br />

con ⎨ − 1 < r < 1<br />

⎪<br />

⎩k<br />

> 0<br />

2<br />

⎧ π ⋅<br />

⎪a<br />

=<br />

⎪<br />

12<br />

⎪ 1<br />

in questo caso ⎨r<br />

=<br />

⎪ 3<br />

⎪ k = 2<br />

⎪<br />

⎩<br />

quindi, l’area occupata dagli infiniti cerchi è:<br />

2<br />

3⋅π<br />

⋅ Λ<br />

C<br />

= 0,2946⋅<br />

<br />

32<br />

2<br />

area % occupata dagli infiniti cerchi rispetto all’intero triangolo equilatero:<br />

2<br />

3⋅π<br />

⋅<br />

100⋅<br />

100⋅ΛC<br />

32 25⋅π<br />

⋅ 3<br />

Λ<br />

C<br />

% = = = 68%<br />

2<br />

ΛT<br />

⋅ 3 2<br />

4<br />

36

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