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Njia ya MaajabuHiari · haipewi alama kamweInaanzia The Pythagorean theorem

Triangles on a ball

Ukurasa huu bado haujatafsiriwa kwa Kiswahili, kwa hiyo unaonyeshwa kwa English.

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Swali

On flat paper the angles of every triangle add up to 180°, and a² + b² = c² in every right triangle. Draw a triangle on a ball instead. Is either rule still true?

Fanya jaribio kabla ya kuendelea kusoma. Hakuna njia mbaya ya kuchunguza.

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Jaribio

Pembetatu kwenye nyuso zilizopinda

Buruta vipeo. Kwenye ndege bapa jumla ya pembe ni 180°; kwenye tufe ni zaidi, na kwenye ndege hipaboliki ni chini.

80.6°66.6°64.7°ABC

Buruta kipeo ili kukisogeza; buruta mahali pengine popote ili kugeuza tufe. Pande ni matao ya duara kuu — njia fupi kabisa juu ya tufe.

  • ∠A80.6°
  • ∠B66.6°
  • ∠C64.7°

∠A + ∠B + ∠C = 212.0°

Ziada ya pembe (jumla − 180°)
32.0° = 0.558 rad
Eneo, limepimwa kwa hesabu za namba
0.558 R²
Sehemu ya tufe zima (eneo ÷ 4πR²)
4.44 %

∠A + ∠B + ∠C − π = Eneo / R²Kwenye tufe lenye nusukipenyo R, ziada kwa radiani ni sawa na eneo likigawanywa kwa R². Eneo linakokotolewa kando, kwa kuunganisha kwa namba (integration), kwa hiyo kulingana kwao ni ukaguzi, si dhana.

∠A + ∠B + ∠C − π = K · Eneo

Ndege bapaK = 0∠A+∠B+∠C = 180°
TufeK = 1/R²∠A+∠B+∠C > 180°
Ndege hipabolikiK = −1∠A+∠B+∠C < 180°

Huu ni mtazamo wa kwanza wa nadharia ya Gauss–Bonnet: kwa pembetatu ambayo pande zake ni jiodesiki (njia fupi kabisa), jumla ya pembe toa π ni sawa na mpindo wote ulio ndani yake. Hapa mpindo K ni uleule kila mahali; kwenye nyuso nyingine K hujumlishwa (huunganishwa) juu ya eneo.

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Vidokezo, ukivitaka

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    Kinachoendelea

    Kinachoendelea

    On a sphere the "straight lines" are great circles — circles whose centre is the centre of the sphere, like the equator. The walk in the first hint makes a triangle with three right angles: its angles add up to 270∘270^\circ.

    Albert Girard found the rule in 1629. On a sphere of radius RR, a triangle with angles α,β,γ\alpha, \beta, \gamma (in radians) has

    α+β+γ−π=areaR2.\alpha + \beta + \gamma - \pi = \frac{\text{area}}{R^2}.

    Our triangle covers one eighth of the sphere, so its area is 18⋅4πR2=π2R2\tfrac18 \cdot 4\pi R^2 = \tfrac{\pi}{2}R^2, and the excess is π2\tfrac{\pi}{2} — exactly the extra 90∘90^\circ.

    Pythagoras fails too. All three sides of that triangle are quarter-circles of the same length cc, so a2+b2=2c2≠c2a^2 + b^2 = 2c^2 \neq c^2. The spherical version of the theorem is

    cos⁡cR=cos⁡aR cos⁡bR,\cos\frac{c}{R} = \cos\frac{a}{R}\,\cos\frac{b}{R},

    and for small triangles it is very close to a2+b2=c2a^2 + b^2 = c^2 — which is why flat geometry works so well for drawing a field or a town, even though the Earth is round.

    The difference between the angle sum and 180∘180^\circ measures how curved a surface is. On a saddle-shaped surface — the hyperbolic plane — triangles have angle sums less than 180∘180^\circ. Gauss, Bolyai, Lobachevsky and Riemann turned these observations into non-Euclidean geometry, and Einstein used curved geometry to describe gravity.

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