Prvo rješenje
Na slici je prikazan poprečni presjek kugle i njoj opisanog uspravnog stošca. Označimo točke kao na slici, te neka je ∣ O D ∣ = r |OD|=r ∣ O D ∣ = r ∣ P C ∣ = R |PC|=R ∣ P C ∣ = R ∣ A P ∣ = h |AP|=h ∣ A P ∣ = h
Kako je ∠ O A D = ∠ C A P \angle OAD=\angle CAP ∠ O A D = ∠ C A P ∠ A D O = ∠ A P C = 90 ∘ \angle ADO=\angle APC=90^\circ ∠ A D O = ∠ A P C = 9 0 ∘ A D O ADO A D O A P C APC A P C
∣ A O ∣ ∣ D O ∣ = ∣ A C ∣ ∣ P C ∣ , \frac{|AO|}{|DO|}=\frac{|AC|}{|PC|}, ∣ D O ∣ ∣ A O ∣ = ∣ P C ∣ ∣ A C ∣ , odnosno
h − r r = h 2 + R 2 R . \frac{h-r}{r}=\frac{\sqrt{h^2+R^2}}{R}. r h − r = R h 2 + R 2 . Nakon kvadriranja dobivamo
h 2 − 2 h r + r 2 r 2 = h 2 + R 2 R 2 , \frac{h^2-2hr+r^2}{r^2}=\frac{h^2+R^2}{R^2}, r 2 h 2 − 2 h r + r 2 = R 2 h 2 + R 2 , odakle slijedi h 2 R 2 − 2 h r R 2 = h 2 r 2 h^2R^2-2hrR^2=h^2r^2 h 2 R 2 − 2 h r R 2 = h 2 r 2
h = 2 r R 2 R 2 − r 2 . h=\frac{2rR^2}{R^2-r^2}. h = R 2 − r 2 2 r R 2 . Promatrani omjer je
V K V S = 4 3 r 3 π 1 3 R 2 π ⋅ h = 4 r 3 R 2 h = 4 r 3 R 2 ⋅ R 2 − r 2 2 r R 2 = 2 r 2 ( R 2 − r 2 ) R 4 = − 2 ( r R ) 4 + 2 ( r R ) 2 . \frac{V_K}{V_S}=\frac{\frac43r^3\pi}{\frac13R^2\pi\cdot h}=\frac{4r^3}{R^2h} =\frac{4r^3}{R^2}\cdot\frac{R^2-r^2}{2rR^2} =\frac{2r^2(R^2-r^2)}{R^4} =-2\left(\frac rR\right)^4+2\left(\frac rR\right)^2. V S V K = 3 1 R 2 π ⋅ h 3 4 r 3 π = R 2 h 4 r 3 = R 2 4 r 3 ⋅ 2 r R 2 R 2 − r 2 = R 4 2 r 2 ( R 2 − r 2 ) = − 2 ( R r ) 4 + 2 ( R r ) 2 . Zato je
V K V S = 1 2 − 2 ( ( r R ) 2 − 1 2 ) 2 , \frac{V_K}{V_S}=\frac12-2\left(\left(\frac rR\right)^2-\frac12\right)^2, V S V K = 2 1 − 2 ( ( R r ) 2 − 2 1 ) 2 , pa traženi maksimum iznosi 1 2 \frac12 2 1 R = r 2 R=r\sqrt2 R = r 2
Drugo rješenje
Označimo točke kao na slici. Neka je ∠ A B C = ∠ A C B = 2 α \angle ABC=\angle ACB=2\alpha ∠ A B C = ∠ A C B = 2 α O O O A B C ABC A B C O C OC O C A C B ACB A C B ∠ O C P = ∠ O C D = α \angle OCP=\angle OCD=\alpha ∠ O C P = ∠ O C D = α
Vrijedi
tan α = ∣ O P ∣ ∣ P C ∣ = r R pa je r = R tan α , \tan\alpha=\frac{|OP|}{|PC|}=\frac rR\quad\text{pa je }r=R\tan\alpha, tan α = ∣ P C ∣ ∣ O P ∣ = R r pa je r = R tan α , tan 2 α = ∣ A P ∣ ∣ P C ∣ = h R pa je h = R tan 2 α . \tan2\alpha=\frac{|AP|}{|PC|}=\frac hR\quad\text{pa je }h=R\tan2\alpha. tan 2 α = ∣ P C ∣ ∣ A P ∣ = R h pa je h = R tan 2 α . Promatrani omjer iznosi
V K V S = 4 3 r 3 π 1 3 R 2 π ⋅ h = 4 r 3 R 2 h . \frac{V_K}{V_S}=\frac{\frac43r^3\pi}{\frac13R^2\pi\cdot h}=\frac{4r^3}{R^2h}. V S V K = 3 1 R 2 π ⋅ h 3 4 r 3 π = R 2 h 4 r 3 . Kako je tan 2 α = 2 tan α 1 − tan 2 α \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha} tan 2 α = 1 − t a n 2 α 2 t a n α
V K V S = 4 ( R tan α ) 3 R 3 tan 2 α = 4 tan 3 α tan 2 α = 4 tan 3 α 2 tan α 1 − tan 2 α = 2 tan 2 α ( 1 − tan 2 α ) . \frac{V_K}{V_S}=\frac{4(R\tan\alpha)^3}{R^3\tan2\alpha}=\frac{4\tan^3\alpha}{\tan2\alpha}=\frac{4\tan^3\alpha}{\frac{2\tan\alpha}{1-\tan^2\alpha}}=2\tan^2\alpha(1-\tan^2\alpha). V S V K = R 3 tan 2 α 4 ( R tan α ) 3 = tan 2 α 4 tan 3 α = 1 − t a n 2 α 2 t a n α 4 tan 3 α = 2 tan 2 α ( 1 − tan 2 α ) . Označimo t = tan 2 α t=\tan^2\alpha t = tan 2 α
V K V S = 2 t ( 1 − t ) = 1 2 − 2 ( t − 1 2 ) 2 , \frac{V_K}{V_S}=2t(1-t)=\frac12-2\left(t-\frac12\right)^2, V S V K = 2 t ( 1 − t ) = 2 1 − 2 ( t − 2 1 ) 2 , pa je traženi maksimalni omjer jednak 1 2 \frac12 2 1
Treće rješenje
Koristimo iste oznake kao u prvom rješenju, te ∣ A B ∣ = ∣ A C ∣ = s |AB|=|AC|=s ∣ A B ∣ = ∣ A C ∣ = s A B C ABC A B C
P A B C = 1 2 ∣ B C ∣ ⋅ ∣ A P ∣ = 1 2 ⋅ 2 R ⋅ h = R h , P_{ABC}=\frac12|BC|\cdot|AP|=\frac12\cdot2R\cdot h=Rh, P A B C = 2 1 ∣ B C ∣ ⋅ ∣ A P ∣ = 2 1 ⋅ 2 R ⋅ h = R h , pa je polumjer njegove upisane kružnice
r = povr s ˇ ina poluopseg = R h 1 2 ( 2 s + 2 R ) = R h s + R . r=\frac{\text{površina}}{\text{poluopseg}}=\frac{Rh}{\frac12(2s+2R)}=\frac{Rh}{s+R}. r = poluopseg povr s ˇ ina = 2 1 ( 2 s + 2 R ) R h = s + R R h . Promatrani omjer je
V K V S = 4 3 r 3 π 1 3 R 2 π ⋅ h = 4 r 3 R 2 h = 4 R 3 h 3 R 2 h ( s + R ) 3 = 4 R h 2 ( s + R ) 3 = 4 R ( s 2 − R 2 ) ( s + R ) 3 = 4 R ( s − R ) ( s + R ) 2 = 4 ( s R − 1 ) ( s R + 1 ) 2 . \begin{aligned} \frac{V_K}{V_S} &=\frac{\frac43r^3\pi}{\frac13R^2\pi\cdot h}=\frac{4r^3}{R^2h}\\ &=\frac{4R^3h^3}{R^2h(s+R)^3}=\frac{4Rh^2}{(s+R)^3}\\ &=\frac{4R(s^2-R^2)}{(s+R)^3}=\frac{4R(s-R)}{(s+R)^2}\\ &=\frac{4\left(\frac sR-1\right)}{\left(\frac sR+1\right)^2}. \end{aligned} V S V K = 3 1 R 2 π ⋅ h 3 4 r 3 π = R 2 h 4 r 3 = R 2 h ( s + R ) 3 4 R 3 h 3 = ( s + R ) 3 4 R h 2 = ( s + R ) 3 4 R ( s 2 − R 2 ) = ( s + R ) 2 4 R ( s − R ) = ( R s + 1 ) 2 4 ( R s − 1 ) . Uvedimo supstituciju t = s R − 1 t=\frac sR-1 t = R s − 1 s R + 1 = t + 2 \frac sR+1=t+2 R s + 1 = t + 2 4 t ( t + 2 ) 2 \frac{4t}{(t+2)^2} ( t + 2 ) 2 4 t ( t − 2 ) 2 ≥ 0 (t-2)^2\ge0 ( t − 2 ) 2 ≥ 0 ( t + 2 ) 2 ≥ 8 t (t+2)^2\ge8t ( t + 2 ) 2 ≥ 8 t
V K V S = 4 t ( t + 2 ) 2 ≤ 4 t 8 t = 1 2 . \frac{V_K}{V_S}=\frac{4t}{(t+2)^2}\le\frac{4t}{8t}=\frac12. V S V K = ( t + 2 ) 2 4 t ≤ 8 t 4 t = 2 1 . Jednakost se postiže ako i samo ako je t = 2 t=2 t = 2 s = 3 R s=3R s = 3 R
