Nakon što je riješio sve zadatke iz testa, učenik je iz dosade počeo crtati. Nacrtao je jednakokračan trokut kojemu je osnovica A 1 A 2 ‾ \overline{A_1A_2} A 1 A 2 a a a x x x A 1 A_1 A 1 α \alpha α A 1 A_1 A 1 A 2 A 3 ‾ \overline{A_2A_3} A 2 A 3 2 3 a \frac23a 3 2 a A 3 A 4 ‾ \overline{A_3A_4} A 3 A 4 2 3 ∣ A 2 A 3 ‾ ∣ \frac23|\overline{A_2A_3}| 3 2 ∣ A 2 A 3 ∣ V 1 , V 2 , … , V n , … V_1,V_2,\ldots,V_n,\ldots V 1 , V 2 , … , V n , … x x x tg α ⋅ x + 5 y − 3 a tg α = 0 \tg\alpha\cdot x+5y-3a\tg\alpha=0 tg α ⋅ x + 5 y − 3 a tg α = 0
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Osnovice nacrtanih trokuta čine geometrijski niz ( a n ) (a_n) ( a n ) 2 3 \frac23 3 2 S n S_n S n
a 1 = a , a 2 = 2 3 a , a 3 = ( 2 3 ) 2 a , … , a n = ( 2 3 ) n − 1 a . a_1=a,\qquad a_2=\frac23a,\qquad a_3=\left(\frac23\right)^2a,\qquad\ldots,\qquad a_n=\left(\frac23\right)^{n-1}a. a 1 = a , a 2 = 3 2 a , a 3 = ( 3 2 ) 2 a , … , a n = ( 3 2 ) n − 1 a . S n = a 1 − ( 2 3 ) n 1 − ( 2 3 ) = 3 a ( 1 − ( 2 3 ) n ) . S_n=a\frac{1-\left(\frac23\right)^n}{1-\left(\frac23\right)}=3a\left(1-\left(\frac23\right)^n\right). S n = a 1 − ( 3 2 ) 1 − ( 3 2 ) n = 3 a ( 1 − ( 3 2 ) n ) . Za apscise traženih vrhova V n V_n V n
x 1 = 1 2 a , x_1=\frac12a, x 1 = 2 1 a , x 2 = a 1 + 1 2 a 2 = a + 1 3 a = 4 3 a , x_2=a_1+\frac12a_2=a+\frac13a=\frac43a, x 2 = a 1 + 2 1 a 2 = a + 3 1 a = 3 4 a , x 3 = a 1 + a 2 + 1 2 a 3 = S 2 + 1 2 a 3 , x_3=a_1+a_2+\frac12a_3=S_2+\frac12a_3, x 3 = a 1 + a 2 + 2 1 a 3 = S 2 + 2 1 a 3 , x 4 = a 1 + a 2 + a 3 + 1 2 a 4 = S 3 + 1 2 a 4 , x_4=a_1+a_2+a_3+\frac12a_4=S_3+\frac12a_4, x 4 = a 1 + a 2 + a 3 + 2 1 a 4 = S 3 + 2 1 a 4 , ⋮ \vdots ⋮ x n = a 1 + a 2 + … + a n − 1 + 1 2 a n = S n − 1 + 1 2 a n = 3 a ( 1 − ( 2 3 ) n − 1 ) + 1 2 ( 2 3 ) n − 1 a \begin{aligned} x_n&=a_1+a_2+\ldots+a_{n-1}+\frac12a_n\\ &=S_{n-1}+\frac12a_n\\ &=3a\left(1-\left(\frac23\right)^{n-1}\right)+\frac12\left(\frac23\right)^{n-1}a \end{aligned} x n = a 1 + a 2 + … + a n − 1 + 2 1 a n = S n − 1 + 2 1 a n = 3 a ( 1 − ( 3 2 ) n − 1 ) + 2 1 ( 3 2 ) n − 1 a x n = 3 a − 5 a 2 ( 2 3 ) n − 1 . (1) x_n=3a-\frac{5a}{2}\left(\frac23\right)^{n-1}.\tag{1} x n = 3 a − 2 5 a ( 3 2 ) n − 1 . ( 1 ) Ordinatu bilo kojeg od vrhova V n V_n V n
tg α = y n a n / 2 \tg\alpha=\frac{y_n}{a_n/2} tg α = a n /2 y n pa je
y n = a 2 ( 2 3 ) n − 1 tg α . (2) y_n=\frac{a}{2}\left(\frac23\right)^{n-1}\tg\alpha.\tag{2} y n = 2 a ( 3 2 ) n − 1 tg α . ( 2 ) Treba još samo pronaći vezu između x x x y y y V n V_n V n
( 2 3 ) n − 1 = 2 y n a ⋅ tg α \left(\frac23\right)^{n-1}=\frac{2y_n}{a\cdot\tg\alpha} ( 3 2 ) n − 1 = a ⋅ tg α 2 y n i uvrstiti u (1). Tada je
x n = 3 a − 5 a 2 ⋅ 2 y n a ⋅ tg α , x_n=3a-\frac{5a}{2}\cdot\frac{2y_n}{a\cdot\tg\alpha}, x n = 3 a − 2 5 a ⋅ a ⋅ tg α 2 y n , odnosno
x n = 3 a − 5 y n tg α . x_n=3a-\frac{5y_n}{\tg\alpha}. x n = 3 a − tg α 5 y n . Pišemo tg α ⋅ x n + 5 y n − 3 a tg α = 0 \tg\alpha\cdot x_n+5y_n-3a\tg\alpha=0 tg α ⋅ x n + 5 y n − 3 a tg α = 0 tg α ⋅ x + 5 y − 3 a tg α = 0 \tg\alpha\cdot x+5y-3a\tg\alpha=0 tg α ⋅ x + 5 y − 3 a tg α = 0
