Smjestimo trokut A B C ABC A B C
Neka je A C = b AC=b A C = b B C = a BC=a B C = a A ( b , 0 ) A(b,0) A ( b , 0 ) B ( 0 , a ) B(0,a) B ( 0 , a ) C ( 0 , 0 ) C(0,0) C ( 0 , 0 ) A B AB A B
y = − a b x + a , y=-\frac abx+a, y = − b a x + a , pa M M M M ( x , − a b x + a ) M\left(x,-\frac abx+a\right) M ( x , − b a x + a )
Računamo udaljenosti:
d 2 ( A , M ) = ( b − x ) 2 + ( − a b x + a ) 2 = b 2 − 2 b x + x 2 + ( a x b ) 2 − 2 a 2 x b + a 2 , d 2 ( B , M ) = x 2 + ( − a b x + a − a ) 2 = x 2 + ( a x b ) 2 , d 2 ( C , M ) = x 2 + ( − a b x + a ) 2 = x 2 + ( a x b ) 2 − 2 a 2 x b + a 2 . \begin{aligned} d^2(A,M)&=(b-x)^2+\left(-\frac abx+a\right)^2\\ &=b^2-2bx+x^2+\left(\frac{ax}{b}\right)^2-\frac{2a^2x}{b}+a^2, \\[4pt] d^2(B,M)&=x^2+\left(-\frac abx+a-a\right)^2\\ &=x^2+\left(\frac{ax}{b}\right)^2, \\[4pt] d^2(C,M)&=x^2+\left(-\frac abx+a\right)^2\\ &=x^2+\left(\frac{ax}{b}\right)^2-\frac{2a^2x}{b}+a^2. \end{aligned} d 2 ( A , M ) d 2 ( B , M ) d 2 ( C , M ) = ( b − x ) 2 + ( − b a x + a ) 2 = b 2 − 2 b x + x 2 + ( b a x ) 2 − b 2 a 2 x + a 2 , = x 2 + ( − b a x + a − a ) 2 = x 2 + ( b a x ) 2 , = x 2 + ( − b a x + a ) 2 = x 2 + ( b a x ) 2 − b 2 a 2 x + a 2 . Slijedi
d 2 ( A , M ) + d 2 ( B , M ) − d 2 ( C , M ) = b 2 − 2 b x + x 2 + ( a x b ) 2 − 2 a 2 x b + a 2 + x 2 + ( a x b ) 2 − x 2 − ( a x b ) 2 + 2 a 2 x b − a 2 = b 2 − 2 b x + x 2 + ( a x b ) 2 . \begin{aligned} d^2(A,M)+d^2(B,M)-d^2(C,M) &=b^2-2bx+x^2+\left(\frac{ax}{b}\right)^2-\frac{2a^2x}{b}+a^2\\ &\quad+x^2+\left(\frac{ax}{b}\right)^2-x^2-\left(\frac{ax}{b}\right)^2+\frac{2a^2x}{b}-a^2\\ &=b^2-2bx+x^2+\left(\frac{ax}{b}\right)^2. \end{aligned} d 2 ( A , M ) + d 2 ( B , M ) − d 2 ( C , M ) = b 2 − 2 b x + x 2 + ( b a x ) 2 − b 2 a 2 x + a 2 + x 2 + ( b a x ) 2 − x 2 − ( b a x ) 2 + b 2 a 2 x − a 2 = b 2 − 2 b x + x 2 + ( b a x ) 2 . Prema uvjetu zadatka
x 2 ( 1 + a 2 b 2 ) − 2 b x + 3 4 b 2 − 1 4 a 2 = 0. x^2\left(1+\frac{a^2}{b^2}\right)-2bx+\frac34b^2-\frac14a^2=0. x 2 ( 1 + b 2 a 2 ) − 2 b x + 4 3 b 2 − 4 1 a 2 = 0. Rješenja su
x 1 , 2 = 2 b ± 4 b 2 − 4 ( 1 + a 2 b 2 ) ( 3 4 b 2 − 1 4 a 2 ) 2 ( 1 + a 2 b 2 ) . x_{1,2}=\frac{2b\pm\sqrt{4b^2-4\left(1+\frac{a^2}{b^2}\right)\left(\frac34b^2-\frac14a^2\right)}}{2\left(1+\frac{a^2}{b^2}\right)}. x 1 , 2 = 2 ( 1 + b 2 a 2 ) 2 b ± 4 b 2 − 4 ( 1 + b 2 a 2 ) ( 4 3 b 2 − 4 1 a 2 ) . Pojednostavimo korijen:
4 b 2 − 4 ( 1 + a 2 b 2 ) ( 3 4 b 2 − 1 4 a 2 ) = b 2 − 2 a 2 + a 4 b 2 = ( b − a 2 b ) 2 . \begin{aligned} &4b^2-4\left(1+\frac{a^2}{b^2}\right)\left(\frac34b^2-\frac14a^2\right)\\ &=b^2-2a^2+\frac{a^4}{b^2}=\left(b-\frac{a^2}{b}\right)^2. \end{aligned} 4 b 2 − 4 ( 1 + b 2 a 2 ) ( 4 3 b 2 − 4 1 a 2 ) = b 2 − 2 a 2 + b 2 a 4 = ( b − b a 2 ) 2 . Slijedi
x 1 , 2 = 2 b ± ( b − a 2 / b ) 2 ( 1 + a 2 / b 2 ) . x_{1,2}=\frac{2b\pm\left(b-a^2/b\right)}{2\left(1+a^2/b^2\right)}. x 1 , 2 = 2 ( 1 + a 2 / b 2 ) 2 b ± ( b − a 2 / b ) . Prvo rješenje daje
x 1 = ( b 2 + a 2 ) / b 2 ( b 2 + a 2 ) / b 2 = b 2 . x_1=\frac{(b^2+a^2)/b}{2(b^2+a^2)/b^2}=\frac b2. x 1 = 2 ( b 2 + a 2 ) / b 2 ( b 2 + a 2 ) / b = 2 b . Zaključujemo da je M M M
Drugo rješenje jest
x 2 = ( 3 b 2 − a 2 ) / b 2 ( b 2 + a 2 ) / b 2 = b ( 3 b 2 − a 2 ) 2 ( b 2 + a 2 ) . x_2=\frac{(3b^2-a^2)/b}{2(b^2+a^2)/b^2}=\frac{b(3b^2-a^2)}{2(b^2+a^2)}. x 2 = 2 ( b 2 + a 2 ) / b 2 ( 3 b 2 − a 2 ) / b = 2 ( b 2 + a 2 ) b ( 3 b 2 − a 2 ) . Mora vrijediti 0 ≤ x 2 ≤ b 0\le x_2\le b 0 ≤ x 2 ≤ b 3 b 2 − a 2 ≥ 0 3b^2-a^2\ge0 3 b 2 − a 2 ≥ 0 a / b ≤ 3 a/b\le\sqrt3 a / b ≤ 3 3 b 2 − a 2 ≤ 2 ( b 2 + a 2 ) 3b^2-a^2\le2(b^2+a^2) 3 b 2 − a 2 ≤ 2 ( b 2 + a 2 ) a / b ≥ 3 / 3 a/b\ge\sqrt3/3 a / b ≥ 3 /3 α = arctan ( a / b ) \alpha=\arctan(a/b) α = arctan ( a / b ) 30 ∘ ≤ α ≤ 60 ∘ 30^\circ\le\alpha\le60^\circ 3 0 ∘ ≤ α ≤ 6 0 ∘ α ≠ 45 ∘ \alpha\ne45^\circ α = 4 5 ∘ α = 45 ∘ \alpha=45^\circ α = 4 5 ∘
M ( b ( 3 b 2 − a 2 ) 2 ( b 2 + a 2 ) , a ( 3 a 2 − b 2 ) 2 ( b 2 + a 2 ) ) . M\left(\frac{b(3b^2-a^2)}{2(b^2+a^2)},\frac{a(3a^2-b^2)}{2(b^2+a^2)}\right). M ( 2 ( b 2 + a 2 ) b ( 3 b 2 − a 2 ) , 2 ( b 2 + a 2 ) a ( 3 a 2 − b 2 ) ) . 
