Algebra + Geometry

Algebra Level 4

12 x 4 − 56 x 3 + 89 x 2 − 56 x + 12 12x^{4}-56x^{3}+89x^{2}-56x+12

Given that the roots of the polynomial above are x 1 , x 2 , x 3 , x 4 x_{1},x_{2},x_{3},x_{4} in increasing order.

Find the area of the quadrilateral with vertices ( ⌊ x 1 ⌋ 2 , ⌊ x 2 ⌋ 2 ) , ( ⌊ x 2 ⌋ 2 , ⌊ x 3 ⌋ 2 ) , ( ⌊ x 3 ⌋ 2 , ⌊ x 4 ⌋ 2 ) , ( ⌊ x 4 ⌋ 2 , ⌊ x 1 ⌋ 2 ) . (\lfloor x_{1} \rfloor^{2},\lfloor x_{2} \rfloor^{2}),(\lfloor x_{2} \rfloor^{2},\lfloor x_{3} \rfloor^{2}),(\lfloor x_{3} \rfloor^{2},\lfloor x_{4} \rfloor^{2}),(\lfloor x_{4} \rfloor^{2},\lfloor x_{1} \rfloor^{2}).

If your answer comes as a b \dfrac{a}{b} where a a and b b are coprime positive integers, submit your answer as a + b a+b .


The answer is 19.

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Akshat Sharda by Akshat Sharda , 21, India

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1 solution

12 x 4 − 56 x 3 + 89 x 2 − 56 x + 12 = 0 Divide both sides by x 2 12 x 2 − 56 x + 89 − 56 x + 12 x 2 = 0 12 ( x + 1 x ) 2 − 56 ( x + 1 x ) + 65 = 0 Let y = x + 1 x ⇒ 12 y 2 − 56 y + 65 = 0 ( 6 y − 13 ) ( 2 y − 5 ) = 0 \begin{aligned} 12x^4-56x^3+89x^2-56x+12 & = 0 \quad \quad \small \color{#3D99F6}{\text{Divide both sides by } x^2} \\ 12x^2-56x+89-\frac{56}{x}+\frac{12}{x^2} & = 0 \\ 12 \left(x+\frac{1}{x}\right)^2 - 56 \left(x+\frac{1}{x}\right) + 65 & = 0 \quad \quad \small \color{#3D99F6}{\text{Let } y = x + \frac{1}{x}} \\ \Rightarrow 12y^2 - 56y + 65 & = 0 \\ (6y-13)(2y-5) & = 0 \end{aligned}

⇒ { x + 1 x = 13 6 ⇒ 6 x 2 − 13 x + 6 = 0 ⇒ ( 3 x − 2 ) ( 2 x − 3 ) = 0 ⇒ x = { 2 3 = x 2 3 2 = x 3 x + 1 x = 5 2 ⇒ 2 x 2 − 5 x + 2 = 0 ⇒ ( 2 x − 1 ) ( x − 2 ) = 0 ⇒ x = { 1 2 = x 1 2 = x 4 \Rightarrow \begin{cases} x+\frac{1}{x} = \dfrac{13}{6} & \Rightarrow 6x^2 -13x+6 = 0 & \Rightarrow (3x-2)(2x-3)=0 & \Rightarrow x = \begin{cases} \dfrac{2}{3} = x_2 \\ \dfrac{3}{2} = x_3 \end{cases} \\ x+\frac{1}{x} = \dfrac{5}{2} & \Rightarrow 2x^2 -5x+2 = 0 & \Rightarrow (2x-1)(x-2)=0 & \Rightarrow x = \begin{cases} \dfrac{1}{2} = x_1\\ \space 2 = x_4 \end{cases} \end{cases}

Therefore the vertices of the quadrilateral are: { ( ⌊ x 1 ⌋ 2 , ⌊ x 2 ⌋ 2 ) = ( ⌊ 1 2 ⌋ 2 , ⌊ 2 3 ⌋ 2 ) = ( 0 , 0 ) ( ⌊ x 2 ⌋ 2 , ⌊ x 3 ⌋ 2 ) = ( ⌊ 2 3 ⌋ 2 , ⌊ 3 2 ⌋ 2 ) = ( 0 , 1 ) ( ⌊ x 3 ⌋ 2 , ⌊ x 4 ⌋ 2 ) = ( ⌊ 3 2 ⌋ 2 , ⌊ 2 ⌋ 2 ) = ( 1 , 4 ) ( ⌊ x 4 ⌋ 2 , ⌊ x 1 ⌋ 2 ) = ( ⌊ 2 ⌋ 2 , ⌊ 1 2 ⌋ 2 ) = ( 4 , 0 ) \begin{cases} ( \lfloor x_1 \rfloor ^2, \lfloor x_2 \rfloor ^2) = \left( \left \lfloor \frac{1}{2} \right \rfloor ^2, \left \lfloor \frac{2}{3} \right \rfloor ^2 \right) = (0,0) \\ ( \lfloor x_2 \rfloor ^2, \lfloor x_3 \rfloor ^2) = \left( \left \lfloor \frac{2}{3} \right \rfloor ^2, \left \lfloor \frac{3}{2} \right \rfloor ^2 \right) = (0,1) \\ ( \lfloor x_3 \rfloor ^2, \lfloor x_4 \rfloor ^2) = \left( \left \lfloor \frac{3}{2} \right \rfloor ^2, \left \lfloor 2 \right \rfloor ^2 \right) = (1,4) \\ ( \lfloor x_4 \rfloor ^2, \lfloor x_1 \rfloor ^2) = \left( \left \lfloor 2 \right \rfloor ^2, \left \lfloor \frac{1}{2} \right \rfloor ^2 \right) = (4,0) \end{cases}

The quadrilateral is as show below and its area = 1 × 3 2 + 1 × 1 + 3 × 4 2 = 17 2 = \dfrac{1\times 3}{2} + 1 \times 1 + \dfrac{3 \times 4}{2} = \dfrac{17}{2}

⇒ a + b = 17 + 2 = 19 \Rightarrow a + b = 17+2 = \boxed{19}

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