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anemone

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MHB

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$\begin{align*}a + b + c &= 1\\ \dfrac{a}{ 1 - a}+\dfrac{b}{1 - b} + \dfrac{c}{1 - c} &= 6ac + 6bc = (a + 1)(b + 1)\end{align*}$

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- MHB
- Thread starter anemone
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- #1

anemone

Gold Member

MHB

POTW Director

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$\begin{align*}a + b + c &= 1\\ \dfrac{a}{ 1 - a}+\dfrac{b}{1 - b} + \dfrac{c}{1 - c} &= 6ac + 6bc = (a + 1)(b + 1)\end{align*}$

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Ick! :sick:

-Dan

-Dan

- #3

Theia

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We have

\[ \begin{cases} a + b + c = 1 \qquad (1) \\ \frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = L \qquad (2) \\ 6c(a + b) = L \qquad (3) \\ (a + 1)(b + 1) = L. \qquad (4) \end{cases} \]

Let's write \( a + b = A. \) Then from (1) \[ c = 1 - A, \]

and from (3) \[ L = 6A(1 - A). \]

Substitute these into (2) and solve the product \( ab = L - A - 1 \) from (4). After simplifying we obtain \[ 36A^4 - 60A^3 + 6A^2 + 19A - 6 =0. \]

Because I like to challenge myself, I completed this to square by solving the cubic equation from the discriminant of quadratic equation... For simplicity I write just few intermediate steps, as there's nothing fancy here:

Put \( A = x + \frac{5}{12}. \Rightarrow \) \[ (16x^2 - 7)^2 = \frac{460}{9} - \frac{608x}{27} \qquad \Rightarrow \] \[ (16x^2 - 7 + q)^2 = 32qx^2 - \frac{608}{27}x + q^2 - 14q + \frac{460}{9} \qquad \Rightarrow \] \[ 729q^3 - 10206q^2 + 37260q - 2888 = 0 \]

Then just write down the solutions: \[ q = \frac{14}{3} + \frac{\sqrt[3]{16\sqrt{287185} - 11408}}{9} + \frac{128}{3\sqrt[3]{16\sqrt{287185} - 11408}} \] \[ x = -\frac{\epsilon \sqrt{2q}}{8} + \frac{\delta}{96\sqrt{3}} \sqrt{-864q + 12096 + \frac{2432\epsilon \sqrt{2q}}{q}}\]

Greek letters take value -1 or 1 to cover all possibilities. Now only \( \epsilon = 1 \) gives real solutions, which we were asked for. \[ A = \frac{10 - 3\sqrt{2q}}{24} + \frac{\delta}{96\sqrt{3}} \sqrt{-864q + 12096 + \frac{2432\sqrt{2q}}{q}}\]

Final solutions: \[ \begin{cases}A \approx -0.5695729\ldots \\ L \approx -5.3639173\ldots \\ c \approx 1.5695729\ldots \\ a,b = \frac{A \pm \sqrt{A^2 - 4(L - A - 1)}}{2} \approx \{ -2.7087187\ldots , 2.1391458\ldots \}\end{cases} \] \[ \begin{cases} A \approx 1.303395799\ldots \\ L \approx -2.37266886\ldots \\ c \approx -0.303395799\ldots \\ a,b \approx \{ -1.60679159\ldots , 2.910187\ldots \} \end{cases} \]

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