Equation Involving Tangent

Short method: With $\tan(180^\circ - y) = \tan(y)$ , we can rewrite the equation as $\dfrac1{x - \tan(20^\circ)} + \dfrac1{x - \tan(140^\circ)} + \dfrac1{x - \tan(80^\circ)} .$ Notice that $\tan(3\times20^\circ) = \tan(3\times80^\circ) = \tan(3\times140^\circ) = \sqrt3$ .

Using the triple angle formula, $\tan(3y) = \dfrac{3\tan (y) - \tan^3(y)}{1-3\tan^2(y)}$ , then the equation $\sqrt3 = \dfrac{3y-y^3}{1-3y^2}$ has roots $\tan(20^\circ), \tan(80^\circ), \tan(140^\circ)$ . Rearranging this equation shows that $y^3 - 3\sqrt3 y^2 - 3y + \sqrt3 = 0$ has roots $\tan(20^\circ), \tan(80^\circ), \tan(140^\circ)$ .

For simplicities sake, let $A,B,C$ denote the values of $\tan(20^\circ), \tan(140^\circ), \tan(80^\circ)$ , respectively.

The given equation can be expressed as $3x^2 - 2x(A+B+C) - (AB + AC +BC) = 0$ . Vieta's formula shows that $A+B+C = 3\sqrt3, AB + AC + BC = -3$ .

Hence, $3x^2 - 6\sqrt3 - 3 = 0$ . With the quadratic formula , $x = \sqrt3 \pm2$ . Our answer is $2 + 3 + 2 +3 =\boxed{10}$ .

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Long method: With $\tan(180^\circ - y) = \tan(y)$ , we can rewrite the equation as $\dfrac1{x - \tan(20^\circ)} + \dfrac1{x - \tan(140^\circ)} + \dfrac1{x - \tan(80^\circ)} .$ For simplicities sake, let $A,B,C$ denote the values of $\tan(20^\circ), \tan(140^\circ), \tan(80^\circ)$ , respectively.

The given equation can be expressed as $\dfrac1{x-A}+\dfrac1{x-B} +\dfrac1{x-C} = 0 \quad \implies \quad 3x^2 - 2x(A+B+C) - (AB + AC +BC) = 0 .$

In order to solve for $x$ , we are left to evaluate the values of $A+B+C$ and $AB+AC+BC$ . Using the compound angle formula, $\sqrt3 = \tan(20^\circ + 140^\circ + 80^\circ) = \dfrac{\tan(20^\circ) + \tan(140^\circ) + \tan(80^\circ) - \tan(20^\circ) \tan(140^\circ) \tan(80^\circ) }{1 - (\tan(20^\circ) \tan(140^\circ) + \tan(20^\circ) \tan(80^\circ) + \tan(140^\circ) \tan(80^\circ) } = \dfrac{A+B+C - ABC}{1 - (AB + AC+ BC)} .$

Now, using the triple angle formula, $\tan(3y) = \tan(y) \tan(60^\circ + y) \tan(60^\circ - y)$ , for $y = 20^\circ$ , we get $\sqrt3 = \tan(60^\circ) = \tan(20^\circ) \tan(40^\circ) \tan(80^\circ) = - \tan(20^\circ) \tan(140^\circ) \tan(80^\circ) = -ABC$ .

Continuing from above, we have $\sqrt3 = \dfrac{ABC + \sqrt3}{1 - (AB + AC+ BC)} .$

We now have to find another relationship between $A+B+C$ and $AB + AC+ BC$ .

Consider the triple angle formula $\cos(3y) = 4\cos^3(y) - 3\cos(y)$ . Then $20^\circ, 100^\circ, 140^\circ$ are roots to the equation $\frac12 = 4\cos^3(y) - 3\cos(y)$ .

Equivalently, $\cos(20^\circ), \cos(100^\circ), \cos(140^\circ)$ are roots to the equation $8z^3 - 6z - 1 = 0$ .

Likewise, $\sec(20^\circ), \sec(100^\circ), \sec(140^\circ)$ are roots to the equation $8/z^3 - 6/z - 1 = 0 \Leftrightarrow z^3 + 6z^2 - 8 = 0$ .

By Vieta's formula ,

$\begin{cases} \sec(20^\circ) + \sec(100^\circ)+ \sec(140^\circ) = 6 \\ \sec(20^\circ) \sec(100^\circ) + \sec(100^\circ) \sec(140^\circ) + \sec(20^\circ) \sec(140^\circ) = 0. \end{cases}$

Thus, $\begin{aligned} \sec^2(20^\circ) + \sec^2(100^\circ)+ \sec^2(140^\circ) &=& 6^2 - 2\cdot 0 = 36 \\ (\sec^2(20^\circ) - 1) + (\sec^2(100^\circ)- 1) + (\sec^2(140^\circ) - 1) &=& 33 \\ \tan^2(20^\circ) + \tan^2(100^\circ)+ \tan^2(140^\circ) &=& 33 \\ \tan^2(20^\circ) + \tan^2(80^\circ)+ \tan^2(140^\circ) &=& 33 \\ A^2 + B^2 + C^2 &=& 33 \\ (A + B + C)^2 - 2(AB + AC + BC) &=& 33 \\ \end{aligned}$

Solving for the two equations $\sqrt3 = \dfrac{ABC + \sqrt3}{1 - (AB + AC+ BC)}$ and $(A + B + C)^2 - 2(AB + AC + BC) = 33$ gives $(A + B+ C, AB + AC + BC) = \left(3\sqrt3,-3\right), \left( -\frac{11}{\sqrt3}, \frac{11}3 \right)$ .

However, note that $A + B + C = \underbrace{\tan(20^\circ)}_{>0} + \underbrace{\tan(80^\circ) - \tan(40^\circ) }_{> 0 } > 0$ . So $(A + B + C, AB + AC + BC) = \left( -\frac{11}{\sqrt3}, \frac{11}3 \right)$ is an extraneous solution.

Hence, $3x^2 - 6\sqrt3 - 3 = 0$ . With the quadratic formula , $x = \sqrt3 \pm2$ . Our answer is $2 + 3 + 2 +3 =\boxed{10}$ .

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Alternatively, we can solve for $A+B+C$ using the following:

$\begin{aligned} A+B+C &=& \tan(20^\circ) + \tan(80^\circ) + \tan(140^\circ) \\ &=& \frac{\sin(20^\circ)}{\cos(20^\circ)} + \frac{\sin(80^\circ)}{\cos(80^\circ)} + \frac{\sin(140^\circ)}{\cos(140^\circ)} \\ &=& \frac{\sin(20^\circ)\cos(80^\circ)\cos(140^\circ) + \cos(20^\circ) \sin(80^\circ)\cos(140^\circ) + \cos(20^\circ)\cos(80^\circ)\cos(140^\circ)}{\cos(20^\circ)\cos(80^\circ) \cos(140^\circ)} \\ \end{aligned}$

Notice that the by using the compound angle formula twice, we have $\begin{aligned} \sin(X+Y+Z) &=& \sin(X+Y) \cos(Z) + \cos(X+Y) \sin (Z) \\ &=& \sin(X) \cos(Y) \cos(Z) + \cos(X) \sin(Y) \sin(Z) + \cos(X) \cos(Y) \sin(Z) - \sin(X) \sin(Y) \sin(Z) \end{aligned}$

Therefore, $A + B+ C = \dfrac{\sin(20^\circ + 80^\circ + 140^\circ) - \sin(20^\circ) \sin(80^\circ) \sin(140^\circ)}{\cos(20^\circ) \cos(80^\circ) \cos(140^\circ)} = \dfrac{\sin(240^\circ)}{\cos(20^\circ) \cos(80^\circ) \cos(140^\circ)} - \underbrace{\tan(20^\circ) \tan(80^\circ) \tan(140^\circ)}_{=-\sqrt3}.$

Using the triple angle formula as shown above, we can deduce that $\cos(20^\circ) \cos(80^\circ) \cos(140^\circ) = -\frac18$ . The rest will follow.