Tangent roots in algebraic equation?

Geometry Level 5

( 1 + 2 x 1 − x 2 ) ( 1 + x 5 − 10 x 3 + 5 x 5 x 4 − 10 x 2 + 1 ) = 2 \left( 1 + \dfrac{2x}{1-x^2} \right) \left( 1 + \dfrac{x^5 - 10x^3 + 5x}{5x^4 - 10x^2 + 1} \right) = 2

The six roots of the equations above are x 1 , x 2 , … , x 6 x_1, x_2, \ldots, x_6 . Given that

x 1 = tan ⁡ b π c , x 2 = tan ⁡ d π e , x 3 = tan ⁡ f π g , x 4 = tan ⁡ h π k , x 5 = tan ⁡ m π n , x 6 = tan ⁡ z π t , x_1 = \tan \dfrac{b \pi }{c} , x_2 = \tan \dfrac{d \pi}e, x_3 = \tan \dfrac{f \pi}g , \\ x_4 = \tan \dfrac{h \pi}k , x_5 = \tan \dfrac{m\pi }n, x_6 = \tan \dfrac{z \pi }t,

where b , c , d , e , f , g , h , k , m , n , z b,c,d,e,f,g,h,k,m,n,z are integers such that gcd ⁡ ( b , c ) = gcd ⁡ ( d , e ) = gcd ⁡ ( f , g ) = gcd ⁡ ( h , k ) = gcd ⁡ ( m , n ) = gcd ⁡ ( z , t ) = 1 \gcd(b,c) = \gcd(d,e) = \gcd(f,g) = \gcd(h,k) = \gcd(m,n) = \gcd(z,t) = 1 .

If all the angles shown above are in the interval ( − π 2 , π 2 ) \left( -\dfrac \pi 2 , \dfrac\pi 2 \right) , compute b + c + d + e + f + g + h + k + m + n + z + t b+c+d+e+f+g+h+k+m+n+z+t .

Inspiration .


The answer is 182.

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Son Nguyen by Son Nguyen , 20, Vietnam

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2 solutions

Zk Lin
Feb 18, 2016

Write c = cos ⁡ t , s = sin ⁡ t c=\cos t, s=\sin t

Note that ( c + i s ) 5 = cos ⁡ 5 t + i sin ⁡ 5 t (c+is)^{5}=\cos 5t+i \sin 5t by De Moivre's theorem.

Expanding ( c + i s ) 5 (c+is)^{5} , we get

( c 5 − 10 s 2 c 3 + 5 s 4 c ) + ( s 5 − 10 s 3 c 2 + 5 s c 4 ) i (c^{5}-10s^{2}c^{3}+5s^{4}c)+(s^{5}-10s^{3}c^{2}+5sc^{4})i

Equating like terms, we find that

cos ⁡ 5 t = c 5 − 10 s 2 c 3 + 5 s 4 c \cos 5t=c^{5}-10s^{2}c^{3}+5s^{4}c

sin ⁡ 5 t = s 5 − 10 s 3 c 2 + 5 s c 4 \sin 5t=s^{5}-10s^{3}c^{2}+5sc^{4}

Therefore, sin ⁡ 5 t cos ⁡ 5 t = s 5 − 10 s 3 c 2 + 5 s c 4 c 5 − 10 s 2 c 3 + 5 s 4 c \frac{\sin 5t}{\cos 5t}=\frac{s^{5}-10s^{3}c^{2}+5sc^{4}}{c^{5}-10s^{2}c^{3}+5s^{4}c} .

Dividing the numerator and denominator by c 5 c^{5} gives

tan ⁡ 5 t = t 5 − 10 t 3 + 5 t 5 t 4 − 10 t 2 + 1 \tan 5t= \frac{t^{5}-10t^{3}+5t}{5t^{4}-10t^{2}+1}

Note also that tan ⁡ 2 t = 2 t 1 − t 2 \tan 2t= \frac{2t}{1-t^{2}}

It is obvious now what we should do. Make the substitution x = tan ⁡ t x=\tan t !

The equation simplifies to:

( 1 + tan ⁡ 2 t ) ( 1 + tan ⁡ 5 t ) = 2 (1+\tan 2t)(1+ \tan 5t)=2

1 + tan ⁡ 5 t + tan ⁡ 2 t + tan ⁡ 5 t tan ⁡ 2 t = 2 1+\tan 5t+\tan2t+\tan 5t \tan 2t=2

1 + sin ⁡ 5 t cos ⁡ 5 t + sin ⁡ 2 t cos ⁡ 2 t + sin ⁡ 5 t sin ⁡ 2 t cos ⁡ 5 t cos ⁡ 2 t = 2 1+ \frac{\sin5t}{\cos5t}+\frac{\sin2t}{\cos2t}+\frac{\sin 5t \sin 2t}{\cos 5t \cos 2t}=2

cos ⁡ 2 t cos ⁡ 5 t + sin ⁡ 5 t cos ⁡ 2 t + sin ⁡ 2 t cos ⁡ 5 t + sin ⁡ 5 t sin ⁡ 2 t cos ⁡ 5 t cos ⁡ 2 t = 2 \frac{\cos 2t\cos 5t+\sin 5t\cos 2t+\sin 2t\cos 5t+\sin 5t\sin 2t}{\cos 5t\cos 2t}=2

cos ⁡ ( 5 t − 2 t ) + sin ⁡ ( 5 t + 2 t ) cos ⁡ 2 t cos ⁡ 5 t = 2 \frac{\cos(5t-2t)+\sin(5t+2t)}{\cos 2t \cos 5t}=2

cos ⁡ ( 5 t − 2 t ) + sin ⁡ ( 5 t + 2 t ) 1 2 ( cos ⁡ 7 t + cos ⁡ 3 t ) = 2 \frac{\cos(5t-2t)+\sin(5t+2t)}{\frac{1}{2}(\cos 7t+ \cos 3t)}=2

cos ⁡ 3 t + sin ⁡ 7 t = cos ⁡ 7 t + cos ⁡ 3 t \cos 3t+\sin 7t=\cos 7t+\cos 3t

tan ⁡ 7 t = 1 \tan 7t=1

The solutions are: (considering principal values of − π 2 ≤ 7 t ≤ π 2 ) \frac{-\pi}{2} \leq 7t \leq \frac{\pi}{2})

7 t = − 11 π 4 , − 7 π 4 , − 3 π 4 , π 4 , 5 π 4 , 9 π 4 , 13 π 4 7t= \frac{-11\pi}{4},\frac{-7\pi}{4}, \frac{-3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}

t = − 11 π 28 , − 7 π 28 , − 3 π 28 , π 28 , 5 π 28 , 9 π 28 , 13 π 28 t= \frac{-11\pi}{28},\frac{-7\pi}{28}, \frac{-3\pi}{28}, \frac{\pi}{28}, \frac{5\pi}{28}, \frac{9\pi}{28}, \frac{13\pi}{28}

x = tan ⁡ t = tan ⁡ ( − 11 π 28 ) , tan ⁡ ( − 7 π 28 ) = − 1 , tan ⁡ ( − 3 π 28 ) , tan ⁡ ( π 28 ) , tan ⁡ ( 5 π 28 ) , tan ⁡ ( 9 π 28 ) , tan ⁡ ( 13 π 28 ) x= \tan t=\tan{(\frac{-11\pi}{28})},\tan{(\frac{-7\pi}{28})}=-1, \tan{(\frac{-3\pi}{28})}, \tan{(\frac{\pi}{28})}, \tan{(\frac{5\pi}{28})}, \tan{(\frac{9\pi}{28})}, \tan{(\frac{13\pi}{28})}

However, we must exclude x = − 1 x=-1 as the answer since 1 − x 2 ≠ 0 1-x^{2} \neq 0 .

Therefore, the desired answer is − 11 − 3 + 1 + 5 + 9 + 13 + 6 ( 28 ) = 182 -11-3+1+5+9+13+6(28)=\boxed{182}

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Moderator note:

Nice observation of the tangent substitution to simplify the expression.

Son Nguyen
Feb 18, 2016

The conditional of the equation is 5 x 4 − 10 x 2 + 1 ≠ 0 5x^4-10x^2+1 \neq 0 and x ≠ ± 1 x\neq \pm 1

Let x = t a n t x=tant and − π 2 < t < π 2 \frac{-\pi }{2}< t< \frac{\pi}{2} the equation can be rewrite as: ( 1 + 2 t a n t 1 − t a n 2 t ) ( 1 + t a n 5 t − 10 t a n 3 t + 5 t a n t 5 t a 4 t − 10 t a n 2 t + 1 ) (1+\frac{2tant}{1-tan^2t})(1+\frac{tan^5t-10tan^3t+5tant}{5ta^4t-10tan^2t+1}) ⇔ ( 1 + 2 t a n t ) ( 1 + 5 t a n t ) = 2 \Leftrightarrow (1+2tant)(1+5tant)=2 ⇔ t a n 2 t + t a n 5 t = 1 − t a n 2 t × t a n 5 t \Leftrightarrow tan2t+tan5t=1-tan2t\times tan5t ⇔ t a n ( 2 t + 5 t ) = 1 \Leftrightarrow tan(2t+5t)=1 ⇔ t a n 7 t = 1 \Leftrightarrow tan7t=1 ⇔ t = π 28 + k π 7 \Leftrightarrow t=\frac{\pi}{28}+\frac{k\pi}{7} So we have six solution: x = t a n − 11 π 28 ; t a n − 3 π 28 ; t a n π 28 ; t a n 5 π 28 ; t a n 9 π 28 ; t a n 13 π 28 x=tan\frac{-11\pi}{28};tan\frac{-3\pi}{28};tan\frac{\pi}{28};tan\frac{5\pi}{28};tan\frac{9\pi}{28};tan\frac{13\pi}{28}

And b + c + d + e + f + g + h + k + m + n + z + t = 182 \LARGE b+c+d+e+f+g+h+k+m+n+z+t=182

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