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2008 iTest Problems/Problem 57

Revision as of 18:16, 13 July 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 57 -- BIG NUMBERS)
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Problem

Let a and b be the two possible values of $\tan\theta$ given that $\sin\theta + \cos\theta = \dfrac{193}{137}$. If $a+b=m/n$, where $m$ and $n$ are relatively prime positive integers, compute $m+n$.

Solution

Let $x = \tfrac{193}{137}$. Multiply both sides by $\tfrac{\sqrt{2}}{2}$ to get \[\frac{\sqrt{2}}{2}\sin\theta + \frac{\sqrt{2}}{2}\cos\theta = \frac{\sqrt{2}}{2}x\] \[\sin(\theta + 45^\circ) = \frac{\sqrt{2}}{2}x\] \[\theta = \sin^{-1} (\frac{\sqrt{2}}{2}x) - 45^\circ \text{ or } 135 - \sin^{-1} (\frac{\sqrt{2}}{2}x)\] Let $y = \tan(\sin^{-1} (\tfrac{\sqrt{2}}{2}x))$. Using the sum formula for tangent, the sum of the two possible values of $\tan \theta$ is \[\left(\frac{y - 1}{1 + y} \right) + \left(\frac{-1 - y}{1 - y} \right)\] \[\left(\frac{y - 1}{1 + y} \right) + \left(\frac{y+1}{y-1} \right)\] \[\left(\frac{y^2 - 2y + 1}{y^2 - 1} \right) + \left(\frac{y^2 + 2y + 1}{y^2 - 1} \right)\] \[\frac{2y^2 + 2}{y^2 - 1}\] If a triangle has opposite side $x\sqrt{2}$ and hypotenuse $2$, by the Pythagorean Theorem, the adjacent side's length is $\sqrt{4-2x^2}$. That means if $\sin(\alpha) = \tfrac{\sqrt{2}}{2}x$, then $\tan(\alpha) = \sqrt{\tfrac{2x^2}{4-2x^2}}$. That means $y = \tan(\sin^{-1} (\tfrac{\sqrt{2}}{2}x)) = \sqrt{\tfrac{2x^2}{4-2x^2}}$. \[\frac{2 \cdot \frac{2x^2}{4-2x^2} + 2}{\frac{2x^2}{4-2x^2} - 1}\] \[\frac{ \frac{4x^2}{4-2x^2} + \frac{8-4x^2}{4-2x^2} }{\frac{2x^2}{4-2x^2} - \frac{4 - 2x^2}{4-2x^2} }\] \[\frac{8}{4x^2-4}\] \[\frac{2}{x^2 - 1}\] \[\frac{2}{(x+1)(x-1)}\] \[\frac{2}{\frac{330}{137} \cdot \frac{56}{137}}\] \[\frac{137 \cdot 137}{330 \cdot 28}\] \[\frac{18769}{9240}\] Thus, $m+n = \boxed{28009}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 56 		Followed by:
Problem 58
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