AIME 1986 · 第 3 题

Solution 1

Since $\cot$ is the reciprocal function of $\tan$:

$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$

Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$

Using the tangent addition formula:

$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}$.

Solution 2

Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$. We only need to find $\tan x \tan y$.

We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$. Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25$.

Similarly, we have $30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}$.

Dividing:

$\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}$. Plugging in to the earlier formula, we have $\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}$.

Solution 3 (less trig required, use of quadratic formula)

Let $a=\tan x$ and $b=\tan y$. This simplifies the equations to:

$$a + b = 25$$ $$\frac{1}{a} + \frac{1}{b} = 30$$ Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$.

We can use substitution to solve the system of equations from above: $b = -a + 25$, so $\frac{1}{a} + \frac{1}{-a + 25} = 30$.

Multiplying by $-a(a-25)$, we get $a + (-a + 25) = -30a(a-25)$, which is $-30a^2 + 750a = 25$. Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$.

Using the quadratic formula gives $a = \frac{150 \pm \sqrt {22380}}{12}$. Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$, which turns out to also be $b = \frac{150 \pm \sqrt {22380}}{12}$, provided that the sign of the radical in $a$ is opposite the one in $b$.

WLOG, assume $a = \frac{150 + \sqrt{22380}}{12}$ and $b = \frac{150 - \sqrt{22380}}{12}$. Multiplying them gives $ab = \frac{22500 - 22380}{144}$ which simplifies to $\frac{5}{6}$.

THe denominator of $\frac{25}{1 - ab}$ ends up being $\frac{1}{6}$, so multiplying both numerator and denominator by 6 gives $\boxed{150}$.

-ThisUsernameIsTaken

Remark: The quadratic need not be solved. The value of $ab$ can be found through Vieta's.