2019 MPFG Problems/Problem 15

Problem

How many ordered pairs $(x, y)$ of real numbers $x$ and $y$ are there such that $-100 \pi \leq x \leq 100 \pi$ , $-100 \pi \leq y \leq 100 \pi$ , $x + y = 20.19$ , and $\tan x + \tan y = 20.19$ ?

Solution 1

According to the $\tan$ angle sum trigonometric identity,

$\tan(x + y) = \frac{\tan x + \tan y}{1 + \tan x \cdot \tan y}$

$\tan 20.19 = \frac{20.19}{1 + \tan x \cdot \tan y}$

$\tan x \cdot \tan y = \frac{20.19}{\tan 20.19} - 1$

The two equations $\tan x \cdot \tan y = \frac{20.19}{\tan 20.19} - 1$ and $\tan x + \tan y = 20.19$ create a set of Vieta's formulas for

$x^2 - 20.19x + \left( \frac{20.19}{\tan 20.19} - 1 \right) = 0,$

whose discriminant $\Delta$ is obviously greater than 0. This indicates that there must be a constant value for the set $(\tan x, \tan y)$ .

Assume that $\tan x > \tan y$ . $\tan x$ is represented by the upper blue line, $\tan y$ is represented by the lower red line.

As we can see, each value of $x$ matches a value of $y$ on the other side of the $y$ -axis. Because $x + y = 20.19$ , which is approximately $6.42 \pi$ , 6 values of $x/y$ close to $-100 \pi$ cannot be taken.

There are $200 - 6 = 194$ values of $(x, y)$ when $\tan x > \tan y$ . Doubling this number, we get $\boxed{388}$ .

~cassphe

~edited by aoum

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2019 MPFG Problems/Problem 15

Problem

Solution 1