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 $tanx + tany = 20.19$ ?

Solution 1

According to the $tan$ angle sum trigonometric identity, $tan(x+y) = \frac{tanx+tany}{1+tanx\cdot tany}$

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

$tanx\cdot tany = \frac{20.19}{tan20.19} - 1$

The two equations $tanx\cdot tany = \frac{20.19}{tan20.19} - 1$ and $tanx + tany = 20.19$ create a set of Vieta's Formula for $x^{2}-20.19+(\frac{20.19}{tan20.19}-1) = 0$ , whose $\delta$ is obviously greater than $0$ . This indicates that there must be a constant value for the set $(tanx,tany)$ .

Assume that $tanx > tany$ . $tanx$ is represented by the upper line, $tany$ is represented by the lower line.

[insert picture]

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 $tanx>tany$ . Doubling this number, we get $\boxed{388}$

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2019 MPFG Problems/Problem 15

Problem

Solution 1