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Difference between revisions of "2019 MPFG Problems/Problem 15"

(Created page with "==Problem== How many ordered pairs <math>(x,y)</math> of real numbers <math>x</math> and <math>y</math> are there such that <math>-100 \pi\leq x \leq 100\pi</math>, <math>-100...")
 
 
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==Problem==
 
==Problem==
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How many ordered pairs <math>(x,y)</math> of real numbers <math>x</math> and <math>y</math> are there such that <math>-100 \pi\leq x \leq 100\pi</math>, <math>-100\pi \leq y \leq 100\pi</math>, <math>x + y = 20.19</math>, and <math>tanx + tany = 20.19</math>?
+
How many ordered pairs <math>(x, y)</math> of real numbers <math>x</math> and <math>y</math> are there such that <math>-100 \pi \leq x \leq 100 \pi</math>, <math>-100 \pi \leq y \leq 100 \pi</math>, <math>x + y = 20.19</math>, and <math>\tan x + \tan y = 20.19</math>?
  
 
==Solution 1==
 
==Solution 1==
−
According to the <math>tan</math> angle sum trigonometric identity, <math>tan(x+y) = \frac{tanx+tany}{1+tanx\cdot tany}</math>
+
According to the <math>\tan</math> angle sum trigonometric identity,  
  
−
<math>tan 20.19 = \frac{20.19}{1 + tan x\cdot tan y}</math>
+
<cmath>
 +
\tan(x + y) = \frac{\tan x + \tan y}{1 + \tan x \cdot \tan y}
 +
</cmath>
  
−
<math>tanx\cdot tany = \frac{20.19}{tan20.19} - 1</math>
+
<cmath>
 +
\tan 20.19 = \frac{20.19}{1 + \tan x \cdot \tan y}
 +
</cmath>
  
−
The two equations <math>tanx\cdot tany = \frac{20.19}{tan20.19} - 1</math> and <math>tanx + tany = 20.19</math> create a set of Vieta's Formula for <math>x^{2}-20.19+(\frac{20.19}{tan20.19}-1) = 0</math>, whose <math>\delta</math> is obviously greater than <math>0</math>. This indicates that there must be a constant value for the set <math>(tanx,tany)</math>.
+
<cmath>
 +
\tan x \cdot \tan y = \frac{20.19}{\tan 20.19} - 1
 +
</cmath>
  
−
Assume that <math>tanx > tany</math>. <math>tanx</math> is represented by the upper line, <math>tany</math> is represented by the lower line.
+
The two equations <math>\tan x \cdot \tan y = \frac{20.19}{\tan 20.19} - 1</math> and <math>\tan x + \tan y = 20.19</math> create a set of [[Vieta's Formulas|Vieta's formulas]] for
  
−
[insert picture]
+
<cmath>
 +
x^2 - 20.19x + \left( \frac{20.19}{\tan 20.19} - 1 \right) = 0,
 +
</cmath>
  
−
As we can see, each value of <math>x</math> matches a value of <math>y</math> on the other side of the <math>y-axis</math>. Because <math>x+y=20.19</math>, which is approximately <math>6.42\pi</math>, 6 values of x/y close to <math>-100\pi</math> cannot be taken.
+
whose discriminant <math>\delta</math> is obviously greater than 0. This indicates that there must be a constant value for the set <math>(\tan x, \tan y)</math>.
  
−
There are <math>200-6=194</math> values of (x,y) when <math>tanx>tany</math>. Doubling this number, we get <math>\boxed{388}</math>
+
Assume that <math>\tan x > \tan y</math>. <math>\tan x</math> is represented by the upper line, <math>\tan y</math> is represented by the lower line.
 +
 
 +
{{image needed}}
 +
 
 +
As we can see, each value of <math>x</math> matches a value of <math>y</math> on the other side of the <math>y</math>-axis. Because <math>x + y = 20.19</math>, which is approximately <math>6.42 \pi</math>, 6 values of <math>x/y</math> close to <math>-100 \pi</math> cannot be taken.
 +
 
 +
There are <math>200 - 6 = 194</math> values of <math>(x, y)</math> when <math>\tan x > \tan y</math>. Doubling this number, we get <math>\boxed{388}</math>.

Latest revision as of 11:36, 16 August 2025

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 line, $\tan y$ is represented by the lower line.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\boxed{388}$.

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