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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==
−
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]
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<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 blue line, <math>\tan y</math> is represented by the lower red line.
 +
 
 +
[[File:Forgot_line.png|710px|center]]
 +
 
 +
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>.
 +
 
 +
~cassphe
 +
 
 +
~edited by aoum

Latest revision as of 06:54, 17 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 blue line, $\tan y$ is represented by the lower red line.

Forgot line.png

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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