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Difference between revisions of "2022 MPFG Problem19"

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Let <math>R</math> be the locus of points <math>P</math> such that <math>P</math> is the intersection of two lines, one of the form <math>Ax + By = 1</math> where <math>(A,B) \in S_-</math> and the other of the form <math>Cx + Dy = 1</math> where <math>(C, D) \in S_+</math>. What is the area of <math>R</math>? Express your answer as a fraction in simplest form.
 
Let <math>R</math> be the locus of points <math>P</math> such that <math>P</math> is the intersection of two lines, one of the form <math>Ax + By = 1</math> where <math>(A,B) \in S_-</math> and the other of the form <math>Cx + Dy = 1</math> where <math>(C, D) \in S_+</math>. What is the area of <math>R</math>? Express your answer as a fraction in simplest form.
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==Solution 1==
  
 
Because <math>Ax+By=1,Cx+Dy=1 ==> (A,B),(C,D)</math> is a solution set of <math>xX+yY=1</math>, which means that the <math>2</math> coordinates are on the line of <math>xX+yY=1</math>.
 
Because <math>Ax+By=1,Cx+Dy=1 ==> (A,B),(C,D)</math> is a solution set of <math>xX+yY=1</math>, which means that the <math>2</math> coordinates are on the line of <math>xX+yY=1</math>.
  
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<math>xX+yY=1 ==> \frac{x}{\frac{1}{x}}+\frac{y}{\frac{1}{y}} = 1</math>
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<math>xX+yY=1 ==> \frac{x}{\frac{1}{X}}+\frac{y}{\frac{1}{Y}} = 1</math>
  
 
<math>S=\int_{x_0}^{x_1} y(x) \,dx</math>
 
<math>S=\int_{x_0}^{x_1} y(x) \,dx</math>
  
 
Let <math>m=\frac{1}{x}</math>.
 
Let <math>m=\frac{1}{x}</math>.
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 +
[[File:2022mpfg19.png|750px|center]]
  
 
<math>\frac{1}{a} = \frac{m-1}{m} ==> a=\frac{m}{m-1} = \frac{\frac{1}{x}}{\frac{1}{x}-1} = \frac{1}{1-x} = \frac{1}{y_1}</math>
 
<math>\frac{1}{a} = \frac{m-1}{m} ==> a=\frac{m}{m-1} = \frac{\frac{1}{x}}{\frac{1}{x}-1} = \frac{1}{1-x} = \frac{1}{y_1}</math>
  
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<math>\frac{b}{2} = \frac{m}{m+1} ==> b=\frac{2m}{m+1} = \frac{\frac{2}{x}}{\frac{1}{x}+1} = \frac{2}{1+x} = \frac{1}{y_1}</math>
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<math>\frac{b}{2} = \frac{m}{m+1} ==> b=\frac{2m}{m+1} = \frac{\frac{2}{x}}{\frac{1}{x}+1} = \frac{2}{1+x} = \frac{1}{y_2}</math>
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<math>\left| y_2 - y_1 \right| = (1-x) -(\frac{1+x}{2}) = (-\frac{3}{2}x+\frac{1}{2})</math>
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and m(<math>\frac{1}{x}</math>) ranging from 3 to infinite <math>==></math>  <math>x_0=0</math> , <math>x_1=\frac{1}{3}</math>
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<math>S = 2\int_{0}^\frac{1}{3} (-\frac{3}{2}x+\frac{1}{2}) \,dx</math>        (times 2 because on both sides）
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<math>=\left. 2(\frac{3}{4}x^2+\frac{1}{2}x)\right|_{0}^{\frac{1}{3}} = 2(-\frac{1}{12} + \frac{1}{6}) = \boxed{\frac{1}{6}}</math>
  
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<math>y_2=y_1 ==> 1-x=\frac{1+x}{2}</math> <math>x_1=\frac{1}{3}</math>
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~cassphe

Latest revision as of 12:26, 3 September 2025

Problem

Let $S_-$ be the semicircular arc defined by \[(x+1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1.\] Let $S_+$ be the semicircular arc defined by \[(x-1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1.\]

Let $R$ be the locus of points $P$ such that $P$ is the intersection of two lines, one of the form $Ax + By = 1$ where $(A,B) \in S_-$ and the other of the form $Cx + Dy = 1$ where $(C, D) \in S_+$. What is the area of $R$? Express your answer as a fraction in simplest form.

Solution 1

Because $Ax+By=1,Cx+Dy=1 ==> (A,B),(C,D)$ is a solution set of $xX+yY=1$, which means that the $2$ coordinates are on the line of $xX+yY=1$.

$xX+yY=1 ==> \frac{x}{\frac{1}{X}}+\frac{y}{\frac{1}{Y}} = 1$

$S=\int_{x_0}^{x_1} y(x) \,dx$

Let $m=\frac{1}{x}$.

2022mpfg19.png

$\frac{1}{a} = \frac{m-1}{m} ==> a=\frac{m}{m-1} = \frac{\frac{1}{x}}{\frac{1}{x}-1} = \frac{1}{1-x} = \frac{1}{y_1}$

$\frac{b}{2} = \frac{m}{m+1} ==> b=\frac{2m}{m+1} = \frac{\frac{2}{x}}{\frac{1}{x}+1} = \frac{2}{1+x} = \frac{1}{y_2}$

$\left| y_2 - y_1 \right| = (1-x) -(\frac{1+x}{2}) = (-\frac{3}{2}x+\frac{1}{2})$

and m($\frac{1}{x}$) ranging from 3 to infinite $==>$ $x_0=0$ , $x_1=\frac{1}{3}$

$S = 2\int_{0}^\frac{1}{3} (-\frac{3}{2}x+\frac{1}{2}) \,dx$ (times 2 because on both sides）

$=\left. 2(\frac{3}{4}x^2+\frac{1}{2}x)\right|_{0}^{\frac{1}{3}} = 2(-\frac{1}{12} + \frac{1}{6}) = \boxed{\frac{1}{6}}$

~cassphe

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