View source for 2022 MPFG Problem19

==Problem==
Let <imath>S_-</imath> be the semicircular arc defined by
<cmath>(x+1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1. </cmath>
Let <imath>S_+</imath> be the semicircular arc defined by
<cmath>(x-1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1. </cmath>

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

==Solution 1==

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

<imath>xX+yY=1 ==> \frac{x}{\frac{1}{X}}+\frac{y}{\frac{1}{Y}} = 1</imath>

<imath>S=\int_{x_0}^{x_1} y(x) \,dx</imath>

Let <imath>m=\frac{1}{x}</imath>.

[[File:2022mpfg19_(1).png|750px|center]]

<imath>\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}</imath>

<imath>\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}</imath>

<imath>\left| y_2 - y_1 \right| = (1-x) -(\frac{1+x}{2}) = (-\frac{3}{2}x+\frac{1}{2})</imath>

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

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

<imath>=\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}}</imath>

~cassphe