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>. | ||
| − | <math>xX+yY=1 ==> \frac{x}{\frac{1}{ | + | <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> | ||
| − | <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> | + | <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> | ||
| − | <math> | + | 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> |
| − | <math>S = 2\int_{0}^\frac{1}{3} (-\frac{3}{2}x+\frac{1}{2}) \,dx</math> | + | <math>S = 2\int_{0}^\frac{1}{3} (-\frac{3}{2}x+\frac{1}{2}) \,dx</math> (times 2 because on both sides） |
<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> | <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> | ||
~cassphe | ~cassphe | ||
Latest revision as of 12:26, 3 September 2025
Problem
Let 
be the semicircular arc defined by
![\[(x+1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1.\]](http://latex.artofproblemsolving.com/9/6/5/96523ed2820abd0475f09931bd1abee7b8b2321b.png)
Let 
be the semicircular arc defined by
![\[(x-1)^2 + (y-\frac{3}{2})^2 = \frac{1}{4} and x \leq -1.\]](http://latex.artofproblemsolving.com/d/5/b/d5b8d189866ef54fff1cf929dd75ca897f0ac850.png)
Let 
be the locus of points 
such that is the intersection of two lines, one of the form 
where 
and the other of the form 
where 
. What is the area of ? Express your answer as a fraction in simplest form.
Solution 1
Because 
is a solution set of 
, which means that the 
coordinates are on the line of .
Let 
.
and m(
) ranging from 3 to infinite 
, 
(times 2 because on both sides）
~cassphe
