https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ssy899&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T11:30:05ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_9&diff=770752008 AIME II Problems/Problem 92016-02-27T06:09:20Z<p>Ssy899: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A particle is located on the coordinate plane at <math>(5,0)</math>. Define a ''move'' for the particle as a counterclockwise rotation of <math>\pi/4</math> radians about the origin followed by a translation of <math>10</math> units in the positive <math>x</math>-direction. Given that the particle's position after <math>150</math> moves is <math>(p,q)</math>, find the greatest integer less than or equal to <math>|p| + |q|</math>.<br />
<br />
__TOC__<br />
== Solution == <br />
=== Solution 1 ===<br />
<br />
Let P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and <math>\theta</math> be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then <math>x'=rcos(\pi/4+\theta)+10 = \sqrt{2}(x - y)/2 + 10</math> and <math>y' = rsin(\pi/4+\theta) = \sqrt{2}(x + y)/2</math>.<br />
Let <math>(x_n, y_n)</math> be the position of the particle after the nth move, where <math>x_0 = 5</math> and <math>y_0 = 0</math>. Then <math>x_{n+1} + y_{n+1} = \sqrt{2}x_n+10</math>, <math>x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10</math>. This implies<br />
<math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>.<br />
Substituting <math>x_0 = 5</math> and <math>y_0 = 0</math>, we have <math>x_8 = 5</math> and <math>y_8 = 0</math> again for the first time. Thus, <math>p = x_{150} = x_6 = -5\sqrt{2}</math> and <math>q = y_{150} = y_6 = 5 + 5\sqrt{2}</math>. Hence, the final answer is<br />
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center><br />
<br />
=== Solution 2 ===<br />
Let the particle's position be represented by a complex number. Recall that multiplying a number by where a is cis<math>\left( \theta \right)</math>. rotates the object in the complex plane by <math>\theta</math> counterclockwise. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to<br />
<center><math>a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10</math></center><br />
where a is cis<math>\left( \theta \right)</math>. By De-Moivre's theorem, <math>\left(cis( \theta \right)^n )</math>=cis<math>\left(n \theta \right)</math>.<br />
Therefore,<br />
<center><math>10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})</math></center><br />
Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is<br />
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_5&diff=708682002 AIME II Problems/Problem 52015-06-25T18:27:31Z<p>Ssy899: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Find the sum of all positive integers <math>a=2^n3^m</math> where <math>n</math> and <math>m</math> are non-negative integers, for which <math>a^6</math> is not a divisor of <math>6^a</math><br />
<br />
== Solution ==<br />
Substitute <math>a=2^n3^m</math> into <math>a^6</math> and <math>6^a</math>, and find all pairs of non-negative integers (n,m) for which <math>(2^n3^m)^{6}</math> is not a divisor of <math>6^{2^n3^m}</math><br />
<br />
Simplifying both expressions:<br />
<br />
<math>2^{6n} \cdot 3^{6m}</math> is not a divisor of <math>2^{2^n3^m} \cdot 3^{2^n3^m}</math><br />
<br />
Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression):<br />
<br />
<math>6n > 2^n3^m</math> OR <math>6m > 2^n3^m</math><br />
<br />
<br />
Using the first inequality <math>6n > 2^n3^m</math> and going case by case starting with n <math>\in</math> {0, 1, 2, 3...}:<br />
<br />
n=0:<br />
<math>0>1 \cdot 3^m</math> which has no solution for non-negative integers m <br />
<br />
n=1:<br />
<math>6 > 2 \cdot 3^m</math> which is true for m=0 but fails for higher integers <math>\Rightarrow (1,0)</math><br />
<br />
n=2:<br />
<math>12 > 4 \cdot 3^m</math> which is true for m=0 but fails for higher integers <math>\Rightarrow (2,0)</math><br />
<br />
n=3:<br />
<math>18 > 8 \cdot 3^m</math> which is true for m=0 but fails for higher integers <math>\Rightarrow (3,0)</math><br />
<br />
n=4:<br />
<math>24 > 16 \cdot 3^m</math> which is true for m=0 but fails for higher integers <math>\Rightarrow (4,0)</math><br />
<br />
n=5:<br />
<math>30 > 32 \cdot 3^m</math> which has no solution for non-negative integers m <br />
<br />
There are no more solutions for higher <math>n</math>, as polynomials like <math>6n</math> grow slower than exponentials like <math>2^n</math>.<br />
<br />
<br />
<br />
Using the second inequality <math>6m > 2^n3^m</math> and going case by case starting with m <math>\in</math> {0, 1, 2, 3...}:<br />
<br />
m=0:<br />
<math>0>2^n \cdot 1</math> which has no solution for non-negative integers n <br />
<br />
m=1:<br />
<math>6>2^n \cdot 3</math> which is true for n=0 but fails for higher integers <math>\Rightarrow (0,1)</math><br />
<br />
m=2:<br />
<math>12>2^n \cdot 9</math> which is true for n=0 but fails for higher integers <math>\Rightarrow (0,2)</math><br />
<br />
m=3:<br />
<math>18>2^n \cdot 27</math> which has no solution for non-negative integers n <br />
<br />
There are no more solutions for higher <math>m</math>, as polynomials like <math>6m</math> grow slower than exponentials like <math>3^m</math>.<br />
<br />
<br />
Thus there ae six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2). <br />
Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is <math>\framebox{042}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=<br />
2002|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_2&diff=708672002 AIME II Problems/Problem 22015-06-25T17:36:26Z<p>Ssy899: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Three [[vertex|vertices]] of a [[cube]] are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?<br />
<br />
== Solution ==<br />
<math>PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}</math><br />
<br />
<math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math><br />
<br />
<math>QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}</math><br />
<br />
So, <math>PQR</math> is an equilateral triangle. Let the side of the cube be <math>a</math>.<br />
<br />
<math>a\sqrt{2}=\sqrt{98}</math><br />
<br />
So, <math>a=7</math>, and hence the surface area is <math>6a^2=\framebox{294}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2002|n=II|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_2&diff=708662002 AIME II Problems/Problem 22015-06-25T17:35:31Z<p>Ssy899: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Three [[vertex|vertices]] of a [[cube]] are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?<br />
<br />
== Solution ==<br />
<math>PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}</math><br />
<br />
<math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math><br />
<br />
<math>QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}</math><br />
<br />
So, <math>PQR</math> is an equilateral triangle. Let the side of the cube be <math>a</math>.<br />
<br />
<math>a\sqrt{2}=\sqrt{98}</math><br />
<br />
So, <math>a=7</math>, and hence the surface area is <math>6a^2=\Box{294}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2002|n=II|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_2&diff=708652002 AIME II Problems/Problem 22015-06-25T17:35:10Z<p>Ssy899: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Three [[vertex|vertices]] of a [[cube]] are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?<br />
<br />
== Solution ==<br />
<math>PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}</math><br />
<br />
<math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math><br />
<br />
<math>QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}</math><br />
<br />
So, <math>PQR</math> is an equilateral triangle. Let the side of the cube be <math>a</math>.<br />
<br />
<math>a\sqrt{2}=\sqrt{98}</math><br />
<br />
So, <math>a=7</math>, and hence the surface area is <math>6a^2=\box{294}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2002|n=II|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_2&diff=681332008 AIME II Problems/Problem 22015-03-03T07:13:39Z<p>Ssy899: </p>
<hr />
<div>== Problem ==<br />
Rudolph bikes at a [[constant]] rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the <math>50</math>-mile mark at exactly the same time. How many minutes has it taken them?<br />
<br />
== Solution ==<br />
Let Rudolf bike at a rate <math>r</math>, so Jennifer bikes at the rate <math>\dfrac 34r</math>. Let the time both take be <math>t</math>. <br />
<br />
Then Rudolf stops <math>49</math> times (because the rest after he reaches the finish does not count), losing a total of <math>49 \cdot 5 = 245</math> minutes, while Jennifer stops <math>24</math> times, losing a total of <math>24 \cdot 5 = 120</math> minutes. The time Rudolf and Jennifer actually take biking is then <math>t - 245,\, t-120</math> respectively. <br />
<br />
Using the formula <math>r = \frac dt</math>, since both Jennifer and Rudolf bike <math>50</math> miles,<br />
<center><cmath>\begin{align}r &= \frac{50}{t-245}\\<br />
\frac{3}{4}r &= \frac{50}{t-120}<br />
\end{align}</cmath></center><br />
Substituting equation <math>(1)</math> into equation <math>(2)</math> and simplifying, we find <br />
<center><cmath>\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\<br />
\frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\<br />
t &= \boxed{620}\ \text{minutes}<br />
\end{align*}</cmath></center><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_1&diff=681322008 AIME II Problems/Problem 12015-03-03T07:06:13Z<p>Ssy899: </p>
<hr />
<div>== Problem ==<br />
Let <math>N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2</math>, where the additions and subtractions alternate in pairs. Find the remainder when <math>N</math> is divided by <math>1000</math>.<br />
<br />
== Solution ==<br />
Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms,<br />
<center><cmath>\begin{align*}<br />
N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\<br />
&= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ <br />
&= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100}<br />
\end{align*}</cmath></center><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=User:Ssy899&diff=63281User:Ssy8992014-09-04T06:06:32Z<p>Ssy899: Replaced content with "
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<div>[[USAJMO]] problems and solutions by year:<br />
*[[2010 USAJMO]]<br />
*[[2011 USAJMO]]<br />
*[[2012 USAJMO]]<br />
*[[2013 USAJMO]]</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=User:Ssy899&diff=48973User:Ssy8992012-10-31T01:10:43Z<p>Ssy899: </p>
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<div><math><math>Insert formula here</math><math><math>Insert formula here</math><math><math>Insert formula here</math><math><math>Insert formula here</math><math><math>Insert formula here</math><math><math>Insert formula here</math><math><math>Insert formula here</math><math>Insert formula here</math></math></math></math></math></math></math></math><math>Insert formula here</math></div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=User:Ssy899&diff=48918User:Ssy8992012-10-25T04:42:50Z<p>Ssy899: Created page with "<math>Insert formula here</math>"</p>
<hr />
<div><math>Insert formula here</math></div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=488922010 AMC 8 Problems/Problem 232012-10-21T23:51:55Z<p>Ssy899: </p>
<hr />
<div>Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<b>Soution</b><br />
According to the pythagorean theorem, The radius of the larger circle is:<br />
<br />
<math>1^2 + 1^2 = \sqrt{2}</math><br />
<br />
Therefore the area of the larger circle is:<br />
<br />
<math>(\sqrt{2})^2\pi = 2\pi </math><br />
<br />
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is:<br />
<br />
<math>1^2\pi=\pi</math><br />
<br />
Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\frac{1}{2}</math>.</div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=488912010 AMC 8 Problems/Problem 232012-10-21T23:51:39Z<p>Ssy899: </p>
<hr />
<div>Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<b>Soution</b><br />
According to the pythagorean theorem, The radius of the larger circle is:<br />
<br />
<math>1^2 + 1^2 = \sqrt{2}</math><br />
<br />
Therefore the area of the larger circle is:<br />
<br />
<math>(\sqrt{2})^2\pi = 2\pi </math><br />
<br />
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is:<br />
<br />
<math>1^2\pi=\pi</math><br />
<br />
Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\frac{1}{2}</math></div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=488902010 AMC 8 Problems/Problem 232012-10-21T23:51:14Z<p>Ssy899: </p>
<hr />
<div>Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<b>Soution</b><br />
According to the pythagorean theorem, The radius of the larger circle is:<br />
<br />
<math>1^2 + 1^2 = \sqrt{2}</math><br />
<br />
Therefore the area of the larger circle is:<br />
<br />
<math>(\sqrt{2})^2\pi = 2\pi </math><br />
<br />
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is:<br />
<br />
<math>1^2\pi=\pi</math><br />
<br />
Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>1\2</math></div>Ssy899https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=488892010 AMC 8 Problems/Problem 232012-10-21T23:45:23Z<p>Ssy899: </p>
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<div>Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<b>Soution</b></div>Ssy899