https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Chenr28&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:28:24ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Radius&diff=101814Radius2019-02-11T12:59:16Z<p>Chenr28: /* Links to other key terms */</p>
<hr />
<div><br />
The '''radius''' of a [[circle]] is the distance from the center to any point on the circle. Identical definitions hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii.<br />
<br />
<br />
The radius of a circle is often denoted using R or r.<br />
<br />
<br />
<asy>size(150);<br />
pair O=origin, P=dir(30);<br />
D(unitcircle);<br />
D(O--P);<br />
MP("O",D(O),NNW);<br />
MP("P",D(P),NE);<br />
MP("r",midpoint(O--P),NNW);</asy><br />
<br />
<br />
== Links to other key terms ==<br />
<br />
<br />
1. diameter: <math> d = 2 r </math><br />
<br />
The diameter is the longest chord is a circle. It is usually expressed by D or d.<br />
<br />
<br />
2. circumference (perimeter): <math> C = 2 \pi r </math><br />
<br />
The circumference is the distance around a circle. It is usually represented by C.<br />
<br />
<br />
3. area: <math> S = \pi r^2 </math><br />
<br />
The area of a circle is the amount of space it occupies. It is usually denoted using S or A.<br />
<br />
<br />
<br />
<br />
By grace.yongqing.yu<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_20&diff=1012722011 AMC 12B Problems/Problem 202019-02-05T14:38:28Z<p>Chenr28: /* Solution 4 (the fastest, easiest, and therefore best way ever) */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> has <math>AB = 13, BC = 14</math>, and <math>AC = 15</math>. The points <math>D, E</math>, and <math>F</math> are the midpoints of <math>\overline{AB}, \overline{BC}</math>, and <math>\overline{AC}</math> respectively. Let <math>X \neq E</math> be the intersection of the circumcircles of <math>\triangle BDE</math> and <math>\triangle CEF</math>. What is <math>XA + XB + XC</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}</math><br />
<br />
==Solutions==<br />
===Solution 1 (Coordinates)===<br />
Let us also consider the circumcircle of <math>\triangle ADF</math>.<br />
<br />
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intersect at <math>P</math>, so <math>P</math> is <math>X</math>.<br />
<br />
The question now becomes calculate the sum of distance from each vertices to the circumcenter. <br />
<br />
We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.)<br />
<br />
Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math><br />
<br />
Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> and passes through <math>(2.5, 6)</math>. <br />
<br />
<math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math><br />
<br />
So <math>X = (7, \frac{33}{8})</math><br />
<br />
and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}</math><br />
<br />
===Solution 2 (Algebra)===<br />
Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. By inspection, we see that <math>XA</math>, <math>XB</math>, and <math>XC</math> are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of <math>3</math>. We can find the circumradius quite easily with the formula <math>\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}</math>, s.t. <math>s=\frac{a+b+c}{2}</math> and R is the circumradius. Since <math>s = \frac{21}{2}</math>:<br />
<br />
<cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath><br />
<br />
After a few algebraic manipulations:<br />
<br />
<math>\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}</math>.<br />
<br />
===Solution 3 (Homothety)===<br />
Let <math>O</math> be the circumcenter of <math>\triangle ABC,</math> and <math>h_A</math> denote the length of the altitude from <math>A.</math> Note that a homothety centered at <math>B</math> with ratio <math>\frac{1}{2}</math> takes the circumcircle of <math>\triangle BAC</math> to the circumcircle of <math>\triangle BDE</math>. It also takes the point diametrically opposite <math>B</math> on the circumcircle of <math>\triangle BAC</math> to <math>O.</math> Therefore, <math>O</math> lies on the circumcircle of <math>\triangle BDE.</math> Similarly, it lies on the circumcircle of <math>\triangle CEF.</math> By Pythagorean triples, <math>h_A=12.</math> Finally, our answer is <cmath>3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}.}</cmath><br />
<br />
<br />
===Solution 4 (the fastest, easiest, and therefore best way ever)===<br />
<br />
After doing a lot of problems with 13-14-15 triangles, you might remember that the circumradius of the triangle is <math>\frac {65} {8}</math>. Because the problem deals with congruent triangles and circumcircles, you can guess that X is the circumcircle. Hence our answer is <cmath>\boxed{\frac{195}{8}.}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_20&diff=1012712011 AMC 12B Problems/Problem 202019-02-05T14:38:02Z<p>Chenr28: /* Solution 4(Really Really Sketch) */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> has <math>AB = 13, BC = 14</math>, and <math>AC = 15</math>. The points <math>D, E</math>, and <math>F</math> are the midpoints of <math>\overline{AB}, \overline{BC}</math>, and <math>\overline{AC}</math> respectively. Let <math>X \neq E</math> be the intersection of the circumcircles of <math>\triangle BDE</math> and <math>\triangle CEF</math>. What is <math>XA + XB + XC</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}</math><br />
<br />
==Solutions==<br />
===Solution 1 (Coordinates)===<br />
Let us also consider the circumcircle of <math>\triangle ADF</math>.<br />
<br />
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intersect at <math>P</math>, so <math>P</math> is <math>X</math>.<br />
<br />
The question now becomes calculate the sum of distance from each vertices to the circumcenter. <br />
<br />
We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.)<br />
<br />
Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math><br />
<br />
Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> and passes through <math>(2.5, 6)</math>. <br />
<br />
<math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math><br />
<br />
So <math>X = (7, \frac{33}{8})</math><br />
<br />
and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}</math><br />
<br />
===Solution 2 (Algebra)===<br />
Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. By inspection, we see that <math>XA</math>, <math>XB</math>, and <math>XC</math> are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of <math>3</math>. We can find the circumradius quite easily with the formula <math>\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}</math>, s.t. <math>s=\frac{a+b+c}{2}</math> and R is the circumradius. Since <math>s = \frac{21}{2}</math>:<br />
<br />
<cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath><br />
<br />
After a few algebraic manipulations:<br />
<br />
<math>\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}</math>.<br />
<br />
===Solution 3 (Homothety)===<br />
Let <math>O</math> be the circumcenter of <math>\triangle ABC,</math> and <math>h_A</math> denote the length of the altitude from <math>A.</math> Note that a homothety centered at <math>B</math> with ratio <math>\frac{1}{2}</math> takes the circumcircle of <math>\triangle BAC</math> to the circumcircle of <math>\triangle BDE</math>. It also takes the point diametrically opposite <math>B</math> on the circumcircle of <math>\triangle BAC</math> to <math>O.</math> Therefore, <math>O</math> lies on the circumcircle of <math>\triangle BDE.</math> Similarly, it lies on the circumcircle of <math>\triangle CEF.</math> By Pythagorean triples, <math>h_A=12.</math> Finally, our answer is <cmath>3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}.}</cmath><br />
<br />
<br />
===Solution 4(the fastest, easiest, and therefore best way ever)===<br />
<br />
After doing a lot of problems with 13-14-15 triangles, you might remember that the circumradius of the triangle is <math>\frac {65} {8}</math>. Because the problem deals with congruent triangles and circumcircles, you can guess that X is the circumcircle. Hence our answer is <cmath>\boxed{\frac{195}{8}.}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=Radius&diff=101270Radius2019-02-05T14:12:16Z<p>Chenr28: </p>
<hr />
<div><br />
The '''radius''' of a [[circle]] is the distance from the center to any point on the circle. Identical definitions hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii.<br />
<br />
<br />
The radius of a circle is often denoted using R or r.<br />
<br />
<br />
<asy>size(150);<br />
pair O=origin, P=dir(30);<br />
D(unitcircle);<br />
D(O--P);<br />
MP("O",D(O),NNW);<br />
MP("P",D(P),NE);<br />
MP("r",midpoint(O--P),NNW);</asy><br />
<br />
<br />
== Links to other key terms ==<br />
<br />
<br />
1. diameter: <math> d = 2 r </math><br />
<br />
The diameter is the longest chord is a circle. It is usually expressed by D or d.<br />
<br />
<br />
2. circumference (perimeter): <math> C = 2 \pi r </math><br />
<br />
The circumference is the distance around a circle. It is usually represented by C.<br />
<br />
<br />
3. area: <math> S = \pi r^2 </math><br />
<br />
The area of a circle is the amount of space it occupies. It is usually denoted using S or A.<br />
<br />
<br />
<br />
<br />
Expanded by grace.yongqing.yu<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_21&diff=1012692011 AMC 12B Problems/Problem 212019-02-05T13:56:46Z<p>Chenr28: /* Short Cut */</p>
<hr />
<div>==Problem==<br />
<br />
The arithmetic mean of two distinct positive integers <math>x</math> and <math>y</math> is a two-digit integer. The geometric mean of <math>x</math> and <math>y</math> is obtained by reversing the digits of the arithmetic mean. What is <math>|x - y|</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70</math><br />
<br />
==Solution==<br />
Answer: (D)<br />
<br />
<math>\frac{x + y}{2} = 10 a+b</math> for some <math>1\le a\le 9 </math>,<math>0\le b\le 9</math>.<br />
<br />
<math>\sqrt{xy} = 10 b+a</math><br />
<br />
<math>100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}</math><br />
<br />
<math>xy = 100b^2 + 20ab + a^2</math><br />
<br />
<math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)</math><br />
<br />
<br />
<br /><br />
<math>|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}</math><br />
<br />
Note that in order for x-y to be integer, <math>(a^2 - b^2)</math> has to be <math>11n</math> for some perfect square <math>n</math>. Since <math>a</math> is at most <math>9</math>, <math>n = 1</math> or <math>4</math><br />
<br />
If <math>n = 1</math>, <math>|x-y| = 66</math>, if <math>n = 4</math>, <math>|x-y| = 132</math>. In AMC, we are done. Otherwise, we need to show that <math>a^2 -b^2 = 44</math> is impossible.<br />
<br />
<math>(a-b)(a+b) = 44</math> -> <math>a-b = 1</math>, or <math>2</math> or <math>4</math> and <math>a+b = 44</math>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no integer solution for <math>a</math>, <math>b</math>.<br />
<br />
==Short Cut==<br />
<br />
We can arrive at <math>|x-y| = 6\sqrt{11(a^2 - b^2)}</math> using the method above. Because we know that <math>|x-y|</math> is an integer, it must be a multiple of 6 and 11. Hence the answer is <math>66.</math><br />
<br />
In addition:<br />
Note that <math>11n</math> with <math>n = 1</math> may be obtained with <math>a = 6</math> and <math>b = 5</math> as <math>a^2 - b^2 = 36 - 25 = 11</math>.<br />
<br />
==Sidenote==<br />
It is easy to see that <math>(a,b)=(6,5)</math> is the only solution. This yields <math>(x,y)=(98,32)</math>. Their arithmetic mean is <math>65</math> and their geometric mean is <math>56</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=20|num-a=22|ab=B}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_21&diff=1012682011 AMC 12B Problems/Problem 212019-02-05T13:54:56Z<p>Chenr28: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The arithmetic mean of two distinct positive integers <math>x</math> and <math>y</math> is a two-digit integer. The geometric mean of <math>x</math> and <math>y</math> is obtained by reversing the digits of the arithmetic mean. What is <math>|x - y|</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70</math><br />
<br />
==Solution==<br />
Answer: (D)<br />
<br />
<math>\frac{x + y}{2} = 10 a+b</math> for some <math>1\le a\le 9 </math>,<math>0\le b\le 9</math>.<br />
<br />
<math>\sqrt{xy} = 10 b+a</math><br />
<br />
<math>100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}</math><br />
<br />
<math>xy = 100b^2 + 20ab + a^2</math><br />
<br />
<math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)</math><br />
<br />
<br />
<br /><br />
<math>|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}</math><br />
<br />
Note that in order for x-y to be integer, <math>(a^2 - b^2)</math> has to be <math>11n</math> for some perfect square <math>n</math>. Since <math>a</math> is at most <math>9</math>, <math>n = 1</math> or <math>4</math><br />
<br />
If <math>n = 1</math>, <math>|x-y| = 66</math>, if <math>n = 4</math>, <math>|x-y| = 132</math>. In AMC, we are done. Otherwise, we need to show that <math>a^2 -b^2 = 44</math> is impossible.<br />
<br />
<math>(a-b)(a+b) = 44</math> -> <math>a-b = 1</math>, or <math>2</math> or <math>4</math> and <math>a+b = 44</math>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no integer solution for <math>a</math>, <math>b</math>.<br />
<br />
==Short Cut==<br />
<br />
We can arrive at <math>|x-y| = 2\sqrt{99(a^2 - b^2)}</math> using the method above. Because we know that <math>|x-y|</math> is an integer, it must be a multiple of 33 because the right side is a multiple of 33. Hence the answer is <math>66.</math><br />
<br />
In addition:<br />
Note that <math>11n</math> with <math>n = 1</math> may be obtained with <math>a = 6</math> and <math>b = 5</math> as <math>a^2 - b^2 = 36 - 25 = 11</math>.<br />
<br />
==Sidenote==<br />
It is easy to see that <math>(a,b)=(6,5)</math> is the only solution. This yields <math>(x,y)=(98,32)</math>. Their arithmetic mean is <math>65</math> and their geometric mean is <math>56</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=20|num-a=22|ab=B}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=Radius&diff=101255Radius2019-02-05T02:37:00Z<p>Chenr28: </p>
<hr />
<div>{{stub}}<br />
<br />
The '''radius''' of a [[circle]] is the distance from the center to any point on the circle. Identical definitions hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii.<br />
<br />
The radius of a circle is often denoted using R or r.<br />
<br />
<asy>size(150);<br />
pair O=origin, P=dir(30);<br />
D(unitcircle);<br />
D(O--P);<br />
MP("O",D(O),NNW);<br />
MP("P",D(P),NE);<br />
MP("r",midpoint(O--P),NNW);</asy><br />
<br />
== See Also ==<br />
* [[Diameter]]<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=Radius&diff=101254Radius2019-02-05T02:08:29Z<p>Chenr28: </p>
<hr />
<div>{{stub}}<br />
<br />
The '''radius''' of a [[circle]] is the distance from the center to any point on the circle. Identical definitions hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii.<br />
<br />
<asy>size(150);<br />
pair O=origin, P=dir(30);<br />
D(unitcircle);<br />
D(O--P);<br />
MP("O",D(O),NNW);<br />
MP("P",D(P),NE);<br />
MP("r",midpoint(O--P),NNW);</asy><br />
<br />
== See Also ==<br />
* [[Diameter]]<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=Radius&diff=101253Radius2019-02-05T02:05:50Z<p>Chenr28: </p>
<hr />
<div>{{stub}}<br />
<br />
The '''radius''' of a [[circle]] is the distance from the center to any point on the circle. Identical definitions hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii.<br />
<br />
<asy>size(115);<br />
pair O=origin, P=dir(30);<br />
D(unitcircle);<br />
D(O--P);<br />
MP("O",D(O),NNW);<br />
MP("P",D(P),NE);<br />
MP("r",midpoint(O--P),NNW);</asy><br />
<br />
== See Also ==<br />
* [[Diameter]]<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_24&diff=1012222009 AMC 12B Problems/Problem 242019-02-04T06:10:23Z<p>Chenr28: /* Better Solution */</p>
<hr />
<div>== Problem ==<br />
For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>?<br />
Note: The functions <math>\sin^{ - 1} = \arcsin</math> and <math>\cos^{ - 1} = \arccos</math> denote inverse trigonometric functions.<br />
<br />
<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math><br />
<br />
== Solution ==<br />
<br />
First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>.<br />
<br />
Hence we get that <math>\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x</math>, thus our equation simplifies to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
<br />
Consider the function <math>f(x) = \sin^{ - 1}(\sin 6x) - x</math>. We are looking for roots of <math>f</math> on <math>[0,\pi]</math>.<br />
<br />
By analyzing properties of <math>\sin</math> and <math>\sin^{-1}</math> (or by computing the derivative of <math>f</math>) one can discover the following properties of <math>f</math>:<br />
<br />
* <math>f(0)=0</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[0,\pi/6]</math>.<br />
* <math>f</math> is decreasing and then increasing on <math>[\pi/6,2\pi/6]</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[2\pi/6,3\pi/6]</math>.<br />
<br />
For <math>x=\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0</math>. Hence <math>f</math> has exactly one root on <math>(0,\pi/6)</math>.<br />
<br />
For <math>x=2\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0</math>. Hence <math>f</math> is negative on the entire interval <math>[\pi/6,2\pi/6]</math>.<br />
<br />
Now note that <math>\forall t: \sin^{-1}(t) \leq \pi/2</math>. Hence for <math>x > 3\pi/6</math> we have <math>f(x) < 0</math>, and we can easily check that <math>f(3\pi/6)<0</math> as well.<br />
<br />
Thus the only unknown part of <math>f</math> is the interval <math>(2\pi/6,3\pi/6)</math>. On this interval, <math>f</math> is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.<br />
<br />
To prove that there are two roots, it is enough to find any <math>x</math> from this interval such that <math>f(x)>0</math>.<br />
<br />
A good guess is its midpoint, <math>x=5\pi/12</math>, where the function <math>\sin^{-1}(\sin 6x)</math> has its local maximum. We can evaluate:<br />
<math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>.<br />
<br />
Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.<br />
<br />
== Better Solution ==<br />
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, <math>x</math> is between that as well. Since <math>x</math> is between <math>0</math> and <math>\pi</math>, combining the two gives <math>x</math> is between <math>0</math> and <math>\pi/2</math>. Now, remove <math>\sin^{ - 1}</math> by taking the sine of both sides. Then, you get <math>sin x = sin 6x</math>.<br />
<br />
From this, either <math>x = 6 x +2 \pi k </math>, or <math> x + 6 x = \pi + 2 \pi k </math>. Solving, and remembering that <math>0<=x<=\pi/2</math>, we get <math>4</math> solutions.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_24&diff=1012212009 AMC 12B Problems/Problem 242019-02-04T06:08:15Z<p>Chenr28: /* Better Solution */</p>
<hr />
<div>== Problem ==<br />
For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>?<br />
Note: The functions <math>\sin^{ - 1} = \arcsin</math> and <math>\cos^{ - 1} = \arccos</math> denote inverse trigonometric functions.<br />
<br />
<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math><br />
<br />
== Solution ==<br />
<br />
First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>.<br />
<br />
Hence we get that <math>\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x</math>, thus our equation simplifies to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
<br />
Consider the function <math>f(x) = \sin^{ - 1}(\sin 6x) - x</math>. We are looking for roots of <math>f</math> on <math>[0,\pi]</math>.<br />
<br />
By analyzing properties of <math>\sin</math> and <math>\sin^{-1}</math> (or by computing the derivative of <math>f</math>) one can discover the following properties of <math>f</math>:<br />
<br />
* <math>f(0)=0</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[0,\pi/6]</math>.<br />
* <math>f</math> is decreasing and then increasing on <math>[\pi/6,2\pi/6]</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[2\pi/6,3\pi/6]</math>.<br />
<br />
For <math>x=\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0</math>. Hence <math>f</math> has exactly one root on <math>(0,\pi/6)</math>.<br />
<br />
For <math>x=2\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0</math>. Hence <math>f</math> is negative on the entire interval <math>[\pi/6,2\pi/6]</math>.<br />
<br />
Now note that <math>\forall t: \sin^{-1}(t) \leq \pi/2</math>. Hence for <math>x > 3\pi/6</math> we have <math>f(x) < 0</math>, and we can easily check that <math>f(3\pi/6)<0</math> as well.<br />
<br />
Thus the only unknown part of <math>f</math> is the interval <math>(2\pi/6,3\pi/6)</math>. On this interval, <math>f</math> is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.<br />
<br />
To prove that there are two roots, it is enough to find any <math>x</math> from this interval such that <math>f(x)>0</math>.<br />
<br />
A good guess is its midpoint, <math>x=5\pi/12</math>, where the function <math>\sin^{-1}(\sin 6x)</math> has its local maximum. We can evaluate:<br />
<math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>.<br />
<br />
Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.<br />
<br />
== Better Solution ==<br />
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, <math>x</math> is between that as well. Since <math>x</math> is between <math>0</math> and <math>\pi</math>, combining the two gives <math>x</math> is between <math>0</math> and <math>\pi/2</math>. Now, remove <math>\sin^{ - 1}</math> by taking the sine of both sides. Then, you get <math>sin x = sin 6x</math>.<br />
<br />
From this, either <math>x = 6x +2\pik</math>, or <math>x+6x = \pi +2\pik</math>. Solving, and remembering that <math>0<=x<=\pi/2</math>, we get <math>4</math> solutions.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_24&diff=1012202009 AMC 12B Problems/Problem 242019-02-04T06:05:09Z<p>Chenr28: /* Better Solution */</p>
<hr />
<div>== Problem ==<br />
For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>?<br />
Note: The functions <math>\sin^{ - 1} = \arcsin</math> and <math>\cos^{ - 1} = \arccos</math> denote inverse trigonometric functions.<br />
<br />
<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math><br />
<br />
== Solution ==<br />
<br />
First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>.<br />
<br />
Hence we get that <math>\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x</math>, thus our equation simplifies to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
<br />
Consider the function <math>f(x) = \sin^{ - 1}(\sin 6x) - x</math>. We are looking for roots of <math>f</math> on <math>[0,\pi]</math>.<br />
<br />
By analyzing properties of <math>\sin</math> and <math>\sin^{-1}</math> (or by computing the derivative of <math>f</math>) one can discover the following properties of <math>f</math>:<br />
<br />
* <math>f(0)=0</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[0,\pi/6]</math>.<br />
* <math>f</math> is decreasing and then increasing on <math>[\pi/6,2\pi/6]</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[2\pi/6,3\pi/6]</math>.<br />
<br />
For <math>x=\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0</math>. Hence <math>f</math> has exactly one root on <math>(0,\pi/6)</math>.<br />
<br />
For <math>x=2\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0</math>. Hence <math>f</math> is negative on the entire interval <math>[\pi/6,2\pi/6]</math>.<br />
<br />
Now note that <math>\forall t: \sin^{-1}(t) \leq \pi/2</math>. Hence for <math>x > 3\pi/6</math> we have <math>f(x) < 0</math>, and we can easily check that <math>f(3\pi/6)<0</math> as well.<br />
<br />
Thus the only unknown part of <math>f</math> is the interval <math>(2\pi/6,3\pi/6)</math>. On this interval, <math>f</math> is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.<br />
<br />
To prove that there are two roots, it is enough to find any <math>x</math> from this interval such that <math>f(x)>0</math>.<br />
<br />
A good guess is its midpoint, <math>x=5\pi/12</math>, where the function <math>\sin^{-1}(\sin 6x)</math> has its local maximum. We can evaluate:<br />
<math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>.<br />
<br />
Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.<br />
<br />
== Better Solution ==<br />
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, x is between that as well. Since x is between 0 and pi, combining the two gives x is between 0 and pi/2. Now, remove the inverse sine by taking the sin of both sides. Then, you get sin x = sin 6x.<br />
<br />
From this, either x = 6x +2pik, or x+6x = pi +2pik. Solving, and remembering that 0<=x<=pi/2, we get 4 solutions.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Chenr28https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_11&diff=1011682001 AMC 12 Problems/Problem 112019-02-03T03:49:41Z<p>Chenr28: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #11]] and [[2001 AMC 10 Problems|2001 AMC 10 #23]]}}<br />
<br />
== Problem ==<br />
<br />
A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?<br />
<br />
<math><br />
\text{(A) }\frac {3}{10}<br />
\qquad<br />
\text{(B) }\frac {2}{5}<br />
\qquad<br />
\text{(C) }\frac {1}{2}<br />
\qquad<br />
\text{(D) }\frac {3}{5}<br />
\qquad<br />
\text{(E) }\frac {7}{10}<br />
</math><br />
<br />
== Solution 1 ==<br />
<br />
Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is <math>\boxed{(\text{D}) \frac {3}{5}}</math>.<br />
<br />
==Solution 2 ==<br />
<br />
Let's assume we don't stop picking until all of the balls are picked. To satisfy this condition, we have to arrange the letters: ''W, W, R, R, R'' such that both ''R's'' appear in the first 4. We find the number of ways to arrange the red balls in the first 4 and divide that by the total ways to chose all the balls. The probability of this occurring is <math>\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Chenr28