https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=J2005&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:25:28ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_6&diff=1061152004 AIME II Problems/Problem 62019-06-05T04:38:16Z<p>J2005: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a [[whole number]] of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the [[ratio]] <math> 3: 2: 1, </math>what is the least possible total for the number of bananas?<br />
<br />
== Solution ==<br />
Denote the number of bananas the first monkey took from the pile as <math>b_1</math>, the second <math>b_2</math>, and the third <math>b_3</math>; the total is <math>b_1 + b_2 + b_3</math>. Thus, the first monkey got <math>\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3</math>, the second monkey got <math>\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3</math>, and the third monkey got <math>\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3</math>. <br />
<br />
Taking into account the ratio aspect, say that the third monkey took <math>x</math> bananas in total. Then,<br />
<br />
<math>x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2 + \frac{11}{48}b_3 = \frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3</math><br />
<br />
Solve this to find that <math>\frac{b_1}{11} = \frac{b_2}{13} = \frac{b_3}{27}</math>. All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</math> is divisible by <math>8</math>, and <math>b_3</math> is divisible by <math>72</math> (however, since the denominator contains a <math>27</math>, the factors of <math>3</math> cancel, and it only really needs to be divisible by <math>8</math>). Thus, the minimal value is when each fraction is equal to <math>8</math>, and the solution is <math>8(11 + 13 + 27) = \boxed{408}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=II|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>J2005https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_1&diff=1051191998 AIME Problems/Problem 12019-04-03T23:40:45Z<p>J2005: /* Solution */</p>
<hr />
<div>== Problem ==<br />
For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math>, <math>8^8</math>, and <math>k</math>? <br />
== Solution ==<br />
It is evident that <math>k</math> has only 2s and 3s in its prime factorization, or <math>k = 2^a3^b</math>. <br />
<br />
*<math>6^6 = 2^6\cdot3^6</math><br />
*<math>8^8 = 2^{24}</math><br />
*<math>12^{12} = 2^{24}\cdot3^{12}</math><br />
<br />
The [[LCM]] of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. <math>[6^6,8^8] = 2^{24}3^6</math>. Therefore <math>12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}</math>, and <math>b = 12</math>. Since <math>0 \le a \le 24</math>, there are <math>\boxed{25}</math> values of <math>k</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1998|before=First question|num-a=2}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_1&diff=1050292019 AMC 10A Problems/Problem 12019-03-30T06:02:55Z<p>J2005: /* Solution */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
==Solution==<br />
The first part can be rewritten as <cmath>2^{0^{1}}=2^{0}=1.</cmath><br />
The second part is <cmath>(1^{1})^{9}=1^{9}=1.</cmath><br />
Adding these up gives <math>\boxed{\textbf{(C) }2}.</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_1&diff=1050282019 AMC 10A Problems/Problem 12019-03-30T05:18:50Z<p>J2005: /* Solution */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
======Solution======<br />
The first part can be rewritten as <cmath>2^{0^{1}}=2^{0}=1.</cmath><br />
The second part is <cmath>(1^{1})^{9}=1^{9}=1.</cmath><br />
Adding these up gives <math>\boxed{\textbf{(C) }2}.</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_1&diff=1050272019 AMC 10A Problems/Problem 12019-03-30T05:18:11Z<p>J2005: /* Solution */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
=Solution=<br />
The first part can be rewritten as <cmath>2^{0^{1}}=2^{0}=1.</cmath><br />
The second part is <cmath>(1^{1})^{9}=1^{9}=1.</cmath><br />
Adding these up gives <math>\boxed{\textbf{(C) }2}.</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1037212019 AMC 10B Problems/Problem 252019-02-22T00:49:35Z<p>J2005: /* See Also */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Thus we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, where <math>f(n)</math> is the number of valid sequences of length <math>n</math>.<br />
<br />
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(5) = 1</math> and <math>f(6) = 2</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose3} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030422019 AMC 10B Problems/Problem 232019-02-16T02:19:01Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}</math><br />
(by Zhen Qin)<br />
(P.S. Will someone please Latex this?)<br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030402019 AMC 10B Problems/Problem 232019-02-16T02:15:03Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}\\<br />
(by Zhen Qin)\\<br />
(P.S. Will someone please Latex this?)</math><br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030392019 AMC 10B Problems/Problem 232019-02-16T02:14:24Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}\\$<br />
(by Zhen Qin)\\<br />
(P.S. Will someone please Latex this?)</math><br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030382019 AMC 10B Problems/Problem 232019-02-16T02:13:45Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}\\$<br />
(by Zhen Qin)</math> <math>\\$<br />
(P.S. Will someone please Latex this?)<br />
<br />
(</math>\LaTeX$ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030372019 AMC 10B Problems/Problem 232019-02-16T02:13:27Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}\\$<br />
(by Zhen Qin)</math>\\$<br />
(P.S. Will someone please Latex this?)<br />
<br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030362019 AMC 10B Problems/Problem 232019-02-16T02:12:12Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>X</math> axis. Then <math>CD</math> would be the perpendicular bisector of <math>AB</math>. Let the center of circle be O. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, that is <math>\frac{OA}{AC} = \frac{AD}{DC}.</math><br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, therefore the slope of CD will be 3. The equation of <math>CD</math> is <math>y-12=3*(x-9)</math>, that is <math>y=3x-15</math>. Let <math>y=0</math>. Then we have <math>x=5</math>, which is the <math>x</math> coordinate of <math>C(5,0)</math>.<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}</math><br />
(by Zhen Qin)<br />
(P.S. Will someone please Latex this?)<br />
<br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030332019 AMC 10B Problems/Problem 232019-02-16T02:05:22Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br />
The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}</math><br />
(by Zhen Qin)<br />
(P.S. Will someone please Latex this?)<br />
<br />
(<math>\LaTeX</math>ed by a pewdiepie subscriber)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030322019 AMC 10B Problems/Problem 232019-02-16T02:04:57Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br />
The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br />
<br />
<math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math><br />
<math>AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}</math><br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math><br />
Therefore <math>OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}</math><br />
Consequently, the area of the circle is <math>pi* OA^2 = pi*\frac{85}{5}</math><br />
(by Zhen Qin)<br />
(P.S. Will someone please Latex this?)<br />
<br />
(<math>\LaTeX</math>)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1030312019 AMC 10B Problems/Problem 232019-02-16T01:56:35Z<p>J2005: /* Solution 3 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br />
<br />
==Problem==<br />
<br />
Points <math>A(6,13)</math> and <math>B(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was <math>(x, 0)</math>. Using Pythagorean Theorem gives <math>x=5</math>.<br />
<br />
Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br />
<br />
<math>2\sqrt{170}x = d * \sqrt{40}</math>, where <math>d</math> represents the distance between circle center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using Pythagorean Theorem on <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle center, we realize that <math>170 + x^2 = 17x^2</math>, at which point <math>x^2 = \frac{85}{8}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2==<br />
First, follow solution 1 and obtain <math>x=5</math>. Label the point <math>(5,0)</math> as point <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line that goes through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. The perpendicular line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. The slope of the perpendicular line is <math>-\frac{1}{13}</math>. The line is hence <math>y=-\frac{x}{13}+\frac{175}{13}</math>. The point <math>(x, 3x-15)</math> lies on this line. Therefore, <math>3x-15=-\frac{x}{13}+\frac{175}{13}</math>. Solving this equation tells us that <math>x=\frac{37}{4}</math>. So the center of the circle is <math>(\frac{37}{4}, \frac{51}{4})</math>. The distance between the center, <math>(\frac{37}{4}, \frac{51}{4})</math>, and point A is <math>\frac{\sqrt{170}}{4}</math>. Hence, the area is <math>\frac{85}{8}\pi</math>. The answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br />
The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br />
<br />
AC=<math>sqrt((6-5)^2+(13-0)^2)=sqrt(170)</math><br />
AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10)<br />
DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160)<br />
Therefore OA=AC*AD/DC=sqrt(85/5)<br />
Consequently, the area of the circle is pi*OA^2=pi*85/5<br />
(by Zhen Qin)<br />
(P.S. Will someone please Latex this?)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
SUB2PEWDS</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=1030272019 AMC 10B Problems/Problem 202019-02-16T01:38:47Z<p>J2005: /* See Also */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br />
==Problem==<br />
As shown in the figure, line segment <math>\overline{AD}</math> is trisected by points <math>B</math> and <math>C</math> so that <math>AB=BC=CD=2.</math> Three semicircles of radius <math>1,</math> <math>\overarc{AEB},\overarc{BFC},</math> and <math>\overarc{CGD},</math> have their diameters on <math>\overline{AD},</math> and are tangent to line <math>EG</math> at <math>E,F,</math> and <math>G,</math> respectively. A circle of radius <math>2</math> has its center on <math>F. </math> The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br />
<cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath><br />
where <math>a,b,c,</math> and <math>d</math> are positive integers and <math>a</math> and <math>b</math> are relatively prime. What is <math>a+b+c+d</math>?<br />
<br />
<asy><br />
size(6cm);<br />
filldraw(circle((0,0),2), grey);<br />
filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br />
filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br />
filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br />
dot((-3,-1));<br />
label("$A$",(-3,-1),S);<br />
dot((-2,0));<br />
label("$E$",(-2,0),NW);<br />
dot((-1,-1));<br />
label("$B$",(-1,-1),S);<br />
dot((0,0));<br />
label("$F$",(0,0),N);<br />
dot((1,-1));<br />
label("$C$",(1,-1), S);<br />
dot((2,0));<br />
label("$G$", (2,0),NE);<br />
dot((3,-1));<br />
label("$D$", (3,-1), S);<br />
</asy><br />
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math><br />
==Solution==<br />
Divide the circle into four parts: The top semicircle: (A), the bottom sector with arc length 120 degrees: (B), the triangle formed by the radii of (A) and the chord: (C), and the four parts which are the corners of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D).<br />
<br />
Area of (A): <math>2\pi</math><br />
<br />
Area of (B): <math>\frac{4\pi}{3}</math><br />
<br />
Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is <math>2\sqrt{3}*1/2=\sqrt{3}</math><br />
<br />
Area of (D): <math>4*1-1/4*\pi*4=4-\pi</math><br />
<br />
Total sum: <math>\frac{7\pi}{3}-\sqrt{3}+4</math><br />
<br />
<math>7+3+3+4=\boxed{17}</math><br />
<br />
For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
SUB2PEWDIEPIE</div>J2005https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=889082004 AMC 10B Problems/Problem 202017-12-14T20:32:59Z<p>J2005: /* Solution (Triangle Areas) */</p>
<hr />
<div>== Problem ==<br />
<br />
<br />
In <math>\triangle ABC</math> points <math>D</math> and <math>E</math> lie on <math>BC</math> and <math>AC</math>, respectively. If <math>AD</math> and <math>BE</math> intersect at <math>T</math> so that <math>AT/DT=3</math> and <math>BT/ET=4</math>, what is <math>CD/BD</math>?<br />
<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} </math><br />
<br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br />
draw(A--B--C--cycle);<br />
draw(A--D);<br />
draw(B--E);<br />
pair T=intersectionpoint(A--D,B--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,N);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,S);<br />
label("$T$",T,2*WNW);<br />
</asy><br />
<br />
== Solution (Triangle Areas) ==<br />
<br />
We use the square bracket notation <math>[\cdot]</math> to denote area.<br />
<br />
WLOG, we can assume <math>[\triangle BTD] = 1</math>. Then <math>[\triangle BTA] = 3</math>, and <math>[\triangle ATE] = 3/4</math>. We have <math>CD/BD = [\triangle ACD]/[\triangle ABD]</math>, so we need to find the area of quadrilateral <math>TDCE</math>.<br />
<br />
Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x = [\triangle TDC]</math>, and <math>y = [\triangle TEC]</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/y=4</math>, and by considering triangles <math>\triangle ATC</math> and <math>\triangle DTC</math>, we obtain <math>(3/4+y)/x=3</math>. Solving, we get <math>x=4/11</math>, <math>y=15/44</math>, so the area of quadrilateral <math>TDEC</math> is <math>x+y=31/44</math>.<br />
<br />
Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math><br />
<br />
== Solution (Mass points) ==<br />
<br />
The presence of only ratios in the problem essentially cries out for mass points.<br />
<br />
As per the problem, we assign a mass of <math>1</math> to point <math>A</math>, and a mass of <math>3</math> to <math>D</math>. Then, to balance <math>A</math> and <math>D</math> on <math>T</math>, <math>T</math> has a mass of <math>4</math>.<br />
<br />
Now, were we to assign a mass of <math>1</math> to <math>B</math> and a mass of <math>4</math> to <math>E</math>, we'd have <math>5T</math>. Scaling this down by <math>4/5</math> (to get <math>4T</math>, which puts <math>B</math> and <math>E</math> in terms of the masses of <math>A</math> and <math>D</math>), we assign a mass of <math>\frac{4}{5}</math> to <math>B</math> and a mass of <math>\frac{16}{5}</math> to <math>E</math>.<br />
<br />
Now, to balance <math>A</math> and <math>C</math> on <math>E</math>, we must give <math>C</math> a mass of <math>\frac{16}{5}-1=\frac{11}{5}</math>.<br />
<br />
Finally, the ratio of <math>CD</math> to <math>BD</math> is given by the ratio of the mass of <math>B</math> to the mass of <math>C</math>, which is <math>\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}</math>.<br />
<br />
== Solution (Coordinates) ==<br />
<br />
Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any <math>\triangle ABC</math>, and we just need to compute it for any single triangle.<br />
<br />
We can choose the points <math>A=(-3,0)</math>, <math>B=(0,4)</math>, and <math>D=(1,0)</math>. This way we will have <math>T=(0,0)</math>, and <math>E=(0,-1)</math>. The situation is <br />
shown in the picture below:<br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(0.8);<br />
pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br />
draw(A--B--C--cycle);<br />
draw(A--D);<br />
draw(B--E);<br />
pair T=intersectionpoint(A--D,B--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,N);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,S);<br />
label("$T$",T,NW);<br />
label("$3$",A--T,N);<br />
label("$4$",B--T,W);<br />
label("$1$",D--T,N);<br />
label("$1$",E--T,W);<br />
<br />
</asy><br />
<br />
The point <math>C</math> is the intersection of the lines <math>BD</math> and <math>AE</math>. The points on the first line have the form <math>(t,4-4t)</math>, the points on the second line have the form <math>(t,-1-t/3)</math>. Solving for <math>t</math> we get <math>t=15/11</math>, hence <math>C=(15/11,-16/11)</math>.<br />
<br />
The ratio <math>CD/BD</math> can now be computed simply by observing the <math>x</math> coordinates of <math>B</math>, <math>C</math>, and <math>D</math>:<br />
<br />
<cmath><br />
\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}<br />
</cmath><br />
<br />
== See also ==<br />
<br />
{{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>J2005