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| − | ==Problem==
| + | #REDIRECT[[2019_AMC_10B_Problems/Problem_23]] |
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| − | 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>?
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| − | <math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }
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| − | \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math>
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| − | ==Solution 1==
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| − | First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.
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| − | Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> is cyclic. Therefore, we can apply Ptolemy's Theorem to give
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| − | <math>2\sqrt{170}x = d \sqrt{40}</math>, where <math>x</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using the Pythagorean Theorem on the triangle formed by the point <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle's center, we find that <math>170 + x^2 = 17x^2</math>, so <math>x^2 = \frac{85}{8}</math>, and thus the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
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| − | ==Solution 2 (coordinate bash)==
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| − | We firstly obtain <math>x=5</math> as in Solution 1. Label the point <math>(5,0)</math> as <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 passing 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>.
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| − | Line <math>AC</math> is <math>y=13x-65</math>. Therefore, the slope of the line perpendicular to <math>AC</math> is <math>-\frac{1}{13}</math>, so its equation is <math>y=-\frac{x}{13}+\frac{175}{13}</math>.
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| − | But notice that this line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. Hence <math>3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}</math>. So the center of the circle is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>.
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| − | Finally, the distance between the center, <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>, and point <math>A</math> is <math>\frac{\sqrt{170}}{4}</math>. Thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
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| − | ==Solution 3==
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| − | The midpoint 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> is the perpendicular bisector of <math>AB</math>. Let the center of the circle be <math>O</math>. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, so <math>\frac{OA}{AC} = \frac{AD}{DC}</math>.
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| − | The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, so the slope of <math>CD</math> is <math>3</math>. Hence, the equation of <math>CD</math> is <math>y-12=3(x-9) \Rightarrow y=3x-15</math>. Letting <math>y=0</math>, we have <math>x=5</math>, so <math>C = (5,0)</math>.
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| − | Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>,
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| − | <math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and
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| − | <math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.
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| − | Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>,
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| − | and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
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| − | ==Solution 4 (how fast can you multiply two-digit numbers?)==
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| − | Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math> for the tangent lines passing <math>A</math> and <math>B</math> respectively. Then the lines normal to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Thus,
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| − | <cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath>
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| − | <cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}</cmath>
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| − | <cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath>
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| − | After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath>
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| − | ==Solution 5 (power of a point)==
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| − | Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.
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| − | ==Video Solution==
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| − | For those who want a video solution: (Is similar to Solution 1)
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| − | https://youtu.be/WI2NVuIp1Ik
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| − | ==Video Solution by TheBeautyofMath==
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| − | https://youtu.be/W1zuqrTlBtU
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| − | ~IceMatrix
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| − | ==See Also==
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| − | {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}
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| − | {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}
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| − | {{MAA Notice}}
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