Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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− | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}} | + | {{duplicate|[[2019 AMC 10B Problems#Problem 23|2019 AMC 10B #23]] and [[2019 AMC 12B Problems#Problem 20|2019 AMC 12B #20]]}} |
==Problem== | ==Problem== | ||
− | 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>? | + | 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>? |
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } | <math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } | ||
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==Solution 1== | ==Solution 1== | ||
− | 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>x=5</math>. | + | 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>. |
− | Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) | + | Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) <math>AOBX</math> is cyclic. |
− | <math>2\sqrt{170} | + | |
+ | Therefore, we can apply [[Ptolemy's Theorem]] to give: | ||
+ | |||
+ | <math>2\sqrt{170}r = d \sqrt{40}</math>, where <math>r</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}r</math>. | ||
+ | |||
+ | Using the Pythagorean Theorem on the right triangle <math>OAX</math> (or <math>OBX</math>), we find that <math>170 + r^2 = 17r^2</math>, so <math>r^2 = \frac{85}{8}</math>, and thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
+ | |||
+ | ===Diagram for Solution 1=== | ||
+ | [[File:Desmos-graph (1).png|900px|caption]] | ||
+ | |||
+ | ~BakedPotato66 | ||
==Solution 2 (coordinate bash)== | ==Solution 2 (coordinate bash)== | ||
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− | ==Solution 4 ( | + | ==Solution 4 (how fast can you multiply two-digit numbers?)== |
− | 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> | + | 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 for the tangent lines passing <math>A</math> and <math>B</math>, respectively, are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math>. Then the lines normal (perpendicular) to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Solving for <math>x</math>, we have |
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<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath> | <cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath> | ||
− | After | + | 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> |
− | |||
+ | ==Solution 5 (power of a point)== | ||
− | + | 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>. | |
− | == | ||
− | + | ==Solution 6 (Similar to #3)== | |
+ | Let the tangent lines from A and B intersect at X. Let the center of <math>\omega</math> be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is <math>(5, 0)</math>. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of <math>AX=\sqrt{170}=\sqrt{17}\sqrt{10}</math>. Using the distance formula on AM, we get that <math>AM=\sqrt{10}</math>. Using the distance formula on MX, we get that <math>MX=4\sqrt{10}</math>. To get AC (the radius of <math>\omega</math>), we use either of these methods: | ||
+ | |||
+ | Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that <math>MC=\frac{\sqrt{17}}{4}</math>. Thus, XC has length <math>XC=\frac{17\sqrt{10}}{4}</math>. Using the Pythagorean Theorem on CAX yields <math>CA=\frac{\sqrt{170}}{4}</math>. | ||
+ | |||
+ | Method 2: Note that CAX and AMX are similar. Thus, <math>\frac{AM}{MX}=\frac{AC}{AX}</math>. Solving for AC yields <math>\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}</math>. | ||
+ | |||
+ | Using the area formula for a circle yields that the area is <math>\frac{85\pi}{8} \longrightarrow \boxed{(C)}</math>. | ||
+ | ~Math4Life2020 | ||
+ | |||
+ | ==Video Solution== | ||
+ | For those who want a video solution: (Is similar to Solution 1) | ||
+ | https://youtu.be/WI2NVuIp1Ik | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/W1zuqrTlBtU | ||
+ | |||
+ | ~IceMatrix | ||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://www.youtube.com/watch?v=sQIWSrio_Hc | ||
− | + | ~The Power of Logic | |
==See Also== | ==See Also== |
Latest revision as of 11:24, 5 February 2022
- The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.
Contents
Problem
Points and lie on circle in the plane. Suppose that the tangent lines to at and intersect at a point on the -axis. What is the area of ?
Solution 1
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is , the Pythagorean Theorem gives . This simplifies to .
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) is cyclic.
Therefore, we can apply Ptolemy's Theorem to give:
, where is the radius of the circle and is the distance between the circle's center and . Therefore, .
Using the Pythagorean Theorem on the right triangle (or ), we find that , so , and thus the area of the circle is .
Diagram for Solution 1
~BakedPotato66
Solution 2 (coordinate bash)
We firstly obtain as in Solution 1. Label the point as . The midpoint of segment is . Notice that the center of the circle must lie on the line passing through the points and . Thus, the center of the circle lies on the line .
Line is . Therefore, the slope of the line perpendicular to is , so its equation is .
But notice that this line must pass through and . Hence . So the center of the circle is .
Finally, the distance between the center, , and point is . Thus the area of the circle is .
Solution 3
The midpoint of is . Let the tangent lines at and intersect at on the -axis. Then is the perpendicular bisector of . Let the center of the circle be . Then is similar to , so . The slope of is , so the slope of is . Hence, the equation of is . Letting , we have , so .
Now, we compute , , and .
Therefore , and consequently, the area of the circle is .
Solution 4 (how fast can you multiply two-digit numbers?)
Let be the intersection on the x-axis. By Power of a Point Theorem, . Then the equations for the tangent lines passing and , respectively, are and . Then the lines normal (perpendicular) to them are and . Solving for , we have
After condensing, . Then, the center of is . Apply distance formula. WLOG, assume you use . Then, the area of is
Solution 5 (power of a point)
Firstly, the point of intersection of the two tangent lines has an equal distance to points and due to power of a point theorem. This means we can easily find the point, which is . Label this point . is an isosceles triangle with lengths, , , and . Label the midpoint of segment as . The height of this triangle, or , is . Since bisects , contains the diameter of circle . Let the two points on circle where intersects be and with being the shorter of the two. Now let be and be . By Power of a Point on and , . Applying Power of a Point again on and , . Expanding while using the fact that , . Plugging this into , . Using the quadratic formula, , and since , . Since this is the diameter, the radius of circle is , and so the area of circle is .
Solution 6 (Similar to #3)
Let the tangent lines from A and B intersect at X. Let the center of be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is . However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of . Using the distance formula on AM, we get that . Using the distance formula on MX, we get that . To get AC (the radius of ), we use either of these methods:
Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that . Thus, XC has length . Using the Pythagorean Theorem on CAX yields .
Method 2: Note that CAX and AMX are similar. Thus, . Solving for AC yields .
Using the area formula for a circle yields that the area is . ~Math4Life2020
Video Solution
For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=sQIWSrio_Hc
~The Power of Logic
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.