Difference between revisions of "2019 AMC 10B Problems/Problem 23"
(→Solution 2) |
Mathkiddie (talk | contribs) (→Solution 5 (power of a point)) |
||
(34 intermediate revisions by 16 users not shown) | |||
Line 1: | Line 1: | ||
− | {{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) } | ||
Line 9: | Line 9: | ||
==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)== | ||
Line 34: | Line 44: | ||
and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
− | |||
− | + | ==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 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 | ||
+ | |||
+ | |||
+ | |||
+ | <cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath> | ||
+ | <cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}</cmath> | ||
+ | <cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==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 20:03, 24 October 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
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.