Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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~The Power of Logic | ~The Power of Logic | ||
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+ | ==Solution 5 (tangent cheese)== | ||
+ | After getting <math>x=5</math>, let <math>C=(5,0)</math>. Get the slopes of the lines <math>AC</math> and <math>BC</math>, namely <math>\frac{13}{6-5}=13</math>, <math>\frac{11}{12-5}=\frac{11}{7}</math>. Then, use tangent angle subtraction to get <math>\tan{2x}=\frac{13-\frac{11}{7}}{1+13*\frac{11}{7}}=\frac{80}{150}=\frac{8}{15}</math>. Then, apply tangent double angle to get <math>\tan{2x}=\frac{8}{15}=\frac{2y}{1-y^2}</math>, where <math>y=tan{x}</math>. Solving, we obtain <math>y=\frac{1}{4}</math>. Then, note that <math>y=R/{BC}</math>, so <math>r=\frac{1}{4}*\sqrt{170}</math>. Finishing off, we obtain <math>A=\pi*r^2=\pi*170/16=\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>. | ||
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+ | ~SigmaPiE | ||
==See Also== | ==See Also== |
Revision as of 09:45, 29 October 2023
- The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.
Contents
[hide]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
Solution 5 (tangent cheese)
After getting , let . Get the slopes of the lines and , namely , . Then, use tangent angle subtraction to get . Then, apply tangent double angle to get , where . Solving, we obtain . Then, note that , so . Finishing off, we obtain .
~SigmaPiE
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.