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>. | ||
+ | |||
+ | ~SigmaPiE | ||
==See Also== | ==See Also== |
Revision as of 10:45, 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
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.