Difference between revisions of "2019 AMC 12B Problems/Problem 20"
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− | + | ==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>? | ||
+ | |||
+ | <math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } | ||
+ | \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math> | ||
+ | |||
+ | ==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>\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) 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 | ||
+ | <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>. | ||
+ | |||
+ | ==Solution 2 (coordinate bash)== | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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>. | ||
+ | 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>. | ||
+ | |||
+ | Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>, | ||
+ | <math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and | ||
+ | <math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>. | ||
+ | |||
+ | Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</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 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, | ||
+ | |||
+ | |||
+ | |||
+ | <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> | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | For those who want a video solution: (Is similar to Solution 1) | ||
+ | https://youtu.be/WI2NVuIp1Ik | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}} | ||
+ | {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 22:36, 8 December 2020
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) defined by the circle's center, ,
, and
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 triangle formed by the point
, either one of
or
, and the circle's center, we find that
, so
, and thus the answer is
.
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 are
and
for the tangent lines passing
and
respectively. Then the lines normal to them are
and
. Thus,
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
.
Video Solution
For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik
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.