Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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==Solution 4 (how fast can you multiply two-digit numbers?)== | ==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>. Thus, 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>. Thus, |
Revision as of 16:04, 1 November 2021
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
. Thus, the lines normal (perpendicular) 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
.
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.