2019 AMC 10B Problems/Problem 23
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
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.