Difference between revisions of "2019 AMC 10B Problems/Problem 23"
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<cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath> | <cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath> | ||
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− | After some condensing, find that <math>x=\frac{37}{4}</math>. Then, the center of circle <math>\omega</math> is <math>(\frac{37}{4}, \frac{53}{4})</math>. Apply distance formula. WLOG, assume you choose <math>A</math> (you choose!). Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{1}{4^2}+\frac{13^2}{4^2}\pi</cmath> | + | |
+ | After some condensing (most tedious step), find that <math>x=\frac{37}{4}</math>. Then, the center of circle <math>\omega</math> is <math>(\frac{37}{4}, \frac{53}{4})</math>. Apply distance formula. WLOG, assume you choose <math>A</math> (you choose!). Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{1}{4^2}+\frac{13^2}{4^2}\pi</cmath> | ||
<cmath>\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}</cmath>. | <cmath>\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}</cmath>. | ||
~BJHHar | ~BJHHar | ||
+ | |||
==Solution 5 (Power of a Point)== | ==Solution 5 (Power of a Point)== | ||
Revision as of 22:25, 25 September 2019
- 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 .
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 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 some condensing (most tedious step), find that . Then, the center of circle is . Apply distance formula. WLOG, assume you choose (you choose!). Then, the area of is
.
~BJHHar
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 .
-bradleyguo
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.