Difference between revisions of "2016 AMC 10A Problems/Problem 21"
(→Solution 2) |
(→Solution 2) |
||
Line 66: | Line 66: | ||
Let the center of the first circle of radius 1 be at (0, 1). | Let the center of the first circle of radius 1 be at (0, 1). | ||
− | Draw the trapezoid <math>PQQ'P'</math> and using Pythagorean | + | Draw the trapezoid <math>PQQ'P'</math> and using Pythagorean Theorem., we get that <math>P'Q' = 2\sqrt{2}</math> so the center of the second circle of radius 2 is at <math>(2\sqrt{2}, 2)</math>. |
− | Draw the trapezoid <math>QRR'Q'</math> and using Pythagorean | + | Draw the trapezoid <math>QRR'Q'</math> and using Pythagorean Theorem., we get that <math>Q'R' = 2\sqrt{2} + 2\sqrt{6}</math> so the center of the third circle of radius 3 is at <math>(2\sqrt{2}+2\sqrt{6}, 3)</math>. |
− | Now, we may use the Shoelace | + | Now, we may use the Shoelace Theorem! |
<math>(0,1)</math> | <math>(0,1)</math> |
Revision as of 03:08, 16 February 2016
Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?
Solution
Notice that we can find in two different ways: and , so
. Additionally, . Therefore, . Similarly, . We can calculate easily because . .
Plugging into first equation, the two sums of areas, .
.
Solution 2
Use the Shoelace Theorem!
Let the center of the first circle of radius 1 be at (0, 1).
Draw the trapezoid and using Pythagorean Theorem., we get that so the center of the second circle of radius 2 is at .
Draw the trapezoid and using Pythagorean Theorem., we get that so the center of the third circle of radius 3 is at .
Now, we may use the Shoelace Theorem!
.
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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.