2016 AMC 10A Problems/Problem 21
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 the Pythagorean Theorem, we get that
so the center of the second circle of radius 2 is at
.
Draw the trapezoid and using the 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.