2019 AMC 10A Problems/Problem 16
- The following problem is from both the 2019 AMC 10A #16 and 2019 AMC 12A #10, so both problems redirect to this page.
Problem
The figure below shows circles of radius
within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius
Solution 1
In the diagram above, notice that triangle and triangle
are congruent and equilateral with side length
. We can see the radius of the larger circle is two times the altitude of
plus
(the distance from point
to the edge of the circle). Using
triangles, we know the altitude is
. Therefore, the radius of the larger circle is
.
The area of the larger circle is thus , and the sum of the areas of the smaller circles is
, so the area of the dark region is
, which implies the answer is
-eric2020
(You also form an equilateral triangle with side length of that goes through the center)~mn28407
Solution 2
We can form an equilateral triangle with side length from the centers of three of the unit circles tangent to the outer circle. The radius of the outer circle is the circumradius of the triangle plus
. By using
or
, we get the radius
.
The shaded region is the area of the outer circle is
Solution 3
Like in Solution 2, we can form an equilateral triangle with side length from the centers of three of the unit circles tangent to the outer circle. We can find the height of this triangle to be
. Then, we can form another equilateral triangle from the centers of the 2nd and 3rd circles in the third row and the center of the bottom circle with side length
. The height of this triangle is clearly
. Therefore the diameter of the large circle is
and the radius is
. The area is thus
. The total area of the 13 circles is
and so the shaded area is
~DBlack2021
Solution 4
In the diagram above, ,
, so
. The larger circle's radius
, so the larger circle's area is
. This minus the combined area of the smaller circles
-SmileKat32
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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.