2022 AIME I Problems/Problem 8
Contents
Problem
Equilateral triangle is inscribed in circle with radius Circle is tangent to sides and and is internally tangent to Circles and are defined analogously. Circles and meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of are the vertices of a large equilateral triangle in the interior of and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of The side length of the smaller equilateral triangle can be written as where and are positive integers. Find
Diagram
~MRENTHUSIASM ~ihatemath123
Solution 1 (Coordinate Geometry)
We can extend and to and respectively such that circle is the incircle of . Since the diameter of the circle is the height of this triangle, the height of this triangle is . We can use inradius or equilateral triangle properties to get the inradius of this triangle is (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is .
Let be the center of the largest circle. We will set up a coordinate system with as the origin. The center of will be at because it is directly beneath and is the length of the larger radius minus the smaller radius, or . By rotating this point around , we get the center of . This means that the magnitude of vector is and is at a degree angle from the horizontal. Therefore, the coordinates of this point are and by symmetry the coordinates of the center of is .
The upper left and right circles intersect at two points, the lower of which is . The equations of these two circles are: We solve this system by subtracting to get . Plugging back in to the first equation, we have . Since we know is the lower solution, we take the negative value to get .
We can solve the problem two ways from here. We can find by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find as they lie on the same vertical, is degrees so we can make use of triangles, and because is the center of triangle . We can draw the diagram as such: Note that . It follows that Finally, the answer is .
~KingRavi
Solution 2 (Euclidean Geometry)
For equilateral triangle with side length , height , and circumradius , there are relationships: , , and .
There is a lot of symmetry in the figure. The radius of the big circle is , let the radius of the small circles , , be .
We are going to solve this problem in steps:
We have is a triangle, and , ( and are tangent), and . So, we get and .
Since and are tangent, we get .
Note that is an equilateral triangle, and is its center, so .
Note that is an isosceles triangle, so
In , Power of a Point gives and .
It follows that . We solve this quadratic equation: .
Since is the circumradius of equilateral , we have .
Therefore, the answer is .
Solution 3 (Simple Geometry)
Let be the center, be the radius, and be the diameter of Let be the radius, are the centers of Let be the desired triangle with side We find using Triangles and – are equilateral triangles with a common center therefore in the triangle
We apply the Law of Cosines to and get
vladimir.shelomovskii@gmail.com, vvsss
Solution 4 (Mixtilinear Incircles)
Let be the center of , be the intersection of further from , and be the center of . Define similarly. It is well-known that the -mixtilinear inradius is , so in particular this means that . Since , it follows by Law of Cosines on that . Then the Pythagorean theorem gives that the altitude of is , so and so the answer is .
~Kagebaka
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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