2022 AIME I Problems/Problem 8

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 pointstwo 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

Solution 1

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:

unitsize(0.3cm);
draw(circle((0,0),18));
pair C = (9 * sqrt(3), -9);
pair B = (-9 * sqrt(3), -9);
pair B2 = (-12 * sqrt(3), -18);
pair C2 = (12 * sqrt(3), -18);

pair A = (0,18);
draw(A--B--C--cycle);
draw(circle((0,-6),12), gray);
draw(circle((3*sqrt(3),3),12), gray);
draw(circle((-3*sqrt(3),3),12), gray);

draw(B--B2,dashed);
draw(C--C2,dashed);
draw(B2--C2,dashed);

dot(B2);
dot(C2);

pair X = (0, 3-sqrt(117));
pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );
pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );
dot(X);
dot(Y);
dot(Z);
dot((0,0));
label("$O$",(0,0),N);

label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
label("$B'$",B2,W);
label("$C'$",C2,E);
label("$X$",X,S);
label("$Y$",Y,E);
label("$Z$",Z,W);

draw(O--X);
draw(O--Y);
draw(O--((sqrt(351)-sqrt(27))/4,((sqrt(117)-3)/2 + (3-\sqrt{117}))/2));


draw(X--Y--Z--cycle, dashed);
 (Error making remote request. Unknown error_msg)

Solution 2

Let bottom left point as the origin, the radius of each circle is , note that three centers for circles are

It is not hard to find that one intersection point lies on since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation , getting that , the length is , leads to the answer

~bluesoul

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.