Olympiad Maths Prep

Track / Stage 5 / 236 of 400 #836 of 2000

Problem 836

AIME late
Geometry Difficulty 5.6 Find the answer

4. On the table lie three balls, touching each other pairwise. The radii of the balls form a geometric progression with a common ratio q1q \neq 1. The radius of the middle one is 2012. Find the ratio of the sum of the squares of the sides of the triangle formed by the points of contact of the balls with the table to the sum of the sides of the triangle formed by the centers of the balls.

Answer: 4024

Official solution

Let the radius of the smaller of the balls be rr, then the radii of the others are rq=2012r q=2012 and r2r^{2}, and the points of contact of the balls with the table are denoted as A,B,CA, B, C respectively. Consider the section of two balls by a plane perpendicular to the table and passing through the centers of these balls (see the figure, for the two smaller balls in size). Then for the balls shown in the figure, we get that OA=r,O1B=rqO A=r, O_{1} B=r q and 0O1=r+rq0 O_{1}=r+r q. Further, O1D=01DDC=01BOA=r(q1)O_{1} D=0{ }_{1} D-D C=0{ }_{1} B-O A=r(q-1). From OO1D\triangle O O_{1} D by the Pythagorean theorem, we get that OD=2rq=ABO D=2 r \sqrt{q}=A B. Similarly, considering the remaining pairs of balls, we get that AC=2rq,BC=2rqqA C=2 r q, B C=2 r q \sqrt{q}, then the desired value is 4r2q+4r2q2+4r2q3r(1+q)+r(q+q2)+r(1+q2)=2rq=22012=4024\frac{4 r^{2} q+4 r^{2} q^{2}+4 r^{2} q^{3}}{r(1+q)+r\left(q+q^{2}\right)+r\left(1+q^{2}\right)}=2 r q=2 \cdot 2012=4024

Recommendations for checking: 2 points for calculating all sides of the triangle formed by the points of contact of the balls.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.