Three circles of equal radius touch one another externally. Circle X encloses all three circles and is internally tangent to each. Circle Y lies in the central gap and is externally tangent to each of the three circles. Find the ratio of the radius of X to the radius of Y.
Answer and solution
Correct answer: C
Solution
Answer: C — (7 + 4√3) : 1
1. The centres of the three equal circles form an equilateral triangle. If each circle has radius r, the side of this triangle is 2r.
2. The distance from the centre of the equilateral triangle to any vertex is 2r/√3.
3. For the larger circle X, its radius is:
4. R = r + 2r/√3
5. For the smaller circle Y, its radius is:
6. S = 2r/√3 − r
7. Therefore,
8. R/S = (1 + 2/√3)/(2/√3 − 1)
9. Multiplying the numerator and denominator by (2/√3 + 1), we get:
10. R/S = (1 + 2/√3)² / (4/3 − 1)
11. Simplifying gives:
12. R/S = 7 + 4√3
13. Answer: (7 + 4√3) : 1.
Why this method works:
The common centre of the outer and inner circles is the centre of the equilateral triangle of the original centres. Its distance d to each original centre is 2r/√3. Internal tangency gives R−r=d; external tangency gives s+r=d.
Independent check and common trap:
Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.
Apply: Centres form an equilateral triangle of side 2r. Centre-to-vertex distance d=2r/√3. Outer/inner radii ratio=(d+r)/(d−r)=7+4√3.
Check: Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.
1. The centres of the three equal circles form an equilateral triangle. If each circle has radius r, the side of this triangle is 2r.
2. The distance from the centre of the equilateral triangle to any vertex is 2r/√3.
3. For the larger circle X, its radius is:
4. R = r + 2r/√3
5. For the smaller circle Y, its radius is:
6. S = 2r/√3 − r
7. Therefore,
8. R/S = (1 + 2/√3)/(2/√3 − 1)
9. Multiplying the numerator and denominator by (2/√3 + 1), we get:
10. R/S = (1 + 2/√3)² / (4/3 − 1)
11. Simplifying gives:
12. R/S = 7 + 4√3
13. Answer: (7 + 4√3) : 1.
Why this method works:
The common centre of the outer and inner circles is the centre of the equilateral triangle of the original centres. Its distance d to each original centre is 2r/√3. Internal tangency gives R−r=d; external tangency gives s+r=d.
Independent check and common trap:
Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.
G Strategy shortcut
Spot: Geometry – Circles.Apply: Centres form an equilateral triangle of side 2r. Centre-to-vertex distance d=2r/√3. Outer/inner radii ratio=(d+r)/(d−r)=7+4√3.
Check: Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.
G Strategy: Spot: Geometry – Circles. Apply: Centres form an equilateral triangle of side 2r. Centre-to-vertex distance d=2r/√3. Outer/inner radii ratio=(d+r)/(d−r)=7+4√3. Check: Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.