AMC 10 · 2015 · #25
Grade 11 geometry-2d
Pick an answer.
The radii themselves are hopeless to track — after six layers there are 65 of them and they are ratios of enormous squares. But the problem does not ask for radii; it asks for 1/√(r), and that is the hint. Tool #1 (Draw a Diagram) puts a radius straight down to the x-axis at each tangency point, turning "these two circles touch" into a right triangle and giving the exact horizontal gap between tangency points. Tool #3 (Eliminate Possibilities) is needed because two different circles are tangent to the x-axis and externally tangent to both parents; the layer count 2^k-1 and the figure pick out the one in the gap. Tool #4 (Introduce a Variable) is the crux: name j(C)=1/√(r(C)), and the tangency relation becomes plain addition, j_new=j_left+j_right. After that no geometry is left. Tool #15 (Organize Information in More Ways) re-reads one layer's total as "twice everything built so far, minus the two end values", and Tool #5 (Look for a Pattern) turns the resulting recursion into a closed form.
Two numbers describe each circle
Two numbers describe each circle.
Drop each radius to the axis and the two centres sit at the corners of a right triangle, so touching becomes one Pythagorean equation.
Dropping each radius to the axis puts the two centres at the corners of a right triangle, so touching becomes one equation.
▸ Why?
Two circles touching on the outside sit exactly the sum of their radii apart, centre to centre.
▸ Why?
That centre-to-centre length is the hypotenuse over the horizontal and vertical gaps.
Pin down which new circle is meant
Touching neighbours gives one equation.
Tangency alone allows two circles; only the one squeezed into the gap keeps the count at one new circle per gap.
9.A-CED.A.3Eliminate PossibilitiesDivide by 2√(r₁r₂z)
Dividing through simplifies it completely.
The square roots all collapse the moment you divide by the square root of all three radii multiplied together.
11.N-RN.A.2Introduce A VariableRename 1/√(r) as j
In the new variable the rule is just addition.
The problem asks for 1/(sqrt r) because that is the one quantity in the picture that simply adds.
9.F-BF.A.1Organize Information In More WaysA layer's total: double, minus the ends
A layer's total doubles, minus the two ends.
Each old circle helps build the circle on its left and the one on its right — except the two outermost, which only have a neighbour on one side.
9.F-IF.A.3Organize Information In More WaysSolve the recursion
That recursion solves in closed form.
Each round nearly triples the running total, and shifting by half of S₀ makes it exactly triple.
11.A-SSE.B.4Look For A PatternPut the numbers in
Substituting gives 143/14, choice (D).
70 · 73=5110=14 · 365, so the 365 built up by six layers cancels straight out.
9.A-SSE.A.2Introduce A VariableCheck the construction really behaves
Rebuilding the layers confirms it.
If the circles ever overlapped or escaped the original pair, the "double minus the ends" count would be wrong — so check that they do not.
9.A-CED.A.3Eliminate PossibilitiesCircles resting on a line add up in 1/√(r), so name that number and the picture turns into a list where every new entry is the sum of its two neighbours.
- Two numbers describe each circle
- Pin down which new circle is meant
- Divide by 2√(r₁r₂z)
- Rename 1/√(r) as j
- A layer's total: double, minus the ends
- Solve the recursion
- Put the numbers in
- Check the construction really behaves