AMC 10 · 2011 · #21
Grade 9 algebrapatternPick an answer.
Tool #11 (Work Backwards): the function is a nest of square roots, and the only fixed fact is the requirement at the innermost layer of the definition, f₁. Writing D_n for the domain of f_n, the recursion lets each domain be pulled back from the one before it, so the whole problem becomes a chain D₁→ D₂→ D₃→… instead of one monstrous inequality. Tool #4 (Introduce a Variable): naming t=√(n²-x) converts a test on x into a construction from t, and the substitution x=n²-t² makes the pullback an exact equality of sets rather than a list of necessary conditions. Tool #9 (Solve an Easier Related Problem): running n=2 and n=3 by hand shows the shape of the recursion before the numbers turn ugly. Tool #5 (Look for a Pattern): the same one-line rule generates every domain, including the induction that shows an empty domain stays empty forever. Tool #14 (Extreme Principle): the problem ends at a boundary case — a domain that touches 0 from below has exactly one usable value, and that single value pins everything afterwards. Tool #6 (Guess and Check): set algebra alone does not prove a number exists, so the surviving candidate is substituted back into the full nested radical.
Write the domain rule exactly
One rule describes every allowed input.
A composition is defined exactly where the inner step lands inside the outer step's domain.
9.F-IF.A.2Work BackwardsOnly the nonnegative part is reachable
Only the nonnegative part is ever reachable.
A square root can only hand the previous function a number that is zero or bigger, so anything negative in the old domain is unusable.
A square root can only hand back a number that is zero or bigger, so anything negative in the old domain is unusable.
▸ Why?
Nothing squared is negative, so the output of a root is bounded below at zero.
▸ Why?
Only the part of the earlier domain that lies at or above zero can ever be reached, so the usable domain shrinks.
Start the chain at f₁
The chain starts as a whole interval.
The outermost square root is the only condition at the start, and it simply caps the input at 1.
8.EE.A.2Solve An Easier Related ProblemRun two steps of the recursion
Two turns of the recursion shift it left.
Each stage subtracts a square that is growing faster than the offset n², so the domains march downward.
9.A-REI.B.3Look For A PatternThe collapse: R₃ is a single point
Then the reachable set collapses to a single point.
An interval that only just touches zero from below leaves exactly one usable value, and that boundary case locks the rest of the chain.
8.EE.A.2Extreme PrinciplePush the single point forward
Pushing it forward gives one last input.
Once the chain is forced to carry the value 16, the next stage must subtract 16²=256 from only 25, and the result falls far below zero.
9.A-CED.A.3Look For A PatternEmpty stays empty
After that everything stays empty.
Nothing can come out of a machine you feed nothing into, so once the chain dies it stays dead.
9.F-IF.A.3Look For A PatternCheck −231 really works, then add
Adding gives -226, choice (A).
A domain argument is only worth as much as one explicit input that survives every square root in the nest.
9.F-IF.A.2Guess And CheckA square root can only ever hand back a number that is zero or bigger, so any negative part of the previous domain is useless — and the moment the usable leftover shrinks to the single value 0, the whole chain is locked onto one number until it runs out.
- Write the domain rule exactly
- Only the nonnegative part is reachable
- Start the chain at f₁
- Run two steps of the recursion
- The collapse: R₃ is a single point
- Push the single point forward
- Empty stays empty
- Check −231 really works, then add