AMC 10 · 2007 · #17
Grade 11 algebraPick an answer.
The five numbers fall into place the moment one thing is known: whether b is above or below 1. Tool #13 (Convert to Algebra) gets that started — dividing by b² shows a and log₁₀b share a sign, so the sign of a and the side of 1 that b lies on are the same fact. The load-bearing move is Tool #4 (Introduce a Variable): setting t=log₁₀b, so b=10^t and b²=100^t, ends the fight between a power of b and a logarithm of b by writing both in terms of one exponent, leaving the single clean relation a=t/100^t. That relation makes the case b ＞ 1 the entire difficulty, and Tool #14 (Extreme Principle) is what kills it: among positive integers the smallest is a whole 1, so t ≥ 100^t — an inequality no real number survives. Tool #9 (Solve an Easier Related Problem) reduces that survival question to the trivial integer fact 100^k ≥ k+1. Tool #11 (Work Backwards) then builds explicit pairs (a,b) so the conclusion is about something real rather than an empty hypothesis, and Tool #15 (Organize Information in More Ways) plus Tool #3 (Eliminate Possibilities) assemble the final ordering and show that two of the five choices could never have been the median under any circumstances.
Divide by b² and read the signs
Rearranging ties the integer's sign to the logarithm's.
When the bottom of a fraction is positive, the top alone decides the sign.
9.A-SSE.A.2Convert To AlgebraName the exponent instead of the number
Naming the exponent instead of the number simplifies everything.
Naming the exponent turns a log-versus-power standoff into a single exponential.
11.F-LE.A.4Introduce A VariableA positive integer is at least a whole 1
A positive integer is at least one, which forces a strong inequality.
There is a gap between 0 and 1 with no integers in it, so "positive integer" means "at least one whole unit".
9.A-CED.A.1Extreme PrincipleBootstrap t past every integer
Repeating it pushes the exponent past every bound, so that branch is empty.
A number forced to be at least its own exponential has to outrun itself forever, and nothing finite can.
11.N-RN.A.1Solve An Easier Related ProblemSo 0 ＜ b ＜ 1, and the order is forced
The surviving branch forces the whole order.
A number below 1 has a reciprocal above 1, so b and 1/b always sit on opposite sides of 1.
A number below one has a reciprocal above one, so the two always sit on opposite sides of one.
▸ Why?
Taking a reciprocal undoes multiplying, so a number and its reciprocal multiply back to exactly one.
▸ Why?
If both were on the same side of one their product would miss one, so the order is forced.
Show the hypothesis is not empty
An explicit pair shows the hypothesis is not empty.
A claim about every solution is worth nothing until you can name one solution.
11.A-REI.D.11Work BackwardsRead off the third smallest
Reading off the middle gives b, choice (D).
With five distinct numbers the median is simply whoever stands third in line.
6.SP.B.5Eliminate PossibilitiesWrite b=10^t and the equation becomes a=t/100^t, a quantity that never reaches 1; since the smallest positive integer is 1, a must be negative, which puts b between 0 and 1 and lands it in the middle of the list.
- Divide by b² and read the signs
- Name the exponent instead of the number
- A positive integer is at least a whole 1
- Bootstrap t past every integer
- So 0 ＜ b ＜ 1, and the order is forced
- Show the hypothesis is not empty
- Read off the third smallest