AMC 10 · 2008 · #24
Grade 9 geometry-2dPick an answer.
The chain is infinite, so nothing can be computed one triangle at a time forever; the goal has to be a formula for A₀A_n. Tool #4 (Introduce a Variable) supplies the frame: write A_n=(x_n,0) and let s_n be the n-th side, and then the two exotic hypotheses collapse into arithmetic. "Equilateral with base on the x-axis" pins the apex over the midpoint of the base at height √(3)/2s_n (tool #1, Draw a Diagram), and "apex on y=√(x)" is just y²=x with y ≥ 0. Together they give one quadratic per step. The quadratic has two readings, though, because A_n could land on either side of A_n-1, and this is exactly where a fast solution cheats: it draws a picture marching rightward and never checks that leftward is impossible. Tool #3 (Eliminate Possibilities) closes that hole, and the reason it closes is the distinctness of the B_n, so the argument has to be made and not waved at. What remains is a recursion carrying a square root, which is unpleasant to iterate. Tool #15 (Organize Information in More Ways) is the pivot of the whole solution: make the square root itself the variable, t_n=√(1+12x_n), and the recursion becomes "add 2". That converts the problem into counting, gives an exact closed form with no pattern-guessing from small cases, and then tool #14 (Extreme Principle) handles the actual question, which is a threshold: the least n crossing a bound. One discipline is built in from the start. Solving the recursion only shows what the chain must look like if it exists; the last algebraic move must be to substitute the closed form back and confirm that this chain is genuinely a legal configuration, distinct points and all.
Put the chain in coordinates
Coordinates locate every base and apex.
An equilateral triangle with its base on a line is completely pinned by that base: the apex must stand over the midpoint at height √(3)/2 times the side.
An equilateral triangle with its base on a line is pinned by that base: the apex stands over the midpoint.
▸ Why?
The apex is equally far from both ends of the base, and those points make up the base's fold line.
▸ Why?
Equal sides face equal angles, so the triangle is a mirror image of itself about that line.
Turn "apex on the curve" into a quadratic
The apex on the curve becomes a quadratic in each step.
"Lies on y=√(x)" is nothing more than y²=x with y ≥ 0, and squaring trades a curve for an equation you can solve.
9.A-CED.A.2Introduce A VariableRule out stepping backwards
Only the forward root is legal, so the steps are determined.
Backing up does not build a new triangle, it redraws the one you just came from — and calling the B_n distinct is precisely what forbids that.
9.A-REI.B.4Eliminate PossibilitiesMake the square root the variable
Making the square root the variable turns it linear.
When the same square root keeps reappearing in a recursion, promote it to being the variable; here doing so turns the whole problem into counting by twos.
9.A-SSE.A.2Organize Information In More WaysRead off the formula and check it is real
That gives a closed form, and the apexes really do land on the curve.
A recursion tells you what the answer would have to be; plugging the formula back in is what tells you the picture actually exists.
9.F-IF.A.3Draw A DiagramFind the first n past 100
Solving for the first position past the bound gives 17, choice (C).
A strictly increasing quantity crosses a threshold exactly once, so two neighbouring values are a complete proof of "least".
9.A-CED.A.1Extreme PrincipleWhen a recursion keeps spitting out the same square root, make that square root your variable: setting t_n=√(1+12x_n) turns this whole chain of triangles into counting by twos.
- Put the chain in coordinates
- Turn "apex on the curve" into a quadratic
- Rule out stepping backwards
- Make the square root the variable
- Read off the formula and check it is real
- Find the first n past 100