AMC 10 · 2010 · #24
Grade 11 algebraPick an answer.
The endpoints of these intervals are roots of a cubic that does not factor, so anyone trying to find them one at a time is stuck. Tool #16 (Change Focus) is the whole idea: the question asks only for a total length, and the total length turns out to be the sum of those roots — and a sum of roots is exactly what Vieta's formulas hand over without ever solving anything. Tool #4 (Introduce a Variable) sets t=x-2010 so the poles become -1,0,1; sliding the picture sideways cannot change any length, and symmetric poles keep the algebra clean. Tool #1 (Draw a Diagram) supplies the shape of f: the poles cut the line into four pieces and on each piece f falls steadily, so each piece can meet the level y=1 at most once. Tool #7 (Identify Subproblems) then treats the four pieces separately, because they do not all behave the same way. Tool #14 (Extreme Principle) handles the piece the eye tends to skip — the far-left one, where every term is negative — and that omission is not cosmetic: the argument only works because the count of contributing pieces is exactly the degree of the cubic. Two things have to be nailed down rather than assumed. First, that each contributing piece supplies exactly one interval, closed at its right end, which is what makes the answer a clean sum. Second, that the three endpoints found are all the roots of the cubic — if even one root were complex, Vieta's sum would include it and would not equal the total length.
Slide the poles to -1,0,1
Shifting puts the poles at three symmetric spots.
Shifting a graph left or right moves the intervals but never stretches them, so the total length is untouched.
9.A-SSE.A.2Introduce A VariableFour pieces, falling on each
The function falls across every piece.
Three falling curves added together still fall, so within one piece the graph can only pass the height 1 going down, and only once.
11.F-IF.C.7Draw A DiagramThe far-left piece gives nothing
The far-left piece stays negative throughout.
Left of every pole all three fractions are negative, so their sum cannot possibly reach 1.
6.NS.C.7Extreme PrincipleEach other piece gives one interval
Each remaining piece yields exactly one interval.
A curve falling without a break from above every height to below every height meets the line y=1 once and only once.
A curve falling without a break from above every height to below it meets the line once and only once.
▸ Why?
A value that only falls can never come back to a height it has already passed.
▸ Why?
So each piece of the graph is matched with exactly one crossing, with none missed and none doubled.
Total length is p+q+r
The pole offsets cancel in the total length.
The three left endpoints add to zero, so all that survives is the sum of the three right endpoints.
7.NS.A.1Change Focus Count The ComplementTurn g(t)=1 into a cubic
The endpoints solve one cubic.
Clearing denominators is safe at a single non-pole point, and it converts a fraction equation into an ordinary cubic.
11.A-APR.D.6Introduce A VariableThese three are all the roots
Those three are all of its roots.
A cubic can hold only three roots, so once three different ones are located the list is closed.
11.A-APR.B.2Identify SubproblemsVieta finishes it
The sum of roots is 3, choice (A).
The t² coefficient of a monic cubic is minus the sum of its roots, which is the one number the problem actually wants.
11.A-APR.B.3Change Focus Count The ComplementThe interval endpoints are ugly roots of a cubic, but their left ends cancel out, so the total length is just the sum of the roots — and a monic cubic tells you that sum straight from its t² coefficient.
- Slide the poles to -1,0,1
- Four pieces, falling on each
- The far-left piece gives nothing
- Each other piece gives one interval
- Total length is p+q+r
- Turn g(t)=1 into a cubic
- These three are all the roots
- Vieta finishes it