AMC 10 · 2017 · #23
Grade 11 algebraPick an answer.
Tool #15 (Organize Information in More Ways): tracking f directly is hopeless, but the three given points are not random — they all sit on the parabola y=x², so I track the difference f(x)-x² instead, and that difference has three known roots. Tool #5 (Look for a Pattern): the values 4, 9, 16 are the squares of 2, 3, 4, and later the three chords all behave the same way with the labels rotated. Tool #4 (Introduce a Variable): the data cannot pin down f completely, so I name the one leftover degree of freedom a and let the sum-24 condition fix it. Tool #7 (Identify Subproblems): each of the three lines becomes its own small factoring problem before the results are added.
Notice the hidden parabola
Each y value is x squared.
Three scattered points become one clean sentence once you see they all obey the same rule y=x².
9.F-IF.A.2Look For A PatternWrite f as a square plus a cubic
So f is a square plus a cubic.
A cubic with three known roots is completely determined up to one scaling number, so the whole problem shrinks to finding a.
11.A-APR.B.2Introduce A VariableTurn meets again into an equation
Subtracting a line leaves three roots.
Where two graphs cross is exactly where their formulas are equal, so an intersection question is a root question.
11.A-REI.D.11Identify SubproblemsFactor out the known roots
Factoring out the two known roots shows the third immediately.
Pulling out the factors you already know collapses a cubic equation into a single linear one.
Pulling out the roots you already know collapses a cubic equation into a single linear one.
▸ Why?
A polynomial vanishing at a root carries that root's linear factor, so the known roots divide out exactly.
▸ Why?
The coefficients already record the sum and product of the roots, so the leftover root is read off directly.
Rotate the labels for AC and BC
All three lines give the same shape of answer.
Each chord uses up two of the three roots, so the leftover root is always the unused one pushed by the same amount -1/a.
11.A-APR.B.3Look For A PatternUse the sum to pin down a
The sum of 24 fixes the last coefficient.
The single unknown scaling number a is exactly what the single extra condition, the sum 24, is there to determine.
9.A-REI.B.3Introduce A VariableEvaluate f at zero
Plugging in zero gives twenty-four fifths.
Once the scaling number is known, the constant term is just the product of the roots times that number.
9.A-SSE.A.1Identify SubproblemsThe three given points all sit on y=x², so studying f(x)-x² instead of f hands you the factored cubic a(x-2)(x-3)(x-4), and the one number a is fixed by the sum 24 — giving f(0)=24/5, choice (D).
- Notice the hidden parabola
- Write f as a square plus a cubic
- Turn meets again into an equation
- Factor out the known roots
- Rotate the labels for AC and BC
- Use the sum to pin down a
- Evaluate f at zero