AMC 10 · 2009 · #23
Grade 9 algebraPick an answer.
The rule g(x) = -f(100 - x) flips the sign of the output and reflects the input across 50. Both flips at once is a half-turn of the whole picture about the point (50, 0), so g's parabola is f's parabola rotated 180 degrees. That single observation is worth more than any expansion: it says the four x-intercepts sit in mirror pairs around 50. Sliding the origin to 50 turns the four unknown intercepts into two, the given gap hands over one of them for free, and the vertex condition becomes one clean equation in the two zeros of a single quadratic. Naming those zeros and comparing them by ratio, not by size, is what makes the unknown leading coefficient drop out.
Read the rule as a half-turn
The rule is a half-turn about a fixed point.
Negating the output and mirroring the input across 50 is one motion, not two: a half-turn of the page around the point (50, 0).
9.F-IF.A.2Visualize Spatial RelationshipsSlide the origin to 50
Sliding the origin there simplifies everything.
Put the origin where the symmetry already is, and the messy 100 disappears from every formula.
9.F-IF.A.2Organize Information In More WaysZeros come in opposite pairs
The roots come in opposite pairs.
A half-turn about the origin sends each intercept to its own negative, so the whole set of intercepts folds onto itself.
A half turn about the centre sends each intercept to its own negative, so the whole set folds onto itself.
▸ Why?
A half turn moves the picture without stretching it, so an intercept must land on another intercept.
▸ Why?
Each intercept is sent to exactly one other, so the four of them pair up with none left over.
Four distinct intercepts pin the labels
Four distinct roots then force the labelling.
Wanting four different intercepts forbids the two collapsing cases, and what survives is one small pair and one large pair, each split between the two parabolas.
9.F-IF.B.4Eliminate PossibilitiesTurn the vertex clue into one equation
The vertex clue becomes one symmetric equation.
Because the second parabola is just the first one turned upside down and backwards, the vertex condition only ever compares p with itself at V and at -V.
9.A-CED.A.1Convert To AlgebraExpand into a symmetric relation
Expanding turns it into a relation between the two roots.
Both sides carry the same factor a, so the unknown steepness cancels and only the shape of the zero pair is left.
9.A-SSE.A.2Introduce A VariableSolve for the ratio of the zeros
Solving gives their ratio.
The condition never fixes how big the parabola is, only the ratio of its two zeros, so one number does all the work.
9.A-REI.B.4Introduce A VariableCheck such a pair really exists
An explicit pair shows the setup really exists.
A rule that says what a number must be is only half a proof until one honest example shows the number can actually happen.
9.A-SSE.B.3Guess And CheckAssemble m, n, and p
Assembling the three numbers gives 752, choice (D).
Once the ratio of the outer pair to the inner pair is known, the given gap of 150 scales straight up to the answer.
9.A-SSE.A.2Introduce A VariableWhen one graph is another one turned upside down, move the origin to the turning point: the four intercepts become two mirror pairs, and the only thing left to find is how much bigger the outer pair is.
- Read the rule as a half-turn
- Slide the origin to 50
- Zeros come in opposite pairs
- Four distinct intercepts pin the labels
- Turn the vertex clue into one equation
- Expand into a symmetric relation
- Solve for the ratio of the zeros
- Check such a pair really exists
- Assemble m, n, and p