AMC 10 · 2025 · #9

Grade 8 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Coordinate Graph Slope as Rate

polynomial-factoringpolynomial-rootssign-analysis convert-to-algebra ↑ Prerequisites: polynomial-factoringquadratic-equations

📏 Long solution 💡 3 insights

Problem

Given f(x) = 100x³ - 300x² + 200x, count how many real numbers a make the shifted graph y = f(x - a) pass through the point (1, 25).

Pick an answer.

(A)

(B)

(C)

(D)

(E)

$\text{more than } 4$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The count of valid a is really a count of intersections, so the load-bearing move is to picture the curve. First Tool #4 (Introduce a Variable): setting t = 1 - a turns "passes through (1,25)" into the clean question "how many real t satisfy f(t) = 25?", and because a = 1 - t is one-to-one the two counts are equal. Then Tool #1 (Draw a Diagram): factor f, mark its roots, and read its shape so the crossings with the horizontal line y = 25 can literally be seen. Finally Tool #14 (Extreme Principle): comparing the hump's peak height and the curve's end behavior against 25 pins down exactly how many crossings occur — the whole answer hinges on those boundary heights.

1STEP 1

Turn the point condition into an equation

Passing through (1, 25) just means the height at x = 1 equals 25, so the whole condition is f(1 - a) = 25.

f(1 - a) = 25

2STEP 2

Rename the input to count cleanly

Let t = 1 - a. The match a = 1 - t is one-to-one, so counting a becomes counting real roots of f(t) = 25.

t = 1 - a → #{a} = #{ t : f(t) = 25 }

3STEP 3

Factor to find the curve's roots

Pull out 100t and factor the quadratic: f(t) = 100t(t-1)(t-2), so the curve is zero at t = 0, 1, 2.

f(t) = 100t(t-1)(t-2)

4STEP 4

Read the sign pattern

With those roots and a positive lead, f is positive only on the hump (0,1) and the rise past t = 2 — the sole places height 25 can occur.

f ＜ 0 on t ＜ 0, f ＞ 0 on (0,1), f ＜ 0 on (1,2), f ＞ 0 on t ＞ 2

5STEP 5

Measure the hump against 25

At the midpoint, f(1/2) = 37.5 tops 25 while f(0) = f(1) = 0, so the hump goes up past 25 and back down: two crossings.

f (1/2) = 100·1/2·(-1/2)(-3/2) = 37.5 ＞ 25

6STEP 6

Count the final rise, then total

Past t = 2 the curve climbs from 0 without bound, crossing 25 once more: 2 + 1 = 3 values of a, choice (C).

2_(0,1) + 1_t ＞ 2 = 3 → (C)

Answer

A hard cap comes for free: f(t)=25 rearranges to the cubic 100t³ - 300t² + 200t - 25 = 0, and a degree-3 polynomial has at most 3 real roots. That alone kills choices (D) 4 and (E) more than 4. We then showed all three roots are real — two on the hump, one past t=2 — and the peak margin is comfortable: 37.5 sits well above 25, so the two hump crossings are genuine and distinct, not a single tangent point. Everything is consistent with exactly 3.

💡Key takeaway

A left-right shift never changes how many times a graph reaches a given height, so just count where the cubic y=f(t) crosses y=25 — and a cubic can cross a flat line at most three times.

Turn the point condition into an equation
Rename the input to count cleanly
Factor to find the curve's roots
Read the sign pattern
Measure the hump against 25
Count the final rise, then total