AMC 10 · 2005 · #8

Grade 9 algebra
quadratic-equationscoordinate-geometryzero-product-property convert-to-algebraextreme-principle ↑ Prerequisites: quadratic-equationscoordinate-geometry
📏 Medium solution 💡 2 insights
Problem
One number controls both a parabola and a line at the same time. Count the real values for which the line passes through the parabola's vertex.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
10
(E)
infinitely many
How to solve
Strategy Convert to Algebra

'Passes through' is a geometric phrase with an exact algebraic translation: the point's coordinates satisfy the equation. So the work splits in two. First find the vertex, which is a minimum question and yields to the extreme principle. Then turn 'the line contains that point' into a single equation in a and count its roots.

1STEP 1

Locate the vertex

The squared term is smallest at zero, so the vertex is on the vertical axis.

y = x² + a² ≥ a², equality at x = 0 ⟹ vertex = (0, a²)
2STEP 2

Translate 'passes through'

Passing through it becomes a single equation in the unknown.

a² = 0 + a ⟺ a² = a
3STEP 3

Solve the equation in a

Factoring, not dividing, keeps both roots: 0 and 1.

a² - a = 0 ⟹ a(a-1) = 0 ⟹ a = 0 or a = 1
4STEP 4

Check both candidates

Both check out, so the count is 2, choice (C).

a = 0: (0,0) ∈ {y = x}; a = 1: (0,1) ∈ {y = x+1}
Answer
2
Test the extreme choices directly. Choice (E) would require a² = a for every a, but a = 2 gives 4 ≠ 2, so it fails. Choice (A) would require no value to work, but a = 0 demonstrably does. Choice (D) would need ten roots from a degree-two equation, which is impossible. A quadratic equation has at most two real roots and here both are real and distinct, so 2 is the largest count available and it is achieved.
💡Key takeaway

'Passes through' always means the same thing: find the point, put its coordinates into the equation, and see what the equation demands.

  • Locate the vertex
  • Translate 'passes through'
  • Solve the equation in a
  • Check both candidates