AMC 10 · 2018 · #21

Grade 8 arithmetic
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
coordinate-geometryquadratic-equationssystems-of-equationssign-analysis convert-to-algebracaseworkbound-inequality-then-enumerate ↑ Prerequisites: coordinate-geometryquadratic-equations
📏 Long solution 💡 3 insights
Problem
Two curves share the xy-plane: the circle x²+y²=a² and the parabola y=x²-a. Both are controlled by the same number a. The question asks which values of a make the curves cross at exactly 3 points.

Pick an answer.

(A)
$a=\frac14$
(B)
$\frac14 < a < \frac12$
(C)
$a>\frac14$
(D)
$a=\frac12$
(E)
$a>\frac12$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Counting where two curves meet is really counting the real solutions of their two equations together, so Tool #4 (Introduce a Variable) leads: the parabola already hands us x²=y+a, which we substitute into the circle to get a single equation in y. Tool #1 (Draw a Diagram) keeps the geometry in view -- a circle with a parabola dipping into it -- so the algebra means something. Tool #7 (Identify Subproblems) splits the job in two: first solve for the y-levels where they meet, then for each y ask how many x-values it gives. Tool #14 (Extreme Principle) finds the exact boundary where one y-level switches from giving 0 points to giving 2 points. Tool #3 (Eliminate Possibilities) turns that boundary into the inequality that matches one answer choice.

1STEP 1

Substitute the parabola into the circle

Rewrite the parabola as x²=y+a and put it into the circle x²+y²=a², clearing x to leave one equation in y: y²+y+a-a²=0.

x²=y+a → y²+y+a-a²=0
2STEP 2

Solve for the two y-levels

The discriminant is the perfect square (2a-1)², so the quadratic gives two clean heights: y=-a or y=a-1.

y=(-1±√((2a-1)²))/2 → y=-a or y=a-1
3STEP 3

The level y=-a always gives exactly one point

At y=-a, x²=y+a becomes x²=0, forcing the single vertex point (0,-a), which lies on the circle for every a.

y=-a: x²=0 → (0,-a) (always 1 point)
4STEP 4

The level y=a-1 gives 0, 1, or 2 points

At y=a-1, x²=y+a becomes x²=2a-1, so this level adds two mirror points exactly when 2a-1 ＞ 0, and none otherwise.

y=a-1: x²=2a-1 → 2 points&2a-1 ＞ 0 ; 1 point&2a-1=0 ; 0 points&2a-1 ＜ 0
5STEP 5

Add the counts and read off a

The guaranteed vertex gives 1 and the mirror pair adds 2 only when a ＞ 1/2, making exactly 3 points — answer (E).

total=1+ 2 ; 0 =3⇔ 2a-1 ＞ 0⇔ a ＞ 1/2 → (E)
Answer
a ＞ 1/2
Test a value past the boundary: at a=1 the levels are y=-a=-1 giving (0,-1), and y=a-1=0 giving x²=2a-1=1, so x=±1 and points (±1,0) -- three points in all, matching a ＞ 1/2. Test the boundary a=1/2: both levels land at y=-1/2 with x=0, a single tangent point, so only 1 -- not 3, which correctly rules out choice (D). A value below, a=1/4, gives x²=2a-1=-1/2 ＜ 0 for the upper level, so again only the bottom point, ruling out (A), (B), and (C). Only a ＞ 1/2 survives, confirming (E).
💡Key takeaway

The parabola's bottom always touches the circle for one point; its two arms only break back out through the circle once a passes 1/2, adding two more for exactly 3 -- choice (E).

  • Substitute the parabola into the circle
  • Solve for the two y-levels
  • The level y=-a always gives exactly one point
  • The level y=a-1 gives 0, 1, or 2 points
  • Add the counts and read off a

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