AMC 10 · 2021 · #20

Grade 10 geometry-2d
parabola-focus-directrixcoordinate-geometryquadratic-equationsvieta-formulas convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometry
📏 Long solution 💡 3 insights
Problem
A parabola has a vertex and a focus. A point on that parabola sits 20 from the focus and 21 from the vertex. More than one parabola can fit this description, so each fitting parabola has its own focal distance. Find the sum of all the focal distances that are possible.

Pick an answer.

(A)
13
(B)
$\frac{40}3$
(C)
$\frac{41}3$
(D)
14
(E)
$\frac{43}3$
How to solve
Strategy Introduce a Variable

Nothing here is numerically pinned down, so Tool #4 (Introduce a Variable) leads: name the wanted length d = FV and note that a parabola with a fixed vertex and axis is completely determined by that one number. Tool #1 (Draw a Diagram) puts V at the origin, F at (0,d), and the directrix at y=-d, which makes every distance in the problem readable from coordinates. Tool #13 (Convert to Algebra) then turns the focus-directrix definition into the equation x²=4dy and turns the two given lengths into two equations. Tool #16 (Change Focus) is the move that keeps the algebra clean: instead of measuring AF to the focus with a square root, measure the equal distance to the directrix, which is a plain subtraction. Two conditions on one point leave a single quadratic in d. Tool #3 (Eliminate Possibilities) tests whether both roots are real parabolas rather than algebraic ghosts, and Tool #9 (Solve an Easier Related Problem) finishes: the question wants the sum of the roots, which the coefficients already carry, so the ugly roots never have to be computed.

1STEP 1

Name the focal distance

Put the vertex at the origin.

V=(0,0), F=(0,d), directrix: y=-d, d > 0
2STEP 2

Get the equation from the definition

The definition gives the equation.

√(x²+(y-d)²) = y+d → x²+y²-2dy+d² = y²+2dy+d² → x² = 4dy
3STEP 3

Measure to the directrix

Measure to the directrix instead.

AF = dist(A, directrix) = y_A + d = 20 → y_A = 20 - d
4STEP 4

Write the other distance

Write the second distance with the formula.

AV² = x_A² + y_A² = 441, x_A² = 4dy_A → 4dy_A + y_A² = 441
5STEP 5

Collapse to a quadratic

It becomes a quadratic in the focal distance.

4d(20-d) + (20-d)² = 441 → 80d-4d²+400-40d+d² = 441 → -3d²+40d-41 = 0 → 3d²-40d+41 = 0
6STEP 6

Check both roots count

Both roots land in the valid range.

Δ = 1600-492 = 1108 > 0; d₁d₂ = 41/3 > 0, d₁+d₂ > 0 → 0 < d₁, d₂ < 20
7STEP 7

Add without solving

The root relations give forty thirds.

d₁ + d₂ = -b/a = --40/3 = 40/3 → (B)
Answer
40/3
Solving the quadratic outright gives d = (40 ± √(1108))/6, roughly 1.119 and 12.214, which do add to about 13.333 = 40/3. Both check numerically: for d ≈ 1.119 the point is y_A ≈ 18.881 and x_A² ≈ 84.50, giving AF = √(84.50 + (18.881-1.119)²) = 20 and AV = √(84.50+18.881²) = 21; for d ≈ 12.214 the point is y_A ≈ 7.786 and x_A² ≈ 380.39, giving AF = 20 and AV = 21 again. The two roots also match the two pictures one expects: a narrow parabola with A high up near the axis, and a wide parabola with A flung far out to the side. Finally, the value 40/3 ≈ 13.33 lies strictly between choices (A) 13 and (D) 14, so anyone who found only one configuration and reported a single root would not land on any listed choice — a sign the problem is deliberately testing that both cases are counted.
💡Key takeaway

When a question asks for the sum of all possible values, aim for one equation in that one unknown — the coefficients of a quadratic hand you the sum of its roots without you ever solving for them.

  • Name the focal distance
  • Write the parabola from its definition
  • Measure to the directrix, not the focus
  • Write AV with the distance formula
  • Collapse it to a quadratic in d
  • Check that both roots count
  • Add the roots without finding them