AMC 10 · 2011 · #14

Grade 11 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
parabola-focus-directrixlaw-of-cosinescoordinate-geometry convert-to-algebra ↑ Prerequisites: parabola-focus-directrix
📏 Medium solution 💡 3 insights
Problem
A line through the focus, square to the axis, cuts a parabola at two points. Find the cosine of the angle at the vertex.

Pick an answer.

(A)
$-\frac{3\sqrt{5}}{7}$
(B)
$-\frac{2\sqrt{5}}{5}$
(C)
$-\frac{4}{5}$
(D)
$-\frac{3}{5}$
(E)
$-\frac{1}{2}$
How to solve
Strategy Introduce a Variable

The problem names no specific parabola, so the temptation is to declare "assume it is y = x²" and compute. That assumes the thing the problem is really asking about: that every parabola gives the same angle. Tool #4 (Introduce a Variable) avoids the assumption entirely. Call the one length in the picture d = VF and refuse to fix a number for it. Tool #1 (Draw a Diagram) adds the directrix, which is what turns the focus-directrix definition into usable distances. Tool #7 (Identify Subproblems) splits the work into three small questions: how far is F from the directrix, how long is AF, and how long is AV. Every length then comes out as a multiple of d, so when the Law of Cosines produces a ratio the d cancels and the answer is proved for all parabolas at once, no "assume" needed. Tool #15 (Organize Information in More Ways) re-encodes the same picture in coordinates and vectors at the end, as an independent check by different machinery.

1STEP 1

Draw the directrix and name one length

The directrix and one length set up the picture.

d = VF = dist(V,l) → dist(F,l) = 2d
2STEP 2

Pin down A and B exactly

The defining property pins both crossing points.

AF = BF = dist(A,l) = 2d, AB = 4d
3STEP 3

Get the two legs VA and VB

The two legs from the vertex come out equal.

AV = BV = √((2d)² + d²) = d√(5)
4STEP 4

Law of Cosines, and watch d vanish

The law of cosines makes the named length vanish.

16d² = 10d² - 10d²cos(∠ AVB) → cos(∠ AVB) = -3/5
5STEP 5

Independent check with vectors

Vectors confirm -3/5, choice (B).

cos(∠ AVB) = (VA · VB)/|VA||VB| = -3p²/5p² = -3/5
Answer
-3/5
The sign is right: from the vertex, the half-angle ∠ AVF has tan(∠ AVF) = AF/FV = 2d/d = 2 ＞ 1, so ∠ AVF ＞ 45° and the full angle ∠ AVB exceeds 90°. An obtuse angle must have negative cosine, which is why every choice is negative. Numerically arccos(-3/5) ≈ 126.87°, and 2arctan(2) ≈ 126.87° — a third route, the double-angle formula cos(2θ) = 2cos²θ - 1 = 2 · 1/5 - 1 = -3/5, lands in the same place. The distractors are readable as specific slips: -2√(5)/5 = -2/√(5) is a half-angle ratio from triangle AFV reported as if it were the full angle's cosine, and -4/5 is what cos(2θ) = (1-tan²θ)/(1+tan²θ) returns if you mistakenly get AF = 3d instead of 2d — that is, if you forget that the vertex is halfway from focus to directrix and put the focus only three vertex-lengths out. Choice -1/2 is the eyeball guess of 120°, close but not the true 126.87°.
💡Key takeaway

Name the one length you have instead of picking a number for it; if it cancels at the end, you have proved the answer for every parabola, not just the one you guessed.

  • Draw the directrix and name one length
  • Pin down A and B exactly
  • Get the two legs VA and VB
  • Law of Cosines, and watch d vanish
  • Independent check with vectors

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