AMC 10 · 2020 · #25

Grade 11 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

pythagorean-identitygeometric-probabilityam-gm-inequalityoptimization convert-to-algebracaseworkextreme-principle ↑ Prerequisites: geometric-probabilitypythagorean-identity

📏 Long solution 💡 4 insights

Problem

Pick one number uniformly from zero up to some positive bound, and another uniformly from zero to one, independently. Consider the probability that the sum of the squares of the sines of pi times each number exceeds one. As the bound ranges over the interval from zero to one, find how large that probability can get.

Pick an answer.

(A)

$\frac{7}{12}$

(B)

$2 - \sqrt{2}$

(C)

$\frac{1+\sqrt{2}}{4}$

(D)

$\frac{\sqrt{5}-1}{2}$

(E)

$\frac{5}{8}$

How to solve

Strategy Draw a Diagram

Tool #13 (Convert to Algebra): the Pythagorean identity turns sin²(π x) + sin²(π y) ＞ 1 into sin(π y) ＞ lvert cos(π x) rvert, one sine against one cosine. Tool #9 (Easier Problem): on [0,1] both sides are values of a monotone piece of the cosine curve, so the trig inequality collapses to the linear condition lvert x - 1/2 rvert + lvert y - 1/2 rvert ＜ 1/2 — no trigonometry survives. Tool #1 (Diagram): that condition is a diamond inscribed in the unit square, so the probability is a ratio of two areas that can be read off a picture. Tool #7 (Subproblems): compute the diamond's area inside the strip 0 ≤ x ≤ a in two cases, a ≤ 1/2 and a ≥ 1/2. Tool #14 (Extreme Principle): the resulting P(a) = 2 - a - 1/2a is maximized by AM-GM, not by scanning. Tool #3 (Eliminate Possibilities): compare the two branches' best values to pick the global maximum.

1STEP 1

Rewrite with the identity

It becomes one inequality.

sin²(π y) ＞ cos²(π x) ⇔ sin(π y) ＞ lvert cos(π x) rvert

2STEP 2

Collapse into distances

The trig collapses into a sum of distances.

lvert x - 1/2 rvert + lvert y - 1/2 rvert ＜ 1/2

3STEP 3

Draw the diamond

A diamond appears inside the square.

Diamond vertices (1/2, 0), (0, 1/2), (1/2, 1), (1, 1/2); area = 1/2

4STEP 4

Read one vertical slice

Read the height of one slice.

g(x) = 2min(x, 1-x), g(1/2) = 1

5STEP 5

Area inside the strip

The bound gives two cases.

P(a) = (2a - a² - 1/2)/a = 2 - a - 1/2a (1/2 ≤ a ≤ 1)

6STEP 6

Minimize the loss

The arithmetic-geometric mean inequality finds it.

a + 1/2a ≥ √(2), equality at a = √(2)/2 → P_max = 2 - √(2)

7STEP 7

Compare the branches

The maximum is two minus root two.

max₀ ＜ a ≤ 1 P(a) = 2 - √(2) ≈ 0.5858 → (B)

Answer

2 - √(2)

Three checks. (1) Bracketing: the diamond fills exactly half the unit square, so P(1) = 1/2; the tent's peak height is 1, so no average can exceed 1. The answer 2 - √(2) ≈ 0.5858 sits between 1/2 and 1 and only modestly above 1/2, exactly the shape of a small gain from trimming the low-payoff right end of the tent. (2) Stationarity cross-check: an average over [0,a] is stationary precisely when the new value equals the running average, i.e. g(a) = P(a). That reads 2(1-a) = 2 - a - 1/2a, which simplifies to a² = 1/2 — the same a = √(2)/2, found by a completely different argument. (3) Decoy separation: the nearest wrong choice is 7/12 ≈ 0.58333, and the exhibited value P(√(2)/2) = 2 - √(2) ≈ 0.585786 is strictly larger, so (A) cannot be the maximum; (C) ≈ 0.6036, (D) ≈ 0.6180, and (E) = 0.625 all exceed the proved upper bound 2 - √(2) and are impossible. Choice (B) confirmed.

💡Key takeaway

The identity sin² + cos² = 1 turns the winning condition into the diamond lvert x - 1/2 rvert + lvert y - 1/2 rvert ＜ 1/2, so P(a) is just diamond-area-inside-the-strip over strip-area = 2 - a - 1/2a, and AM-GM says stop at a = √(2)/2 for a maximum of 2 - √(2).

Rewrite with the Pythagorean identity
Collapse the trig into distances
Draw the diamond in the square
Read one vertical slice
Area of the strip, two cases
Minimize the leftover by AM-GM
Compare the two branches