AMC 10 · 2018 · #22

Grade 8 geometry-2d

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

geometric-probabilityarea-circlespolygon-inequalitylaw-of-cosines convert-to-algebracomplementary-countingcasework ↑ Prerequisites: geometric-probability

📏 Long solution 💡 3 insights

Problem

Pick two numbers x and y at random, each somewhere between 0 and 1. Treat x, y, and 1 as the three side lengths of a triangle. What is the chance that this triangle is obtuse (has one angle bigger than 90°)?

Pick an answer.

(A)

$text{ 0.21}$

(B)

$text{ 0.25}$

(C)

$text{ 0.29}$

(D)

$text{ 0.50}$

(E)

$text{ 0.79}$

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

How to solve

Strategy Draw a Diagram

Tool #13 (Convert to Algebra): the words "triangle" and "obtuse" hide two clean inequalities in x and y — the triangle inequality and the Pythagorean inequality — so the first job is to turn the conditions into algebra. Tool #1 (Draw a Diagram): once x and y are two coordinates, every outcome is a point in the unit square, and the favorable outcomes fill a region; the probability is just that region's area divided by the square's area 1. Tool #16 (Change Focus / Count the Complement): the favorable region is awkward to integrate directly, but it equals a quarter circle with one triangle sliced off, so we compute it by subtraction instead of head-on.

1STEP 1

When do the lengths form a triangle?

Since x ≤ 1 and y ≤ 1, side 1 is the longest, so the only triangle condition that can fail is x + y ＞ 1.

x + y ＞ 1

2STEP 2

When is the triangle obtuse?

The largest angle faces side 1, and by the Pythagorean inequality it is obtuse exactly when x² + y² ＜ 1.

1² ＞ x² + y² ⟹ x² + y² ＜ 1

3STEP 3

Draw the sample space and the winning region

Each outcome is a point in the unit square, whose area is 1, so the probability is the winning area above x + y = 1 and inside x² + y² ＜ 1.

{(x,y) : x + y ＞ 1 and x² + y² ＜ 1} ⊂ [0,1]²

4STEP 4

Area of the quarter circle

Inside the square, everything with x² + y² ＜ 1 is a quarter of a radius-1 circle, so its area is π/4.

quarter-circle area = 1/4π (1)² = π/4

5STEP 5

Slice off the triangle below the line

The quarter circle still holds a right triangle of area 1/2 below the chord; removing it leaves π/4 - 1/2.

favorable area = π/4 - 1/2

6STEP 6

Evaluate and pick the closest choice

Since the square's area is 1, the favorable area is the probability itself: π/4 - 1/2 ≈ 0.2854, which rounds to about 0.29 — choice (C).

π/4 - 1/2 ≈ 0.2854 ≈ 0.29 → (C)

Answer

text{ 0.29}

The result 0.2854 sits safely between 0 and 1, as any probability must. It is reasonable that being obtuse is fairly uncommon: we already need x + y ＞ 1 just to form a triangle (a 1/2 chance), and on top of that the corner facing side 1 must be pried open, which trims the region further. The trap answers each match a wrong stopping point: 0.50 is the bare triangle condition, 0.79 ≈ π/4 forgets to remove the triangle, and 0.21 = π/4 - 0.58-style miscounts come from subtracting the wrong piece. Only π/4 - 1/2 ≈ 0.29 survives, matching (C).

💡Key takeaway

When two random numbers set up a shape, draw them as a point in a square and turn the chance into the area of the region that wins.

When do the lengths form a triangle?
When is the triangle obtuse?
Draw the sample space and the winning region
Area of the quarter circle
Slice off the triangle below the line
Evaluate and pick the closest choice