AMC 10 · 2002 · #18

Grade 9 probability
geometric-probabilityperpendicular-bisectorarea-trapezoidcoordinate-geometry convert-to-algebracomplementary-counting ↑ Prerequisites: geometric-probabilitycoordinate-geometry
📏 Long solution 💡 3 insights
Problem
A point P is picked uniformly at random from the rectangle whose corners are (0,0), (2,0), (2,1), (0,1). Find the probability that P is nearer to the origin than to the point (3,1).

Pick an answer.

(A)
$\frac 12$
(B)
$\frac 23$
(C)
$\frac 34$
(D)
$\frac 45$
(E)
1
How to solve
Strategy Draw a Diagram

Because the point is uniform, the probability is just an area ratio, so the real task is to draw the boundary between "nearer the origin" and "nearer (3,1)" and measure what it cuts off — that is Tool #1 (Draw a Diagram). To find that boundary exactly, Tool #4 (Introduce a Variable) names a generic point (x,y) and writes both distances, and Tool #13 (Convert to Algebra) turns the phrase "is closer to" into an inequality; comparing squared distances collapses that messy square-root comparison into a single straight-line inequality, so the boundary is a line and the two pieces are polygons whose areas are easy. Finally, the piece nearer (3,1) is a clean trapezoid while the piece nearer the origin is a five-sided blob, so Tool #16 (Change Focus / Count the Complement) says: measure the easy piece and subtract.

1STEP 1

Name the point and both distances

Distances are never negative, so compare their squares and skip the roots.

P nearer (0,0) ⇔ x² + y² < (x-3)² + (y-1)²
2STEP 2

Turn the comparison into a line

The squared terms cancel, reducing the condition to the line 3x + y < 5.

x² + y² < x² - 6x + 9 + y² - 2y + 1 ⟺ 6x + 2y < 10 ⟺ 3x + y < 5
3STEP 3

Draw the divider across the rectangle

That line crosses the rectangle from (5/3, 0) to (4/3, 1), cutting off a trapezoid.

3x + y = 5 → (5/3, 0) and (4/3, 1)
4STEP 4

Measure the easier piece

The far piece is a trapezoid of area 1/2, the easier of the two to measure.

A_far = (1/3 + 2/3)/2 · 1 = 1/2
5STEP 5

Convert areas into the probability

So the near piece is 3/2 out of area 2, giving probability 3/4, choice (C).

Pr = (2 - 1/2)/2 = 3/2/2 = 3/4 (C)
Answer
3/4
Spot-check the rule 3x + y < 5 at the four corners. At (0,0): 0 < 5, nearer the origin, which is obvious since P is the origin. At (2,1): 3(2) + 1 = 7 > 5, nearer (3,1), also obvious since (2,1) is only √(1² + 0²) = 1 away from (3,1) but √(5) from the origin. At (0,1) and (2,0): 1 < 5 and 6 > 5, matching the picture. The rectangle sits mostly on the origin's side, so the answer should exceed 1/2 but fall short of 1 — and 3/4 does, ruling out (A) and (E). A fine grid sweep over the rectangle counting how many sample points satisfy x² + y² < (x-3)² + (y-1)² returns about 0.75, confirming the exact value.
💡Key takeaway

The points tied between two fixed points always form a straight line, so "who is closer" questions turn into cutting a region with a line and comparing areas.

  • Name the point and both distances
  • Turn the comparison into a line
  • Draw the divider across the rectangle
  • Measure the easier piece
  • Convert areas into the probability