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2003 AMC 10A Problem 12

2003 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Geometric Probability · difficulty 2 of 5 · about 2 min · position P11-15

A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$ , $(4,0)$ , $(4,1)$ , and $(0,1)$ . What is the probability that $x<y$ ?

A)
$\frac{1}{8}$
B)
$\frac{1}{4}$
C)
$\frac{3}{8}$
D)
$\frac{1}{2}$
E)
$\frac{3}{4}$

Answer

Key insight

Inside the 4-by-1 rectangle, x < y only in the small triangle above the line y = x, which has area 1/2 out of 4.

Solution

The rectangle has area $4 \cdot 1 = 4$ . The condition $x < y$ describes the points above the line $y = x$ .

Inside the rectangle, $y$ never exceeds $1$ , so $x < y$ forces $x < 1$ as well. The line $y = x$ enters at $(0,0)$ and leaves the rectangle at $(1,1)$ ; the region above it is the right triangle with vertices $(0,0)$ , $(1,1)$ , $(0,1)$ , with legs of length $1$ and area $\tfrac{1}{2}$ .

The probability is $\dfrac{1/2}{4} = \dfrac{1}{8}$ .

The answer is $\boxed{\textbf{(A)}\ \frac{1}{8}}$ .

Why this works

A uniformly random point turns probability into area: favorable area over total area. Sketching the boundary line and noting where it exits the region is the whole problem; most of the wide rectangle lies to the right of $x = 1$ , where $x < y$ is impossible.

The trap

Comparing the triangle to the unit square it sits in (area 1) and answering 1/2, instead of to the whole rectangle of area 4.

Common mistakes

Comparing the triangle to the unit square it sits in (area 1) and answering 1/2, instead of to the whole rectangle of area 4.
Assuming the line $y = x$ splits the rectangle into two equal halves, which is only true for a square.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)