AMC 10 · 2009 · #23

Grade 11 probability
Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region
complex-numbersgeometric-probabilityarea-trianglescomplementary-counting convert-to-algebraidentify-subproblemssymmetry-argument ↑ Prerequisites: complex-numbersgeometric-probability
📏 Long solution 💡 3 insights
Problem
A point is picked uniformly from a square in the complex plane and multiplied by a fixed complex number. Find the probability the product lands back in the square.

Pick an answer.

(A)
$\frac12$
(B)
$\frac23$
(C)
$\frac34$
(D)
$\frac79$
(E)
$\frac78$
How to solve
Strategy Convert to Algebra

The condition "the product is in S" is a statement about a complex number, so the first move is to write out the product in a + bi form and read off what it demands of x and y. That turns one vague requirement into two clean inequalities, and because S tests the real and imaginary parts separately, the translation runs both ways: the inequalities are not just necessary but sufficient. From there it is a pure area question, and the failing set is easier to measure than the winning set, so I count the complement. The one place this argument can quietly break is the last subtraction: removing four corner pieces is only legal if those pieces do not overlap. I check that explicitly by pushing x and y to their extreme allowed values, which pins each bad piece inside its own quadrant.

1STEP 1

Multiply out and split the condition

Multiplying out splits the condition into two inequalities.

(3/4 + 3/4i)(x + iy) = 3/4(x-y) + 3/4(x+y)i in S ⇔ |x-y| ≤ 4/3 and |x+y| ≤ 4/3
2STEP 2

Turn it into a picture of areas

Together they cut the square with a diamond.

P = frac{area(S ∩ {|x| + |y| ≤ 4/3})}{4}
3STEP 3

The failures are four corner triangles

The failures are four corner triangles.

each corner triangle = 1/2 · 2/3 · 2/3 = 2/9
4STEP 4

Check the four bad pieces never overlap

They never overlap, so their areas add.

x + y ＞ 4/3, x ≤ 1, y ≤ 1 → x ＞ 1/3 and y ＞ 1/3
5STEP 5

Subtract, then divide

Subtracting and dividing gives 7/9, choice (C).

(4 - 4 · 2/9)/4 = 28/9/4 = 28/36 = 7/9
Answer
7/9
The size is believable. The multiplier has modulus |3/4 + 3/4i| = 3√(2)/4 ≈ 1.06, so multiplication rotates by 45° and stretches lengths by only about six percent. Most of the square should survive, so the answer must be a little under 1; 7/9 ≈ 0.78 fits, while 1/2 would be far too harsh. A second check bounds the answer from the other side: the tilted square |x| + |y| ≤ 4/3 has area 2(4/3)² = 32/9 ≈ 3.56, so the overlap must be less than both 32/9 and 4, and 28/9 ≈ 3.11 is. A third check recomputes the octagon straight from its eight vertices (± 1, ±1/3) and (±1/3, ± 1) by the shoelace formula, which returns 28/9 again. Finally, the disjointness in Step 4 is not a formality: it fails once the threshold drops below 1, and then the answer would not be 4 minus four separate triangles.
💡Key takeaway

Multiplying by a complex number just spins and stretches the whole plane, so "stays inside the square" becomes two straight-line conditions, and after that it is only an area you can trim at the corners.

  • Multiply out and split the condition
  • Turn it into a picture of areas
  • The failures are four corner triangles
  • Check the four bad pieces never overlap
  • Subtract, then divide

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