AMC 10 · 2024 · #7

Grade 12 geometry-2d
vector-additionsymmetry-argumentequal-spacingmedian-of-trianglecoordinate-geometry symmetry-argumentpattern-recognitioneasier-related-problem ↑ Prerequisites: pythagorean-theoremcoordinate-geometryisosceles-right-triangle
📏 Medium solution 💡 3 insights
Problem
Triangle ABC has a right angle at B and legs BA = BC = √(2). The 2024 points P₁, P₂, …, P₂₀₂₄ lie on the hypotenuse AC and chop it into equal pieces, with AP₁ = P₁P₂ = … = P₂₀₂₃P₂₀₂₄ = P₂₀₂₄C. Add the 2024 vectors BP₁, BP₂, …, BP₂₀₂₄ into one vector, and find the length of that vector.

Pick an answer.

(A)
1011
(B)
1012
(C)
2023
(D)
2024
(E)
2025
How to solve
Strategy Organize Information in More Ways

Adding 2024 vectors one after another is hopeless, so the entire problem is about choosing a better order to add them in. Tool #15 (Organize Information in More Ways) is that move: instead of sweeping from P₁ to P₂₀₂₄, pair the outermost two points, then the next two inward, and so on. Each such pair straddles the midpoint M of AC, so each pair adds up to the very same vector 2BM, and 2024 different vectors collapse into many copies of one. Tool #1 (Draw a Diagram) supplies coordinates with B at the origin, which is what makes the symmetry checkable instead of merely believable. Tool #4 (Introduce a Variable) names the index i so the pairing rule can be stated once and reused. Tool #9 (Solve an Easier Related Problem) is the safety net: rerun the argument with 2 points instead of 2024 and confirm the pattern before trusting it at scale.

1STEP 1

Put the right angle at the origin

With B at the origin, A and C sit on the axes.

B = (0,0), A = (0, √(2)), C = (√(2), 0)
2STEP 2

The hypotenuse has length 2

Pythagoras gives AC equals 2.

AC² = (√(2))² + (√(2))² = 2 + 2 = 4 ⟹ AC = 2
3STEP 3

2024 points make 2025 gaps

There is one more gap than point.

P_i = A + i/2025 (C - A) = (i√(2)/2025, √(2) - i√(2)/2025), i = 1, 2, …, 2024
4STEP 4

Pair the ends inward

Every pair sums to twice the midpoint.

(BP_i + BP₂₀₂₅-i)/2 = BM ⟹ BP_i + BP₂₀₂₅-i = 2 BM
5STEP 5

1012 identical pairs

The total becomes 2024 times the median vector.

S = Σ_i=1²⁰²⁴ BP_i = 1012 · (2 BM) = 2024 BM
6STEP 6

The median to the hypotenuse is 1

That median is half the hypotenuse, namely 1.

|BM| = AC/2 = 2/2 = 1
7STEP 7

Take the length

2024 times 1 is 2024.

|S| = |2024 BM| = 2024 · |BM| = 2024 · 1 = 2024 ⟹ (D)
Answer
2024
Recompute with no pairing at all. With B at the origin, Σ_i=1²⁰²⁴ BP_i = Σ_i=1²⁰²⁴(A + i/2025(C - A)) = 2024 A + 1/2025·(2024 · 2025)/2(C - A) = 2024 A + 1012 (C - A) = 1012 (A + C). In coordinates that is 1012 (√(2), √(2)), of length 1012√(2 + 2) = 1012 · 2 = 2024. Same answer — and notice the 2025 cancelled, so the number of pieces never mattered, only the symmetry. A second check settles the size: the hypotenuse lies on the line x + y = √(2), whose distance from the origin is 1, so every single BP_i has a shadow of length exactly 1 along the direction of BM. Those shadows add to 2024 and the sideways parts cancel in symmetric pairs, so the length is exactly 2024 and not one of the neighbouring choices. The distractors are the off-by-one traps: 1012 is the number of pairs before doubling, 2025 is the number of pieces mistaken for the number of points, and 1011 and 2023 come from miscounting the pairing.
💡Key takeaway

Never add 2024 vectors one at a time — pair the first with the last, because every such pair lands on twice the midpoint vector, and here that midpoint vector has length exactly 1.

  • Put the right angle at the origin
  • The hypotenuse has length 2
  • 2024 points make 2025 gaps
  • Pair the ends inward
  • 1012 identical pairs
  • The median to the hypotenuse is 1
  • Take the length