Problem 8 · AMC 8 Stretch
Core
Geometry & Measurement
Counting & Probability
pigeonholevisual-representation
Five points are placed inside an equilateral triangle with sides of length 1. Show that at least 2 of the points are less than \(\tfrac12\) apart.
Show answer
Answer: two points less than 1/2 apart
Show hints
Hint 1 of 4
Cut the big triangle into smaller equal triangles, like cutting the square into smaller squares.
Still stuck? Show hint 2 →
Hint 2 of 4
Connect the midpoints of the three sides. This makes 4 smaller equilateral triangles. What is the side length of each?
Still stuck? Show hint 3 →
Hint 3 of 4
Each small triangle has side \(\tfrac12\) — those are your 4 boxes. You have 5 points.
Show solution
Approach: Pigeonhole — 5 points into 4 side-1/2 triangles
- Connect the midpoints of the three sides of the big triangle. This splits it into 4 smaller equilateral triangles, each with side length \(\tfrac12\). These 4 small triangles are our boxes.
- Drop the 5 points into the 4 small triangles. Since \(5 > 4\), some small triangle holds at least 2 points.
- Inside any triangle, the farthest apart two points can be is the length of its longest side; here that is \(\tfrac12\).
- So the two points in the same small triangle are less than \(\tfrac12\) apart.
Mark:
· log in to save