Problem 20 · 2018 Math Kangaroo
Hard
Geometry & Measurement
careful-counting
In a regular 2018-sided polygon the vertices are numbered 1 to 2018 in order. Two diagonals are drawn: one joins vertices 18 and 1018, the other joins vertices 1018 and 2000. How many vertices do the three resulting polygons have?
Show answer
Answer: A — 38, 983, 1001
Show hints
Hint 1 of 2
The two diagonals share vertex 1018, cutting the polygon into three pieces.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the vertices on each piece inclusively (shared endpoints belong to both adjoining pieces).
Show solution
Approach: count the vertices on each arc inclusively
- Diagonal 18–1018 bounds the arc 18, 19, …, 1018: that is \(1018-18+1 = 1001\) vertices.
- Diagonal 1018–2000 bounds the arc 1018, …, 2000: that is \(2000-1018+1 = 983\) vertices.
- The third piece runs 2000, …, 2018, 1, …, 18 together with vertex 1018: \(19 + 18 + 1 = 38\) vertices.
- So the three polygons have 38, 983, 1001 vertices.
Mark:
· log in to save