Track / Stage 5 / 136 of 400 #736 of 1964

Problem 736

AIME late

Geometry Difficulty 5.4 Find the answer

# Problem 4. (3 points)

Inside a pentagon, 1000 points were marked and the pentagon was divided into triangles such that each of the marked points became a vertex of at least one of them. What is the smallest number of triangles that could have resulted?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

# Answer: 1003

## Solution:

The sum of the angles of a pentagon is $540^{\circ}$ . The sum of the angles at each internal point is either $180^{\circ}$ , if it lies on the side of one of the triangles and is the vertex of several others, or $360^{\circ}$ , if angles of triangles are located on all sides. That is, the sum of the angles at each internal point is at least $180^{\circ}$ .

Therefore, for a thousand points, the sum of the angles of all triangles is at least $540^{\circ} + 1000 \cdot 180^{\circ} = 1003 \cdot 180^{\circ}$ . For this, we need at least 1003 triangles, since the sum of the angles in each is $180^{\circ}$ .

An example where the value 1003 is achieved is constructed as follows: first, divide the pentagon into three triangles using diagonals emanating from one of the vertices. Then, mark 1000 points on one of the diagonals and connect them to the opposite vertex of the triangle for which this diagonal is a side. The triangle will be divided into 1001 parts, giving us another 1000 new triangles.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.