Welcome back! Today, we finally answer Tuesday’s Brain Twister challenge:

## Question

In regular Polygon Q the ratio of the length of the longest line that can be drawn between two vertices to the length of any one side is greater than 2.

A

The number of sides of Polygon Q

B

6

## Answer and Explanation

A good way to approach this problem is to assume that polygon Q is a square. That way, you can figure out the ratio of the vertices and any side. If we assume the length of a square is 1, then the longest possible line between two vertices and that side would be would be √2:1, or about 1.4.

If we assume Polygon Q is a regular hexagon, then, using the properties of an equilateral triangle (a hexagon is broken up into six equilateral triangles), we get a length of 2 for the vertex and length of 1 for the side. Therefore, Polygon Q has to have more than six sides. But be careful not to choose 6, since the question says “greater than 2”. And as we can infer, from working with the square and hexagon, that this ratio increases the greater the number of sides we have.

Therefore, the answer must be (A).

Thanks for playing!

Hi Chris, thanks for the solution.

However, i tried for regular pentagon. If i draw a longest line between two vertices, i get an isosceles triangle with 108 deg opposite to longer side and 38 degrees each opposite to 2 sides of pentagon. So lengths are always in proportion to angles opposite to corresponding sides. In this case, line drawn is more than the double of the sides. Because 108 is more thane the double of 38. Please throw some light on this.

thanks

subbu

I did the calculations for pentagon and if we take side as 1 then the diagonal comes as 1.6666