Sevens between tens

Probability Level 5

There is an equilateral triangular lattice of points distance one unit apart, bounded by an equilateral triangle having 10 points on each side (see image). How many segments having the length of 7 units can be formed with the vertices of the lattice?

The answer is 36.

1 solution

Maria Kozlowska
May 19, 2015

There are $6 \times 3 + 6 \times 3 = \boxed{36}$ segments of length 7.

Note: In the Cartesian plane, points of the lattice have coordinates $(\frac{m}{2} , \frac{n \sqrt{3}}{2} )$ where $m , n$ are both even or both odd integers. We are looking for a vector $u: |u| =7$ . $m^2 + 3n^2 = 196$ $\Rightarrow (|m|,|n|) = (13,3), (11,5), (14,0), ( 7,7), (2,8)$

how 6 * 3 is coming ?can u explain the picture in detail .. can u show why only (m/2 , n*root(3) / 2) is coming ?

Tarit Goswami - 3 years, 9 months ago

The picture shows 6 segments at an angle, there is a set of those in each corner of the triangle so 6 * 3. There are also 6 horizontal segments, 3 at the bottom line, 2 in the middle and 1 on the third line. There is a set of those at each side of the triangle so 6 * 3.

Maria Kozlowska - 3 years, 9 months ago

Sevens between tens

The answer is 36.

1 solution

0 pending reports