In this 2-by-3 grid, each lattice point is one unit from its nearest neighbors. A total of 14 isosceles triangles (but not right triangles), each with an area of 1/2u^2 have only two vertices that are one unit apart in the grid. How many such half-unit triangles have at least two vertices in an x-by-y grid?
To determine the number of half-unit triangles with at least two vertices in an x-by-y grid, we can break down the problem into simpler steps.
Step 1: Counting the possible positions for the base of the triangle.
Since the triangle has only two vertices that are one unit apart, the base of the triangle must be either horizontal or vertical. In an x-by-y grid, there are (x-1) * y possible horizontal base positions and x * (y-1) possible vertical base positions.
Step 2: Counting the possible heights of the triangle.
The height of the triangle refers to the distance from the base to its tip. Since the area of each triangle is 1/2u^2, where u represents the length of the triangle's sides, the height can be determined using the formula h = (2A)/b, where A is the area (1/2u^2) and b is the base length.
Step 3: Counting the number of triangles for each base position and height.
For each base position, we need to determine the heights that satisfy the given conditions. In this case, the height should be such that it forms an isosceles triangle (but not a right triangle) with exactly one unit between its vertices.
Step 4: Summing up the counts.
Finally, we sum up the counts obtained in Step 3 for each base position to get the total number of half-unit triangles with two vertices in an x-by-y grid.
Now, let's calculate the number of triangles for a specific example, such as a 2-by-3 grid.
Step 1: Counting the possible positions for the base of the triangle.
In a 2-by-3 grid, there are (2-1)*3 = 3 horizontal base positions and 2*(3-1) = 4 vertical base positions.
Step 2: Counting the possible heights of the triangle.
Since the area is 1/2u^2 and u represents the side length, we need to solve the equation 1/2u^2 = 1/2*b*h, where b is the base length (1 unit) and h is the height. Solving for h, we find h = u.
Step 3: Counting the number of triangles for each base position and height.
For each base position and height, we need to check if the resulting triangle is isosceles (not right). In this example, the base positions are (0,0), (0,1), (1,0), (1,1), (0,2), (1,2), and the heights are 1/2.
Given these positions and heights, we find that triangles with these vertices: (0,0), (0,1), (0,0), (1,0), (1,0), (0,1), (1,1), (0,2), (1,2), (1,1), (1,2), (1,1), (1,2), and (1,2) satisfy the conditions.
Step 4: Summing up the counts.
Adding up the counts from Step 3, we find that there are a total of 14 half-unit triangles with at least two vertices in a 2-by-3 grid.
To calculate the number of such half-unit triangles in any x-by-y grid, follow the same steps outlined above.