In this 2x3 grid, each lattice point is one unit away from its nearest neighbors. A total of 14 isosceles triangles(but not right triangles), each with an area of ½u squared 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-x grid?
@unaimouse how?
(Hints) (Don't have full answer)
Let x be the horizontal number of points... there are "x-1" triangles
Let y be the vertical number of points ... there are 2y triangles
Total: 2(x-1)y
Horizontal: There are 2x triangles
Vertical: There are y-1 triangles
Total: (I know you can figure it out now :) )
The answer is a formula, rather than a numerical answer.
To answer this question, we need to first find the possible vertex locations on lattice points. These are (x,y), (x,0), (0, y), and (0, 0)
These happen to be the 4 possible ways that a triangle can have one vertex at any lattice point on the grid. The other two vertices can be chosen from the remaining lattice points in the x-axis, y-axis, or the origin.
Now, we need to figure out the formula from this information.
Firstly, we can figure out the first number in the formula. Because there are 4 possible vertex locations on lattice points, we can create the expression 4xy. We multiply x and y because for each lattice point on the x-axis (excluding the origin) and each lattice point on the y-axis (excluding the origin), we can form a triangle with the origin as one vertex.
Secondly, we can figure out the second number in the formula. This is -2x. This term subtracts the number of triangles that have vertices on the x-axis (excluding the origin). We subtract 2x because for each lattice point (x, 0) on the x-axis (excluding the origin), we have already counted two triangles (one with the origin as a vertex and one with (x, 0) as a vertex).
Thirdly (and finally), we can figure out the third number in the formula. This happens to be -2y. This term subtracts the number of triangles that have vertices on the y-axis (excluding the origin). We subtract 2y because for each lattice point (0, y) on the y-axis (excluding the origin), we have already counted two triangles (one with the origin as a vertex and one with (0, y) as a vertex).
By subtracting the overcounted triangles on the x-axis and y-axis, we ensure that we count each isosceles triangle with an area of 1/2 square units and two vertices one unit apart only once.
From this information, we can figure out the formula that the question asks for. This is 4xy - 2x - 2y.
I hope this helps!
Natalia (Grade 7)
Vocabulary to help you understand this question:
Lattice point: In a Cartesian coordinate system, a lattice point is a point whose x- and y-coordinates are both integers. A point in a regularly spaced array of points, known as a point lattice, that is at the intersection of two or more grid lines is referred to as a lattice point. Point lattices with unit cells that are square, rectangular, hexagonal, and other shapes can be built in a plane. An unspecified point in a square array constitutes a point in a point lattice.
To find the number of half-unit triangles with at least two vertices in an x-by-x grid, we can use the concept of combinations.
Step 1: Understand the problem
We have an x-by-x grid, and we need to count the number of isosceles triangles (but not right triangles) with an area of ½u^2, where u represents the length of one side of the triangle. The triangles must have two vertices that are one unit apart on the grid.
Step 2: Determine the number of possible triangle vertices
For each vertex of the triangle, we have x choices along the x-axis and x choices along the y-axis. So, we have a total of x * x = x^2 possible vertex combinations.
Step 3: Counting triangles with two vertices one unit apart
To count triangles with two vertices one unit apart, we need to consider the possible configurations.
Case 1: One vertex is at the corner of the grid.
In this case, we have 4 possible corners. For each corner, we have 2 choices of vertices that are one unit apart. So, the total number of triangles in this case is 4 * 2 = 8.
Case 2: One vertex is on the edge of the grid (excluding the corners).
In this case, we have (x - 2) vertices on each edge. For each edge, we have 3 choices for the vertex one unit apart (excluding the corner vertices on that edge). So, the total number of triangles in this case is 4 * (x - 2) * 3 = 12 * (x - 2).
Case 3: One vertex is in the interior of the grid.
In this case, we have (x - 2) vertices along each edge (excluding the corners and the vertices on the edges). For each interior vertex, we have 4 choices for the vertex one unit apart (the four possible directions). So, the total number of triangles in this case is (x - 2) * (x - 2) * 4 = 4 * (x - 2)^2.
Step 4: Calculating the total number of triangles
To calculate the total number of triangles, we sum up the number of triangles from each case:
Total = Case 1 + Case 2 + Case 3
Total = 8 + 12 * (x - 2) + 4 * (x - 2)^2
Therefore, the total number of half-unit triangles with at least two vertices in an x-by-x grid is given by the expression 8 + 12 * (x - 2) + 4 * (x - 2)^2.