Greatest Area

Visually played around with Heron's formula a bit, and realized that if you've skewed a triangle's shape by moving one of its three points (say 'A'), then that triangle's area will remain ${\color{#20A900}unchanged}$ as long as that point is moved along a path parallel to the line created by the two other points, (the line 'B-C'). Here's a gif I made:

Click here for the interactive version.

In this week's problem, triangle E's area will remain unchanged as long as its top-most point is shifted horizontally anywhere along the third row of dots.

Similarly, triangle C's area will remain unchanged as long as its bottom-most point is moved anywhere along the second row of grey dots.

Triangles B and D are a bit tougher to see, but let's say the bottom left grey dot is the the origin (0,0) of a coordinate system x,y. If you squint (or bring out a ruler) you'll see that in triangle B for example, its right-most dot can be moved diagonally along any of the grey dots with the 'coordinates' (1,0), (2,1), (3,2), (4,3).

So, because all triangles in this week's problem are essentially the same as triangle A but with their shape changed by shifting one of the points along a line parallel to the two other points, all the triangles have the same area.

A trivial consequence of Pick's theorem perhaps (as others have mentioned), but I found it's still a nice visual reminder of what the theorem actually entails.

EDIT: Yes, you guys are correct -- there was no need for me to mess around with Heron's formula. I could have just used good old Area=(base)(height)(0.5) , and since neither base nor height is changing, its area will of course stay the same. To be honest, I was just playing around in Geogebra and happened to find something I thought was cool, so I figured I'd share it with the community. Didn't expect this to get up-voted to one of the top solutions. As a way of thanking you for all your up-votes I have made another animation for you (screenshot below, gif file size was too large). I may have gone a bit overboard with the animations in this new one, but all it's really showing is that none of the triangles change their base or height, and therefore their area doesn't change.

Click here for the interactive version.

Each triangle has zero interior lattice points and three boundary lattice points (the three vertices), so by Pick's Theorem , the area of each triangle is 1/2.

It is easy to see that triangles A, C, E, & F all have base and height of 1 unit and area of $\frac12$ square units. For triangles B & D, I labelled the vertices of the triangles and other relevant points in the diagram. Starting with triangle B, I divided the triangle into 2 sections by drawing BD and extending it to E. $\triangle BCF \sim \triangle BDE. \quad \therefore \frac{BE}{DE}=\frac{BF}{CF}=\frac23=\frac{1}{1\frac12}\quad \implies DE=1\frac12 \implies BD=\frac12.$ So the upper and lower parts of triangle B both have a base of $\frac12$ and a height of 1 and an area of $\frac14$ giving a total area of $\frac12$ . Now for triangle D, the method is similar. Divide the triangle into 2 by drawing GI. $\triangle HGI \sim \triangle HKJ. \quad \therefore \frac{HI}{GI}=\frac{HJ}{KJ}=\frac34=\frac{1}{1\frac13}\quad \implies GI=1\frac13 \implies GL=\frac13.$ So the upper and lower parts of triangle D both have a base of $\frac13$ and heights of 1 and 2 respectively. The total area is again $\frac12$ . Thus $\boxed{\text{all the triangles have the same area.}}$

Each triangle has the same area equal to 0.5 units . it can be calculated like this

A, C, E, and F all have a base and height of 1. B and D have a base of √2 and height of √2/2 (create a line connecting the 2 closest vertices and see that the 3rd is half a diagonal distance away from the line.) All have an area of 1/2.

The area of a 2D parallelogram can be calculated by treating two of the sides as vectors $\mathbf{a}$ and $\mathbf{b}$ , and then computing $|a_x b_y - a_y b_x|$ . The area of the triangle is half that of of the parallelogram, $A=\frac{1}{2}|a_x b_y - a_y b_x|$ . Since all these triangles are on a grid, it's easy to compute these components:

• A: Starting from the bottom left corner, we identify $\mathbf{a}_A=(1,1)$ and $\mathbf{b}_A=(0,1)$ . Then the area is $A_A=\frac{1}{2}|1\cdot 1 - 1 \cdot 0| = \frac{1}{2}$ .
• B: We have $\mathbf{a}_B=(3,2)$ and $\mathbf{b}_B=(1,1)$ . So, $A_B=\frac{1}{2}|3\cdot 1 - 2 \cdot 1| = \frac{1}{2}$ .
• C: Here, $\mathbf{a}_C=(3,1)$ and $\mathbf{b}_C=(2,1)$ . So, $A_C=\frac{1}{2}|3 \cdot 1 - 1 \cdot 2| = \frac{1}{2}$ .
• D: Here, $\mathbf{a}_D=(3,2)$ and $\mathbf{b}_D=(4,3)$ . So, $A_D=\frac{1}{2}|3 \cdot 3 - 2 \cdot 4| = \frac{1}{2}$ .
• E: Starting from the top left, $\mathbf{a}_E=(2,-1)$ and $\mathbf{b}_E=(1,-1)$ . So, $A_E=\frac{1}{2}|2 \cdot (-1) - 1 \cdot (-1)| = \frac{1}{2}$ .

This turns out not to be the most elegant way to solve this, since we can apparently use Pick's Theorem, but this method has broader applications. For a three dimensional triangle, the area of the triangle is given by a vector cross product $\frac{1}{2}|\mathbf{a}\times\mathbf{b}|$ - we can see the above formula is a special case of a cross product where the third component is zero.

An almost identical calculation is also very important in real-time 3D computer graphics, where it's used to compute an 'edge function' and 'barycentric coordinates' determining whether a particular pixel is inside a triangle, and if so, how to interpolate quantities defined on the vertices in a perspective-correct way. In a rasterisation process, after the vertex shader has run, you have the positions of each vertex projected onto 2D plane of the screen. So, much as in the above case, you have the x and y components of the edges as vectors.

Suppose you have a point $\mathbf{p}$ and you want to know if it's inside a triangle with vertices $\mathbf{v}_1$ , $\mathbf{v}_2$ , $\mathbf{v}_3$ . For each edge - take for example the edge 12 - you can find two vectors $\mathbf{p}-\mathbf{v}_1$ and $\mathbf{v}_2-\mathbf{v}_1$ , and then calculate $(p_x - v_{1x}) (v_{2y}-v_{1y}) - (p_y-v_{1y}) (v_{2x}-v_{2y})$ - this is twice the 'signed area' of a triangle defined by the two vertices and the point, and it is positive if the point is on one side of the edge, and negative if it is on the other. By computing this for all three edges, you know the point is inside the triangle iff all three edge functions are positive. Moreover, these signed area functions turn out to be exactly what you need for finding, for example, the texture coordinates at this point in a perspective-correct way.

For more information on that and the rasterisation algorithm in general, see Scratchapixel's wonderful tutorial .

As others have said, A, C, E, & F all have base 1 and height 1. For the other 2, treat the 45 degree side as the base. Now moving the 3rd point parallel to that maintains the height. You can move those point to where they also make 1 by 1 triangles, thus the same area.

I just guessed.I had this feeling that all triangle are same because points A,B and C are three points and triangles are stretched at one of these points.

There are zero interior points[I(P)] in each of the triangles and 3 boundary points[B(P)] or vertices in each of the triangles. Therefore by Pick's Formula, Area of each triangle =I(P)+(B(P)/2)-1=0+(3/2)-1=1/2 Therefore all the triangles have the same area...

You can shear triangles C and E about their horizontal sides, and B and D about their sides angled at $45^\circ$ , so as to produce a triangle congruent to A in each case. Since shearing preserves area, it follows that triangles B, C, D and E all have the same area as A.

You can simply use base x height /2 for the area of triangles A, C and E to get 1/2 for all of them.

For triangles B and D you can extend the triangles and by working out the total area - added area, you get the original area of the triangle.

If you were to label all the points on the grid as A to F horizontally and 1 to 4 vertically from the bottom left corner, (like a chessboard) the extension can be better expressed. First I will start with triangle B. By leaving points A1 and D3 and creating a new point C3, you can form a triangle with area 1x 2 /2 = 1 as the total area. The added area (B2, C3, D3) is mathematically similar to triangle E and has an area of 1 x 1 /2 = 1/2. Therefore by taking away 1/2 from 1 we get the area of triangle B to be 1/2.

The same principal can be applied to triangle D. By creating a triangle with points B1, C1 and F4 we get (1x3 /2 ) - (1x2 /2) = 1/2 as the area of triangle D. Thus all the triangles have the same area.

Reuben - age 14

According to Pick’s Theorem, the area of a lattice polygon is $\frac{B}{2} + I - 1$ , where $B$ and $I$ are the numbers of boundary and interior points respectively. Each one of the given triangles has area $A = \frac{3}{2} + 0 - 1 = 0.5$ square units.

Just create parallelograms using the triangles as one half and you will notice that the triangles that form the parallelograms from A,C,F, and E all have height and base 1. Since the area of the triangle is half the area of the parallelogram you get 1/2. You could then calculate distance for B and D.

My solution was to have my friend go on his computer and tell me the answers btw I am the highest level in everything in my class 🤣

For triangles A, C, and E, take the horizontal side as the base. Then they all have bases and heights of 1, so area 1/2. For B and D, take the side that's a single diagonal as the base. It has length sqrt(2). The heights are then 1/2 of a diagonal, so length 1/sqrt(2), and the areas are again 1/2.

By assuming points to be in a Cartesian plane area of each triangle =1/2

Treat the dots as points in a xy plane. Find coordinates for the vertices of the triangles. Use determinants or any other method to calculate area

Please help me find the areas of B and D!