Home > Geometry, Number theory, Programming, Uncategorized > Project Euler – Problem 264

## Project Euler – Problem 264

Today I managed to solve problem 264 from Project Euler. This is my highest rating problem until now: 85%. You can click the link for the full text of the problem. The main idea is to find all triangles ABC with vertices having integer coordinates such that

• the circumcenter O of each of the triangles is the origin
• the orthocenter H (the intersection of the heights) is the point of coordinates (0,5)
• the perimeter is lower than a certain bound

I will not give detailed advice or codes. You can already find a program online for this problem (I won’t tell you where) and it can serve to verify the final code, before going for the final result. Anyway, following the hints below may help you get to a solution.

The initial idea has to do with a geometric relation linking the points A, B, C, O and H. Anyone who did some problems with vectors and triangles should have come across the needed relation at some time. If not, just search for important lines in triangles, especially the line passing through O and H (and another important point).

Once you find this vectorial relation, it is possible to translate it in coordinates. The fact that points A, B, C are on a circle centered in O shows that their coordinates satisfy an equation of the form $x^2+y^2=n$, where $n$ is a positive integer, not necessarily a square… It is possible to enumerate all solutions to the following equation for fixed $n$, simply by looping over $x$ and $y$. This helps you find all lattice points on the circle of radius $\sqrt{n}$.

Once these lattice points are found one needs to check the orthocenter condition. The relations are pretty simple and in the end we have two conditions to check for the sum of the x and y coordinates. The testing procedure is a triple loop. We initially have a list of points on a circle, from the previous step. We loop over them such that we dont count triangles twice: i from 1 to m, j from i+1 to m, k from j+1 to m, etc. Once a suitable solution is found, we compute the perimeter using the classical distance formula between two points given in coordinates. Once the perimeter is computed we add it to the total.

Since the triple loop has cubic complexity, one could turn it in a double loop. Loop over pairs and construct the third point using the orthocenter condition. Then just check if the point is also on the circle. I didn’t manage to make this double loop without overcounting things, so I use it as a test: use double loops to check every family of points on a given circle. If you find something then use a triple loop to count it properly. It turns out that cases where the triple loop is needed are quite rare.

So now you have the ingredients to check if on a circle of given radius there are triangles with the desired properties. Now we just iterate over the square of the radius. The problem is to find the proper upper bound for this radius in order to get all the triangles with perimeter below the bound. It turns out that a simple observation can get you close to a near optimal bound. Since in the end the radii get really large and the size of the triangles gets really large, the segment OH becomes small, being of fixed length 5. When OH is very small, the triangle is almost equilateral. Just use the upper bound for the radius for an equilateral triangle of perimeter equal to the upper bound of 100000 given in the problem.

Using these ideas you can build a bruteforce algorithm. Plotting the values of the radii which give valid triangles will help you find that you only need to loop over a small part of the radii values. Factoring these values will help you reduce even more the search space. I managed to  solve the problem in about 5 hours in Pari GP. This means things could be improved. However, having an algorithm which can give the result in “reasonable” time is fine by me.