Detailed mathematical approach

Problem Summary

Problem 557 asks for \(S(n)\), the sum of the total areas \(T\) of all valid cutting-triangle configurations with \(T \le n\). After the geometry is reduced, every candidate can be encoded by positive integers \((a,b,c,d)\), and the condition \(b \le c\) is imposed so that mirror-image cases are not counted twice.

The total area is not approximated numerically; it is forced by an exact rational formula. That turns the task into a Diophantine search: find all integer triples \((a,b,c)\) for which the formula becomes integral, recover \(d\), and add the resulting total area \(T\).

A naive search over four positive integers would be far too slow for \(n=10000\). The successful idea is to eliminate the geometric unknowns first and then use the resulting inequalities to prune the arithmetic search very aggressively.

Mathematical Approach

Let \(S(n)\) be the sum of all admissible total areas \(T\) with \(T \le n\). The key step is to replace the geometric picture by exact integer conditions.

Step 1: Eliminate the geometry

Once the two auxiliary inner areas are solved away, the total area of a candidate configuration is determined by

$T=\frac{a(a+b)(a+c)}{a^2-bc}.$

So a valid cut must satisfy

$a^2-bc\gt 0,\qquad T\in \mathbb{Z}_{\ge 1},\qquad d=T-a-b-c\in \mathbb{Z}_{\ge 1},\qquad b\le c.$

This reduction is the heart of the solution: the geometric existence condition becomes a rational expression plus positivity and integrality constraints.

Step 2: Reduce the search to \((a,b,c)\)

Because the four piece areas add up to the whole triangle, we have

$T=a+b+c+d.$

Therefore \(d\) does not need its own loop; once \(a\), \(b\), and \(c\) are known, \(d\) is determined by

$d=T-a-b-c.$

Substituting the formula for \(T\) gives

$d=\frac{bc(2a+b+c)}{a^2-bc}.$

Hence whenever \(a^2-bc\gt 0\) and \(T\) is an integer, \(d\) is automatically positive; and since \(d=T-a-b-c\), it is automatically integral as well. The implementations still test \(d\gt 0\) explicitly as a final safety check.

Step 3: Derive tight bounds for \(b\)

Several inequalities cut down the search before any divisibility test is attempted.

Since \(b\le c\) and \(bc\lt a^2\), we get \(b^2\lt a^2\), so

$b\le a-1.$

Also, from \(T=a+b+c+d\le n\), together with \(c\ge b\) and \(d\ge 1\), we obtain

$a+2b+1\le n,$

hence

$b\le \left\lfloor\frac{n-a-1}{2}\right\rfloor.$

There is a third bound. For fixed \(a\) and \(b\), the total area is strictly increasing as a function of \(c\) on the region \(bc\lt a^2\), because

$\frac{dT}{dc}=\frac{a^2(a+b)^2}{(a^2-bc)^2}\gt 0.$

So the smallest possible total for that pair occurs at \(c=b\):

$T_{\min}=\frac{a(a+b)^2}{a^2-b^2}=\frac{a(a+b)}{a-b}.$

Requiring \(T_{\min}\le n\) yields

$b\le \left\lfloor\frac{a(n-a)}{a+n}\right\rfloor.$

The implementation takes the minimum of these three upper bounds and only scans \(b\) in that reduced range.

Step 4: Derive tight bounds for \(c\)

Now fix \(a\) and \(b\). Because \(a^2-bc\gt 0\), we may rearrange \(T\le n\) without changing the direction of the inequality:

$\frac{a(a+b)(a+c)}{a^2-bc}\le n.$

Collecting the terms involving \(c\) gives

$c\bigl(a(a+b)+nb\bigr)\le a^2(n-a-b),$

so

$c\le \left\lfloor\frac{a^2(n-a-b)}{a(a+b)+nb}\right\rfloor.$

Two additional constraints are used simultaneously. From \(bc\lt a^2\),

$c\le \left\lfloor\frac{a^2-1}{b}\right\rfloor.$

From \(d\ge 1\) and \(T\le n\),

$c\le n-a-b-1.$

The final upper bound for \(c\) is the minimum of these three values. If that minimum is smaller than \(b\), then there is no admissible \(c\) for the current pair \((a,b)\).

Step 5: Test integrality and accumulate \(S(n)\)

For each surviving triple \((a,b,c)\), the implementations test whether the denominator divides the numerator:

$a^2-bc \mid a(a+b)(a+c).$

If so, then

$T=\frac{a(a+b)(a+c)}{a^2-bc}$

is an integer. The program keeps the candidate only if \(T\le n\), computes

$d=T-a-b-c,$

and confirms that \(d\gt 0\). Every valid configuration contributes its full total area \(T\) to \(S(n)\).

Worked Example: \(n=55\), \(a=20\), \(b=2\)

For this pair, the bounds on \(b\) are satisfied, so we move on to \(c\). The inequality from \(T\le n\) gives

$c\le \left\lfloor\frac{20^2(55-20-2)}{20(20+2)+55\cdot 2}\right\rfloor=\left\lfloor\frac{13200}{550}\right\rfloor=24.$

The bound from \(bc\lt a^2\) gives

$c\le \left\lfloor\frac{20^2-1}{2}\right\rfloor=199,$

and the condition \(d\ge 1\) together with \(T\le 55\) gives

$c\le 55-20-2-1=32.$

Therefore only \(2\le c\le 24\) must be checked. At the endpoint \(c=24\),

$T=\frac{20\cdot 22\cdot 44}{20^2-2\cdot 24}=\frac{19360}{352}=55,$

so

$d=55-20-2-24=9.$

This produces the valid quadruple \((20,2,24,9)\). Another valid quadruple with the same total area is \((22,8,11,14)\), matching the small checkpoint data used by the implementations.

How the Code Works

The C++, Python, and Java implementations all follow the same arithmetic strategy. They loop over \(a\), derive the tight admissible range for \(b\), derive the tight admissible range for \(c\) for each remaining pair \((a,b)\), and only then perform the divisibility test that determines whether \(T\) is integral.

The C++ implementation is the main high-performance solver. For large \(n\) it partitions the outer range of \(a\) into chunks, processes those chunks on multiple threads, and sums one partial total from each worker at the end. It also checks a few small known cases before evaluating the full target input.

The Python implementation simply launches the compiled C++ solver and returns its numeric output, so it inherits the same mathematics and the same performance characteristics. The Java implementation reproduces the same bounded integer search directly with integer arithmetic. No floating-point geometry is required anywhere in the pipeline.

Complexity Analysis

Define

$B(a)=\min\left(a-1,\left\lfloor\frac{n-a-1}{2}\right\rfloor,\left\lfloor\frac{a(n-a)}{a+n}\right\rfloor\right),$

and

$C(a,b)=\min\left(\left\lfloor\frac{a^2(n-a-b)}{a(a+b)+nb}\right\rfloor,\left\lfloor\frac{a^2-1}{b}\right\rfloor,n-a-b-1\right).$

Then the number of candidate triples examined by the arithmetic search is

$\sum_{a=1}^{n-3}\sum_{b=1}^{B(a)} \max\bigl(0,\ C(a,b)-b+1\bigr).$

This is much smaller than a naive search over all positive quadruples \((a,b,c,d)\). A coarse worst-case bound is still \(O(n^3)\) time, because after eliminating \(d\) the algorithm is essentially a three-variable integer search, but the derived inequalities remove most impossible cases before the expensive integrality test is reached.

Each surviving candidate needs only a constant amount of arithmetic: a few multiplications, one divisibility test, and a couple of comparisons. Memory usage is \(O(1)\) for the sequential search, apart from the running total. The multithreaded C++ version adds only \(O(p)\) extra storage for \(p\) worker partial sums, so the method remains extremely light on memory.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=557
2. Diophantine equations: Wikipedia — Diophantine equation
3. Triangle area background: Wikipedia — Area of a triangle
4. Floor function and integer bounds: Wikipedia — Floor and ceiling functions
5. Constrained integer search: Wikipedia — Integer programming