<< problem 504 - Square on the Inside | Prime triples and geometric sequences - problem 518 >> |

# Problem 510: Tangent Circles

(see projecteuler.net/problem=510)

Circles A and B are tangent to each other and to line L at three distinct points.

Circle C is inside the space between A, B and L, and tangent to all three.

Let r_A, r_B and r_C be the radii of A, B and C respectively.

Let S(n) = sum{r_A + r_B + r_C}, for 0 < r_A <= r_B <= n where r_A, r_B and r_C are integers.

The only solution for 0 < r_A <= r_B <= 5 is r_A = 4, r_B = 4 and r_C = 1, so S(5) = 4 + 4 + 1 = 9.

You are also given S(100) = 3072.

Find S(10^9).

# My Algorithm

A simple `bruteForce()`

function easily verifies S(100) (and even manages to solve S(1000))

but then falls apart due to rounding problems and - more import - performance issues.

However, it reveals a few things:

- most solutions are multiples of (4,4,1)

- and in general: for each solution all of its multiples are solutions, too

- if I remove these multiples, then the basic solutions are squares only

Just a few days I was in a museum and saw some ancient drawings which helped me to get great insights into problem 199

(which is not really solved yet: I still have some severe performance issues but they aren't related to the geometric solution).

Anyway, I learnt that these kind of problems have been around for ages and were analyzed by Appolonius more than 2000 years ago (see en.wikipedia.org/wiki/Problem of Apollonius).

Renè Descartes found a great equation (see en.wikipedia.org/wiki/Descartes' theorem) if four circles touch each other and their curvature is k = 1 / r

(1) (k_1 + k_2 + k_3 + k_4)^2 = 2 (k_1^2 + k_2^2 + k_3^2 + k_4^2)

Unfortunately, I don't have four but three circles and a line ...

and fortunately, if I set k = 0 then the formula solves the case of three circles and a line, too !

Now the whole equation simplifies to

(2) (k_A + k_B + k_C)^2 = 2 (k_A^2 + k_B^2 + k_C^2)

If I solve for k_C then

(3) sqrt{k_C} = sqrt{k_A} + sqrt{k_B}

And with regards to the radii instead of the curvatures (remember: k = 1 / r)

(4) dfrac{1}{sqrt{r_C}} = dfrac{1}{sqrt{r_A}} + dfrac{1}{sqrt{r_B}}

As I said before, all solutions are squares (or multiples of squares).

If I assume a = sqrt{r_A} and b = sqrt{r_B} and c = sqrt{r_C} then:

(5) dfrac{1}{c} = dfrac{1}{a} + dfrac{1}{b}

Solved for c

(6) c = dfrac{ab}{a+b}

My program iterates over all a and b where b < a (that's different from the problem statement but it doesn't affect the result).

If a^2 b^2 is a multiple of a^2 + b^2 then I have found another c^2.

That means I have found a triple (a^2, b^2, c^2) = (r_A, r_B, r_C).

It's possible that the multiples of squares are squares, too.

Therefore I have to make sure that they are coprime (gcd(a,b,c) = 1) otherwise they would be counted twice.

The sum of all multiples of r_A + r_B + r_C is based on the triangular numbers (see `triangle()`

).

If m is the number of multiples of a basic solution:

(7) m = dfrac{limit}{r_A} → because I defined r_A >= r_B

(8) sum_{k=1..m}{k} = dfrac{m(m+1)}{2}

(9) sum_{k=1..m}{k (r_A + r_B + r_C)} = (r_A + r_B + r_C) dfrac{m(m+1)}{2}

## Alternative Approaches

After submitting my solution I saw in the Project Euler forum that you can heavily re-arrange the `for`

-loops such that the solution is found in a few milliseconds.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo 100 | ./510`

Output:

*Note:* the original problem's input `1000000000`

__cannot__ be entered

because just copying results is a soft skill reserved for idiots.

*(this interactive test is still under development, computations will be aborted after one second)*

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <cmath>
// ---------- taken from my toolbox ----------

// greatest common divisor

template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}
// ---------- brute force solution (not used anymore) ----------

// try all combinations of a,b,c

unsigned int bruteForce(unsigned int limit)
{
// accept solution if left side of the equation of Descartes theorem
// doesn't differ by more than epsilon from the right side
const double Epsilon = 0.0000000000001;
// sum of all valid a,b,c
unsigned int result = 0;
for (unsigned int a = 1; a <= limit; a++)
for (unsigned int b = 1; b <= a; b++)
for (unsigned int c = 1; c <= b; c++)
{
// curvatures
double k1 = 1.0 / a;
double k2 = 1.0 / b;
double k3 = 1.0 / c;
// both sides of the equation
double leftSide = (k1 + k2 + k3) * (k1 + k2 + k3);
double rightSide = 2 * (k1*k1 + k2*k2 + k3*k3);
// acceptable error ?
if (fabs(leftSide - rightSide) < Epsilon)
{
auto sum = a + b + c;
//std::cout << a << "+" << b << "+" << c << "=" << sum << std::endl;
result += sum;
}
}
return result;
}
// ---------- relevant code ----------

// compute triangular number

unsigned long long triangle(unsigned int n)
{
// sum of 1+2+3+...+n
// see https://en.wikipedia.org/wiki/Triangular_number
unsigned long long nn = n;
return nn * (nn + 1) / 2;
}
typedef long long Number;
unsigned long long evaluate(unsigned int limit)
{
// all multiples of the basic 4+4+1 solution
unsigned long long result = 0;
// all multiples of a^2, b^2, c^2
// where 1/c2 = (1/a2 + 1/b2) + 2sqrt(1/a2 * 1/b2)
// => sqrt(1/c2) = sqrt(1/a2) + sqrt(1/b2)
// => 1/c = 1/a + 1/b
// => c = a*b / (a+b)
for (unsigned long long a = 1; a*a < limit; a++)
for (unsigned long long b = 1; b <= a; b++)
{
// numerator = a*a * b*b
auto a2 = a*a;
auto b2 = b*b;
auto numerator = a2 * b2;
// denominator = (a+b)^2
auto denominator = (a+b) * (a+b);
// c^2 must be an integer
auto c2 = numerator / denominator;
// no remainder left by the previous divison ?
if (c2 * denominator != numerator)
continue;
// all values must be coprime
if (gcd(gcd(a2, b2), c2) == 1)
{
auto multiples = limit / (unsigned int)a2;
auto sum = (a2 + b2 + c2) * triangle(multiples);
result += sum;
}
}
return result;
}
int main()
{
unsigned int limit = 1000000000;
std::cin >> limit;
//std::cout << bruteForce(limit) << std::endl;
std::cout << evaluate (limit) << std::endl;
return 0;
}

This solution contains 16 empty lines, 27 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 4.2 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL`

)

See here for a comparison of all solutions.

**Note:** interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without `-DORIGINAL`

.

# Changelog

September 8, 2017 submitted solution

September 8, 2017 added comments

# Difficulty

Project Euler ranks this problem at **30%** (out of 100%).

# Links

projecteuler.net/thread=510 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# Heatmap

*Please click on a problem's number to open my solution to that problem:*

green | solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too | |

yellow | solutions score less than 100% at Hackerrank (but still solve the original problem easily) | |

gray | problems are already solved but I haven't published my solution yet | |

blue | solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much | |

orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |

red | problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too |

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I scored 13,386 points (out of 15600 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort !!!

<< problem 504 - Square on the Inside | Prime triples and geometric sequences - problem 518 >> |