# Fermat points and parameterizing the 120 degree integer triangle (Project Euler 143)

Problem 143 of Project Euler is an interesting geometry problem, and one notorious for having one of the lowest solvers to time released ratios of all the problems on the site: under 600 solvers in over 3 years as time of writing.

So what is the problem?

The problem is quite simple. We have a triangle $ABC$, with all angles smaller than $120^\circ$. Point $T$ is inside the triangle such that $p+q+r$ is at its minimum ($T$ can always be determined uniquely).

We define a Torricelli triangle as one where $a$, $b$, $c$, $p$, $q$, $r$ are all integers.

Generate all Torricelli triangles with $p + q + r \leq 120000$.

### The Fermat point

In triangle $ABC$, with $P$ being the point such that $PA+PB+PC$ is minimized, $P$ is called the Fermat point of triangle $ABC$.

We can use Ptolemy’s theorem to locate the Fermat point in the triangle.

Let $D$ be a point such that $BCD$ is equilateral, then construct the circumcircle of $BCD$:

We now show that $P$ must be at the intersection of $AD$ and the circumcircle of $\triangle BCD$.

Assume that $P$ does not lie on the circumcircle of $\triangle BCD$. Then, according to Ptolemy’s theorem:

$BP \times CD + PC \times BD > PD \times BC$

Equality only occurs when $PCDB$ is cyclic. Since $BCD$ is equilateral, we have $BC = CD = BD$ so we can divide it out to get

$PB + PC > PD$

Adding $PA$ to both sides:

$PA + PB + PC > PA + PD$

Now if $P$ did lie on the circumcircle of $\triangle BCD$, we would have $PA + PB + PC = PA + PD$, so $P$ must lie on the circumcircle of $\triangle BCD$ in order to be optimal.

So we know $P$ is on the circle. But now we assume that $P$ is not on $AD$. As we had before, $PA + PB + PC = PA + PD$.

Next if $P$ is not on $AD$, then $AP + PD > AD$, or by substitution,

$PA + PB + PC > AD$

Of course if $P$ were actually on $AD$, then $PA + PB + PC = AD$, and the sum $PA+PB+PC$ would be optimal.

This proves the optimal place for $P$ to be on the intersection of $AD$ and the circumcircle of $\triangle BCD$.

If we repeat this for the other three sides, we notice that $P$ is the intersection of the three circumcircles, and also the intersection of $AD$, $BE$, and $CF$:

Further, we see that quadrilaterals $BPCD$, $APCE$, and $FAPB$ are all cyclic. As $\angle BDC = 60^\circ$, $\angle BPC = 120^\circ$, and similarly $\angle APC = 120^\circ$ and $\angle APB = 120^\circ$.

### Designing an algorithm

We have enough information now to start working on an algorithm. Let us come back to the previous diagram:

So knowing that the central angles are all $120^\circ$, we can apply the cosine law ($\cos 120^\circ = -\frac{1}{2}$):

$a^2 = q^2 + r^2 - 2qr \cos 120$

$a^2 = q^2 + r^2 +qr$

A similar formula applies to sides $b$ and $c$. We call $(x,y)$ a pair if $x^2 + xy + y^2$ is a perfect square.

We have found a Torricelli triangle if for three integers $p$, $q$, $r$, all of $(p,q)$, $(p,r)$, $(q,r)$ are all pairs.

This leaves us with an outline of an algorithm:

1. Generate all pairs $(x,y)$ under the limit (with $x+y < 120000$ and $x^2 + xy + y^2$ being a square)
2. Sort the list of pairs and index them (to be explained in step 3)
3. For each pair $(a,b)$ in the list, search through the list to check if there exists some $c$ where $(a,c)$ is a pair and $(b,c)$ is a pair. We index the list to drastically improve searching time.
4. If an $(a,b,c)$ triple is found and $a+b+c \leq 120000$, then mark $a+b+c$ as found. This is easily implemented as a bit array of size 120000 with all bits initially 0 and flipped to 1 when a sum is encountered.
5. Add up the sums and output

Step 1 takes up most of the time in this algorithm. To do this by ‘brute force’, we need an outer loop from 1 up to 120000, then another inner loop again from 1 to 120000 (actually a bit less), which is essentially $120000^2$ operations, which is in the billions. An implementation of this algorithm would take about 3 minutes.

### Parameterizing the a^2 + ab + b^2 = c^2 equation

It is well known that all primitive triples of the form $a^2 + b^2 = c^2$ can be generated by $a = m^2 - n^2$, $b = 2mn$, $c = m^2 + n^2$ where $m$ and $n$ are coprime and are of opposite parity.

Can a similar parameterization be found for the equation $a^2 + ab + b^2$ equation?

Letting $x = \frac{a}{c}$, and $y = \frac{b}{c}$ and dividing by $c^2$, we get

$x^2 + xy + y^2 = 1$

which is the equation of an ellipse.

Originally we wanted to find integer solutions to the equation $a^2 + ab + b^2 = c^2$. This is equivalent to finding all rational points on the ellipse $x^2 + xy + y^2 = 1$:

It is easy to see why: if $x$ and $y$ are rational points such that $x^2 + xy + y^2 = 1$ then $x = \frac{a}{c}$ and $y = \frac{b}{c}$ where $a$, $b$, $c$ are positive integers, and we arrive back at the original form of the equation.

Also in order for $a$, $b$, $c$ to be positive, we will only consider points on the ellipse in the first quadrant, with both $x$ and $y$ being positive. We do this by choosing a point on the ellipse, and from there combine the equations of the ellipse and the line.

Let us choose the point $(0,-1)$ to be the first point of the line. Then the equation of the line is $y = tx-1$, where $t$ is valid only if it is positive and greater than 1.

Substituting into the equation of the ellipse:

$x^2 +tx^2 - x + t^2 x^2 - 2tx + 1 = 1$

Simplifying:

$x (t^2+t+1) = 2t + 1$

$x = \frac{2t+1}{t^2 + t + 1}$

Now evaluating for $y$:

$y = t (\frac{2t+1}{t^2 + t + 1})-1 = \frac{t^2-1}{t^2+t+1}$

Notice now that for $x$ and $y$ to be rational, we just need the slope $t$ to be rational. So we can write $t$ as $\frac{m}{n}$ where $m$ and $n$ are positive integers and $m>n$.

Simplifying the expressions for $x$ and $y$:

$x = \frac{2 \frac{m}{n} + 1}{(\frac{m}{n})^2 + \frac{m}{n} + 1} = \frac{2mn+n^2}{m^2 + mn + n^2}$

$y = \frac{(\frac{m}{n})^2 - 1}{(\frac{m}{n})^2 + \frac{m}{n} + 1} = \frac{m^2 - n^2}{m^2 + mn + n^2}$

Since we defined $x = \frac{a}{c}$ and $y = \frac{b}{c}$, we have a valid triple if $a = 2mn+n^2$, $b = m^2 - n^2$, and $c = m^2 + mn + n^2$.

If $m$ and $n$ are not coprime, then neither will $a$, $b$, and $c$ be coprime as they will share a common divisor. So in order for $(a,b,c)$ to be primitive we will need $m$ and $n$ to be coprime.

Still this isn’t quite sufficient. Consider what would happen if $m \equiv n (\mod 3)$. We have $a = 2mn+n^2$ which can be written as $n(n-m+3m)$ which is divisible by 3. Next $b = (m+n) (m-n)$ which is divisible by 3. Finally, $c = m^2 + mn + n^2$ which can be written as $(m-n)^2 + 3mn$, again divisible by 3! So if $m \equiv n \mod 3$, then the resulting triple is not primitive!

But it turns out that we can ignore all cases where $m \equiv n \mod 3$. Given a triple $(2mn+n^2, m^2-n^2, m^2+mn+n^2)$, with $m \equiv n \mod 3$, it is possible to find an identical parameterization giving $(\frac{2mn+n^2}{3}, \frac{m^2-n^2}{3}, \frac{m^2 + mn + n^2}{3})$.

As $m \equiv n mod 3$, we have $m+2n \equiv 0 \mod 3$ and also $m-n \equiv 0 \mod 3$. Thus we can have $u$ and $v$ where $u$ and $v$ are positive integers:

$u = \frac{m+2n}{3}$

$v = \frac{m-n}{3}$

Multiplying by 3, $3u = m+2n$, and $3v = m-n$. Combining the two and substituting we get

$m = u+2v$

$n = u-v$

If we substitute $u$ and $v$ for $m$ and $n$ in the triple $(2mn+n^2, m^2-n^2, m^2+mn+n^2)$, we get:

$2mn + n^2 = 3 (u^2-v^2)$

$m^2 - n^2 = 3(2uv + v^2)$

$m^2 + mn + n^2 = 3 (u^2 + uv + v^2)$

So this proves that when $m \equiv n \mod 3$, the case is nonprimitive and has already been covered by a smaller (possibly primitive) case. We are done.

### Implementation in C++

Now using the new parameterization scheme, we can generate all pairs in about $n$ operations instead of $n^2$ operations. So we can loop up to $\sqrt{120000}$, which cuts down the time down to about 2 seconds:

#include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
using namespace std;

// Limit and square root of limit
#define L 120000
#define LSQ 347
typedef long long int int64;

int64 gcd(int64 a, int64 b)
{
int64 c = a % b;
while(c != 0)
{
a = b;
b = c;
c = a % b;
}
return b;
}

int main()
{
// Store pairs in here
vector< pair<int64,int64> > pairs;

// Use the parameterization
for(int64 u=1; u<LSQ; u++){
for(int64 v=1; v<u; v++){
if(gcd(u,v) != 1) continue;
if((u-v) % 3 == 0) continue;
int64 a = 2*u*v + v*v;
int64 b = u*u - v*v;
if(a+b > L) break;
// From coprime pairs make composite pairs
for(int k=1; k*(a+b)<L; k++){
pairs.push_back(make_pair(k*a,k*b));
pairs.push_back(make_pair(k*b,k*a));
}
}
}

// Sort pairs list
sort(pairs.begin(),pairs.end());

// Create index
int index[L];
for(int i=0; i<L; i++) index[i] = -1;
for(int i=0; i<pairs.size(); i++){
if(index[pairs[i].first] == -1)
index[pairs[i].first] = i;
}

// Which sums have been reached?
bool sums[L];
for(int i=0; i<L; i++) sums[i] = false;

// Iterate through all pairs
for(int i=0; i<pairs.size(); i++){
int64 a = pairs[i].first;
int64 b = pairs[i].second;

// Construct vectors for indices
vector<int64> va;
vector<int64> vb;

// Fetch indices
int ia = index[a];
int ib = index[b];

while(ia<pairs.size()){
pair<int64,int64> next = pairs[ia];
if(next.first != a) break;
va.push_back(next.second);
ia++;
}

while(ib<pairs.size()){
pair<int64,int64> next = pairs[ib];
if(next.first != b) break;
vb.push_back(next.second);
ib++;
}

// Find common objects between va and vb
for(int ja=0; ja<va.size(); ja++){
for(int jb=0; jb<vb.size(); jb++){
if(va[ja]==vb[jb]){
// Potential c found
int64 c = va[ja];
if(a+b+c<L) sums[a+b+c] = true;
}
}
}
}

// Tally up sums
int64 s = 0;
for(int i=0; i<L; i++)
if(sums[i]) s+=i;

printf("%lld\n",s);
return 0;
}

# Socks in a Drawer

I’ve began studying some probability and statistics for fun, and I’m using a book called Fifty Challenging Problems in Probability by Mosteller:

Printing math textbooks is a great way to use up the school’s printing credits at the end of the year.

The first problem is about socks in a drawer. At first glance it looks really, deceptively simple. But when I start doing the problem I find that it’s a bit trickier than I expected.

Solutions to the problems are provided at the back of the book. For this problem, I find that the book’s solution is really ugly and unintuitive. Thus I’m going to give an alternate solution here.

### Problem (my version)

A drawer contains red and black socks. I randomly take a sock out of the drawer. I then randomly take another sock out of the drawer.

The probability that both are red is exactly $\frac{1}{2}$. Determine how many socks of each color there are in the drawer.

### Solution

Let r be the number of red socks in the drawer, and b be the number of black socks. Then the total number of socks in the drawer is r+b.

The probability that a sock you randomly take is red would be

$\frac{r}{r+b}$

Now the probability that the next sock you take would also be red is

$\frac{r-1}{r+b-1}$

This is because there is one less red sock in the drawer, and at the same time there is one less sock (of any color) in the drawer.

The probability that both socks will be red is the two probabilities multiplied:

$P = \frac{1}{2} = \frac{r}{r+b} \cdot \frac{r-1}{r+b-1}$

Now we’re solving the equation to find possible values of r and b. There are two variables, and one equation. However, both variables have to be positive integers. This type of equation is called a diophantine equation.

We begin by simplifying the equation into a quadratic.

$\begin{array}{rcl} \frac{r(r-1)}{(r+b)(r+b-1)} &=& \frac{1}{2} \\(r+b)(r+b-1)&=& 2r(r-1) \\ r^2+2rb+b^2-r-b&=&2r^2-2r \\ r^2+(-2b-1)r+(b-b^2)&=&0 \end{array}$

Now using the quadratic formula to solve for r:

$r = \frac{(2b+1) + \sqrt{(-2b-1)^2-4(b-b^2)}}{2}$

It’s not necessary to consider the negative root of the quadratic. Simplifying the quadratic further, we get

$r = \frac{(2b+1)+ \sqrt{8b^2+1}}{2}$

Here we can make a substitution. Let $k = \sqrt{8b^2+1}$. Then,

$r = \frac{(2b+1)+ k}{2}$

In this equation r will be an integer as long as b is an integer and k is an odd integer.

We are now finding integer solutions to the equation

$k = \sqrt{8b^2+1}$

..where k is also odd.

Simplifying, we get:

$\begin{array}{rcl} k^2 &=& 8b^2+1 \\ k^2-8b^2&=&1 \end{array}$

At this point we can drop the condition that k is odd, because if k is even then left hand side is even, but the right hand side is 1. As a result, k cannot be even if it’s a solution to this equation.

We’ve basically converted this problem into a form of Pell’s equation, which generally has this form:

$x^2 - Dy^2 = 1$

..where D is a constant and x and y are unknown integers. Our instance of Pell’s equation is the case where D=8.

Fortunately for us, the Pell’s equation is well studied. We can use existing results in number theory to help us with this problem.

By trial and error, it can be found that the smallest nontrivial solution, or the fundamental solution to the equation is $(k,b) = (3,1)$. Once the fundamental solution to a Pell type equation has been found, all other solution pairs can be generated. And there are infinitely many such pairs.

Generating solution pairs can be done with this recurrence relation:

$\begin{array}{rcl} k_{i+1} &=& k_1 k_i + D b_1 b_i \\ b_{i+1} &=& k_1 b_i + b_1 k_i \end{array}$

It’s simple to write a computer program to generate these pairs. For instance, the first ten pairs of $(k,b)$ are:

(3,1)
(17,6)
(99,35)
(577,204)
(3363,1189)
(19601,6930)
(114243,40391)
(665857,235416)
(3880899,1372105)
(22619537,7997214)

All that is left is to substitute values of b to get integer values of r. The first ten pairs of $(r,b)$ are:

(3,1)
(15,6)
(85,35)
(493,204)
(2871,1189)
(16731,6930)
(97513,40391)
(568345,235416)
(3312555,1372105)
(19306983,7997214)

We are done; we can use the same recurrence relation to generate arbitrarily large sock combinations.