Diophantine equation interesting problem from geometry

Question

Today I came up to an interesting problem that i would like to share with the ME community to the enjoy of the reader. The initial problem (which has $2$ almost equivalent statements) was about integer sides in a triangle. This problem is (not very surprisingly) equivalent to a bit challenging problem about Diophantine-equations

Statement $1$

Let $\triangle ABC$ be an isosceles triangle with $\overline{BC}=\overline{CA}$. Consider the height $AD$ over side $BC$, with $D$ being the base of the height. Find all triangles such that $\overline{BD}, \overline{DC}, \overline{AB}, \overline{AD} \in \mathbb{N}$ (or $\in \mathbb{Z}$ if you consider directed lenghts)

Statement $2$

Let $\Omega$ be a circle with center $O$. Consider a radius $OA$, and a point $B \in OA$. Denote $C$ the point that results from the intersection of $\Omega$ with a perpendicular line to $OA$ through $B$. Find all possible configurations such that $\overline{OB}, \overline{BA}, \overline{BC}, \overline{CA} \in \mathbb{N}$ (or $\in \mathbb{Z}$ if you consider directed lenghts)

Hint

Letting $u=\overline{OB}, v = \overline{BA}$, the problem es equivalent to solve the following system of Diophantine equations:\begin{align}(u+v)^2 - u^2 = v(2u +v)= x^2 \\(u+v)^2 - u^2 + v^2 = 2v(u+v) =z^2\end{align} Now, $z$ is even, so $z=2y$ and we have to solve the following system\begin{align} v(2u +v)= x^2 \\ v(u+v) = 2y^2 \end{align}

Answer 1

Hints: As can be seen in figure AB, AD and DB are Pythagorean triples such that:

$DB=2 k+1$, $AD=2k(k+1)$ and $AB=2k(k+1)+1$; $k\in N$

Similarly AD , DC and AC follow the same relation. That is we can always construct such triangles with required constrains. This helps you to solve the equation you found.

Answer 2

Fact: If $(a,b,c)\in\Bbb{N}^3$ satisfies $a^2+b^2=c^2$ then, after interchanging $a$ and $b$ if necessary, there exist $k,m,n\in\Bbb{N}$ with $\gcd(m,n)=1$ and $m>n>0$ and $mn$ even such that $$a=k(m^2-n^2),\qquad b=2kmn,\qquad c=k(m^2+n^2).$$ Such triples are known as Pythagorean triples, and this parametrization has been known since antiquity. A triple with $\gcd(a,b,c)=1$ is called primitive, and of course has $k=1$ and $a$ odd.

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Suppose $ABC$ is a triangle with $AB,AD,BD,CD\in\Bbb{N}$, where $D$ is as described in Statement 1. Then $AC=BC=BD+CD$, so also $AC,BC\in\Bbb{N}$, and so $(AC,AD,CD)$ and $(AB,AD,BD)$ are two Pythagorean triples. Note that $\gcd(AC,AD,CD)$ divides $AD$ and also $AC-CD$, where $$AC-CD=BC-CD=BD,$$ and hence $\gcd(AC,AD,CD)$ also divides $AB$ because $AD^2+BD^2=AB^2$. This shows that $\gcd(AC,AD,CD)$ divides $\gcd(AB,AD,BD)$, and so the configuration $ABCD$ can be scaled down so that the triple $(AC,AD,CD)$ is a primitive Pythagorean triple.

Next note that at most one of the legs of a Pythagorean triple is odd. So if $AD$ is odd then $BD$ and $CD$ are even, and hence $AC=BC=BD+CD$ is even. But this contradicts the fact that $$AD^2+CD^2=AC^2.$$ Hence $AD$ is even and so there exist $m,n,r,s,t\in\Bbb{N}$ with $\gcd(m,n)=\gcd(s,t)=1$ and $m>n>0$ and $s>t>0$, and $mn$ and $st$ even, such that \begin{eqnarray*} CD&=&m^2-n^2,&\qquad AD&=&2mn,&\qquad AC&=&m^2+n^2,\\ BD&=&r(s^2-t^2),&\qquad AD&=&2rst,&\qquad AB&=&r(s^2+t^2).\\ \end{eqnarray*} Then $AC$ and $CD$ are odd, so $BD=BC-CD=AC-CD$ is even. Of course $s^2-t^2$ is odd because $\gcd(s,t)=1$ and $st$ is even, and so it follows that $r$ is even, say $r=2w$. Then $$2n^2=AC-CD=BD=2w(s^2-t^2).$$ We also see that $AD=2mn=4wst$, and hence $$2w(nst)=n(4wst)=n(2mn)=m(2n^2)=2mw(s^2-t^2),$$ which shows that $$2nst=m(s^2-t^2).$$ Because $m$ and $n$ are coprime this shows that $m$ divides $2st$. Similarly, because $s$ and $t$ are coprime and $s^2-t^2$ is odd, this shows that $2st$ divides $m$. That means $m=2st$ and hence $n=s^2-t^2$. Plugging this back into our original parametrization shows that \begin{eqnarray*} CD&=&m^2-n^2&\qquad AD&=&2mn&\qquad AC&=&m^2+n^2\\ &=&(2st)^2-(s^2-t^2)^2&\qquad&=&2(2st)(s^2-t^2)&\qquad&=&(2st)^2+(s^2-t^2)^2\\ &=&-s^4+6s^2t^2-t^4&\qquad&=&4s^3t-4st^3&\qquad&=&s^4+2s^2t^2+t^4\\ BD&=&r(s^2-t^2),&\qquad AD&=&2rst,&\qquad AB&=&r(s^2+t^2).\\ \end{eqnarray*} Comparing the expressions for $AD$, it is clear that $r=2(s^2-t^2)$.

Conversely, it is a matter of rote algebra to verify that $$(-s^4+6s^2t^2-t^4)^2+(4s^3t-4st^3)^2=(s^4+2s^2t^2+t^4)^2,$$ which means that for any choice of $r,s,t\in\Bbb{N}$ with $\gcd(s,t)=1$ and $s>t>0$ and $st$ even these parametrizations yield a solution. To also meet the requirement that $CD>0$ we must have $m>n$, which is equivalent to $2st>s^2-t^2$, or put differently $$(s-t)^2<t^2.$$ Of course $s$ and $t$ are both nonnegative, so this is in turn equivalent to $s<2t$. In conclusion:

A triangle $ABC$ with $AC=BC$ has $AB,AD,BD,CD\in\Bbb{N}$, with $D$ as defined in statement 1, if and only if after scaling there exist coprime $s,t\in\Bbb{N}$ with $2t>s>t>0$ and $st$ even, such that \begin{eqnarray*} AB&=&2(s^4-t^4),\\ AC=BC&=&s^4+2s^2t^2+t^4,\\ AD&=&4st(s^2-t^2),\\ BD&=&2(s^2-t^2)^4,\\ CD&=&-s^4+6s^2t^2-t^4. \end{eqnarray*}