Is there a $\triangle ABC$, with $D$ on $BC$, such that $AB$, $BC$, $CA$, $AD$, $BD$, $CD$ are all integers, but only one is even?

Question

This is just a question I came up with while working through another problem.

If we have a triangle $ABC$ and we draw a cevian from $A$ to $D$ on side $BC$, is it possible to assign integer values to lengths $AB$, $BC$, $AC$, $AD$, $BD$ or $CD$ such that only one of the listed lengths has an even valued length? If it is possible, what are these lengths?

I have done some parts of this problem on my own. I figured out that this even length could only be $BC$, $CD$, or $BD$, with $BD$ and $CD$ being the same case.

With $BC$ being the only even length I computed for $BC$ using Stewart's Theorem and got $$BC = \frac{(AC^2)(BD)+(AB^2)(CD)}{(BD)(CD)+(AD)^2}$$ where every length but $BC$ is odd. The equation seems to work but I don't know how to obtain its integer solutions.

Similarly with $BD$/$CD$ case, I applied Stewart's theorem and got $$BD=\frac{(AB^2)(CD)-(BC)(AD^2)}{(CD)(BC)-(AC)^2}$$ Once again this seems to work, but I don't know how to obtain its integer solutions.

Pretty much all I need help is verifying whether or not my work is right, and if it is how to find the integer solutions of my provided equations. If there is a better way or my work has flaws please let me know.

Answer 1

Too long for a comment but here is food for thought.

It would appear, incorrectly, that we can find what you need if we join two dissimilar Pythagorean triples on a matching side-A. This side and be any odd number greater than one. We begin with Euclid' formula here shown as.

$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$ If we solve the A-function for k, we can find triples that match a length "A" by testing a finite range of m-values to see which ones yield integers.

\begin{equation} A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2} \end{equation} The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$. $$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$ $$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$

If these two are joined at side-A, we have an acute triangle with sides $$(17,113,120)$$ with a cevian $(15)$ going from the $(17,113)$ vertex to the. 120-side, splitting it into lengths of $8$ and $112$. Unfortunately, this yields $3$ even lengths. If we were to join the even sides of two triples such as

$$(5,12,13)\qquad (35,12,37)$$ we would have $(13,37,40)$ with $40$ split into $(5,35)$. That would still leave two even lengths: the base $(40)$ and the cevian $(12)$. This probably eliminates Pythagorean triples from the solution. There are methods of finding integer cevians in a right triangle from a non-right vertex to an adjacent side but the question remains about the split-length of the target side being integers.

Answer 2

A simple solution found by brute force (by setting $AD=AB$):

and three more interesting ones found by exhaustive search:

Answer 3

According to Heron's formula, area and semi-perimeter are given as follows.
Sides a,b and c.
$Area=\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$

We consider the case of $AB=AD$.
Let $a=AB, b=AC, c=BC, a=AD, c1=BD, c2=DC$.

Let apply Heron's formula to triangle $(ABC, ABD, ADC)$.

Triangle $ABC$:
$s=\frac{a+b+c}{2}$
$Area=\sqrt{s(s-a)(s-b)(s-c)}$
$c2=c-c1$

Triangle $ABD$:
$s1=\frac{2a+c1}{2}$
$Area1=\sqrt{s1(s1-a)(s1-a)(s1-c1)}$

Triangle $ADC$:
$s2=\frac{a+b+c2}{2}$
$Area2=\sqrt{s2(s2-a)(s2-b)(s2-c2)}$

$Area=Area1+Area2.$
\begin{split} \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} &= c1\sqrt{(2a+c1)(2a-c1)} \\ &+ \sqrt{(c1-a-b-c)(a-b-c+c1)(a-b+c-c1)(a+b-c+c1)} \end{split}

By simple algebra, we know that equation $Area-Area1-Area2=0$ has a factor $-cc1+a^2-b^2+c^2$.
Thus problem was reduced to finding the integer solutions of $-cc1+a^2-b^2+c^2=0$.
We found the integer solutions of $-cc1+a^2-b^2+c^2=0$ with $(a,b,c)<30$.
Jack D'Aurizio's solutions are included.
Similarly, we can apply this method to general case.

            [a  b  c  c1 c2] 

            [ 3  7  8  3  5]
            [ 5  7  8  5  3]
            [ 5  9  8  1  7]
            [ 5 13 16  7  9]
            [ 5 23 24  3 21]
            [ 7 15 16  5 11]
            [ 7 17 16  1 15]
            [ 7 19 24 11 13]
            [ 9 11  8  3  5]
            [ 9 15 16  7  9]
            [ 9 17 16  3 13]
            [ 9 21 24  9 15]
            [11 13 16 13  3]
            [11 17 24 17  7]
            [11 19 16  1 15]
            [11 23 24  7 17]
            [11 25 24  3 21]
            [13 15  8  1  7]
            [13 17 24 19  5]
            [13 23 24  9 15]
            [13 25 24  5 19]
            [15 21 24 15  9]
            [15 27 24  3 21]
            [17 19 24 21  3]
            [17 23 16  1 15]
            [17 29 24  1 23]
            [19 21 16 11  5]
            [19 23 24 17  7]
            [19 25 24 13 11]
            [23 29 24 11 13]
            [25 29 24 15  9]
            [27 29 16  9  7]