analytic geometrygeometrytriangles
I recently got fascinated by this question:
Does there exist a scalene triangle $\triangle ABC$ where $AB, AC, $ and $BC$ are integers and $\triangle ABC$ has an integer height?
There is some ambiguity around where the 'height' is but I assume that it is perpendicular from any of the sides.
Cheers!
EDIT: Right-angled triangles are not permitted, I think the question wants a generalisation, if it exists…
No, none of the lengths can be prime.
Let me write $d,e,f$ for $h_a,h_b,h_c$, then twice the area is $$ 2\mathrm{Area} = ad=be=cf $$
Furthermore let $c=x+y$ with $x,y$ possibly signed lengths satisfying $x^2+f^2=a^2$ and $y^2+f^2=b^2$, so $x,y$ are (possibly signed) integers.
Consider the case in which a side length is prime, wlog let it be $c$.
Then since $c>d$, $c\nmid d$, but $c\mid ad$ so we must have $a=tc$, and similarly $c>e$ so $b=uc$ for integers $t,u$. Since $a,b,c$ make a triangle $a<b+c$ and $b<a+c$ we have $t<u+1,u<t+1\implies t=u$ and the triangle must be isosceles with $a=b$.
Then $x=y=c/2$, but $a^2=f^2+x^2$ requires $x$ to be an integer, so $c$ is even and can only be prime if $c=2$. But then $a^2=f^2+1$ which is impossible for positive integers $a,f$, so there cannot be such a triangle with a prime side.
Consider the case in which a height is prime, wlog let it be $f$.
If $f\mid a$ then $f\mid x$ and $(x/f)^2+1=(a/f)^2$ which is impossible for nonzero integers $x/f,a/f$. So $f\nmid a$ and $f\mid ad \implies d=tf$. Similarly $f\nmid b$ and so $e=uf$. Then from $ad=be=cf$ we have $c=ta=ub$. Wlog $a\ge b$, then from $c<a+b$ we have $(t-1)a<b \implies t=1$ and the triangle must be isosceles with $a=c$.
Then with $c=ub,e=uf$ and by symmetry the altitude bisecting $b$ $$ (b/2)^2+e^2=c^2 \\ (b/2)^2+u^2f^2=u^2b^2 \\ f^2=\left(\frac{b}{2u}\right)^2(4u^2-1) $$ Since $f$ is an integer and $\gcd(2u,4u^2-1)=1$ we must have $2u\mid b$. Since we have already shown $f\nmid b$ and by assumption $f$ is prime, then $\gcd(f,b)=1$ and we must have $b/2u=1$. Then $f^2=4u^2-1$, but this is impossible for positive integers $u,f$, so there cannot be such a triangle with prime height.
Let $A'$ be the midpoint of $BC$; let $B'$ be the midpoint of $AC$. Let $M$ be the intersection point of the medians. We will find the lengths of the medians $AA'$ and $BB'$ from the fact that the medians are divided by the intersection point $M$ forming the ratio $2/1$, that is, $$ {AM\over A'M} = {BM\over B'M} = 2. $$ Suppose that $${1\over3}AA'=MA'=x, \qquad {1\over3}BB'=MB'=y.$$
By the Pythagorean theorem, from the right triangles $AMB'$ and $BMA'$ we have the system of equations $$ 4x^2 + y^2 = 9, $$ $$ x^2 + 4y^2 = 16. $$ Solving these equations we find $$ x = {2\over\sqrt3}, \qquad y = \sqrt{11\over3}. $$ Therefore, the medians are $$ AA' = 3x = {2\sqrt3}, \qquad BB' = 3y = \sqrt{33}. $$ We can find the third side, $AB$, from the right triangle $AMB$: $$ AB^2 = AM^2 +BM^2 = 4x^2 + 4y^2 = 20. $$ So we now know all tree sides of triangle $ABC$: $$ a = BC = 8, \qquad b=AC=6, \qquad c = AB = \sqrt{20}. $$ Finally, we can find the area of $ABC$ using an alternative form of Heron's formula: $$ \mbox{Area}(ABC)=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =4\sqrt{11}\approx13.266. $$
Best Answer
A nice example is
and if you multiply these numbers by $195$ you get the more impressive example with all three heights integers: