Question: In triangle ABC, all three side lengths are integers and angle A = 60 degrees. The point D lies on the side AC so that BC = BD and DC = 2. Find the smallest possible area of the triangle ABC.
I know that because BC=BD, one of the inner triangles is isosceles. However, I do not understand how this knowledge can help me solve the problem.
Best Answer
The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us another Diophantine constraint. Draw the altitude $BH$ from $B$ onto $AC$:
Since $\triangle BCD$ is isosceles, $H$ is the midpoint of $CD$, so $CH = 1$; by looking at the right triangle $\triangle BHC$, we get $\cos \angle C = \frac{CH}{BC} = \frac1a$. The Law of Cosines tells us that $$c^2 = a^2 + b^2 - 2b.$$
Substituting the first equation into the second yields $$c^2 = (b^2 + c^2 - bc) + b^2 - 2b \implies bc = 2b^2 - 2b \implies c=2(b-1),$$ and substituting this value of $c$ back into the first equation tells us that $$a^2 = b^2 + (2b-2)^2 - b(2b-2) \implies a^2 - 3(b-1)^2 = 1,$$ which is a Pell equation. The simplest solution to $x^2-3y^2=1$ is $x=2,y=1$, and all other solutions are obtained by taking the coefficients of $1$ and $\sqrt3$ in $(2 + \sqrt3)^k$ for some $k$.
If we use $x=2$ and $y=1$, then this gives us $a=b=c=2$, which is certainly a solution to the problem: in this case, $\triangle ABC$ is equilateral and $D=A$. This triangle has area $\sqrt 3$.
If you don't like this solution, because it's a somewhat degenerate case, then $(2+\sqrt3)^2 = 7+4\sqrt3$ yields $x=7$ and $y=4$ as the next solution. In this case, $a=7$, $b=5$, and $c=8$, which is the triangle I used for the diagram. By Heron's formula, $\triangle ABC$ has area $10 \sqrt 3$.
One of these, depending on whether you find the equilateral solution acceptable or not, is the solution with the smallest area, but we can find infinitely many triangles with integer sides that satisfy the conditions, by taking higher powers of $2+\sqrt 3$ to solve Pell's equation.