, 2020

We first observe the following properties.

Lemma 1: For any integer-point triangle whose area is an integer, exactly 1 or 3 of its sides have midpoints that are integer points.
Proof. By Pick's theorem, a triangle with integer area has an even number of integer points lying on its edges. Suppose $P$ is an integer point on some side $AB$ of the triangle. Let $M$ be the midpoint; then the reflection $P'$ of $P$ about $M$ is also an integer point (since $A + B = P + P'$), so the total number of integer points among the three vertices and the three midpoints must be even, hence an odd number of midpoints are integer points, as claimed. □

Lemma 2: There is no integer-point equilateral triangle.
Proof. The area of an integer-point triangle must be rational. But suppose there exists an integer-point equilateral triangle with side length $a$; then its area is $\frac{\sqrt{3}}{4}a^2$, and by the Pythagorean theorem $a^2$ is an integer, so the area is not rational, a contradiction. □

Lemma 3: If an isosceles integer-point triangle has area a power of 2, then its base is parallel to the $x$-axis, the $y$-axis, or the line $x = y$ or $x = -y$.
Proof. Suppose otherwise. Let the triangle be $ABC$ with base $BC$. Reflect $B$ about $A$ to get $B'$, and reflect $C$ about $A$ to get $C'$, obtaining a rectangle $BCB'C'$. This rectangle's area is also a power of 2. Suppose $BC = \sqrt{a^2 + b^2}$. If $p \mid a^2 + b^2$, then if the area of $BCB'C'$ is an integer it is also a multiple of $p$, so $a^2 + b^2 = 2^n$. Let $g = \gcd(a, b)$; then
$$g^2 \left( \left( \frac{a}{g} \right)^2 + \left( \frac{b}{g} \right)^2 \right) = 2^n$$
However, it is well known that if two numbers are not both even, their sum of squares is not divisible by 4, so $\left(\frac{a}{g}\right)^2 + \left(\frac{b}{g}\right)^2$ can only take the value 1 or 2. Hence $a = b = g$ or $ab = 0$, which correspond exactly to the cases mentioned in the lemma, as claimed. □

Lemma 4: There is no acute integer-point triangle with area $\frac{1}{2}$.
Proof. Suppose triangle $ABC$ has area $\frac{1}{2}$, where $\angle A$ is the largest angle and $b = AC, c = AB$; then $\angle A \ge \frac{\pi}{3}$.
$$\frac{1}{2}bc \sin A = \frac{1}{2} \Rightarrow bc = \frac{1}{\sin A} \le \frac{2}{\sqrt{3}}$$
Since $\sqrt{2} > \frac{2}{\sqrt{3}}$, we must have $b = c = 1$, but this clearly cannot hold, as claimed. □

Returning to the original problem, let us call an acute integer-point triangle of area $2^n$ an $n$-good triangle. Suppose there are $a_n$ scalene $n$-good triangles and $b_n$ isosceles $n$-good triangles. Furthermore, call a good triangle whose three side-midpoints are all integer points an even good triangle, and one with only one integer-point midpoint an odd good triangle.

Now, let us consider, separately for scalene and isosceles good triangles, a mapping that produces an $n$-good triangle from an $(n-1)$-good triangle:
- For a scalene $(n-1)$-good triangle $ABC$, consider a corresponding map $f(\Delta ABC, AB)$ sending it to a scalene $n$-good triangle, with the following rule: without loss of generality assume $CA < CB$; reflect $A$ about $C$ to get $A'$, and take the resulting triangle to be $ABA'$.
- For an isosceles $n$-good triangle $ABC$ with $AB = AC$, consider the corresponding map $g(\Delta ABC)$, defined as follows: if $AB > BC$, reflect $C$ about $B$ to get $C'$, and take the resulting triangle to be $ACC'$; if $AB < BC$, reflect $A$ about $B$ to get $A'$, and take the resulting triangle to be $ACA'$.

Now, running over all $(n-1)$-good triangles and their three sides, and applying $f, g$, we construct a total of $3a_{n-1} + b_{n-1}$ $n$-good triangles, but among these some $n$-good triangles may be constructed multiple times, so we need to count how many times each $n$-good triangle is produced. Consider the following three cases:
1. The $n$-good triangle is a scalene even triangle: connecting the midlines separately yields 3 $(n-1)$-good triangles, so each such triangle is produced 3 times. Moreover, taking the midpoint triangle of a scalene even $n$-good triangle gives a bijection onto all scalene $(n-2)$-good triangles, so in total these are produced $3a_{n-2}$ times.
2. The $n$-good triangle is a scalene odd triangle: connecting to the unique integer midpoint yields one $(n-1)$-good triangle, so each is produced once; subtracting the number of scalene even triangles gives that the number of scalene odd $n$-good triangles is $a_n - a_{n-2}$.
3. The $n$-good triangle is isosceles: regardless of whether the midpoint of the base is an integer point, connecting the midlines does not yield an acute triangle, so if it is produced at all it must be via a leg. Hence only even isosceles good triangles are produced, each exactly once, and by an argument similar to before, these are produced a total of $b_{n-2}$ times.

Combining the above discussion, we obtain the relation
$$3a_{n-1} + b_{n-1} = 3a_{n-2} + (a_n - a_{n-2}) + b_{n-2} \implies a_n - 3a_{n-1} + 2a_{n-2} + (b_{n-2} - b_{n-1}) = 0 \quad (1)$$

Now we compute $b_n$, splitting into two cases:
1. Even $n$-good triangles: there are $b_{n-2}$ of these.
2. Odd $n$-good triangles: let the triangle be $ABC$ with $AB = AC$, and let $M$ be the midpoint of $BC$, where Lemma 1 shows that $M$ must be an integer point. Since the triangle is acute, $MA > MB$; by Lemma 3, together with the fact that the area of $\triangle AMB$ is a power of 2, it is easy to see that $\frac{MA}{MB} = 2^k$, so the midpoint of $MA$ is an integer point. Applying Lemma 1 to $\triangle AMB$, we see that the midpoint of $AB$ being an integer point is equivalent to the midpoint of $BM$ being an integer point. By Lemma 3 again, all possible triangles for which the midpoint of $BM$ is not an integer point are those with $BC = 2$ and $BC = 2\sqrt{2}$, and it is easy to verify that both of these satisfy the required conditions.

Therefore, $b_n = b_{n-2} + 2$; since the base of a good isosceles triangle with integer area must be $2^n$ or $2^n\sqrt{2}$, we can deduce that $b_0 = 0, b_1 = 1$, so
$$b_n = n$$

Substituting this result back into equation (1):
$$a_n - 3a_{n-1} + 2a_{n-2} - 1 = 0 \quad (2)$$
Rearranging gives
$$(a_n + n) - 3(a_{n-1} + (n-1)) + 2(a_{n-2} + (n-2)) = 0$$
Since the characteristic equation $x^2 - 3x + 2 = 0$ has roots 1, 2, this second-order recurrence has general solution $a_n + n = p2^n + q$.

Since an integer-point triangle must always have at least one side whose midpoint is an integer point, if there existed a scalene $0$-good triangle then there would have to exist an acute integer-point triangle of area $\frac{1}{2}$. By Lemma 4 this is impossible, so $a_0 = 0$.
Similarly, if there existed a scalene $1$-good triangle, then there would have to exist a $0$-good triangle, but there is none, so
$a_1 = 0$.

In summary, $a_0 + 0 = 0, a_1 + 1 = 1$; substituting gives $p + q = 0, 2p + q = 1$, solving gives $p = 1, q = -1$, and hence
$$a_n = 2^n - 1 - n$$

The answer sought is therefore
$$a_{2020} + b_{2020} = 2^{2020} - 1 - n + n = 2^{2020} - 1$$