Problem 1850

Let $ABC$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_1$, $B_1$, $C_1$ be the projections of $P$ onto the sides $BC$, $CA$, and $AB$, respectively. We seek the locus of points $P$ such that $AA_1$, $BB_1$, and $CC_1$ are concurrent and $\angle PAB + \angle PBC + \angle PCA = 90^\circ$.

To solve this, we consider the possible locations for $P$. The only possible points are the incenter, circumcenter, and orthocenter of $\triangle ABC$. We can verify that all three points satisfy the given conditions using Ceva's Theorem and Trigonometric Ceva's Theorem.

Suppose $P$ is in the locus. Define $x_1 = \angle PAB$, $x_2 = \angle PBC$, $x_3 = \angle PCA$, $y_1 = \angle PAC$, $y_2 = \angle PBA$, and $y_3 = \angle PCB$. By Trigonometric Ceva's Theorem, we have:
$$\sin x_1 \sin x_2 \sin x_3 = \sin y_1 \sin y_2 \sin y_3.$$

Next, we observe that $AC_1 = PA \cos x_1$, and similarly for the other five segments. By Ceva's Theorem, we have:
$$\cos x_1 \cos x_2 \cos x_3 = \cos y_1 \cos y_2 \cos y_3.$$

Since the sum of all six angles is $\pi$, we get:
$$x_1 + x_2 + x_3 = y_1 + y_2 + y_3 = \frac{\pi}{2}.$$

Conversely, if $P$ satisfies these three conditions, then $P$ is in the locus (since Ceva's and Trigonometric Ceva's Theorems are if-and-only-if statements).

In fact, we can prove that if $P$ satisfies the conditions, then $\{x_1, x_2, x_3\} = \{y_1, y_2, y_3\}$. Note that all six angles are in $(0, \pi/2)$, so it suffices to show $\{\tan x_1, \tan x_2, \tan x_3\} = \{\tan y_1, \tan y_2, \tan y_3\}$.

First, note that $\tan x_1 \tan x_2 \tan x_3 = \tan y_1 \tan y_2 \tan y_3$ by dividing the sine equation by the cosine equation. Next, note that:
$$0 = \cos(x_1 + x_2 + x_3) = \cos x_1 \cos x_2 \cos x_3 - \sin x_1 \cos x_2 \cos x_3 - \cos x_1 \sin x_2 \cos x_3 - \cos x_1 \cos x_2 \sin x_3.$$

Dividing by $\cos x_1 \cos x_2 \cos x_3$ (which is positive) and rearranging, we get:
$$\tan x_1 \tan x_2 + \tan x_2 \tan x_3 + \tan x_3 \tan x_1 = 1.$$

The same identity holds for $y_1, y_2, y_3$ as well, so:
$$\tan x_1 \tan x_2 + \tan x_2 \tan x_3 + \tan x_3 \tan x_1 = \tan y_1 \tan y_2 + \tan y_2 \tan y_3 + \tan y_3 \tan y_1.$$

Now, note that:
$$1 = \sin(x_1 + x_2 + x_3) = \sin x_1 \cos x_2 \cos x_3 + \cos x_1 \sin x_2 \cos x_3 + \cos x_1 \cos x_2 \sin x_3 - \sin x_1 \sin x_2 \sin x_3.$$

Dividing by $\cos x_1 \cos x_2 \cos x_3$ and rearranging, we get:
$$\frac{1}{\cos x_1 \cos x_2 \cos x_3} + \tan x_1 \tan x_2 \tan x_3 = \tan x_1 + \tan x_2 + \tan x_3.$$

However, the same identity holds for $y_1, y_2, y_3$, and the left-hand side doesn't change when we replace $x_1, x_2, x_3$ with $y_1, y_2, y_3$. Thus:
$$\tan x_1 + \tan x_2 + \tan x_3 = \tan y_1 + \tan y_2 + \tan y_3.$$

Thus, the three symmetric sums of $\{\tan x_1, \tan x_2, \tan x_3\}$ and $\{\tan y_1, \tan y_2, \tan y_3\}$ are equal, which means that $\{\tan x_1, \tan x_2, \tan x_3\} = \{\tan y_1, \tan y_2, \tan y_3\}$ and thus $\{x_1, x_2, x_3\} = \{y_1, y_2, y_3\}$.

We now consider cases based on $x_1$:

Case 1: $x_1 = y_1$. Then $(x_2, x_3) \neq (y_3, y_2)$, so $x_2 = y_2$ and $x_3 = y_3$. This implies that $P$ lies on each angle bisector, so $P = I$ (the incenter).

Case 2: $x_1 = y_2$. Then $x_2 \neq y_1$, so $x_2 = y_3$ and $x_3 = y_1$. This implies that $PA = PB = PC$, so $P = O$ (the circumcenter).

Case 3: $x_1 = y_3$. Then $x_3 \neq y_1$, so $x_3 = y_2$ and $x_2 = y_1$. Then, we see that $\angle A + \angle BPC = x_1 + y_1 + (180^\circ - x_2 - y_3) = 180^\circ$, so the reflection of $P$ over $BC$ lies on $(ABC)$. This implies that $P$ lies on $(BHC)$, and similarly it lies on $(AHB)$ and $(CHA)$, so $P = H$ (the orthocenter).

We have exhausted all cases for $x_1$, so the locus of points $P$ is the set of the incenter, circumcenter, and orthocenter of $\triangle ABC$.

The answer is: $\boxed{\text{the incenter, circumcenter, and orthocenter of } \triangle ABC}$.