At what angle does the hypotenuse of a right-angled triangle with a perimeter of 2 units appear when viewed from a point on the internal angle bisector of the right angle, $\sqrt{2}$ units away from the vertex?
I. solution. Let the right angle vertex of the $ABC$ triangle be $C$. Then, by extending the sides $CA$ and $CB$, we can place the triangle inside a unit square $CGPF$. Since the length of the diagonal $CP$ is $\sqrt{2}$, $P$ is precisely the point we are looking for, and we need to determine the angle $BPA$. (Note: it can be proven that the legs of the $ABC$ triangle are shorter than 1 unit.)

![](https://cdn.mathpix.com/cropped/2024_05_02_41e6535f1215e0ffc035g-1.jpg?height=456&width=501&top_left_y=344&top_left_x=801)

Let the sides of the $ABC$ triangle be denoted by $a, b,$ and $c$ in the usual way, and let the foot of the perpendicular from $P$ to $AB$ be $E$, and let $PE = x$ (see the figure). The area of the square is 1, so

$$
\frac{x c}{2} + \frac{a b}{2} + \frac{1-a}{2} + \frac{1-b}{2} = 1
$$

Since the perimeter of the triangle is 2, we have

$$
c = 2 - a - b
$$

Substituting this into the previous equation, we get

$$
x = \frac{a + b - ab}{2 - a - b}
$$

By the Pythagorean theorem, $a^2 + b^2 = c^2 = (2 - a - b)^2$. Squaring and simplifying this equation, we get $4 - 4a - 4b + 2ab = 0$, from which $2 - a - b = a + b - ab$. Substituting this into the previous equation for $x$, we get $x = 1$.

Since $GP = EP = 1$, the right triangles $PGB$ and $PEB$ are congruent. Similarly, the congruence of triangles $PEA$ and $PFA$ can be shown. This implies that $\angle GPC = \angle BPE$ and $\angle EPA = \angle APF$. Therefore, the angle $\angle BPA$ is half of the right angle $\angle GPF$, i.e., $\angle BPA = \angle BPE + \angle EPA = 45^\circ$.

Note. From the equality $PG = PE = PF$, we can conclude that point $P$ is the center of the circle circumscribed around the hypotenuse of the triangle. It is known that in a right triangle, the radius of the circle circumscribed around the hypotenuse is equal to half the perimeter, which is 1 unit in this case. This statement can essentially be proven using the reasoning in the above solution. Furthermore, the lengths of the tangents drawn from an external point to a circle are equal, so $PGBE$ and $PEAF$ are kites, from which the solution to the problem can also be obtained.

II. solution. Using the diagram from the previous solution, let $\angle APF = \alpha$ and $\angle GPC = \beta$. The equation $4 - 4a - 4b + 2ab = 0$ derived in the previous solution can be rearranged as

$$
a = \frac{2 - 2b}{2 - b} = 2 - \frac{2}{2 - b}
$$

Then $AF = 1 - b$ and $BG = 1 - a = \frac{2}{2 - b} - 1$. Furthermore,

$$
\tan \alpha = \frac{1 - b}{1}, \quad \text{and} \quad \tan \beta = \frac{\frac{2}{2 - b} - 1}{1}
$$

Using the tangent addition formula,

$$
\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{(1 - b) + \left(\frac{2}{2 - b} - 1\right)}{1 - (1 - b)\left(\frac{2}{2 - b} - 1\right)} = \frac{(1 - b)(2 - b) + b}{(2 - b) - (1 - b)b} = 1
$$

Thus, $\tan(\alpha + \beta) = 1$, which implies $\alpha + \beta = 45^\circ$. Since $\angle BPA = 90^\circ - (\alpha + \beta) = 45^\circ$, the measure of the desired angle is $45^\circ$.

Note. Using trigonometric functions, another solution can be obtained by first applying the cosine rule to the sides $PB$ and $PA$ of triangles $PBC$ and $CAP$ (from which $PB^2 = a^2 - 2a + 2$ and $PA^2 = b^2 - 2b + 2$ follow), and then applying the cosine rule again to the side $BA$ of triangle $PBA$ (which gives $\cos \angle APB = \frac{1}{\sqrt{2}}$ as the final result). Some solved this problem using coordinate geometry, in which case the measure of angle $\angle APB$ can be calculated from the dot product of vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$.

III. solution. Using the notation from the previous solutions. Rotate the triangle $PGB$ $90^\circ$ counterclockwise around $P$. Since $PG = PF$, $\angle GPF = 90^\circ$, and $\angle PGB + \angle PFA = 180^\circ$, after the rotation, we get triangle $PFH$, whose vertex $H$ lies on the line $AF$.

![](https://cdn.mathpix.com/cropped/2024_05_02_41e6535f1215e0ffc035g-2.jpg?height=409&width=555&top_left_y=576&top_left_x=774)

Since $PA = PA$, $PB = PH$ (due to the rotation), and $AH = AF + FH = (1 - b) + (1 - a) = 2 - a - b = c = AB$, the triangles $APB$ and $APH$ are congruent, which implies $\angle BPA = \angle APH$. By the rotation, $\angle BPH = 90^\circ$, so $\angle BPA = \frac{90^\circ}{2} = 45^\circ$.