A right-angled triangle has legs of length 126 and 168 units. What is the perimeter of the triangle determined by the feet of the angle bisectors?
Let's denote the vertices of the triangle in the usual way, and let \( AC = 126 \), \( BC = 168 \). From the Pythagorean theorem, we have \( AB = 210 \).

The angle bisector from \( A \) intersects \( CB \) at \( A_1 \), the angle bisector from \( B \) intersects \( AC \) at \( B_1 \), and the angle bisector from \( C \) intersects \( AB \) at \( C_1 \). We want to determine the perimeter of the triangle \( A_1 B_1 C_1 \).

It is known that the angle bisector divides the opposite side in the ratio of the adjacent sides, for example, \( CA_1 : A_1 B = AC : AB \). Let \( CA_1 = x \), \( CB_1 = y \), and \( BC_1 = z \), then

\[
x : (168 - x) = 126 : 210, \quad y : (126 - y) = 168 : 210, \quad \text{and} \quad z : (210 - z) = 168 : 126.
\]

From these, we get \( x = 63 \), \( y = 56 \), and \( z = 120 \).

The length of the side \( A_1 B_1 \) can be easily calculated from the right triangle \( A_1 B_1 C \):

\[
A_1 B_1 = \sqrt{63^2 + 56^2} = \sqrt{7105}
\]

The lengths of \( B_1 C_1 \) and \( A_1 C_1 \) can be calculated using the cosine rule in triangles \( AB_1 C_1 \) and \( BA_1 C_1 \), respectively, but we need to know the cosines of \( \angle CAB = \alpha \) and \( \angle CBA = \beta \).

We know that \( \cos \alpha = \frac{126}{210} \). Let's write the cosine rule for \( B_1 C_1 \):

\[
(B_1 C_1)^2 = 70^2 + 90^2 - 2 \cdot 70 \cdot 90 \cdot \frac{126}{210} = 5440
\]

Thus, \( B_1 C_1 = \sqrt{5440} \).

Similarly, from the triangle \( BA_1 C_1 \) and \( \cos \beta = \frac{168}{210} \):

\[
(A_1 C_1)^2 = 105^2 + 120^2 - 2 \cdot 105 \cdot 120 \cdot \frac{168}{210} = 5265
\]

Thus, \( A_1 C_1 = \sqrt{5265} \).

Finally, the perimeter we are looking for is

\[
K = \sqrt{7105} + \sqrt{5440} + \sqrt{5265} \approx 230.61 \text{ units. }
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_b6e867b02c924d468fc1g-1.jpg?height=529&width=703&top_left_y=1427&top_left_x=711)