Two congruent right-angled isosceles triangles (with baselength 1) slide on a line as on the picture. What is the maximal area of overlap?

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=287[/img]
To find the maximal area of overlap between two congruent right-angled isosceles triangles with base length 1, we need to consider two cases: when the overlapping area forms a triangle and when it forms a pentagon.

### Case 1: The Overlapping Area is a Triangle

1. Let \( x \) be the length of the segment along the base where the two triangles overlap.
2. The base of the overlapping triangle is \( 1 - x \).
3. Since the triangles are congruent right-angled isosceles triangles, the height of the overlapping triangle is \( \frac{1 - x}{2} \).
4. The area \( A \) of the overlapping triangle is given by:
   \[
   A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (1 - x) \times \frac{1 - x}{2} = \frac{1}{4} (1 - x)^2
   \]
5. To find the critical points, we differentiate \( A \) with respect to \( x \):
   \[
   \frac{dA}{dx} = \frac{d}{dx} \left( \frac{1}{4} (1 - x)^2 \right) = \frac{1}{4} \times 2(1 - x)(-1) = -\frac{1}{2} (1 - x)
   \]
6. Setting the derivative equal to zero to find the critical points:
   \[
   -\frac{1}{2} (1 - x) = 0 \implies x = 1
   \]
7. The second derivative test:
   \[
   \frac{d^2A}{dx^2} = \frac{d}{dx} \left( -\frac{1}{2} (1 - x) \right) = \frac{1}{2}
   \]
   Since the second derivative is positive, \( x = 1 \) is a minimum point.

### Case 2: The Overlapping Area is a Pentagon

1. Let \( x \) be the length of the segment along the base where the two triangles overlap.
2. The pentagon can be split into two trapezoids.
3. The height \( h \) of each trapezoid is \( \frac{1 - x}{2} \).
4. The bases of the trapezoids are \( a = x \) and \( b = \frac{1 + x}{2} \).
5. The area \( A \) of the pentagon is:
   \[
   A = 2 \times \left( \frac{1}{2} \times h \times (a + b) \right) = h \times (a + b) = \frac{1 - x}{2} \times \left( x + \frac{1 + x}{2} \right)
   \]
6. Simplifying the expression:
   \[
   A = \frac{1 - x}{2} \times \left( x + \frac{1 + x}{2} \right) = \frac{1 - x}{2} \times \frac{3x + 1}{2} = \frac{1}{4} (1 - x)(3x + 1)
   \]
7. To find the critical points, we differentiate \( A \) with respect to \( x \):
   \[
   \frac{dA}{dx} = \frac{d}{dx} \left( \frac{1}{4} (1 - x)(3x + 1) \right) = \frac{1}{4} \left( (1 - x) \cdot 3 + (3x + 1) \cdot (-1) \right) = \frac{1}{4} (3 - 3x - 3x - 1) = \frac{1}{4} (2 - 6x)
   \]
8. Setting the derivative equal to zero to find the critical points:
   \[
   \frac{1}{4} (2 - 6x) = 0 \implies 2 - 6x = 0 \implies x = \frac{1}{3}
   \]
9. Substituting \( x = \frac{1}{3} \) back into the area formula:
   \[
   A = \frac{1}{4} (1 - \frac{1}{3})(3 \cdot \frac{1}{3} + 1) = \frac{1}{4} \left( \frac{2}{3} \cdot \frac{4}{3} \right) = \frac{1}{4} \cdot \frac{8}{9} = \frac{2}{9}
   \]
10. The second derivative test:
    \[
    \frac{d^2A}{dx^2} = \frac{d}{dx} \left( \frac{1}{4} (2 - 6x) \right) = -\frac{3}{2}
    \]
    Since the second derivative is negative, \( x = \frac{1}{3} \) is a maximum point.

Conclusion:
The maximal area of overlap is achieved in Case 2 when \( x = \frac{1}{3} \), and the maximal area is \( \frac{2}{9} \).

The final answer is \( \boxed{\frac{2}{9}} \).