For each $a$, $1<a<2$, the graphs of functions $y=1-|x-1|$ and $y=|2x-a|$ determine a figure. Prove that the area of this figure is less than $\frac13$.
1. **Identify the points of intersection:**
   - The function \( y = 1 - |x - 1| \) is a V-shaped graph with a vertex at \( (1, 1) \). It intersects the x-axis at \( x = 0 \) and \( x = 2 \).
   - The function \( y = |2x - a| \) is also a V-shaped graph with a vertex at \( x = \frac{a}{2} \). It intersects the y-axis at \( y = a \) and the x-axis at \( x = \frac{a}{2} \).

2. **Determine the points of intersection:**
   - To find the points of intersection, we solve \( 1 - |x - 1| = |2x - a| \).
   - For \( x \leq 1 \), \( 1 - (1 - x) = x \) and \( |2x - a| = a - 2x \).
     \[
     x = a - 2x \implies 3x = a \implies x = \frac{a}{3}
     \]
     Substituting \( x = \frac{a}{3} \) into \( y = 1 - |x - 1| \):
     \[
     y = 1 - \left| \frac{a}{3} - 1 \right| = 1 - \left| \frac{a - 3}{3} \right| = \frac{a}{3}
     \]
     So, one point of intersection is \( \left( \frac{a}{3}, \frac{a}{3} \right) \).

   - For \( x \geq 1 \), \( 1 - (x - 1) = 2 - x \) and \( |2x - a| = 2x - a \).
     \[
     2 - x = 2x - a \implies 3x = a + 2 \implies x = \frac{a + 2}{3}
     \]
     Substituting \( x = \frac{a + 2}{3} \) into \( y = 1 - |x - 1| \):
     \[
     y = 1 - \left| \frac{a + 2}{3} - 1 \right| = 1 - \left| \frac{a - 1}{3} \right| = \frac{4 - a}{3}
     \]
     So, another point of intersection is \( \left( \frac{a + 2}{3}, \frac{4 - a}{3} \right) \).

3. **Identify the vertices of the quadrilateral:**
   - The vertices of the quadrilateral are \( \left( \frac{a}{3}, \frac{a}{3} \right) \), \( (1, 1) \), \( \left( \frac{a + 2}{3}, \frac{4 - a}{3} \right) \), and \( \left( \frac{a}{2}, 0 \right) \).

4. **Calculate the area of the quadrilateral:**
   - Using the Shoelace formula for the area of a quadrilateral with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), and \((x_4, y_4)\):
     \[
     \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
     \]
     Substituting the vertices:
     \[
     \text{Area} = \frac{1}{2} \left| \frac{a}{3} \cdot 1 + 1 \cdot \frac{4 - a}{3} + \frac{a + 2}{3} \cdot 0 + \frac{a}{2} \cdot \frac{a}{3} - \left( \frac{a}{3} \cdot 1 + 1 \cdot \frac{a + 2}{3} + \frac{4 - a}{3} \cdot \frac{a}{2} + 0 \cdot \frac{a}{3} \right) \right|
     \]
     Simplifying the expression:
     \[
     \text{Area} = \frac{1}{2} \left| \frac{a}{3} + \frac{4 - a}{3} + \frac{a^2}{6} - \left( \frac{a}{3} + \frac{a + 2}{3} + \frac{a(4 - a)}{6} \right) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{a + 4 - a + a^2}{6} - \left( \frac{a + a + 2 + a(4 - a)}{6} \right) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{4 + a^2}{6} - \frac{2a + 2 + 4a - a^2}{6} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{4 + a^2 - 6a + a^2}{6} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{2a^2 - 6a + 4}{6} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{a^2 - 3a + 2}{3} \right|
     \]
     \[
     = \frac{1}{6} \left| a^2 - 3a + 2 \right|
     \]
     \[
     = \frac{1}{6} \left| (a - 1)(a - 2) \right|
     \]

5. **Find the maximum area:**
   - Since \( 1 < a < 2 \), \( (a - 1) \) and \( (a - 2) \) are both negative, making their product positive.
   - The maximum value of \( (a - 1)(a - 2) \) in the interval \( 1 < a < 2 \) is achieved as \( a \to 2^- \):
     \[
     \lim_{a \to 2^-} (a - 1)(a - 2) = 0
     \]
   - However, the supremum of the area is:
     \[
     \frac{1}{6} \left| (a - 1)(a - 2) \right| \leq \frac{1}{6} \cdot 1 = \frac{1}{6}
     \]

6. **Conclusion:**
   - The area of the figure is less than \( \frac{1}{3} \).

The final answer is \( \boxed{ \frac{1}{3} } \)