Given triangle $ ABC$ of area 1. Let $ BM$ be the perpendicular from $ B$ to the bisector of angle $ C$. Determine the area of triangle $ AMC$.
1. **Identify the given elements and their properties:**
   - Triangle \(ABC\) has an area of 1.
   - \(BM\) is the perpendicular from \(B\) to the angle bisector of \(\angle C\).
   - We need to determine the area of triangle \(AMC\).

2. **Extend \(BM\) to meet \(CA\) at \(D\):**
   - Since \(BM\) is perpendicular to the angle bisector of \(\angle C\), \(BM\) is also perpendicular to \(CD\) (where \(D\) is the point where the angle bisector intersects \(CA\)).

3. **Analyze the properties of \(CM\):**
   - \(CM\) is both an angle bisector and an altitude of \(\triangle BCD\).
   - This implies that \(\triangle BMC\) is isosceles with \(BM = CM\).

4. **Determine the area of \(\triangle AMC\):**
   - Since \(BM\) is perpendicular to \(CD\), \(\triangle BMC\) is a right triangle.
   - The area of \(\triangle BMC\) is \(\frac{1}{2} \times BM \times CM\).

5. **Relate the area of \(\triangle AMC\) to the area of \(\triangle ABC\):**
   - Since \(BM\) is perpendicular to \(CD\) and \(CM\) is the angle bisector, \(\triangle BMC\) is half of \(\triangle ABC\).
   - Therefore, the area of \(\triangle BMC\) is \(\frac{1}{2}\) of the area of \(\triangle ABC\), which is \(\frac{1}{2} \times 1 = \frac{1}{2}\).

6. **Calculate the area of \(\triangle AMC\):**
   - Since \(\triangle AMC\) shares the same base \(AC\) and height from \(M\) as \(\triangle BMC\), the area of \(\triangle AMC\) is also \(\frac{1}{2}\) of the area of \(\triangle ABC\).

The final answer is \(\boxed{\frac{1}{2}}\).