Let $ABC$ be a triangle with $\angle A = 60^o$. Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to ℓ. Given that $AB = 20$ and $AC = 22$, compute $XY$ .
1. **Identify the given information and set up the problem:**
   - We have a triangle \(ABC\) with \(\angle A = 60^\circ\).
   - Line \(\ell\) intersects segments \(AB\) and \(AC\) and splits triangle \(ABC\) into an equilateral triangle and a quadrilateral.
   - Points \(X\) and \(Y\) are on \(\ell\) such that lines \(BX\) and \(CY\) are perpendicular to \(\ell\).
   - Given \(AB = 20\) and \(AC = 22\), we need to compute \(XY\).

2. **Set up the coordinate system:**
   - Place \(A\) at \((0, 0)\).
   - Since \(\angle A = 60^\circ\), place \(B\) at \((20, 0)\) and \(C\) at \((11, 11\sqrt{3})\).

3. **Determine the coordinates of \(X\) and \(Y\):**
   - Since \(BX\) and \(CY\) are perpendicular to \(\ell\), and \(\ell\) is parallel to the base of the equilateral triangle, \(X\) and \(Y\) must lie on a line parallel to \(BC\).
   - The height of the equilateral triangle is \(\frac{\sqrt{3}}{2} \times 22 = 11\sqrt{3}\).

4. **Calculate the coordinates of \(X\) and \(Y\):**
   - Since \(X\) and \(Y\) are on the line parallel to \(BC\) and at the same height, we can place \(X\) at \((1, 11\sqrt{3})\) and \(Y\) at \((22, 11\sqrt{3})\).

5. **Compute the distance \(XY\):**
   - The distance between \(X\) and \(Y\) is the horizontal distance between their x-coordinates.
   - \(XY = |22 - 1| = 21\).

The final answer is \(\boxed{21}\).