In right triangle $ ABC$ with right angle at $ C$, $ \angle BAC < 45$ degrees and $ AB \equal{} 4$.  Point $ P$ on $ AB$ is chosen such that $ \angle APC \equal{} 2\angle ACP$ and $ CP \equal{} 1$.  The ratio $ \frac{AP}{BP}$ can be represented in the form $ p \plus{} q\sqrt{r}$, where $ p,q,r$ are positive integers and $ r$ is not divisible by the square of any prime.  Find $ p\plus{}q\plus{}r$.
1. **Define the angles and segments:**
   Let $\angle ACP = \theta$ and $\angle APC = 2\theta$. Also, let $AP = x$ and $PB = 4 - x$ because $AB = 4$. Let $AC = y$.

2. **Apply the Law of Sines in $\triangle ACP$:**
   \[
   \frac{\sin \theta}{x} = \frac{\sin 2\theta}{y}
   \]
   Since $\sin 2\theta = 2 \sin \theta \cos \theta$, we can rewrite the equation as:
   \[
   \frac{\sin \theta}{x} = \frac{2 \sin \theta \cos \theta}{y}
   \]
   Simplifying, we get:
   \[
   \cos \theta = \frac{y}{2x}
   \]

3. **Draw a parallel line and use similar triangles:**
   Draw a line parallel to $AC$ through $P$ and let it intersect $CB$ at $D$. Since $AC \parallel PD$, $\angle ACP \cong \angle CPD$, so $\angle CPD = \theta$. In $\triangle CPD$, we have:
   \[
   \cos \theta = \frac{DP}{CP} = \frac{DP}{1}
   \]
   Therefore, $DP = \cos \theta$.

4. **Use similarity of triangles:**
   $\triangle BPD \sim \triangle BAC$, so:
   \[
   \frac{BP}{PD} = \frac{BA}{AC}
   \]
   Substituting the known values, we get:
   \[
   \frac{4 - x}{\cos \theta} = \frac{4}{y}
   \]
   Using $\cos \theta = \frac{y}{2x}$ from step 2, we substitute:
   \[
   \cos \theta = \frac{y(4 - x)}{4}
   \]

5. **Equate the two expressions for $\cos \theta$:**
   \[
   \frac{y}{2x} = \frac{y(4 - x)}{4}
   \]
   Simplifying, we get:
   \[
   \frac{1}{2x} = \frac{4 - x}{4}
   \]
   Cross-multiplying, we obtain:
   \[
   4 = 2x(4 - x)
   \]
   Simplifying further:
   \[
   4 = 8x - 2x^2
   \]
   Rearranging into a standard quadratic form:
   \[
   2x^2 - 8x + 4 = 0
   \]
   Dividing by 2:
   \[
   x^2 - 4x + 2 = 0
   \]

6. **Solve the quadratic equation:**
   Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -4$, and $c = 2$:
   \[
   x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}
   \]
   Since $AP = x$ must be positive and less than 4, we take $x = 2 + \sqrt{2}$.

7. **Find the ratio $\frac{AP}{BP}$:**
   \[
   \frac{AP}{BP} = \frac{x}{4 - x} = \frac{2 + \sqrt{2}}{4 - (2 + \sqrt{2})} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}}
   \]
   Rationalizing the denominator:
   \[
   \frac{2 + \sqrt{2}}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{(2 + \sqrt{2})^2}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{4 + 4\sqrt{2} + 2}{4 - 2} = \frac{6 + 4\sqrt{2}}{2} = 3 + 2\sqrt{2}
   \]

8. **Sum the integers $p, q, r$:**
   The ratio $\frac{AP}{BP}$ is $3 + 2\sqrt{2}$, so $p = 3$, $q = 2$, and $r = 2$. Therefore:
   \[
   p + q + r = 3 + 2 + 2 = 7
   \]

The final answer is $\boxed{7}$