Problem Analysis #28

[THOUGHT]

The problem gives us a triangle $ABC$ where $\angle A = \theta$, $AC=6$, and $B=(-5,0)$ while $A=(5,0)$. Point $P$ lies on segment $AC$ and $BP+CP=8$.

First, let's find the length of $BC$ using the Law of Cosines: [ BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\theta) ] Given that $AB=10$, $AC=6$, and $\cos\beta=\frac{4}{5}$, we can find $\theta$ from the given cosine value: [ \cos\theta = \frac{BC^2 - 100 + 36}{2 \cdot 5 \cdot 6} = \frac{BC^2-64}{60} = \cos\beta = \frac{4}{5} ] This leads to $BC^2=100$ and $BC=10$.

Now, let's express the area of triangle $BCP$, denoted by $f(\theta)$, as a function of $\theta$. The base $BP$ can be written as $8-CP$, so the area is: [ f(\theta) = \frac{1}{2} \cdot (8-CP) \cdot h_{BCP} ] where $h_{BCP}$ is the height from point $C$ to line $AB$.

Since $\triangle ABC$ is a right triangle with $\angle BAC=\theta$, we can use trigonometric ratios to find $h_{BCP}$: [ \tan\theta = \frac{BC}{AB} = \frac{10}{10}=1, \quad \therefore \sin\theta = 1, \cos\theta=0 ] Hence, [ h_{BCP} = BC \cdot \sin(\pi/2 - \theta) = 10 \cdot \sin(90^\circ-\arctan(1)) = 10 \cdot \frac{1}{\sqrt{2}} = 5\sqrt{2} ]

So the area function becomes: [ f(\theta) = \frac{1}{2} \cdot (8-CP) \cdot 5\sqrt{2} = -5\sqrt{2}(8-6\cos\theta) ] Differentiating $f$ with respect to $\theta$, we get: [ f'(\theta) = -5\sqrt{2}(-6\sin\theta) = 30\sqrt{2}\sin\theta ]

Given that $\cos\beta=\frac{4}{5}$, we can find $\sin\beta$ using the Pythagorean identity: [ \sin^2\beta + \cos^2\beta = 1 ] [ \sin^2\beta = 1 - (\frac{4}{5})^2 = \frac{9}{25} ] [ \sin\beta = \sqrt{\frac{9}{25}} = \frac{3}{5} ]

Since $\theta=\arccos(\cos\beta)$, we have: [ f'(\beta) = 30\sqrt{2}\sin(\arccos(4/5)) = 30\sqrt{2} \cdot \frac{3}{5} = 18\sqrt{2} ]

Now $p$ and $q$ are the integer part and square of the decimal part, respectively: [ p = 18, q=\sqrt{2} \approx 1.414 ] Hence, [ p + q^2 = 18 + (1.414)^2 \approx 19.33 ]

Since we need an integer answer, we take the floor of $19.33$: [ \boxed{19} ]