This problem states that on $△ABC$, $AB=3$, $AC=1$, $AD=1$ $∠BAC=θ$. It says that there are two points that fullfill the condition $DP_0=\frac{1}{3}BC$ and asks for the range of $\cos \theta$ that is needed to fullfill the condition. These images are given:
The problem says that by looking at image#2, it can be known that the conditions for P to exist are $0\leq AP<AP_0$ and the explanation of the problem says that this is because $P$ is between $A$ and $P_0$ but I don't understand this. How can I get the conditions for $P$ to exist only by looking at the second image?
Applying law of cosine in $\triangle ABC$,
$BC^2 = 3^2 + 1^2 - 2 \cdot 3 \cos \theta = 10 - 6 \cos \theta$
Now in $\triangle ADP$, if $AP = x$,
$DP^2 = \frac{BC^2}{9} = 1^2 + x^2 - 2 x \cos \theta$
i.e $ \ \displaystyle \frac{10 - 6 \cos \theta}{9} = 1 - \cos^2 \theta + (x- \cos \theta)^2$
$(1 - 3 \cos \theta)^2 = 9 (x-\cos\theta)^2$
$1 - 3 \cos \theta = \pm (3x - 3 \cos \theta)$
So, $x = \frac{1}{3} $ or $x = \frac{6\cos\theta - 1}{3}$
Range of $\theta$ so we have two points on line segment $AC$ meeting the condition $DP = \frac{BC}{3}$,
we must have $0 \lt x \lt 1 \implies \frac{1}{6} \lt \cos\theta \lt \frac{2}{3}$.
Also note that when $\cos \theta = \frac{1}{3}$, we have only one point $P$ where $DP = \frac{BC}{3}$.