To find the minimum value of $ PC + \frac{1}{2}PA $ geometrically, we can use the following approach:
Identify Points and Segments:
- Given points: $A(1, 0)$, $B(5, 0)$, and $C(0, 5)$.
- Circle centered at $B$ with radius $2$.
- Let $P$ be a point on the circle.
- Select point $ F $ on segment $ AB $ such that $ BF = 1 $.
Triangles and Similarity:
- Since $ PB = 2 $ and $ BF = 1 $: $$ \frac{BF}{PB} = \frac{1}{2} $$
- Triangles $ \triangle PBF $ and $ \triangle ABP $ are similar because: $$ \frac{PB}{AB} = \frac{2}{4} = \frac{1}{2} = \frac{BF}{PB} $$ Therefore, by AA similarity ($\angle PBF = \angle ABP$), we have: $$ \triangle PBF \sim \triangle ABP $$
Proportional Segments:
- Since the triangles are similar: $$ \frac{PF}{PA} = \frac{BF}{PB} = \frac{1}{2} \implies PF = \frac{1}{2}PA $$
Inequality and Minimum Value:
- From the above proportional relationship: $$ PC + \frac{1}{2}PA = PC + PF \geq CF $$
- When $ C $, $ P $, and $ F $ are collinear, $ PC + \frac{1}{2}PA $ is minimized, and this minimum value is the length of segment $ CF $.
Calculate ( CF ):
- Since $ O $ is the origin, $ OC = 5 $ and $ OF = OB - 1 = 5 - 1 = 4 $: $$ CF = \sqrt{OC^2 + OF^2} = \sqrt{5^2 + 4^2} = \sqrt{41} $$
Thus, the minimum value of $ PC + \frac{1}{2}PA $ is: $$ \sqrt{41} $$ Is this correct?
