9-10th grade olympiad problem

Question

So first things first, I apologize for my bad English in advance.
In one math olympiad (for 9-10th grade) there was a problem:
There is given a rectangle $ ABCD $ such that $AB=1$ and $BC=2$. There is a point $P$ on diagonal $BD$ and a point $Q$ on $BC$. Find the lowest possible value of $CP+PQ$.
So what I tried to do was to prove that only if PQ $\bot$ BC we get the least possible value of PQ. Therefore, we get right-angled $\bigtriangleup CPQ$, where $\angle CQP=90°$. By using Pythagorean theorem, we get that $CP+PQ=\sqrt{CQ^2+PQ^2}+PQ$. I tried to find out if there is any dependence on CQ and PQ. Apparently, there is, such that $PQ=\frac{2-CQ}{2}$. I tried to write that in $\sqrt{CQ^2+PQ^2}+PQ$ and what I got was $\sqrt{1,25CQ^2-CQ+1}+1-\frac{CQ}{2}$. What I did next was one of the dumbest decisions I could have possibly made there. Since I didn't know how to calculate the minimum value of this expression of $CP+PQ$, I tried to make $\sqrt{1,25CQ^2-CQ+1}=1$, which was pretty logical (so we would get a rational answer), but absolutely not proved. So from this equation I got that $CQ=\frac{4}{5}$ and so $CP+PQ=\sqrt{1,25CQ^2-CQ+1}+1-\frac{CQ}{2}=1,6$ (by the way, $1,6$ IS the correct answer). So what I wanted to ask you was, is there any smart way to find the minimum value of $\sqrt{1,25CQ^2-CQ+1}+1-\frac{CQ}{2}$ (this is basically all I needed to get full 7 points for this problem).

Answer 1

Let $C'$ and $Q'$ be the reflectional images of $C$ and $Q$ respectively in $BD$. Then $Q'$ lies on $BC'$ and $PQ'=PQ$. It is equivalent to find the smallest possible value of $CP+PQ'$.

For a fixed $Q$ (and hence a fixed $Q'$), $CP+PQ'$ is the smallest when $CQ'$ is a straight line. So what we have to find is the minimum distance between $C$ and $BC'$. This minimum distance is equal to the perpendicular distance from $C$ to $BC'$, i.e.

$$BC\cdot\sin\angle CBC'=2\sin2\angle CBD=4\sin\angle CBD\cos\angle CBD=4\left(\frac{1}{\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right)=\frac{8}{5}$$

Answer 2

Consider a cartesian coordinate system with$$A(0;0),B(1;0),C(1;2),D(0;2)$$ then we get $$P(x;-2x+2)$$ and $$Q'(1;-2x+2)$$ and we get $$CP=\sqrt{(1-x)^2+(-2x+2-2)^2}$$ and $$PQ'=\sqrt{(1-x)^2}$$ and we obtain $$CP+PQ'=\sqrt{5x^2-2x+1}+1-x$$ with $$0\le x\le 1$$ and finally we obtain $$\sqrt{5x^2-2x+1}+1-x\geq \frac{5}{8}$$ if and only if $$\left(x-\frac{2}{5}\right)^2\geq 0$$