So first things first, I apologize for my bad English in advance.
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).

• Why is $\angle CQP=90^{\circ}$? Commented Apr 22, 2018 at 9:38
• Yes, this was absolutely unproved in my solution, though it was correct (we get the least possible value of PQ only and only if $\angle CQP=90°$, and the problem asks us to find the least possible value of $PQ+CP$. To be mentioned, 1 out of 7 points was for noticing the fact that $\angle CQP$ must be $90°$ Commented Apr 22, 2018 at 9:44
• Guys, I fully understand that you are making good and proper solutions to this problem (thank you very much for this, by the way), but please, could you instead tell me how should I have found the minimum value of this expression of $CP+PQ$? Thank you very much! Commented Apr 22, 2018 at 10:25

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.
$$BC\cdot\sin\angle CBC'=2\sin2\angle CBD=4\sin\angle CBD\cos\angle CBD=4\left(\frac{1}{\sqrt{5}}\right)\left(\frac{2}{\sqrt{5}}\right)=\frac{8}{5}$$
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 $$\left(x-\frac{2}{5}\right)^2\geq 0$$