Sliding a semicircle within a rectangle

Question

Let ABCD be a rectangle with sides sizes of $1$ and $CD=AB=\sqrt{2}$. And let a semicircle with center $O$ of radius $1$ be such that its diameter is on the $AB$-produced so $CD$ will be tangent to the circle at point $Q$. $AB$ is always within the diameter of the semicircle. So semicircle cuts AC and BD at points P and R respectively. How to prove that the angle $POR$ always is at least (or maybe exactly) $\pi/2$ ?

PS - I am trying to prove a theorem relating "lattice point covering property" and I reduced the proof to the question I asked above; so $POR$-angle to be $\ge \pi/2$ will finish the proof. Please help!

Answer 1

As stated in TonyK's comment, the rectangle should be $ABDC$. (The result does not hold for $ABCD$.)

Note that $PR\ge\sqrt{2}$ and $\displaystyle \cos\angle POR=\frac{1^2+1^2-PR^2}{2(1)(1)}$.

We can conclude that $\cos\angle POR\le 0$ and hence $\angle POR\ge\dfrac\pi2$.

Answer 2

This is the situation:

We want to show that angle $POR$ is $\ge \pi/2$, or alternatively that $\theta+\varphi\le\pi/2$. This will be the case if and only if $$\cos(\theta+\varphi)=\cos\theta\cos\varphi-\sin\theta\sin\varphi\ge 0$$

Plugging in the values of these trigonometric ratios from the diagram gives us $$xy\ge zw$$

Both sides are positive, so we can square both sides:

$$x^2y^2\ge z^2w^2$$

By Pythagoras, $z^2=1-x^2$ and $w^2=1-y^2$, so this is

$$x^2y^2\ge (1-x^2)(1-y^2)$$ Rearranging, $$x^2+y^2\ge 1$$

So we have reduced the problem to showing that if $x+y=\sqrt 2$, then $x^2+y^2\ge 1$. And this follows from

$$x^2+y^2=x^2+(\sqrt 2-x)^2=2x^2-2\sqrt 2x+2=2(x-\frac12\sqrt 2)^2+1\ge 1$$

with equality if and only if $x=y=\frac12\sqrt 2$.