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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!

  • @TonyK, because AB=√2 but diameter is 2. –  Jun 11 '19 at 15:26
  • Let ABCD be a rectangle with sides sizes of 1 and CD=AB=√2. In other words, it's a unit square ACBD. :) – lurker Jun 11 '19 at 15:59
  • 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. I'm struggling with this part. AB is a diagonal of the unit square. CD is the other diagonal, orthogonal to AB. If you have a semicircle with diameter on AB, I can't get CD to be tangent to the circle and have AB be fully contained on the circle (or semicircle) diameter. I think I need a picture... – lurker Jun 11 '19 at 16:04
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    @lurker: Now that you mention it, I suppose the question does admit of that interpretation. But it all makes sense if you understand $ABCD$ (or more properly $ABDC$) to be a $1\times\sqrt 2$ rectangle. – TonyK Jun 11 '19 at 16:27
  • @TonyK ah that was it. I was seeing CD as the other diagonal for some reason. Makes sense now! – lurker Jun 11 '19 at 17:24

2 Answers2

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This is the situation:

enter image description here

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 $x=y=\frac12\sqrt 2$.

TonyK
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  • Since $PR\ge\sqrt{2}$, $\cos\angle POR=\frac{1^2+1^2-PR^2}{2(1)(1)}\le \frac{1^2+1^2-(\sqrt{2})^2}{2(1)(1)}=0$. $\angle POR \ge\frac\pi2$. – CY Aries Jun 11 '19 at 16:46
  • @CYAries: Ha! I suspected I was going the long way round. You should post that as an answer. – TonyK Jun 11 '19 at 16:53
  • I posted my answer, taking advantage of your nice figure. – CY Aries Jun 11 '19 at 16:59
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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 $\angle POR\ge\dfrac\pi2$.

CY Aries
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