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A movie screen $10$ m high is mounted on a wall $5$ m above the floor. Find the distance from the movie screen at which a point on the floor subtends the largest angle.

Let $A=(0,5)$ and $B=(0,15).$

I want to solve this by finding the circle though $A$ and $B$ which is tangent to the $x$-axis and then explain why this is the solution using Euclidean geometry. I may be wrong because I'm just guessing, but is the problem equivalent to finding the radius of this circle?

And even if that was the case, how would I find that radius? I know $(0,5)$ and $(0,15)$ are on this circle so if the centre is at $(h,k)$ then we have $h^2 + (5-k)^2=r^2\;(1)$ $ h^2 + (15-k)^2=r^2\;(2).$ We are also given that the circle is tangent to the $x$-axis, so it has a point $(b,0),$ which means that $(b-h)^2+k^2=r^2\;(3).$ From the first two equations, I have that $20k=200\Rightarrow k=10.$ So using $(1)$ and $(3),$ $$r^2=h^2+25\Rightarrow (b-h)^2=h^2-75\Rightarrow b^2-2hb+75=0.$$ But then how would I find an exact value for $h,$ which would give me the value of $r$?

Edit: I'll reward anyone with the bounty as long as they show that drawing the given circle maximizes the angle. Could someone provide a more thorough, short, and concise answer that is purely geometrical???

Also, I already know how to solve this from a calculus-based approach.

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  • $\begingroup$ I won't comment on whether the two problems are equivalent (because I don't know) but I will comment: If you want to make a circle that passes through $(0,5)$ and $(0,15)$, its centre point has to have $y$-coordinate $y=10$. If you also want it to be tangent to the $x$-axis, it has to have a radius of equal to the $y$-coordinate, meaning the the radius is $10$. Additionally, notice that this doesn't mean that the tangent point is at $x=10$... $\endgroup$
    – Matti P.
    Oct 28, 2019 at 7:33
  • $\begingroup$ desmos.com/calculator/g496zhzrwh $\endgroup$
    – Matti P.
    Oct 28, 2019 at 7:35
  • $\begingroup$ @MattiP. I wanted to solve this problem using a different method than the previous one, so that’s why I posted it separately. $\endgroup$
    – user717371
    Oct 28, 2019 at 13:14
  • $\begingroup$ @MattiP why does drawing the circle maximize the angle???? $\endgroup$
    – user717371
    Nov 1, 2019 at 0:24
  • $\begingroup$ @GerardWilliamson this problem is a specific case for the Regiomontanus' angle maximization problem. $\endgroup$
    – user718615
    Nov 1, 2019 at 2:37

3 Answers 3

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First a picture of the problem.enter image description here

We already recognize (see same question previously, now deleted) that we could maximize $\theta$ from the relationship \begin{equation} f(b)=\tan^{-1}\left(\frac{15}{b}\right)-\tan^{-1}\left(\frac{5}{b}\right)\label{eq:Scr01} \end{equation} To maximize, we set $df(b)/db$= 0, and solve for $b.$ \begin{equation} \frac{df(b)}{db}=\frac{-10b^{2}+750}{b^{4}+250b^{2}+5625}=0\label{eq:Scr02} \end{equation} \begin{equation} b=\pm5\sqrt{3}\label{eq:Scr03} \end{equation} It is assumed that we can do the calculus. Now, given $b,$$\ \theta$ is computed to be \begin{equation} \theta=f(b)=.5236\ rads=30^{\circ}.\label{eq:Scr04} \end{equation} However, we want to find $\theta$ with more intuition and less calculus. Imagine that the point $(b,0)$ is very far away from the screen. At that distance, $\theta$ will be very small. Likewise, if $(b,0)$ is very close to the screen, then $\theta$ will again be very small. That implies that at some point between 0 and infinity, that $\theta$ will be maximized. Now suppose that the screen top and bottom represent two points on a circle, and that a third point represents the eyes of an observer and that we measure $A$ and $B$ such that the third point is on our $x$-axis. We draw two chords, $B$ to $A$ and $A$ to $(b,0).$ The Circumcenter Theorem says that perpendicular bisectors of these two chords intersect at the center of the circle made by these 3 points. The perpendicular bisector of the vertical chord is the line \begin{equation} y=10.\label{eq:Scr5} \end{equation} The other chord has slope \begin{equation} chord\ slope=\frac{\Delta y}{\Delta x}=\frac{5}{-b}\label{eq:Scr06} \end{equation} and thus its bisector has slope $b/5.$ Therefore, using the point slope form of a line with centerpoint $(h,10),$we get the equation for that midpoint bisector to be \begin{equation} y-10=\frac{b}{5}(x-h).\label{eq:Scr07} \end{equation} Solving these two equations simultaneously we get \begin{equation} 10-10=\frac{b}{5}(x-h) \end{equation} Since $b\ne0,$we can conclude that $x=h$ and since the $x$ in question is distance $b,$ we also have $x=h=b.$ Now since $h=b$ and $b$ is on the circle, then the circle must be tangent to the $x$-axis, which establishes the circle radius as 10. \begin{equation} radius=10 \end{equation} Then the circle equation, using point $A=(0,5)$ is \begin{equation}(0-h)^{2}+(5-10)^{2}=10^{2} \end{equation}

and \begin{equation} h=b=\sqrt{75}\ =5\sqrt{3} \end{equation}

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  • $\begingroup$ @GerardWilliamson There's no reason to get aggressive here. I'm also puzzled by this. $\endgroup$
    – Matti P.
    Nov 1, 2019 at 6:31
  • $\begingroup$ I do think that the intuition is not obvious, which is probably why it is not popular. But, the reason that the point is maximized by the circle is that it is minimized when going either way away from the tangent. $\endgroup$
    – Narlin
    Nov 1, 2019 at 10:50
  • $\begingroup$ you know, I should've set the reason of the bounty to be "Reward an existing answer." I'd like to reward @Narlin for his amazing answer! $\endgroup$
    – user717371
    Nov 1, 2019 at 13:07
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Consider this definition of circle: Locus of all points that substend same angle with respect to a line segment.

The region outside the circle denotes those points that make less angle and inside region denotes those that make more angle with respect to line segment. (comment if you want elaboration.)

So for your question, we need the smallest circle that passes through AB and still has at least one point common to x-axis. The answer would be a circle tangent to x-axis.

There you have it.

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Fig 1 Condition for minimum subtended angle on x-axis

$$ y = \angle BPO -\angle APO= \tan^{-1}\frac{b}{x}-\tan^{-1}\frac{a}{x} \tag1 $$

Differentiate w.r.t. x and set to zero

$$ y'(x)= \frac{-b}{x^2+b^2} -\frac{-a}{x^2+a^2}=0 \tag2 $$ $$ \frac{ax^2+ab^2-bx^2-a^2b}{(x^2+a^2) (x^2+b^2)}=0 \tag3 $$

Further simplify

$$ x=\sqrt{ab}, y=0 \tag4 $$

Figs 1 & 2

Fig 2. Condition for tangency seen as double root or coincidental cutting points through zero discriminant

The center must be on perpendicular line $y= (a+b)/2$ due to equality of radii at C.

$$ (x-h)^2 +[y-\big(\frac{a+b}{2} \big)] ^2 =h^2 +\big(\frac{a+b}{2}\big)^2 \tag5 $$ Put $y=0$ $$ (x-h)^2 + [\big(\frac{a+b}{2} \big)] ^2=h^2+\big(\frac{a-b}{2} \big)] ^2 \tag6 $$ Simplifying $$ x^2- 2 x\,h +ab =0 \tag7 $$ In order that this quadratic equation should have coincident double root its discriminant $ \Delta =0$ $$ 4h^2= 4 ab,\, h=\sqrt{ab},y=0 \tag8 $$

Two Separate & Independent conditions but they are shown to produce the same point $ P(\sqrt{ab},0) $

  1. Yet a third independent condition that takes care of tangency normality and equal distances from $(A,B)$ in one shot is to use known property of parabolic loci ( here we take A and B as foci and x-axis as the directrix) and find out intersection center of circle as C.

Locus of focal point A $$ (y-a)^2 + x^2 = y^2 \rightarrow y= \frac{x^2+a^2}{2a} \tag9 $$

Similarly focal point B locus

$$ y= \frac{x^2+b^2}{2b} \tag{10} $$

To find intersection equate the RHS_s to obtain after simplification:

$$ x=\sqrt{ab}, y= \frac{a+b}{2}, \quad (=h)\tag{11}$$ at parabola intersection

enter image description here

and pedal line foot of of perpendicular on x-axis tangency point P directly below is

$$ x=\sqrt{ab}, y= 0, R= CA =10 \tag{12} $$

The parabolas are drawn as above. Also did not use numerical values as symbolic use of symbols is better than numerical calculation when you want to look at two situations.

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