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Art Gallery Problem


Date: 11/30/97 at 21:14:00
From: Emily Cloutier
Subject: The Art Gallery Problem

A room in an art gallery contains a picture you are interested in 
viewing. The picture is two meters high and is hanging so the bottom 
of the picture is one meter above your eye level. How far from the 
wall on which the picture is hanging should you stand so that the 
angle of vision occupied by the picture is a maximum?  

What is the maximum angle? How much advantwge would a person 20 cm 
taller have in viewing the picture? The art gallery is 6 meters wide. 

I have looked in the library and researched "maximizing angle of 
vision," but I can not find anything about it. I realize that the 
closer the person is the larger the angle is, but that is not always 
logical. How can I solve this? Basically, where is the place that 
the angle of vision would be the best?


Date: 12/01/97 at 02:42:39
From: Doctor Pete
Subject: Re: The Art Gallery Problem

Hi,

What the problem means when it says "angle of vision" is this:  
Draw an imaginary vertical line through the picture, cutting it in 
half.  Then where this line intersects the top edge of the picture, 
call this point A. Similarly, where this line intersects the bottom 
edge of the picture, call this point B. Now, say your eye is a third 
point located 1 meter below point B and x meters away from the wall; 
call this point O. Then the "angle of vision" the problem refers to is 
the angle AOB.

Now, what does this mean?  If x = 0, that is, if your face is planted 
right on the wall, then points A, B, and O will be collinear, and 
angle AOB will equal 0; obviously you wouldn't get a very good view of 
the picture. On the other hand, if x = 100000, which would be 
somewhere 100 km away from the wall the picture is hanging from, it's 
pretty obvious you wouldn't see much either, assuming you had a clear 
view. So there must be some value of x between 0 and "infinity" at 
which angle AOB is maximized; this is the maximum "angle of vision."

To find this, we would require calculus.  Draw a 2-dimensional diagram 
of points A, B, and O, as well as point C, which is 1 meter directly 
below B; then triangle AOC is right, with angle ACO = 90 degrees.  Let 
angle AOC = p, and angle BOC = q.  Then notice that p - q = angle AOB.  
But
     
  Tan[p] = (2+1)/x,  and   Tan[q] = 1/x.

Therefore,

     p-q = ArcTan[3/x] - ArcTan[1/x]
         = ArcTan[(3/x - 1/x)/(1+(3/x)(1/x))]
         = ArcTan[2x/(x^2+3)].

So Tan[AOB] = 2x/(x^2+3). Since we wish to find the value x such that 
AOB is maximum, we take the derivative of AOB with respect to x. But 
notice that if AOB is maximized, Tan[AOB] is also maximized (why?).  
So to save us work, we simply take the derivative of 2x/(x^2+3), 
instead of ArcTan[2x/(x^2+3)] (again, why?).  So we set

     (x^2+3)(2) - (2x)(2x)
     --------------------- = 0,
            x^2 + 3

from which we find that 2x^2 - 6 = 0, or x = {Sqrt[3], -Sqrt[3]}.  
Since x > 0, we take the positive root, x = Sqrt[3] meters. This is 
our optimum viewing distance. I leave it to you to find the angle of 
vision at this distance.

Finally, say you had someone 20 cm = 0.2 m taller. Then mark a point D 
between B and C, such that BD = 0.8, and CD = 0.2. Also mark point Q 
such that BDQ is a right angle, and QD = y. Then solve the similar 
problem for y. What is the viewing distance and angle?

One last thing:  in one step of my calculations, I used the formula

     ArcTan[a] + ArcTan[b] = ArcTan[(a+b)/(1-ab)].

This is called the arctangent addition formula, and requires proof.  
To see this, notice that the tangent addition formula says

     Tan[A+B] = (Tan[A] + Tan[B])/(1-Tan[A]Tan[B]).

Then simply let a = Tan[A], b = Tan[B], and take the arctangent of 
both sides to get the result.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Trigonometry

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