Art Gallery ProblemDate: 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/ |
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