# Assignment 6: Picture Viewing

## Installing a Picture or a TV

The purpose of this write up is to find the maximum angle for viewing something on a wall. The assignment asks about a picture, 4 feet high and 4 feet wide, affixed to the wall so that the lower edge is 2 feet about a person's eye. Horizontally, we want to center ourselves in the middle. We're worried about the vertical direction. At what distance from the picture is viewing angle greatest?

While most of us don't worry about viewing pictures, a very similar situation arises if someone is trying to install a flat screen TV on a wall. While I like the top of my TV at eye height, many people prefer to look upwards, as in a movie theater.

Here, we'll use three different types of mathematics to find the maximum angle. First, we'll measure the angle at various distances. Second, we'll use trigonometry to create an equation for the inclination, then calculus to maximize that angle. Finally, we'll use properties of circles and angles from geometry.

## Measurement Approach

The direct approach, one I suspect almost all non-mathematicians would use, would be to compare distances with an angle measuring tool. A school protractor costs less than a dollar. A more precise device, such as this digital angle finder, costs a bit more. In either event, we could walk backwards at eye height and measure the angle from lower edge to upper edge, searching for the best view.

We can simulate this on the computer, which I did in a GeoGebra applet. The picture is on the wall, with endpoints in light red. Eye position is the big blue dot on the horizontal axis. Click on the dot and drag it along the axis to change the angle.

Click through to picture.html to try the GeoGebra applet if you like.

## Trigonometry and Calculus Approach

For practical use, the measurement approach is likely good enough. Even if we're off by a half foot, at 4 feet, the angle is still 29.74 degrees, less than 1% off the maximum. Nevertheless, we can find an exact solution using trigonometry. The wall and the distance, the two axes, form a right angle. The angle we want is \alpha . If we call the distance from the wall x, we can use trigonometry to get two relationships. Recall that the tangent of an angle equals the opposite side divided by the adjacent side.

\tan (\beta) = \frac{2}{x} ~~~~ \; \; \; \; \tan (\alpha + \beta) = \frac{6}{x}

We can then convert this to arc-tangent, yielding \beta = \arctan (2/x) and \alpha + \beta = \arctan (6/x) . Substituting and subtracting yields \alpha = \arctan (6/x) - \arctan(2/x) . If we know calculus, we can take the derivative, which after some work yields \frac{48 - 4 x^2}{x^4 + 40 x^2 + 144} . Finding the maximum means setting this derivative to zero and solving for x. Eventually we reach x^2 = 12 . The distance that gives maximum angle \alpha is x = \sqrt{12} \approx 3.46 .

I skipped some steps in the derivation above, such as the derivative of an arctangent. If you know calculus, you can complete the problem. If not, but you do know geometry, the upcoming third approach is another way to find the answer.

## Geometry Approach

Geometrically, given any three points, we can construct a circle that includes these three points. This circle is called the circumcircle. Our angle \alpha is formed from three points: top, eye, and bottom. Because all three points are on the circle, the angle has a special name, an inscribed angle.

The key theorem we need was proved around 300 BC, in Euclid's Elements, Book III, Proposition 20. It says that the size of the inscribed angle is half the size of the central angle. On the picture, blue inscribed angle \alpha is half the size of green central angle \delta . This means that if we want to maximize \alpha , we can maximize the central angle \delta .

How do we do that? Since the top and bottom points are fixed, we make \delta bigger by getting point C closer to the painting. If we could, we'd put C right by the painting and get an angle close to 180 degrees. Unfortunately, we can't do that, because C is the center of a circle. We need to make the circle as small as possible. What's restricting us?

The restriction is that the eye point must be on the circle. The distance from center C to our eye is drawn as a black line. Since it goes from the center to a point on the edge, it's a radius. We want to shorten the black line, which will make the circle smaller and increase the central angle.

Because the edges of the painting are fixed, our central angle will always be largest if we vertically place C midway between the top and bottom points. Here, that's 4 feet above the ground. More technically, the perpendicular bisector between the top and bottom points will be horizontal because the line between top and bottom is vertical, always splitting the difference. I drew the perpendicular bisector as a dashed line. How do we get from the dashed line to our eye?

The shortest distance from a point to a line is a straight line segment. We need to move the eye point directly below C, so a vertical line connects them. This line will be exactly four feet long. The radius of the minimal circle is 4 feet.

To find the distance from the wall, we need to rely on the angle bisector. Since it literally bisects the green angle \delta , the angle from the bottom through C along the bisector is half the central angle. As Euclid proved, half the central angle equals the inscribed angle \alpha . The bisector is also literally perpendicular, forming a right triangle. The distance from the bottom of the picture to the center is the radius, 4 feet. The distance from the bottom to B is half of the picture, or 2 feet. The final side, BC, the thick blue line, is the distance x, which we can find via the Pythagorean theorem.

2^2 + x^2 = 4^2

This yields the same result as through calculus and measurment, in feet x = \sqrt{12} \approx 3.46 .