# The Golden Christmas Tree

Last time, Alexandria Jones and her family were on their way to Uncle William’s tree farm to find the perfect Christmas tree, and Dr. Jones taught us about the Golden Section:

$The \; Golden \; Section \; ratio$

|———————A———————|————B————|

$A \; is \; to \; B \; as \; \left(A + B \right) \; is \; to \; A, \; or . . .$

$\frac{A}{B} = \frac{A + B}{A} = \: ?$

I gave you three algebra puzzles to solve. Did you try them?

• What is the exact value of the Golden Section ratio?

• If a 7-foot tree will fit in the Jones family’s living room, allowing for the tree stand and for a star on top, how wide will the tree be?
• Approximately how much surface area will Alex and Leon have to fill with lights and ornaments?

## Math Adventurer’s Rule: Figure It Out for yourself

Whenever I give a problem in an Alexandria Jones story, I will try to post the answer soon afterward. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. So if you have not worked these problems yet, go back to the original post. Figure them out for yourself — and then check the answers just to prove that you got them right.

## The Golden Section Ratio

To find the Golden Section ratio, remember that the length of the line does not matter. That means that we can pick a line of any convenient length. We will make our problem easier by choosing B = 1. Then the Golden Section ratio A/B will be the same as the length of A:

|———————A———————|————B————|

$\frac{A}{B} = \frac{A}{1} = A$

And that will make our Golden Section equation look like this:

$\frac{A}{B} = \frac{A + B}{A}$

$\frac{A}{1} = \frac{A + 1}{A}$

Multiply both sides of the equation by A:

${A}^{2} = A + 1$

Now put it into standard quadratic form:

${A}^{2} - A - 1 = 0$

That looks better, doesn’t it? Unfortunately, this equation cannot be factored, so we will have to use the dreaded Quadratic Formula. Finally we get:

$A = \frac{1 + \sqrt{5}}{2} \approx 1.618$

Answer: So the Golden Section ratio, which compares the long section A to the short section B (or the whole line to the long section), is approximately 1.618.

## How Wide Was the Christmas Tree?

Now for the next part of the puzzle: What size was the Jones family’s Christmas tree? If the tree was pruned to the Golden Section ratio, then the height H compared to the width W should be about 1.618:

$\frac{H}{W} = 1.618$

We can solve this equation for W:

$H = 1.618 \times W$

$W = \frac {H}{1.618} = \frac {7}{1.618} = 4.3$

Since we are being overtly American and measuring in feet, we need to handle decimals with care. One foot is 12 inches, and 0.3 will be about 1/3 of a foot, so…

Answer #1: For a 7-foot tall tree, W is about 4 feet 4 inches.

If, however, the tree was trimmed to Grandpa Jones’s rule of thumb, then:

$H = 1.5 \times W$

$W = \frac {H}{1.5} = \frac {7}{1.5} = 4.7$

Answer #2: For a 7-foot tall tree, that gives a width of 4 feet 8 inches.

The Golden Section makes for a slimmer tree. But with such a small difference, you can see why Grandpa preferred the simpler rule.

## Time to Trim the Tree

Finally, we asked how much space Alex and Leon had to fill with lights and ornaments. To calculate the surface area of the tree, we will have to make an approximation. A right circular cone is the closest geometric shape to a Christmas tree, and the surface area of a cone is:

$A = \pi r h$

[Click on image to enlarge.]

The radius r is easy, since it is just half the width of our tree. But be careful! That h is not the height of the cone, but the slant height: the length from top to bottom along the cone’s outside surface. To find the slant height, we have to use the Pythagorean Theorem:

${h}^{2} = {H}^{2} + {r}^{2}$

$h = \sqrt{{H}^{2} + {r}^{2}}$

Which gives us the following formula:

$A = \pi \times \left(\frac{W}{2} \right) \times \sqrt{{H}^{2} + {\left(\frac{W}{2} \right)}^{2}}$

Answer: Using our values of H and W from above, the Golden Section tree would have a surface area of almost 50 square feet. Grandpa’s slightly fatter tree would have an area of 54.5 square feet.

If you got either of these answers, congratulations — you are a master of algebra!

## To Be Continued…

Read all the posts from the November/December 1998 issue of my Mathematical Adventures of Alexandria Jones newsletter.