This is for those who worked on the puzzle that I posted two weeks ago.
OK, so you gave it a shot and finally figured it out. Good for you. Let's see, did you do it the same way that I did? When I tried to solve this problem, one thing that I did was draw a lot of lines on this triangle. I drew bisectors and perpendiculars and arcs and lots of other stuff and looked for a thousand relationships that would give me the answer. Finally, for reasons I can't remember, I constructed a equilateral triangle on one of the lines from the corner to the middle. Then I drew a line from the tip of the new equilateral triangle to the tip of the old equilateral triangle and formed some new triangles. It looked like this:
It was then that I noticed that angle α, cut out the same piece of the little equilateral triangle (aaa) as it did the big equilateral triangle (xxx). In both cases, it leaves the angle 60 - α. The magnitude isn't so important as the fact that they are the same. Now we go back to 10th grade geometry and remember that two triangles are congruent (identical) if they have the same side-angle-side (SAS) pattern. So we see that the external triangle (a,60-α,x) is the same as the internal triangle (a,60-α,x). And that means that the last line that I drew in the new construction is going to be the same length as "c". So now we're done. All we have to do now is fill in the math.
When I first did this, I didn't have the internet and I didn't know the "Law of Cosines". But I did know that if you have a triangle and you know all the sides, there must be a way to figure out the angles. By doing a bunch of construction, I spent a day coming up with the law on my own only to find out that any real math nerd knew it already and it was clearly laid out in my own CRC Handbook of Standard Math Tables. Anyhow, if you have a triangle with sides a, b and c, then the angle (γ) opposite c is given by:
So now the internal triangle axc can now be described. The angle opposite x in triangle abx is γ + 60 and by the law of cosines again:
And now we plug in for γ:
OK, the formula for the area of an equilateral triangle is:
And then swapping out the X squared we get finally:
So I haven't even bothered to calculate the answer with a, b and c equal to 200, 300 and 400 ft., respectively. But maybe I will someday.
I will mention that I have always been bothered by that arccos and cos stuff in the answer. You have to have a calculator or some tables to arrive at the answer and I never liked that. Recently, while I was drawing up these pictures, I came across another solution. Of course, it's related to the construction above. It uses Heron's formula. You might have seen it on the wikipedia page on triangles. Heron's formula gives the area of a scalene triangle with sides a, b and c and one form is:
OK, here goes. As you did in the first step, construct two more equilateral triangles using the b line and the c line. What you will end up with looks like this:
The greens are the constructed equilateral triangles and the pink are 3 identical abc triangles. Now you can calculate the total area of this hexagonal blob as being equal to the three different equilateral triangles and three times the area of the abc triangle. Now look at it a different way:
In this view, you see that in the process you have managed to reproduce all the internal triangles on the outside and therefore the area of the triangle xxx is half the area of the hexagonal blob. So the area of triangle xxx, which was the original question, can also be given as:
And that's enough of that. If you're wondering how I made the equations, go to CodeCogs and open up the LaTex Equation Editor
