## Thursday, December 13, 2012

### Inner triangle area solution

See the original puzzle

There's a more elegant way to do this puzzle, and a less elegant way.  First, I'll show you the less elegant way.

We can associate the points with vectors t, u, and v.  This is useful because you can express any point on the line between t and u as x*t + (1-x)*u for some value of x.  See spoiler image.

Anyway, you can go through the algebra and calculate that point A is at (4/7*t+3/7*u+1/7*v).  The rest will be left as an exercise to the reader because I'm going to move on to the more elegant solution.

There's a more elegant solution involving center of mass.  Let's suppose that there are weights at each of the corners of the triangle, and that the center of mass is at point A.  How much weight must you place at point A?

See spoiler image

You have to place weights with ratios 4:2:1 on points t, u, and v respectively.  If you only consider the weights on t and u, then their center of mass is at B, with weight 6.  And so we can conclude that point A is 6/7ths of the way from point v to B.

The rest is simple geometry.  Triangle tAB is 1/7th the area of tvB, which is 1/3rd the area of tuv.  So triangle tvA is 6/21 of the entire area.  The same argument can be applied to triangles tuC and uvE.

If you sum up tvA, tuC, and uvE, the remaining area is 1/7th of the total, so that is our answer.

#### 1 comment:

miller said...

I realize I'm abusing some notation here by mixing points and vectors.