Vector+Trig Bash (meh)

by math_explorer, Sep 21, 2010, 10:41 AM

Quote:

We've already done Part 1 without Lagrange multipliers.
I finally figured out Part 2 with vectors and trig, except that to think of the solution you more or less have to know the answer beforehand, and it doesn't seem to work easily for obtuse triangles, although I'm sure it could be bashed out. Still, it took me a while. Maybe I'll fix it later.

Click to reveal hidden text

The next one is 167583, and I'll just quote it in its entirety.
April wrote:
Let $ P$ and $ P'$ be two points in interior of the given triangle $ ABC$. Let $ D$, $ E$, $ F$ be the traces on sides $ BC$, $ CA$, $ AB$, respectively with the lines through $ P$ and parallel to $ P'A$, $ P'B$, $ P'C$. Let $ D'$, $ E'$, $ F'$ be the traces on sides $ BC$, $ CA$, $ AB$, respectively with the lines through $ P'$ and parallel to $ PA$, $ PB$, $ PC$.
Prove that $ AD$, $ BE$, $ CF$ are concurrent if and only if $ AD'$, $ BE'$, $ CF'$ are concurrent.

I hope "trace" means what it seems to mean.

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