Re: [EMHL] Another Problem (locus)
- I wrote:
>>Let P be a point on the plane of a triangle ABC, and A'B'C' its cevianPaul Yiu wrote:
>>From B', C' draw parallels to BC intersecting AB, AC at Ba, Ca, respectively.
>>Let A* be the intersection of B'C'and BaCa. Similarly define B*,C*.
>> / \
>> C' Ca
>> / \
>> / A* \
>> Ba B'
>> / P \
>> / \
>>Which is the locus of P such that A*B*C*, ABC are in perspective?
>[...] the perspector is always the centroid.Hmm.....
>The desmic system is
>(u(v+w):vw:vw) & (::) & (::) & (u^2(v+w):v^2(w+u):w^2(u+v) \\
>(uvw:v^2(w+u):w^2(u+v)) & (::) & (::) & (1:1:1)
> Desmon & (u:v:w) \\
> Harmon & (u(-\frac 1u+\frac 1v+ \frac 1w)::)
Now, after Paul's solution, I recall this well-known exercise in Euclidean
If the non-parallel sides of a trapezium meet at X, and its diagonals at Y,
then XY bisects the parallel sides.
Let's see it in the figure of the problem above:
/ A* \
Assume that AA* intersects BaB' at D.
Then by Ceva theorem we get:
DBa CaB' C'A
--- x ---- x ----- = - 1
DB' CaA C'Ba
We have C'Ca // BaB', therefore D is the midpoint of B'Ba,
and AA* bisects every parallel to it.
And since BC is parallel to BaB', the AA* is the a-median of ABC etc
Note that the points A',B',C' in the problem do not have to form
a triangle in perspective with ABC (they can be arbitrarily chosen
on the triangle sides)