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.
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The triangle is
.
Suppose the vertices of the Miquel's triangle, from point
, are
,
,
.
Let
be the midpoints of
respectively. Let
be the circumcenter.
Let
,
,
. We have
iff
. Let
be the length of
.
Finally, let
be the area of
.
We want to prove that the area of
, which happens to be
, is larger than the area of
. Equivalently, we can prove that
, using brackets for area.
Divide both sides by
and using some simple ratios with the Law of Sines area formula, our goal becomes
, cyclically summing over
.
Expand the terms and we find we have to prove that
.
Now, finally, the vectors come in: we find that
and similarly for
and
. It's well known that
where
is the angle between the two vectors. Therefore
.
To simplify things, let's let
,
,
. So
.
We must prove that
(that dot is scalar multiplication) is non-positive. We can multiply it by
without changing its sign, so our target is
.
Observe now that
, so
, so
. Plugging in, we must prove
is non-positive. So, rearranging the terms, we get
.
is non-negative, and
is non-positive for acute triangles, so the sum is non-positive.

Suppose the vertices of the Miquel's triangle, from point




Let



Let







Finally, let


We want to prove that the area of



![$3s/4 \leq [AEF] + [BFD] + [CDE]$](http://latex.artofproblemsolving.com/3/2/a/32ab7261927aa5d0ae7a40c5287ec165920261f9.png)
Divide both sides by



Expand the terms and we find we have to prove that

Now, finally, the vectors come in: we find that






To simplify things, let's let




We must prove that



Observe now that







The next one is 167583, and I'll just quote it in its entirety.
April wrote:
Let
and
be two points in interior of the given triangle
. Let
,
,
be the traces on sides
,
,
, respectively with the lines through
and parallel to
,
,
. Let
,
,
be the traces on sides
,
,
, respectively with the lines through
and parallel to
,
,
.
Prove that
,
,
are concurrent if and only if
,
,
are concurrent.























Prove that






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