Geometry Dash
by djmathman, Aug 13, 2015, 1:15 AM
Being removed from official math contests, I now essentially have the freedom to practice math whenever I want without the pressure of competition. (Technically, I've always studied math without an organized study regimen, but the extra freedom is a new plus.) As a result, I'm able to practice math a little... unconventionally.
One of these new ways came through v_Enhance's new GeoGuessr handout here. I've never really explored the mixtillinear incircle configuration before to that extent, so I tried an experiment: my goal was to see how many of the listed facts I could prove/recall the proof to within the span of an hour. It was actually quite a fun exercise!
Just for kicks, here are my mostly-unedited sketches of my proofs. Edits are denoted in parentheses.
Gogogo
Also triple speed ship in Electrodynamix is so hard ahhh
One of these new ways came through v_Enhance's new GeoGuessr handout here. I've never really explored the mixtillinear incircle configuration before to that extent, so I tried an experiment: my goal was to see how many of the listed facts I could prove/recall the proof to within the span of an hour. It was actually quite a fun exercise!
Just for kicks, here are my mostly-unedited sketches of my proofs. Edits are denoted in parentheses.
Gogogo
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
where
. (The past equality comes from
as from #5.) Hence
is cyclic as desired.
Partial Problem 10
Nine out of 11 problems in the hour - I'd say that's not bad
For completeness, here are solutions to 10 and 11 - I solved these the next day after a few minutes in the car.
Problem 10
Problem 11
(Sorta mem) Homothety.
, etc.
(More clarification: the homothety sending the mixtillinear incircle (which I shall denote
) to the circumcircle sends
to the tangent to
parallel to
, which touches at
. Thus
is taken to
.)

(More clarification: the homothety sending the mixtillinear incircle (which I shall denote







Problem 2
(Mem) Pascal on
yields
,
,
collinear, and
angle bis.
midpt.






Problem 3
Let
be this top point. Note that
(easiest way to prove this is to note both lines are
to
), and sim.
, so
is a
-gram. Thus
passes through the midpoint of
. But by #1 we know
,
homothetic, so
,
, and this midpoint are collinear. Thus
. Similar reasonings - homothety - prove the other statements, and
follows by homothety wrt
.
















Problem 4
(Mem)
-inversion sends
, done.


Problem 5
Let
. Note that
is refl. of
across
and
is refl. of
, so
passes through refl. of
across (this line). Call this point
. Then
, so
.











Problem 6
Similar to #3;
and
so
.



Problem 7
It's well known that
(both isoc. and have vert. angles
), so from #3
. This implies
cyclic. Similar reasoning applies for
.





Problem 8
Note that since
is tangent to
,
. This implies
is tangent to
. Similar reasoning goes for
and
.







Problem 9




Partial Problem 10
It suffices to prove that
, since then the homothety sending
to
sends
to
and
to itself. This I could not do in the allotted hour.
(This is wrong
)






(This is wrong

Nine out of 11 problems in the hour - I'd say that's not bad

For completeness, here are solutions to 10 and 11 - I solved these the next day after a few minutes in the car.
Problem 10
Let
and
. Note that
and
, so
is cyclic. Thus
. But then
as well. Thus
, which implies
.






![\[\angle C_1Y_1C=\angle(M_CM_B, BC)=\frac12(\widehat{M_BC}-\widehat{M_CB})=\frac12\left(\widehat{M_BA}-\widehat{M_BC}\right)=\dfrac12\widehat{XA}\]](http://latex.artofproblemsolving.com/1/7/8/178516b2d4b33126e9fe4949e93b8522b763ad9a.png)


Problem 11
Homothety centered at
sends
to the bottom point of
. But by #3
passes through the top point, and
, so
also passes through the bottom point. QED






Also triple speed ship in Electrodynamix is so hard ahhh
This post has been edited 2 times. Last edited by djmathman, Aug 13, 2015, 1:17 AM
Reason: fixed typo
Reason: fixed typo