Geo practice
by MP8148, Jul 24, 2024, 9:03 PM
I tried some ISL geometry problems:
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I tried G5 for a while but it looked kinda hard so I gave up. I didn't look at the rest of the problems.
Also, I passed the driver's test somehow, even though I'm horrible driver and I drive into a wall recently while parking
G1 wrote:
Let
be a convex pentagon such that
. Suppose that the midpoint of
is the circumcenter of triangle
. Let
be the circumcenter of triangle
.
Prove that line
passes through the midpoint of segment
.






Prove that line


The first thing I did is let
, to get a circle with diameter
, and then the condition nicely becomes
is a parallelogram.
After this I got stuck for a bit (yes I didnt see the similar triangles shhh). I noticed that this can probably be solved by complex numbers with
as unit circle, and decided to give up and look at the solutions.



After this I got stuck for a bit (yes I didnt see the similar triangles shhh). I noticed that this can probably be solved by complex numbers with

G2 wrote:
Let
be a triangle with
let
be the circumcircle of
and let
be its radius. Point
is chosen on
such taht
and point
is the foot of the perpendicular from
to
. Ray
mets
again at
. Point
is chosen on line
such that
and
lie on a line in that order. Finally, let
be a point satisfying
and
. Prove that
lies on
.























I got a pretty messy diagram, but fortunately I realize the conclusion is equivalent to
so
can be completely eliminated.
Then I looked at the weird length condition
. I realized that
, and so
. Then I realized that in fact
is the circumcenter of
by angle chase, which kills the problem.


Then I looked at the weird length condition





G3 wrote:
Let
be a cyclic quadrilateral with
. Let
be the midpoint of the arc
not containing
. Suppose there is a point
inside
such that
and
.
Prove that lines
, and
are concurrent.









Prove that lines


This looked pretty hard, but turns out to actually be pretty straightforward. I extended
,
to meet
at
,
, so one of the angle conditions becomes
. The other angle condition immediately gives
, so there are suddenly lots of isosceles trapezoids -> lots of parallel lines.
I noticed that if
, then
and
are homothetic. The center of the homothety is
which lies on
, and Pascal's means
lies on this line too.







I noticed that if






G4 wrote:
Let
be an acute-angled triangle with
. Let
be the circumcircle of
. Let
be the midpoint of the arc
of
containing
. The perpendicular from
to
meets
at
and meets
again at
. The line through
parallel to
meets line
at
. Denote the circumcircle of triangle
by
. Let
meet
again at
. Prove that the line tangent to
at
meets line
on the internal angle bisector of
.



























Apparently this is also IMO P2, which I solved last year, but I forgot about it already (I did feel this is kinda familiar though).
The first thing I did (while drawing the diagram) is eliminate
, cuz
is just
. Then angle chasing gives the tangent at
is
where
is the point on
with
. After struggling with more random angle chase, I noticed that
and
are homothetic, and that finishes the problem.
I thought the homothety is pretty nice (the 18 people who upvoted TheUltimate123 should agree).It's pretty funny cuz I solved this problem last year with a very different method (I constructed some circles and made the concurrency point the radical center).
The first thing I did (while drawing the diagram) is eliminate










I thought the homothety is pretty nice (the 18 people who upvoted TheUltimate123 should agree).It's pretty funny cuz I solved this problem last year with a very different method (I constructed some circles and made the concurrency point the radical center).
I tried G5 for a while but it looked kinda hard so I gave up. I didn't look at the rest of the problems.
Also, I passed the driver's test somehow, even though I'm horrible driver and I drive into a wall recently while parking
