Number Line DDIT Proof (Mostly delusional)
by YaoAOPS, Nov 14, 2024, 7:30 AM
We prove the following
Claim
. An involution on a conic is projection through some point.
Proof: Well-known, but take two pairs of the involution and their intersection is said point.
Let
be the polar of 
with respect to 
, which by Dual Brokard's lies on 
and 
. Since 
lies on , it follows that 
passes through as well.
Claim
. 
lie on .
Proof: Replace with any line through . Then take homography that maps 
to a Rhombus, the result follows by symmetry.
Claim
. 
are coconic.
Proof: Let
. Then 
lie on and thus 
lies on the conic. Now, let 
be the tangency points from to 
. We get that 
form a pencil involution 
.
also lies on this but I haven't been bothered to prove this yet.
Claim
. Suppose an involution 
on maps has fixed points 
and lies on 
. The 
is a pencil involution.
Proof: The lies on condition just means its well-defined by Brokard's, finish by "Euclidean constructions."
Numberline DDIT wrote:
Let real quadrilateral have inconic and let 
. Let be a point on , and let the tangency points from to be 
. Then if we define 
as the distance function from to , then
![\[
(d(A), d(C)), (d(B), d(D)), (d(E), d(F)), (d(J), d(K))
\]](//latex.artofproblemsolving.com/e/7/9/e797a0a51a000ba1aa4eaaf14f50999d699c8d64.png)
is an involution on the real line 
(notably, 
).
have inconic 
and let 
. Let 
be a point on 
, and let the tangency points from to be 
. Then if we define 
as the distance function from to , then![\[
(d(A), d(C)), (d(B), d(D)), (d(E), d(F)), (d(J), d(K))
\]](http://latex.artofproblemsolving.com/e/7/9/e797a0a51a000ba1aa4eaaf14f50999d699c8d64.png)
is an involution on the real line 
(notably, 
).Claim
. An involution on a conic is projection through some point.Proof: Well-known, but take two pairs of the involution and their intersection is said point.
Let
be the polar of with respect to , which by Dual Brokard's lies on 
and 
. Since lies on , it follows that 
passes through as well.Claim
. 
lie on .Proof: Replace
with any line through . Then take homography that maps to a Rhombus, the result follows by symmetry.Claim
. 
are coconic.Proof: Let
. Then 
lie on and thus 
lies on the conic. Now, let 
be the tangency points from to 
. We get that 
form a pencil involution 
. also lies on this but I haven't been bothered to prove this yet.Claim
. Suppose an involution 
on maps has fixed points 
and lies on 
. The 
is a pencil involution.Proof: The
lies on condition just means its well-defined by Brokard's, finish by "Euclidean constructions."
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