A line through the incenter $I$ of triangle $ABC$, perpendicular to $AI$, intersects $AB$ at $P$ and $AC$ at $Q$. Prove that the circle tangent to $AB$ at $P$ and to $AC$ at $Q$ is also tangent to the circumcircle of triangle $ABC$.
Problem
Source: 1-st Taiwanese Mathematical Olympiad 1992
Tags: geometry, circumcircle, incenter, geometry unsolved
TomciO
12.01.2007 20:09
It is just very easy converse of the problem from here.
vanu1996
18.10.2013 12:47
I think something error in this problem,(Draw these two circles as large as possible).
jayme
18.10.2013 12:52
Dear Mathlinkers, this circle is the A-mixtilinear incircle of ABC. (when perpendicular at I) You can see http://jl.ayme.pagesperso-orange.fr/ vol. 4 a new mixtilinear incircle adventure I, p. 10 Sicerely Jean-Louis
vanu1996
18.10.2013 13:04
Oh sorry,I think two different circles tangent at $P$ and $Q$,I make mistake to read the question.
vanu1996
18.10.2013 13:19
let $\omega$ be a circle which is tangent to the sides $AB,AC$ at $X,Y$ and circumcircle of $ABC$ at $T$.Draw $TX$ and $TY$ and they meet the circumcircle at $P$ and $Q$,note that $P$ and $Q$ are the midpoint of the arcs $AB$ and $AC$.Apply pascal to $BACPTQ$ and we see that $X,I,Y$ are collinear.Hence we get the result.
parmenides51
05.09.2024 15:04
This is the construction of A- mixtilinear incircle I though I had found an inversion solution , but failed
Let (I,r) be the incircle of ABC, tangent at AB,AC at points K,L resp.
Since AI^2 = AK x AP = AL x AQ, (1)
We do the inversion (A,AI)
Line AB, AC remain fixed as they both pass through A
Line BC inverts to circle (AB''C') where B',C' are the inverses of B, C
A -> A
K -> P
L -> Q
Line B' C' inverts of circle (ABC)
Line B'C' is tangent to incircle *
Circle tangent to AB,AC , BC, B'C' inverts to circle tangent to inverses of AB, AC, BC, B'C' which are AB, AC, (AB'C') and (ABC)
But because of (1), the iinverse of the incircle is the circle tangent to both AB,AC at points P,Q resp.
* What remains to be proved