Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2001 Iran NMO (Round 3) Problems/Problem 5

Revision as of 15:47, 27 April 2008 by Boy Soprano II (talk | contribs) (problem and solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

In triangle $ABC$, let $I$ be the incenter and $I_a$ the excenter opposite $A$. Suppose that $\overline{II_a}$ meets $\overline{BC}$ and the circumcircle of triangle $ABC$ at $A'$ and $M$, respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$. Let lines $NI$ and $NI_a$ intersect the circumcircle of triangle $ABC$ again at $S$ and $T$, respectively. Prove that $S$, $T$, and $A'$ are collinear.

Solution

[asy] size(300); defaultpen(1); pair A=(1,2.7), B=(0,0), C=(3,0); path O=circumcircle(A,B,C); pair I=incenter(A,B,C); pair BB=A+5(B-A); pair II=bisectorpoint(BB,B,C); pair Ia=IntersectionPoint(B--(B+6(II-B)),A--(A+5(I-A))); pair Aa=IntersectionPoint(A--Ia,B--C); pair M=IntersectionPoint(A--Ia,O,1); pair NN=bisectorpoint(A,C,M); pair N=IntersectionPoint(C--(5(NN-C)+C),O,1); pair S=IntersectionPoint(N--(N+3(I-N)),O,1); pair T=IntersectionPoint(N--(2Ia-N),O,1); draw(O); draw(A--B--C--A--Ia); draw(Ia--N--S); draw(S--T,dashed); draw(B--I--C--Ia--B,dashed); draw(circumcircle(I,B,C),dotted); draw(circumcircle(I,T,S),dotted); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$I$",I,NE); label("$I_a$",Ia,SE); label("$A'$",Aa,SE); label("$M$",M,SW); label("$N$",N,W); label("$S$",S,E); label("$T$",T,SW); [/asy]

We will use directed angles mod $\pi$, and directed arcs mod $2\pi$.

Since $\widehat{MN}\equiv \widehat{NA}$, it follows that \[\angle I_a TS \equiv \angle NTS \equiv \tfrac{1}{2} (\widehat{NA} + \widehat{AS}) \equiv \tfrac{1}{2} (\widehat{MN} + \widehat{AS}) \equiv \angle MIS \equiv \angle I_a IS .\] It follows that quadrilateral $I_aITS$ is cyclic.

On the other hand, $\angle IBI_a \equiv \angle ICI_a \equiv \tfrac{\pi}{2}$, so quadrilateral $IBCI_a$ is cyclic.

Now, since $II_a$ is the radical axis of the circumcircles of $I_aITS$ and $IBCI_a$, $BC$ is the radical axis of the circumcircles of $IBCI_a$ and $ABC$, and $TS$ is the circumcircle of $ABC$ and $I_aITS$, these three lines concur at $A'$, as desired. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_Iran_NMO_(Round_3)_Problems/Problem_5&oldid=25626"