Let $r,R$ and $r_a$ be the radii of the incircle, circumcircle and A-excircle of the triangle $ABC$ with $AC>AB$, respectively. $I,O$ and $J_A$ are the centers of these circles, respectively. Let incircle touches the $BC$ at $D$, for a point $E \in (BD)$ the condition $A(IEJ_A)=2A(IEO)$ holds. Prove that \[ED=AC-AB \iff R=2r+r_a.\]
Problem
Source: Turkey TST 2014 Day 1 Problem 3
Tags: geometry, circumcircle, parallelogram, geometry proposed
12.03.2014 20:38
let $D'$ is tangency point of $(J_A)$, $M$ is midpoint of the side $BC$ and $P$ is midpoint of the smaller arc $BC$ of $(O)$. We have the point $P$ is midpoint of $IJ_A$. Hence \[ A(IEO)=\frac{A(IEJ_A)}{2}=A(IEP) \] and $EI$ passes through midpoint of $OP$(let us it's $N$). Obviously, $E$ lies on the ray $CB$(since $AC>AB$ and $D\in ]BC[ $ we get that $B$ lies on the segment $EC$). Let $X$ is midpoint of $J_AE$. (1) if $ED=AC-AB$ then $ED=DD'=2DM$ and $EI=2IN$. So $I $ is centroid of the $OEP$ and since $D$ is midpoint of $ED'$ $ \Rightarrow $ $X,D,I$ are collinear and $2DX=J_AD'=r_a$. Thus $INPX$ is parallelogram and $ R=OP=2NP=2IX=2(ID+DX)=2r+r_a $. (2) if $R=2r+r_a$ then let $I'\in EN$ such that $XI'\perp BC$ and since $XP\parallel I'N$ we get $I'NPX$ is parallelogram, $ XI'=PN=\frac{R}{2}=r+\frac{r_a}{2} $. Let $XI'\cap ED'=T$. Then $ XT=\frac{r_a}{2} $ $ \Rightarrow $ $TI'=r=ID$. So $I'\equiv I$ and $D\equiv T$ is midpoint of $ED'$, $ED=DD'=AC-AB$.
15.03.2014 15:02
I don't understand how is $ E $ defined... can somebody help me?
15.03.2014 15:38
anonymouslonely wrote: I don't understand how is $ E $ defined... can somebody help me? $E \in ]BC[$ means it is on the segment $BC$, but not $B$ or $C$.
15.03.2014 16:49
emregirgin35 wrote: anonymouslonely wrote: I don't understand how is $ E $ defined... can somebody help me? $E \in ]BC[$ means it is on the segment $BC$, but not $B$ or $C$. It actually means it is on the line BC, but not onto the segment! Best regards, sunken rock
15.03.2014 23:06
sunken rock wrote: emregirgin35 wrote: anonymouslonely wrote: I don't understand how is $ E $ defined... can somebody help me? $E \in ]BC[$ means it is on the segment $BC$, but not $B$ or $C$. It actually means it is on the line BC, but not onto the segment! Best regards, sunken rock Oh, actually I wanted to write $]BD[$ (as $(BD)$, they mean the same thing here). But as a result of typo I wrote $]BC[$ . Luckily, the statements are true if $E \in [DB$ and $AC > AB$ guarantees that $C \notin [EB]$. So the way that I wrote the problem is true, and does not alter the solution. But, in order to protect the originality I've edited the post as $(BD)$.
09.10.2014 20:04
What does A(IEJ)=2A(IEO) mean?
10.10.2014 08:23
alibagheri wrote: What does A(IEJ)=2A(IEO) mean? Here it means that the area of one triangle is double the area of the other one! Best regards, sunken rock