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Inscribed circle centre problem
01-06-2011, 10:26 PM
Post: #1
Inscribed circle centre problem
[float=right]     [/float]Given three circles $(O_1), (O_2), (O)$, (see the figure) with $(O_1)$ contact to $(O_2)$ at point $T$, $(O_1)$ contact to $(O)$ at point $E$, $(O_2)$ contact to $(O)$ at point $F$, $BC$ tangent to $(O_1)$ and $(O_2)$ at points $P, Q$ respectively, $AT$ tangent to $(O_1)$ and $(O_2)$ at point $T$, And $A, B, C$ lie on the circle $(O)$.

Prove that $T$ is the inscribed circle centre
of $\triangle ABC$.


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01-06-2011, 10:39 PM
Post: #2
RE: inscribed circle center problem
   
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01-06-2011, 10:52 PM
Post: #3
RE: inscribed circle center problem
The circles $\odot (O)$ and $\odot (O_1)$ are homothetic at $E$, while $\odot (O)$ and $\odot (O_2)$ are homothetic at $F$, consequently $EP$ and $FQ$ concur at $D$, the midpoint of the arc $BC$ of the circle $\odot (O)$.
The inversion centered $D$ and power $BD^2$ transforms the circle $\odot (O)$ into the line $BC$, so it sends $E$ to $P$ and $F$ to $Q$, hence keeping the circles $\odot(O_1)$ and $\odot (O_2)$ unchanged.

Consequently, the following equality holds: $BD^2=DE\cdot DP=DF\cdot DQ$ (1); these relations show that $D$ belongs to the radical axis of the circles $\odot (O_1)$ and $\odot (O_2)$, i.e. $DE\cdot DF=DT^2$ (2).

(1) and (2) implies $BD=DT=DC$ which proves that $T$ is the incenter of $\triangle ABC$.    
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01-22-2011, 10:29 AM
Post: #4
RE: Inscribed circle centre problem
   
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