Circum-Incentral Triangle

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Circum-Incentral Triangle

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Nine Point Circle

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#1 h Dec 31, 2018, 12:44 PM • 3 Y

Y by srijonrick, Adventure10, Mango247

Here are some trivial properties of circum-incentral triangle that connect to the excentral triangle, Prove them,
So basically these are 2 cents from my side to all the AOPSers and wish you Happy New Year!!

$\text{(1) } \text{ If } M_A,M_B,M_C \text{ are circumcenters of } \Delta BIC , \Delta AIC , \Delta AIB, \text{ then, } \Delta M_AM_BM_C \text{ is the circum-incentral triangle w.r.t to } \Delta ABC$
$\text{(2) } M_A,M_B,M_C \text{ lie on circle with circumcenter of } \Delta ABC \text{ as the center }$
$\text{(3) } A,B,C \text{ are the reflections of } I \text{ over } M_BM_C, M_AM_C , M_AM_B$
$\text{(4) } I \text{ is the orthocenter of } \Delta M_AM_BM_C$
$\text{(5) Let } S_9 \text{ be the mid-point of } I \text{ and } O, \text{ then, } S_9 \text{ is the nine-point center of } \Delta M_AM_BM_C$
$\text{(6) } \text{ Homothety } \mathcal{X} \equiv 2 \text{ at } I , \mathcal{X}: M_A \rightarrow I_A \text{ sends } \Delta M_AM_BM_C \text{ to } \Delta I_AI_BI_C$
$\text{(7) } I \text{ is the orthocenter of } \Delta I_AI_BI_C$
$\text{(8) } \Delta ABC \text{ is the orthic triangle of } \Delta I_AI_BI_C$
$\text{(9) } \Delta M_AM_BM_C \text{ is the euler tiangle of } I_AI_BI_C$
$\text{(10) } O \text{ is the nine-point center of } \Delta I_AI_BI_C$
$\text{(11) Let } A',B',C' \text{ be the mid-points of } AI,BI,CI , \text{ then } \Delta A'B'C' \text{ is the orthic triangle of } \Delta M_AM_BM_C$
$\text{(12) } S_9 \text{ is the circumcenter of } \Delta A'B'C'$
$\text{(13) If } E \text{ is the circumcenter of } \Delta I_AI_BI_C , \text{ then, } IO=OE$
$\text{(14) } I,S_9,O,E \text{ are collinear }$
$\text{(15) } \left(\Delta A'B'C' \right) \text{ is the nine-point circle of } \Delta M_AM_BM_C$

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I42857

#2 h Jan 6, 2019, 6:09 PM • 2 Y

Y by AlastorMoody, Adventure10

$(1)$ Nothing to show here.

$(2)$ We have $\widehat{BM_AC}=\widehat{BM_AI}+\widehat{IM_AC}=2\cdot\widehat{ICB}+2\cdot\widehat{IBC}=2\bigg(\frac{\gamma}{2}+\frac{\beta}{2}\bigg)=\gamma+\beta$ . Then $\widehat{BAC}+\widehat{BM_AC}=\alpha+\beta+\gamma=\pi$ and $ABM_AC$ is cyclic. Similarly, $BCM_BA$ and $CAM_CB$ are cyclic; so the circumcenter of $M_AM_BM_C$ is the same of $ABC$ .

$(3)$ Since $M_AB=M_AC$ , $M_A$ lies on the perpendicular bisector of $BC$ and is the midpoint of arc $BC$ , so $A,$ $I$ and $M_A$ are collinear. Similarly $B$ , $I$ and $M_B$ and $C$ , $I$ and $M_C$ are collinear.
Since $AM_CBM_ACM_B$ is cyclic, we have $\widehat{AM_BM_C}=\widehat{ACM_C}=\widehat{ACI}=\frac{\gamma}{2}=\widehat{ICB}=\widehat{IM_BM_C};$ similarly we find $\widehat{AM_CM_B}=\widehat{IM_CM_B}$ and so $AM_BM_C\cong IM_BM_C$ for the ASA criterion. So $I$ is the reflection of $A$ on $M_BM_C$ . The other two situations are identical.

$(4)$ From $(3)$ and the fact that $A,$ $I$ and $M_A$ are collinear we find $AI\perp M_BM_C$ and similarly for the other two vertexes. So $I$ is the ortocenter of triangle $M_AM_BM_C$ .

$(5)$ Since $I$ is the orthocenter of $M_AM_BM_C$ and $O$ its circumcenter, their midpoint $S_9$ is its nine-point center.

$(6)$ $BI_A$ is an external bisector of $\widehat{CBA},$ so $\widehat{I_ABI}=\widehat{I_ABC}+\widehat{CBI}=\frac{\pi}{2}$ , and similarly $\widehat{I_ACI}=\frac{\pi}{2}$ . So $IBI_AC$ is cyclic and $II_A$ is a diameter offf its circumscribed circle. But this circle has $M_A$ as center, so $II_A=2IM_A$ . From this follows the thesis.

$(7)$ We have $\widehat{I_AAI_B}=\widehat{I_AAB}+\widehat{BAI_B}=\frac{\pi}{2}$ , so $I_AA\perp I_BI_C$ . From a similar argument follow $I_BB \perp I_CI_A$ and $I_CC \perp I_AI_B$ . So $I$ is the orthocenter of $I_AI_BI_C$ .

$(8)$ This is a direct consequence of $(7)$ .

$(9)$ From $(6)$ we know that $I_AI=2I_AM_A$ , $I_BI=2I_BM_B$ and $I_CI=2I_CM_C$ , so the thesis follows immediatly.

$(10)$ $A$ , $B$ and $C$ are the feet of the perpendiculars in $I_AI_BI_C,$ so their circumcenter $O$ is the nine-point center of this triangle.

$(11)$ From $(3)$ we have that $A'=M_BM_C\cap M_AI$ ; since $I$ is the orthocenter of $M_AM_BM_C$ the thesis follows.

$(12)$ This is a direct consequence of $(5)$ : since $S_9$ is nine-point center of $M_AM_BM_C$ , it is circumcircle of its orthic triangle $A'B'C'$ .

$(13)$ From $(10)$ we have that $O$ is nine-point center of $I_AI_BI_C$ . So it is the midpoint of its orthocenter $I$ and its circumcenter $E$ .

$(14)$ For $(13)$ $I$ , $O$ , $E$ are collinear; for $(5)$ $I,$ $S_9$ , $O$ are collinear. So we have $I,$ $S_9$ , $O$ and $E$ collinear.

$(15)$ This is a direct consequence of $(11)$ .

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#3 h Jan 20, 2019, 7:03 AM • 2 Y

Y by Adventure10, Mango247

I42857 wrote:

$(3)$ Since $M_AB=M_AC$ , $M_A$ lies on the perpendicular bisector of $BC$ and is the midpoint of arc $BC$ , so $A,$ $I$ and $M_A$ are collinear. Similarly $B$ , $I$ and $M_B$ and $C$ , $I$ and $M_C$ are collinear.
Since $AM_CBM_ACM_B$ is cyclic, we have $\widehat{AM_BM_C}=\widehat{ACM_C}=\widehat{ACI}=\frac{\gamma}{2}=\widehat{ICB}=\widehat{IM_BM_C};$ similarly we find $\widehat{AM_CM_B}=\widehat{IM_CM_B}$ and so $AM_BM_C\cong IM_BM_C$ for the ASA criterion. So $I$ is the reflection of $A$ on $M_BM_C$ . The other two situations are identical.

Would you like to explain the following part which you mentioned in your answer of 3 ?

Quote:

$\widehat{AM_BM_C}=\widehat{ACM_C}=\widehat{ACI}=\frac{\gamma}{2}=\widehat{ICB}=\widehat{IM_BM_C}$

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#4 h Jan 20, 2019, 7:41 AM • 1 Y

Y by Adventure10

My Solution to (3): It follows from (2) that, $BM_A=IM_A$ and $BM_C=IM_C$ , and angle chasing shows that, $\angle M_CM_AI=\frac{C}{2}$ and $\angle BIM_A=\frac{A+B}{2} \implies M_AM_C \perp BI \implies \text{ that } M_AM_C \text{ bisects } BI$

This post has been edited 1 time. Last edited by AlastorMoody, Jan 20, 2019, 7:41 AM

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