Incircle and circumcircle

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Incircle and circumcircle

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Source: tst- Greece 2019

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#1 h Sep 23, 2019, 12:06 PM • 2 Y

Y by mathematicsy, Adventure10

Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$ . Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$ .Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$ .( $AN$ is a diametre of the circle $\Gamma$ ).

This post has been edited 1 time. Last edited by stergiu, Sep 23, 2019, 12:08 PM
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#2 h Sep 23, 2019, 12:09 PM • 3 Y

Y by Pluto1708, Mathasocean, Adventure10

Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$ . Inverting about the incircle gives that $S$ maps to $R$ , so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.

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#3 h Sep 23, 2019, 3:46 PM • 1 Y

Y by Adventure10

TheDarkPrince wrote:

Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$ . Inverting about the incircle gives that $S$ maps to $R$ , so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.

Just a simple question to this nice solution:

Why $S$ goes to $R$ and not to another point $Q$ ,of the circle $(AI)$ ? Thank you !

( Ok ! The circle $(A,B,C)$ has invers the Euler circle of triangle $DEF$ and $S$ belongs to this circle.Allmost obvious, but ....)

This post has been edited 3 times. Last edited by stergiu, Sep 23, 2019, 5:41 PM
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jayme

#5 h Sep 24, 2019, 8:15 AM • 2 Y

Y by Adventure10, Mango247

Dear Matlinkers,

http://www.artofproblemsolving.com/community/c6h614584

Sincerely
Jean-Louis

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#6 h Sep 24, 2019, 5:04 PM • 2 Y

Y by Adventure10, Mango247

Lets complex bash
Set the incircle as the unit circle with incentre as the origin and let $d = 1$ .

Then $a = \frac{2ef}{e + f} , b = \frac{2f}{1 + f} , c = \frac{2e}{1 + e}$
$\therefore \frac{a - b}{\overline{a - b}} = -f^2$ .

$nb \perp ba \implies -\frac{a - b}{\overline{a - b}} = \frac{c - n}{\overline{c - n}} = f^2$

$\therefore \overline{n} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$

Similarly $\overline{n} = \frac{(1+e)n-2e+2e^2}{e^2(1+e)}$

$\therefore \frac{(1+e)n-2e+2e^2}{e^2(1+e)} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$
After simplifying we obtain that
$n = (1 + e + f - ef)k$

where $k = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
Now we also obtain that $\overline{k} = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
So $\overline{k} = k \implies k \in\mathbb{R}$
Now also note that $s = \frac{1}{2}(1 + e + f - ef)$
So $\frac{n}{s} = 2k \in \mathbb{R}$
So $N$ , $I$ and $S$ are collinear.

This post has been edited 4 times. Last edited by Euler365, Sep 25, 2019, 10:53 AM

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gambi

#8 h Oct 17, 2021, 2:05 PM

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Let $L$ be the midpoint of arc $BAC$ in $\Gamma$ , and let $M$ be the antipode of $L$ in $\Gamma$ .
Let $K$ be the second intersection point of $MD$ with $\Gamma$ .

Claim 1. $K$ is the Miquel point of $BFEC$ .
Let $\Psi$ be the inversion around $(BIC)$ . By the Incenter Lemma, we get that such inversion is centered in $M$ . Clearly,
$\left\{\begin{array}{lll} \Psi ((ABC))=BC \\ \Psi (KM)=KM \\ \end{array} \right. \Longrightarrow \Psi (K)=D$ Therefore,
$MD\cdot MK=MI^2 \Longrightarrow (KID) \thickspace \text{tangent to} \thickspace IM$ Hence
$\measuredangle DKI=\measuredangle DIM=\measuredangle LMA=\measuredangle LKA$ But then
$90^o=\measuredangle MKL=\measuredangle DKL=\measuredangle DKI+\measuredangle IKL=\measuredangle LKA+\measuredangle IKL=\measuredangle IKA$ This way $K\in (AEIF)$ and so Claim 1 is proved.

Claim 2. $K,S,I$ are collinear.
Right triangles $\triangle FSD$ and $\triangle IDC$ are similar because
$\measuredangle DFS=\measuredangle DFI+\measuredangle IFS=\measuredangle DBI+\measuredangle IAE=90^o-\measuredangle ICD$ Analogously, right triangles $\triangle ESD$ and $\triangle IDB$ are similar.
From these two similarities, we get
$\left\{\begin{array}{lll} \frac{FS}{SD}=\frac{ID}{DC} \\ \\ \frac{SE}{SD}=\frac{ID}{BD} \\ \end{array} \right. \Longrightarrow \frac{FS}{SE}=\frac{BD}{DC} \qquad (\star)$ From Claim 1, we get that $K$ is the center of the spiral similarity sending $BC$ to $FE$ . But from $(\star)$ , we get that such spiral similarity also sends $D$ to $S$ .
Hence $\triangle KFS$ and $\triangle KBD$ are similar. Consequently,
$\measuredangle FKS=\measuredangle BKD=\measuredangle BKM=\measuredangle BAM=\measuredangle FAI=\measuredangle FKI$ Therefore, $K,S,I$ are collinear and so Claim 2 is proved.

Finally, from Claim 1 we have $K\in (AEFI) \Longrightarrow IK\perp KA$ .
But, in light of Claim 2, this implies that lines $ISK$ and $AK$ , are perpendicular, so ray $SI$ will intersect $\Gamma$ at $N$ , the antipode of $A$ .

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#9 h Apr 10, 2025, 1:46 AM

Y by

Easiest bash exercise

$(DEF)\in\mathbb{S}^1~D=d~E=e~F=f$
$A=\frac{2ef}{e+f}$
$B=\frac{2fd}{f+d}$
$C=\frac{2de}{d+e}$

$O=\frac{2def(d+e+f)}{(d+e)(e+f)(f+d)}$

$N=2O-A=\frac{2ef(d^2+de+df-ef)}{(d+e)(e+f)(f+d)}$

$S=\frac12(d+e+f-\frac{ef}{d})$

$\frac{s-i}{n-i}=\frac{(d+e+f-\frac{ef}{d})(d+e)(e+f)(f+d)}{4ef(d^2+de+df-ef)}=\frac{(d+e)(e+f)(f+d)}{4def}\in\mathbb{R}$

This post has been edited 2 times. Last edited by Sadigly, Apr 10, 2025, 1:56 AM

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