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geometry problem with many circumcircles
Melid   0
7 minutes ago
Source: own
In scalene triangle $ABC$, which doesn't have right angle, let $O$ be its circumcenter. Circle $BOC$ intersects $AB$ and $AC$ at $A_{1}$ and $A_{2}$ for the second time, respectively. Similarly, circle $COA$ intersects $BC$ and $BA$ at $B_{1}$ and $B_{2}$, and circle $AOB$ intersects $CA$ and $CB$ at $C_{1}$ and $C_{2}$ for the second time, respectively. Let $O_{1}$ and $O_{2}$ be circumcenters of triangle $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2}$, respectively. Prove that $O, O_{1}, O_{2}$ are collinear.
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Melid
7 minutes ago
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interesting geo config (2/3)
Royal_mhyasd   2
N 12 minutes ago by Ilikeminecraft
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
2 replies
Royal_mhyasd
Yesterday at 11:36 PM
Ilikeminecraft
12 minutes ago
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interesting geo config (1\3)
Royal_mhyasd   1
N 17 minutes ago by Ilikeminecraft
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
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Royal_mhyasd
Yesterday at 11:18 PM
Ilikeminecraft
17 minutes ago
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Incircle and circumcircle
stergiu   6
N Apr 10, 2025 by Sadigly
Source: tst- Greece 2019
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 $\Gamma$).
6 replies
stergiu
Sep 23, 2019
Sadigly
Apr 10, 2025
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Incircle and circumcircle
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Source: tst- Greece 2019
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stergiu
1648 posts
stergiu
#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
Reason: explanation
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TheDarkPrince
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TheDarkPrince
#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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stergiu
1648 posts
stergiu
#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
Reason: correction
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jayme
9801 posts
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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Euler365
143 posts
Euler365
#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
82 posts
gambi
#8 h  Oct 17, 2021, 2:05 PM
Y by
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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Sadigly
229 posts
Sadigly
#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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