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Tangent circles
oVlad   6
N Sep 15, 2024 by Helixglich
Source: Russian TST 2018, Day 9 P2 (Group NG), P4 (Groups A & B)
The point $K{}$ is the middle of the arc $BAC$ of the circumcircle of the triangle $ABC$. The point $I{}$ is the center of its inscribed circle $\omega$. The line $KI$ intersects the circumcircle of the triangle $ABC$ at $T{}$ for the second time. Prove that the circle passing through the midpoints of the segments $BC, BT$ and $CT$ is tangent to the circle which is symmetric to $\omega$ with respect to $BC$.
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oVlad
Mar 30, 2023
Helixglich
Sep 15, 2024
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Tangent circles
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Source: Russian TST 2018, Day 9 P2 (Group NG), P4 (Groups A & B)
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oVlad
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oVlad
#1 h  Mar 30, 2023, 4:41 PM • 2 Y
Y by NO_SQUARES, Rounak_iitr
The point $K{}$ is the middle of the arc $BAC$ of the circumcircle of the triangle $ABC$. The point $I{}$ is the center of its inscribed circle $\omega$. The line $KI$ intersects the circumcircle of the triangle $ABC$ at $T{}$ for the second time. Prove that the circle passing through the midpoints of the segments $BC, BT$ and $CT$ is tangent to the circle which is symmetric to $\omega$ with respect to $BC$.
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hef4875
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hef4875
#2 h  Apr 3, 2023, 6:04 AM
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Bump for this
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nguyenducmanh2705
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nguyenducmanh2705
#4 h  Feb 11, 2024, 2:07 PM
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Bump....
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JafoNoksy
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JafoNoksy
#5 h  May 6, 2024, 9:48 PM • 1 Y
Y by NO_SQUARES
In file you can see my solution.
Attachments:
something.pdf (383kb)
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NO_SQUARES
1133 posts
NO_SQUARES
#6 h  May 8, 2024, 7:01 PM • 1 Y
Y by JafoNoksy
JafoNoksy wrote:
In file you can see my solution.

Very nice solution and problem!
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math_comb01
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math_comb01
#7 h  Aug 18, 2024, 2:34 PM
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We claim this circle is ninepoint circle reflected over $MI$ which finishes by feurbach tangency, and so it suffices to show that $MI$ passes through the midpoint of $AT'$ call it $W$.It suffices to show $WMI$ are collinear and $W$ is on NPC, the latter follows by homoethety of $2$ at $A$ and the former, let nagel cevian meet farther incircle at $L$ then well known $AL \parallel MI$, so suffices to show reflection of $T'$ over $I$ lies on nagellian but follows as $MT'$ is refl of nagellian in $I$.
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Helixglich
113 posts
Helixglich
#8 h  Sep 15, 2024, 12:57 PM
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Perhaps a more reasonable solution. Let $T$' be the reflection of $T$ over $BC$. Now take homothety at $T$' with factor $2$. This leads to proving that this homothetized circle is tangent to the circumcircle. Work with classical incentre setup.
$t = \frac{k-p}{k\overline{p}-1} = -\frac{2abc+a^2b+a^2c}{2a+b+c}$. Now we compute $t' = b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c}$ and hence $-i' = 2(ab+bc+ca)+b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c} = \frac{(2a+b+c)(ab+bc+ca)^2}{a(2bc+ab+ac)}$.
Hence we conclude that the power of $I$' with respect to the circumcircle is $i'\overline{i'} = \frac{(a+b+c)^2(ab+bc+ca)^2}{a^2b^2c^2} = (\overline{i}i)^2$ and thus we are done as this clearly implies that $Pow(I',(ABC)) = (2Rr)^2$.
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