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A well-known geo configuration revisited
Tintarn   6
N 7 minutes ago by Primeniyazidayi
Source: Dutch TST 2024, 2.1
Let $ABC$ be a triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $D$ be the reflection of $A$ in $B$ and let $E$ the reflection of $A$ in $C$. Let $M$ be the midpoint of segment $DE$. Show that the tangent to $\Gamma$ in $A$ is perpendicular to $HM$.
6 replies
Tintarn
Jun 28, 2024
Primeniyazidayi
7 minutes ago
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incircle geometry
Tuguldur   0
an hour ago
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The diagonals $AC$ and $BD$ meet at $E$. The rays $CB$ and $DA$ meet at $F$. Prove that the line through the incenters of $\triangle ABE$ and $\triangle ABF$ and the line through the incenters of $\triangle CDE$ and $\triangle CDF$ meet at a point lying on $\omega$.
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Tuguldur
an hour ago
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geometry
realitywarper_888   2
N Oct 17, 2022 by PROF65
Let $\Delta ABC$ be a triangle. $E, F$ are points such that $EA \perp AB, EB \perp BC, FA \perp AC, FC \perp BC$. $M$ is the midpoint of $BC$, $L$ lies on $BC$ such that $AL$ is the symmedian of $\Delta ABC$. Perpendicular bisector of $LM$ cut $AM$ at $K$. Prove that $(KLM)$ touches two circles with diameters $BE$ and $CF$
2 replies
realitywarper_888
Oct 16, 2022
PROF65
Oct 17, 2022
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realitywarper_888
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realitywarper_888
#1 h  Oct 16, 2022, 1:41 PM
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Let $\Delta ABC$ be a triangle. $E, F$ are points such that $EA \perp AB, EB \perp BC, FA \perp AC, FC \perp BC$. $M$ is the midpoint of $BC$, $L$ lies on $BC$ such that $AL$ is the symmedian of $\Delta ABC$. Perpendicular bisector of $LM$ cut $AM$ at $K$. Prove that $(KLM)$ touches two circles with diameters $BE$ and $CF$
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on_gale
65 posts
on_gale
#2 h  Oct 16, 2022, 9:59 PM • 2 Y
Y by realitywarper_888, Mango247
Let $X, Y$ be the second points of intersection of $AL, AM$ with $(ABC)$.
Let $Z$ be a point in $AL$ such that $XK \| ZM$.

Using inversion with center $A$ and mapping $B$ to $C$, we conclude that the inverses of $(EAB), (FAC)$ are the lines $CD, BD$ respectively.
Obviously, $(X, M)$ and $(L, Y)$ are pairs of inverses. Also, since $AZ \cdot AK = AX \cdot AM$, we see that $Z$ is the inverse of $K$.

Therefore, it suffices to prove that $(XYZ)$ is tangent to the lines $BD$ and $CD$.
And that's equivalent to prove that $(XYZ)$ and $(ABC)$ are homothetic. But we already know that $Y$ is the reflection of $X$ over $DM$.

So we just have to prove that $\dfrac{DA}{DX} = \dfrac{DX}{DZ}$ which is the same as saying $DX^2 = DA \cdot DZ$.

Also, since $(A, X; B, C) = -1$, then $\angle{XML} = \angle{LMK} = \angle{KLM}$ so $LK \| XM$.

And that's it, the rest is just bashing:
Let $p = AK$, $q = KM$. Then $AL = pr$, $LX = qr$ for some $r$.

Now, we'll use that $(A, X; L, D) = -1$ to say that $XD = \dfrac{qr(p+q)}{p-q}$.
Furthermore, we have $\dfrac{AX}{XZ} = \dfrac{AK}{KM}$ implying that $XZ = \dfrac{qr(p+q)}{p}$.

So $DZ = DX - XZ = \dfrac{q^2r(p+q)}{p(p-q)}$. And now we just check it:

$DA \cdot DZ = (\dfrac{pr(p+q)}{p-q}) \cdot (\dfrac{q^2r(p+q)}{p(p-q)}) = \dfrac{q^2r^2(p+q)^2}{(p-q)^2} = DX^2$
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PROF65
2016 posts
PROF65
#3 h  Oct 17, 2022, 12:36 AM • 1 Y
Y by realitywarper_888
Let $O_1,O_2 , S$ the centers of $(AEB)$, $(AFC)$ , the intersection of $BC$ and $O_1O_2$; $T,T'$ the tangency points of $(AEB),(AFC)$ with their common tangent other than $BC$ :
it s clear that $S$ is the ex-similicenter of $(AEB),(AFC)$ and $M$ is on their radical axis:
then $SA^2=SB.SC$ so $AS$ is tangent to $(ABC)$ hence $(S,L,B,C)=-1$ implies $ ML.MS=MB^2$ hence the inversion with center $M$ and power $MB^2$ swaps $S$ with $ L$ ; $TT'$ with $(I)$, the common tangent circle to $(O_1),(O_2)$ passing through $L,M$ ; $K'$ with the midpoint of $TT'$ say $G$ where $K'=(I)\cap AM$.
Whence $ MK'L\sim GSM$ but $ SG =SM$ hence $K'M=K'L$ which means $K=K'$ and $ (KLM)$ touches the two circles.
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