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Yet another way to reflect H to the circumcircle!
Tintarn   1
N 35 minutes ago by bin_sherlo
Source: Baltic Way 2024, Problem 14
Let $ABC$ be an acute triangle with circumcircle $\omega$. The altitudes $AD$, $BE$ and $CF$ of the triangle $ABC$ intersect at point $H$. A point $K$ is chosen on the line $EF$ such that $KH\parallel BC$. Prove that the reflection of $H$ in $KD$ lies on $\omega$.
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Tintarn
an hour ago
bin_sherlo
35 minutes ago
No more topics!
A projectional vision in IGO
Shayan-TayefehIR   6
N 3 hours ago by X.Allaberdiyev
Source: IGO 2024 Advanced Level - Problem 3
In the triangle $\bigtriangleup ABC$ let $D$ be the foot of the altitude from $A$ to the side $BC$ and $I$, $I_A$, $I_C$ be the incenter, $A$-excenter, and $C$-excenter, respectively. Denote by $P\neq B$ and $Q\neq D$ the other intersection points of the circle $\bigtriangleup BDI_C$ with the lines $BI$ and $DI_A$, respectively. Prove that $AP=AQ$.

Proposed Michal Jan'ik - Czech Republic
6 replies
Shayan-TayefehIR
Thursday at 7:59 PM
X.Allaberdiyev
3 hours ago
A projectional vision in IGO
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Source: IGO 2024 Advanced Level - Problem 3
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Shayan-TayefehIR
104 posts
#1
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In the triangle $\bigtriangleup ABC$ let $D$ be the foot of the altitude from $A$ to the side $BC$ and $I$, $I_A$, $I_C$ be the incenter, $A$-excenter, and $C$-excenter, respectively. Denote by $P\neq B$ and $Q\neq D$ the other intersection points of the circle $\bigtriangleup BDI_C$ with the lines $BI$ and $DI_A$, respectively. Prove that $AP=AQ$.

Proposed Michal Jan'ik - Czech Republic
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math_comb01
633 posts
#2 • 1 Y
Y by ehuseyinyigit
Sketch
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VicKmath7
1373 posts
#3
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I didn't like it very much, because it's basically two not much related problems combined into one problem (or at least my solution makes it look like that). Anyway, we show that $AP=AI=AQ$, refer to the first image for $AQ=AI$ and to the second for $AP=AI$.
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bin_sherlo
485 posts
#4 • 2 Y
Y by SomeonesPenguin, egxa
Part $I$: $AI=AQ$.
Proof of Part $I$: We have $I_AI.I_AA=I_AB.I_AC=I_AD.I_AQ$ hence $Q,A,I,D$ are concyclic. Also since $-1=(DC,DA;DI_A,DI)=(DC,DA;DQ,DI)$ and $\measuredangle ADC=90$, we observe that $\measuredangle QDA=\measuredangle ADI$ which implies $AQ=AI$.$\square$

Part $II$: $AI=AP$.
Proof of Part $II$: First we present a lemma.
Lemma: $ABC$ is a triangle with $AB=AC$ and altitude $AD$. Let $P$ be an arbitrary point on $BC$ where $I_1,I_2$ are the incenters of $\triangle PAB$ and $\triangle PAC$. Prove that $I_1,I_2,P,D$ are concyclic.
Proof: This is a special case of Serbia 2018 P1 and it can be proved by the method of moving points with rotations centered at $A$ and $D$.$\square$
Let $K\in BI$ with $AI=AK$ We will show that $\measuredangle I_CDK=\measuredangle I_CBK=90$. Let $C'$ be the reflection of $C$ over $D$. $CI\cap AD=F,BI_C\cap C'F=S$. By above lemma, since $S$ and $I$ are the incenters of triangles $\triangle AC'B$ and $\triangle ACB$ we see that $S,B,I,D$ are concyclic. Apply DDIT on quadrilateral $SI_CIK$ to get that $(\overline{DS},\overline{DI}),(\overline{DI_C},\overline{DK}),(\overline{DB},\overline{DF})$ is an involution. Note that $\measuredangle BDF=90=\measuredangle SDI$ hence this involution must be rotating $90$ degrees. Thus, $\measuredangle I_CDK=90$ as desired.$\blacksquare$
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Reason: typo
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SomeonesPenguin
98 posts
#5
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This is also very easy using barycentric coordinates.

Claim: $AQ=AI$.

Proof: Notice that $AI_cBI$ is cyclic so by PoP we have \[I_aA\cdot I_aI=I_aB\cdot I_aI_c=I_aD\cdot I_aQ\]So $DIAQ$ is cyclic. Now note that $-1=(I_a,I;AI\cap BC,A)$ and $\angle ADC=90^\circ$, so $DA$ is the angle bisector of $\angle QDI$ so $AQ=AI$.

Now we use barycentric coordinates to show that $AP=AI$. Set $ \triangle ABC$ as the reference triangle.

\begin{align*}
    D&=(0:a^2+b^2-c^2:a^2+c^2-b^2)\\
    I_c&=(a:b:-c)
\end{align*}
The equation of the circle $(I_cBD)$ is \[-a^2yz-b^2zx-c^2xy+(x+y+c)(ux+vy+wz)=0\]Now since $B$ lies on this circle we get $v=0$. Plugging $D$ and canceling a factor of $a^2(a^2+c^2-b^2)$ gives $w=(a^2+b^2-c^2)/2$. Finally, plugging $I_c$ in the equation yields \[ua-wc=-abc\implies ua+wc=2wc-abc=c(a^2+b^2-c^2-ab)\]
Now let $P=(a:t:c)$. Plugging $P$ in the equation and canceling a factor of $c(a-b+c)$ gives \[t=\frac{a^2}{b+c}-c\]Therefore \[P=\left(\frac{b+c}{a+b+c},\frac{a^2-bc-c^2}{a(a+b+c)},\frac{bc+c^2}{a(a+b+c)}\right)\]So $\overrightarrow{AP}$ has displacement vector \[\left(\frac{-a}{a+b+c},\frac{a^2-bc-c^2}{a(a+b+c)},\frac{bc+c^2}{a(a+b+c)}\right)\]Finally, plugging this into the length formula gives

\begin{align*}
    (a+b+c)^2\cdot AP^2&=-(a^2-bc-c^2)(bc+c^2)+b^2(bc+c^2)+c^2(a^2-bc-c^2)\\
    &=-a^2bc+2b^2c^2+b^3c+c^3b
\end{align*}
The displacement vector of $\overrightarrow{AI}$ is \[\left(\frac{-b-c}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)\]So by the distance formula \[(a+b+c)^2\cdot AI^2=-a^2bc+2b^2c^2+b^3c+c^3b\]And this concludes the proof. $\blacksquare$
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egxa
158 posts
#6 • 1 Y
Y by bin_sherlo
Another way for $AI=AP$ is in complex plane $P=a^2+b^2+\frac{b^2c}{a}+bc+ba$
This post has been edited 2 times. Last edited by egxa, Yesterday at 8:37 AM
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X.Allaberdiyev
93 posts
#7
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One can easily prove that $AIDQ$ is cyclic by PoP. Next, harmonic bundles imply that $AD$ bisects $QDI$. Then, $AQ=AI$. Then, $\angle QAI=\angle IDI_a=2\angle QDB=2\angle QPB$. Hence, $A$ is the center $(QPI)$. So, we are done.
This post has been edited 2 times. Last edited by X.Allaberdiyev, 3 hours ago
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