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Source: 1st TASIMO, Day1 Problem1

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NJAX

#1 h May 18, 2024, 8:57 AM • 3 Y

Y by dmusurmonov, navi_09220114, ItsBesi

Let $ABC$ be a triangle with $AB<AC$ and incenter $I.$ A point $D$ lies on segment $AC$ such that $AB=AD,$ and the line $BI$ intersects $AC$ at $E.$ Suppose the line $CI$ intersects $BD$ at $F,$ and $G$ lies on segment $DI$ such that $FD=FG.$ Prove that the lines $AG$ and $EF$ intersect on the circumcircle of triangle $CEI.$
Proposed by Avan Lim Zenn Ee, Malaysia

This post has been edited 3 times. Last edited by NJAX, May 18, 2024, 1:37 PM

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#2 h May 18, 2024, 11:13 AM • 3 Y

Y by VicKmath7, NJAX, Math_.only.

Very suitable IMO1 angle chasing problem!

Denote $\angle ABC = \beta$ , $\angle ACB = \gamma$ . Observe that $\angle ABD = \frac{\beta + \gamma}{2}$ due to $AB = AD$ , so $\angle IBD = \angle ICD = \frac{\gamma}{2}$ and $BIDC$ is cyclic. Next, $FD = FG$ implies $\angle GFB = \gamma = \angle BCD = \angle EID = \angle EIG$ , so $BIGF$ is also cyclic. We now bravely claim that the second intersection point $T$ of the circumcircles of $CEI$ and $BIGF$ lies on both $AG$ and $EF$ , which would complete the proof.

On one hand, $\angle IFT = \angle IBF = \angle ICE = \angle ITE$ , so $E$ , $F$ , $T$ are collinear. On the other hand, $\angle FTG = \angle FIG = \frac{\beta - \gamma}{2}$ , so the collinearity of $A$ , $G$ , $T$ is now equivalent to $\angle ETA = \frac{\beta-\gamma}{2}$ , i.e. (as $\angle CTE = \angle CIE = \frac{\beta + \gamma}{2}$ to $\angle CTA = \beta$ - that is, $ABTC$ to be cyclic. However, $\angle BTI + \angle CTI = \angle BFI + 180^{\circ} - \angle CEI = \frac{\beta}{2} + \angle IBC + \angle ECB = \beta + \gamma = 180^{\circ} - \angle BAC$ and so we are done.

Remark

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This post has been edited 2 times. Last edited by Assassino9931, Jun 1, 2024, 11:44 PM

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#3 h Jun 4, 2024, 10:54 PM • 2 Y

Y by NJAX, Math_.only.

Let $X$ be the Miquel point of complete quadrilateral $IEDF$ .

Claim: $BIDC$ is concyclic
$\angle BID=180^{\circ}-2\angle BIA=\angle DCB$ Claim: $E$ , $F$ , and $X$ are collinear
It is sufficient to show that both $XF$ and $XE$ bisect $\angle IXD$ . $\angle IXF=\angle IBF=\angle FCD=\angle FXD$ $\angle IXE=\angle ICE=\angle DBE=\angle DXE$ Claim: $X$ lies on the circumcircle of $ABC$
$\angle BXC=\angle FIE+\angle FDE=\angle FBC+\angle ACB+\angle ABD=180^{\circ}-\angle BAC$ Claim: $BGIFX$ is concyclic
$\angle IGF=\angle IDF=\angle IBF$ Claim: $A$ , $G$ , and $X$ are collinear
$\angle GXB=\angle GFB=\angle ACB=\angle AXB$

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This post has been edited 2 times. Last edited by sami1618, Jun 5, 2024, 2:18 PM

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#5 h Jun 9, 2024, 4:19 PM

Y by

My solution is a bit different (Lovely problem by the way :love:

:love:

)

Let $\angle BAC=\alpha , \angle ABC=\beta , \angle ACB=\gamma , \odot (ABC)=\Gamma , \odot(CEI)=\Omega , EF \cap \Omega =\{X\}$
So $\alpha +\beta+\gamma=180 , \angle ABE=\angle CBE=\frac{\beta}{2} , \angle ACI=\angle BCI=\frac{\gamma}{2}$

$\textbf{Claim:}$ Points $B,D,I$ and $C$ -are concyclic
$\textbf{Proof:}$ From the isosceles triangle $\triangle ABD$ we have:

$\angle ABD+ \angle ADB+\angle BAD=180 \implies 2\angle ABD +\alpha =180 \implies \angle ABD=90-\frac{\alpha}{2}=\frac{\beta+\gamma}{2}$ $\implies \frac{\beta+\gamma}{2}=\angle ABD=\angle ABE+\angle EBD=\frac{\beta}{2}+\angle EBD \implies \angle EBD=\frac{\gamma}{2} \implies \angle ICD=\frac{\gamma}{2}=\angle EBD \equiv IBD \implies$
$\angle ICD=\angle IBD \implies$ Points $B,D,I$ and $C$ -are concyclic $\square$ . Let $\odot(BDIC)=\sigma$

$\textbf{Claim:}$ Points $F,D,X$ and $C$ -are concyclic
$\textbf{Proof:}$

$\angle FDC \equiv \angle BDC \stackrel{\sigma}{=} \angle BIC=180-\angle EFC \stackrel{\Omega}{=} 180-\angle EXC \equiv 180- \angle FXC \implies \angle FDC=180-\angle FXC \implies$
$\angle FDC+\angle FXC =180 \implies$ Points $F,D,X$ and $C$ -are concyclic $\square$ . Let $\odot(FDXC)=\omega_1$

$\textbf{Claim:}$ Points $B,I,F$ and $X$ -are concyclic
$\textbf{Proof:}$ By Power Of The Point Theorem (POP) we have:

$EI \cdot EB=Pow(E,\sigma)=ED \cdot EC=Pow(E,\omega_1)=EF \cdot EX \implies EI \cdot EB= EF \cdot EX$ which by the converse of POP means that Points $B,I,F$ and $X$ -are concyclic $\square$ . Let $\odot(BIFX)=\omega_2$

$\textbf{Claim:}$ $G \in \omega_2$
$\textbf{Proof:}$ Since $FD=FC$ we get $\angle FGD=\angle FDG$ so:

$180-\angle IGF=\angle FGD=\angle FDG \equiv \angle BGI \stackrel{\sigma}{=} \angle BCI =\frac{\gamma}{2}=\angle EBC \equiv IBF \implies 180-\angle IGF =\angle IBF \implies$
$\angle IGF+\angle IBF=180 \implies$ Points $B,I,F$ and $G$ -are concyclic $\iff G \in \odot(BIF) \iff G \in \omega_2 \square$

$\textbf{Claim:}$ $X \in \Gamma \iff X \in \odot (ABC)$
$\textbf{Proof:}$

$\angle BXC=\angle BXF+\angle FXC \stackrel{\omega_1}{=} \angle BXF + (180-\angle FDC) \stackrel{\omega_2}{=} (180-BIF)+(180-\angle FDC)=$
$=360-\angle BIF-\angle FDC \equiv 360-\angle BIC-\angle BDC \stackrel{\sigma}{=} 360-\angle BDC-\angle BDC=360-2\angle BDC=360-2(180-\angle BDA)=$
$=360-360+2\angle BDA=2\angle BDA=2\angle ABD=2 \cdot \frac{\beta+\gamma}{2}=\beta+\gamma \implies \angle BXC=\beta+\gamma$

So $\angle BAC+\angle BXC=\alpha +(\beta+\gamma)=\alpha+\beta+\gamma=180 \implies \angle BAC+\angle BXC =180 \implies$
Points $A,B,C$ and $X$ -are concyclic $\iff X \in (ABC) \iff X \in \Gamma \square$

$\textbf{Claim:}$ Points $A-G-X$ are collinear
$\textbf{Proof:}$

$\angle GXB \stackrel{\omega_2}{=} \angle GFB=180-\angle GFD \stackrel{\triangle GFD}{=} \angle FGD+\angle FDG \stackrel{FD=FG}{=} \angle FDG+\angle FDG=2\angle FDG \equiv 2\angle BDI \stackrel{\sigma}{=} 2\angle BCI=2 \cdot \frac{\gamma}{2}=\gamma=$
$=\angle ACB \stackrel{\Gamma}{=} \angle AXB \implies \angle GXB=\angle AXB \implies$ Points $A-G-X$ are collinear $\blacksquare$

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This post has been edited 1 time. Last edited by ItsBesi, Jun 9, 2024, 4:22 PM
Reason: typo

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117 posts

#6 h Jan 26, 2025, 8:49 AM

Y by

sami1618 wrote:

Let $X$ be the Miquel point of complete quadrilateral $IEDF$ .

Claim: $BIDC$ is concyclic
$\angle BID=180^{\circ}-2\angle BIA=\angle DCB$ Claim: $E$ , $F$ , and $X$ are collinear
It is sufficient to show that both $XF$ and $XE$ bisect $\angle IXD$ . $\angle IXF=\angle IBF=\angle FCD=\angle FXD$ $\angle IXE=\angle ICE=\angle DBE=\angle DXE$ Claim: $X$ lies on the circumcircle of $ABC$
$\angle BXC=\angle FIE+\angle FDE=\angle FBC+\angle ACB+\angle ABD=180^{\circ}-\angle BAC$ Claim: $BGIFX$ is concyclic
$\angle IGF=\angle IDF=\angle IBF$ Claim: $A$ , $G$ , and $X$ are collinear
$\angle GXB=\angle GFB=\angle ACB=\angle AXB$

Thing is real, if you could not notice that $X$ lies on the $ABC$ .

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