Prove concyclic and tangency

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Prove concyclic and tangency

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Source: Japan Olympiad Finals 2014, #4

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202 posts

#1 h May 17, 2014, 7:34 AM • 6 Y

Y by itslumi, KrazyGuy, mathleticguyyy, Adventure10, Mango247, ItsBesi

Let $\Gamma$ be the circumcircle of triangle $ABC$ , and let $l$ be the tangent line of $\Gamma$ passing $A$ . Let $D, E$ be the points each on side $AB, AC$ such that $BD : DA= AE : EC$ . Line $DE$ meets $\Gamma$ at points $F, G$ . The line parallel to $AC$ passing $D$ meets $l$ at $H$ , the line parallel to $AB$ passing $E$ meets $l$ at $I$ . Prove that there exists a circle passing four points $F, G, H, I$ and tangent to line $BC$ .

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#2 h May 17, 2014, 9:48 AM • 4 Y

Y by mathematiculperson, Mathematicsislovely, Adventure10, Mango247

let parallel line to $AC$ passing $D$ meet $BC$ at $X$ .
$\frac{BD}{AD}=\frac{BX}{CX}=\frac{AE}{EC}\Rightarrow XE\left | \right |AB,$
$E,X,I$ in same line.
$\widehat{HAB}=\widehat{ACB}=\widehat{HXB}\Rightarrow$ $AXBH$ is cycle $\Rightarrow$ $HD.DX=AD.BD=DF.DG\Rightarrow$ $HFXG$ is cycle.similar we find $FXGI$ is cycle. $\Rightarrow$ $HFXGI$ is cycle( $o$ ).
$\widehat{AIX}=\widehat{ACB}=\widehat{HXB}$ $\Rightarrow$ $o$ is tangent to $BC$ at $X$ .
done

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#3 h Jun 15, 2014, 10:41 AM • 3 Y

Y by mathematiculperson, Adventure10, Mango247

Let $HD \cap BC = P$ . So, $\frac{BP}{PC}=\frac{BD}{DA}=\frac{AE}{EC} \implies EP \parallel AB \implies P \in EI$ .
Now, $\measuredangle CPI = \measuredangle CBA = \measuredangle IAC \implies EI.EP = EA.EC = EF.EG \implies I \in \odot PFG$ . Similarly, $H \in \odot PFG$ .
$\therefore FGHIP$ cyclic. Now, $\measuredangle CPI = \measuredangle PBA = \measuredangle PHI \implies CP, i.e., BC$ touches $\odot FGHI$ at $P$ .

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#4 h Oct 2, 2014, 10:05 AM • 2 Y

Y by mathematiculperson, Adventure10

Let $DX$ be the line parallel to $AC$ with $X$ on $BC$ .Then $\frac{AE}{EC}=\frac{DB}{AD}=\frac{BX}{XC} \implies EX \parallel AB$ .Thus the parallels meet at the same point on $BC$ .Now we see that $\angle{HXB}=C$ and since $HI$ is the tangent at $A$ , $\angle{HAB}=C$ .Thus points $A,H,B,X$ ar concyclic.So $FD \cdot DG=AD \cdot DB=DH \cdot DX$ or points $F,H,G,X$ concyclic.Similarly points $G,I,F,X$ are concyclic.Thus $FHIGX$ is a cyclic pentagon.Now note that $\angle{HIX}=\angle{HAB}=C=\angle{HXB}$ or $\odot{HFXGI}$ is tangent to $BC$ at $X$ ,as desired.

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#5 h Feb 21, 2020, 9:44 AM • 2 Y

Y by aopsuser305, Adventure10

not really adding anything new, but here it is.

Let $X$ be the point on $\overline{BC}$ satisfying $\tfrac{BD}{DA} = \tfrac{AE}{EC} = \tfrac{BX}{XC}$ . Then $\overline{XD} \parallel \overline{AC}$ so $X$ lies on line $HD$ ; likewise $X$ lies on line $IE$ .

Now from
$\measuredangle HXB = \measuredangle ACB = \measuredangle HAB$ we deduce $HAXB$ is cyclic, so from $DH \cdot DX = DA \cdot DB = DF \cdot DG$ we obtain that $HFXG$ is cyclic. Likewise $IFXG$ is cyclic.

To finish, we need to show that $(HIX)$ is tangent to $\overline{BC}$ : this follows as
$\measuredangle HIX = \measuredangle HAB = \measuredangle HXB.$

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#6 h Apr 4, 2020, 6:04 PM • 1 Y

Y by mkomisarova

Nice problem. The key step is to prove that $HD$ and $EI$ intersects on $BC$ .
Let $X$ be intersection of $HD$ and $EI$ . It is obvious that $ADXE$ is parallelogram. Therefore $\frac{BD}{DA}=\frac{AE}{EC}=\frac{BD}{XE}=\frac{DX}{EC}$ . We conclude that $\triangle BDX \sim \triangle XEC$ . As a result $\angle DXB = \angle ECX$ . Note that $\angle EXC=180 -\angle A -\angle DXB$ and $\angle A = \angle DXE$ . Therefore $\angle BXC = 180$ and it means that $X$ lies on $BC$ .
The rest is easy. Note that $\angle HAB=\angle ACB= \angle HXB$ . This implies that $HAXB$ is cyclic. As a result $AD \cdot DB=HD \cdot DX$ , but $AD \cdot DB=FD \cdot DG$ , therefore $HD \cdot DX=FD \cdot DG$ . As a result $HFXG$ is cyclic. Similarly we get that $IGXF$ is cyclic. We conclude that $HFXGI$ is cyclic.
We claim that the circumcircle of $HFXGI$ is tangent to $BC$ at $X$ . This is obvious, since $\angle BXH=\angle ACB=\angle HIX$ .

This post has been edited 1 time. Last edited by Kimchiks926, Apr 4, 2020, 8:07 PM
Reason: typo

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#7 h Aug 27, 2020, 4:45 AM • 7 Y

Y by itslumi, srijonrick, Mathematicsislovely, v4913, HamstPan38825, Mango247, MS_asdfgzxcvb

Solution from OTIS office hours:

We start with the following critical definition: Define $X$ on $BC$ such that $\frac{BD}{DA} = \frac{BX}{XC} = \frac{AE}{EC}.$ Then $ADXE$ is obviously a parallelogram, so that $X = \overline{HD} \cap \overline{IE}$ .
$[asy] pair A = dir(110); pair B = dir(210); pair C = dir(330); filldraw(unitcircle, invisible, blue); draw(A--B--C--cycle, blue); real r = 0.7; pair D = r*B+(1-r)*A; pair E = r*A+(1-r)*C; pair X = r*B+(1-r)*C; pair H = extension(X, D, A, dir(90)*A+A); pair I = extension(X, E, A, dir(90)*A+A); draw(H--I--X--cycle, red); filldraw(circumcircle(H, I, X), invisible, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(210)); dot("$E$", E, dir(45)); dot("$X$", X, dir(270)); dot("$H$", H, dir(H)); dot("$I$", I, dir(I)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 unitcircle 0.1 lightcyan / blue A--B--C--cycle blue ! real r = 0.7; D = r*B+(1-r)*A R210 E = r*A+(1-r)*C R45 X = r*B+(1-r)*C R270 H = extension X D A dir(90)*A+A I = extension X E A dir(90)*A+A H--I--X--cycle red circumcircle H I X 0.1 lightred / red */ [/asy]$

Claim: The circumcircle of $\triangle HIX$ is tangent to $X$ at $BC$ .
Proof. Because $\measuredangle HIX = \measuredangle HAB = \measuredangle ACB = \measuredangle HXB$ . $\blacksquare$

Claim: Points $D$ and $E$ lie on the radical axis of $(HIX)$ and $(ABC)$ .
Proof. We saw already $\measuredangle HAB = \measuredangle ACB = \measuredangle HXB$ so quadrilateral $HAXB$ is cyclic. Then $D$ is the radical center of $(HIX)$ , $(ABC)$ , $(HAXB)$ . $\blacksquare$

Thus $F$ and $G$ are the two intersections of the circles, as needed.

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#8 h Oct 19, 2020, 7:09 PM • 1 Y

Y by Mango247

Let $M, N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ , let $E' \neq E$ be on $\overline{AC}$ such that $NE = NE'$ , and let $T = \overline{HD} \cap \overline{EI}$ . The ratio condition implies that $\overline{DE'} \parallel \overline{BC}$ ; in particular, this implies that $\overline{DE}$ is bisected by $\overline{MN}$ . As $ADTE$ is a parallelogram, it follows that $\overline{AT}$ is bisected by $\overline{MN}$ as well, implying that $T$ lies on $\overline{BC}$ .

Now, we have
$\begin{align*} \angle HAD = \angle HAB = \angle BCA = \angle BTH, \end{align*}$ implying that $HATB$ is cyclic. By Power of a Point, we now have
$\begin{align*} DH \cdot DT = DA \cdot DB = DF \cdot DG, \end{align*}$ implying that $FTGH$ is cyclic. Similarly, $FTGI$ is cyclic, so $FGHIT$ is cyclic. To finish, we have
$\begin{align*} \angle HIT = \angle HAD = \angle BTH, \end{align*}$ implying that $\overline{BC}$ is tangent to $(FGHIT)$ . We are done! $\Box$

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#9 h Oct 21, 2020, 1:50 PM

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Let lines $\overline{HD}$ and $\overline{BC}$ meet at a point $T$ . Now note that $TC/TB = DA/DB = EC/EA$ , so $T$ lies on $\overline{IE}$ . Note that $\angle HAB = \angle ACB = \angle DTB = \angle HTB$ , so $AHBT$ is cyclic. Similarly, $AICT$ is cyclic. Now $HD \cdot TD = AD \cdot BD = FD \cdot GD$ , which implies that $HFTG$ is cyclic. By symmetry $IGTF$ is also cyclic. Now $\angle HTB = \angle HAB = \angle HIT$ , so $\overline{BTC}$ is tangent to $(HIGTF)$ , as desired.

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#10 h Nov 9, 2020, 11:02 PM

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Heavy 2020 APMO 1 vibes.

Let $X_1$ be the point on $BC$ such that $DX_1 \parallel AC$ and $X_2$ be the point on $BC$ such that $EX_1 \parallel AB$ . We see that $\frac{BX_1}{CX_1} = \frac{BD}{DA} = \frac{AE}{EC} = \frac{BX_2}{CX_2}$ hence $X_1 = X_2$ are the same point. Let this point be $X$ , on $BC$ such that $ADXE$ is a parallelogram. Since $\angle HAB = \angle ACB = \angle HXB \text{ and } \angle IAC = \angle ABC = \angle IXC$ we see that $HAXB$ and $IAXC$ are cyclic. Now through PoP we get that $FD \cdot GD = BD \cdot AD = HD \cdot XD$ $FE \cdot GE = AE \cdot CE = IE \cdot XE$ thus $HFXG$ and $IGXF$ are cyclic and thus $HFXGI$ are all cyclic. Indeed, since $\angle BXH = \angle BCA = \angle XIH$ , this circle is tangent to $BC$ , as desired. $\blacksquare$

Attachments:

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#11 h Nov 23, 2020, 9:45 AM

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Wonderful problem!

Claim. $DH \cap EI \in BC$ .
Proof. To prove this, we use phantom point. Suppose that $X \in BC$ such that $DX \parallel AC$ and $E' \in AC$ such that $XE' \parallel AB$ . We will prove that $E \equiv E'$ . Now, notice that we have $\triangle BDX \sim \triangle XE'C$ and $AEX'D$ being a parallelogram. Therefore, we must have
$\frac{BD}{DX} = \frac{XE'}{E'C} \Rightarrow \frac{BD}{AE'} = \frac{AD}{E'C} \Rightarrow \frac{BD}{AD} = \frac{AE'}{E'C}$ However, $E \in AC$ such that $\frac{AE}{EC} = \frac{BD}{AD} = \frac{AE'}{E'C}$ , which forces $E = E'$ .

Now, notice that $(HIX)$ tangent to $BC$ at $X$ since $\measuredangle HIX = \measuredangle HXB$ . It suffices to prove that $F, G \in (HIX)$ .
Claim. $AICX$ and $AXBH$ is cyclic.
Proof. Simple angle chasing yields $\measuredangle AIX = \measuredangle HAB = \measuredangle ACB \equiv \measuredangle ACX$ . Similarly, we have $\measuredangle AHX = \measuredangle IAC = \measuredangle ABC \equiv \measuredangle ABX$ .

Claim. $DE$ is the radical axis of $(HXI)$ and $(ABC)$ .
Proof. Radical axis theorem on $(HIX), (ABC), (AHBX)$ gives us $D = AB \cap HX$ lies on radical axis of $(HXI)$ and $(ABC)$ . Similarly, radical axis theorem on $(HIX), (ABC), (AICX)$ gives us $E = AC \cap IX$ lies on radical axis of $(HIX)$ and $(ABC)$ . Therefore, $DE$ is the radical axis of $(HXI)$ and $(ABC)$ .

Now, notice that $(HIX)$ must intersect $(ABC)$ at two points, both of which lies in its radical axis. However, we know that $DE \cap \Gamma = F,G$ , and therefore, $F,G \in (HIX)$ as well.

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#12 h Dec 13, 2020, 6:08 AM

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First interpret the length conditions as $P$ lying on $\overline{BC}$ , which can be done with some simple ratio manipulations to get $\frac{BD}{PD} = \frac{PE}{CE}$ , which is $\triangle BDP \sim \triangle PEC$ .

We continue with angle chase: $\measuredangle HAB = \measuredangle ACB = \measuredangle HPB,$ implying $HABP$ cyclic. Similarly, $AICP$ is cyclic. Then, $\angle HIP = \angle AIP = \angle ACP = \angle HPB,$ thus $(HIP)$ is tangent to $\overline{BC}$ at $P$ . Next, redefine $F'$ and $G'$ to be $(PHI) \cap (ABC)$ . However, from $HABP$ cyclic, we get $HD \cdot PD = AD \cdot BD$ , aka $D$ lies on the radical axis of $(PHI)$ and $(ABC)$ , and similarly for $E$ . Thus $\{F', G'\} = \{F, G\}$ and we are done.

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#13 h Jan 7, 2021, 5:46 PM

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Let $P= EI \cap BC$ . Since $EP \parallel AB \implies \frac{CP}{BP}= \frac{CE}{EA}= \frac{AD}{BD} \implies PD \parallel AC \implies P$ , $D$ and $H$ are collinear.

Now, observe that since $AI$ is tangent to $\Gamma$ and $AC \parallel PH$ , $\angle ABC= \angle CAI= \angle PHA \implies AHBP$ is cyclic $\implies DH.DP=DA.DB=DF.DG \implies HFPG$ is cyclic. Similarly, $IGPF$ is cyclic $\implies HFGI$ is cyclic. Moreover, $\angle IPC= \angle ABC= \angle PHI \implies$ since $P \in (HFGI) \implies BC$ is tangent to $HFGI$ at $P$ , as desired.

$\blacksquare$

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#14 h Feb 20, 2021, 11:11 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let the line through $D$ parallel to $\overline{AC}$ intersect $\overline{BC}$ at $X.$ The given ratio condition implies that $\overline{EX}\parallel\overline{AB}.$

$\textbf{Key Claim: }$ $B,H,A,X$ are concyclic.

$\emph{Proof: }$ Since $\overline{HA}$ is tangent to $\Gamma,$ we have $\angle BAH = \angle BCA = \angle BXH.$ Thus, quadrilateral $BHAX$ is cyclic, as desired. $\blacksquare$

Now, by power of a point, we have $\text{Pow}_{(XHI)}D=DH\cdot DX=DA\cdot DB=\text{Pow}_{\Gamma}D.$ Therefore, $D$ lies on the radical axis of $(XHI)$ and $\Gamma.$ We can similarly show that $E$ lies on this radical axis, so pentagon $FGHIX$ is cyclic.

Finally, note that $\angle IXC = \angle ABX = \angle AHX=\angle IXH.$ Hence, $(FGHIX)$ is tangent to $\overline{BC}$ at $X,$ as needed.

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#15 h Aug 16, 2021, 1:43 PM

Y by

Very cool problem - same as the solutions above, yet, I may or may not have written out the PoP rather than quoting Radical Axes.
Solution

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#16 h Aug 16, 2021, 6:31 PM

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Because of the given ratio condition, $HD$ and $EI$ meet on $BC$ , call it $J$ .

By angle chase, $HAJB$ and $AICJ$ are cyclic.

By PoP through $D$ , $DF \cdot DG = DA \cdot DB = DH \cdot DJ$ , so $HFJG$ is cyclic.

By PoP through $E$ , $EG \cdot EF = EA \cdot EC = EI \cdot EJ$ , so $IGJF$ is cyclic.

So, $HFJGI$ is cyclic, and $\angle DJB = \angle ACB = \angle AIJ$ , hence $(HFGJI)$ is tangent to $BC$ at $J$ . We are done.

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#17 h Aug 19, 2021, 7:43 AM

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Claim : $HD \cap IE=J \in BC$
Proof. Let $HD \cap BC=J_1, IE \cap BC = J_2$ . We have
$\frac{BJ_1}{J_1C} =\frac{BD}{DA}$ and $\frac{CJ_2}{J_2B}=\frac{CE}{EA}$ Therefore $\frac{BJ_1}{J_1C} \times \frac{CJ_2}{J_2B}=\frac{BD}{DA} \times \frac{AD}{DB}=1 \implies \frac{BJ_1}{J_1C}=\frac{BJ_2}{J_2C}$ Finally, $J_1 \equiv J_2$ , as desired.

Back to the main problem, Note that $AHBJ, AJCI$ is cyclic. Now, by PoP, see that
$DF \times DG =DA \times DB= DH \times DJ$ Therefore $FGJH$ is cyclic. Applying PoP is a very similar way, we see that $GIFJ$ is cyclic. This implies that $FGJHI$ is cyclic, proving the first part.
For the second part, note that
$\begin{align*} \angle{HJB} &= \angle{ACB} \\ &=\angle{JIA} \end{align*}$ As desired $\blacksquare$

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L567

#18 h Aug 26, 2021, 9:31 AM • 1 Y

Y by starchan

Let $X$ be on $BC$ such that $DX || AC$ , by the length condition, we have $EX || AB$ too, so $ADXE$ is a parallelogram and $H,D,X$ and $I,E,X$ are collinear.

Note that $(HIX)$ is tangent to $BC$ since $\angle HIX = \angle AIE = 180 - \angle IAE - \angle IEA = 180 - \angle ABC - \angle BAC = \angle ACB = \angle HXB$

Now, note that $AHXB$ and $AICX$ are cyclic and so by PoP, we get $AD.DB = XD.DH$ and $AE.EC = EX.EI$ which means $D,E$ lie on the radical axis of $(HIX)$ and $(ABC)$ , so $F,G \in (HIX)$ too

So, there does indeed exist a circle passing through $F,G,H,I$ , and tangent to $BC$ at $X$ , as desired. $\blacksquare$

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3217 posts

#19 h Dec 25, 2021, 5:46 AM

Y by

Illegally cute

Denote the given ratio as $r$ , and let $X$ be the point on $BC$ such that $\frac{BX}{XC}=r$ . We claim that $X$ is the desired tangency point. Note that triangles $BXD,XCE$ are similar, which gives that $X=IE\cap HD$ .

Claim: $(AIX)$ is tangent to $BC$ .

Proof: Because $AC\parallel HX$ , we have $\angle EXC=\angle B=\angle IAC=\angle IHX$ .

Now, by since $\triangle XEC\sim\triangle AEI$ we have $EA\cdot EC=EI\cdot EX$ and similar for $D$ , hence $\overline{DE}$ is the radical axis of $(ABC),(AIX)$ and it follows that $H,G$ are precisely $(AIX)\cap (ABC)$ .

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#20 h Dec 27, 2021, 6:48 AM

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Let $J=IE\cap BC$ so we have :
$BD : DA=AE : EC=BJ : CJ$
and so $H,D,J$ are collinear
note that $\angle{HAB}=\angle{ACB}=\angle{HJB}$
$\implies AHBJ$ is cyclic
$\implies DH.DJ=DA.DB=DF.DG$
$\implies FHGJ$ is cyclic
and similarly $FIGJ$ is cyclic
and so $FHIGJ$ is cyclic and we have :
$\angle{IHJ}=\angle{ABC}=\angle{IJC}$
So $(FHIGJ)$ is tangent to $BC$ .

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#21 h Jan 8, 2022, 3:59 PM

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Let $X$ be the point on $BC$ such that $\frac{BX}{XC} = \frac{BD}{DA}=\frac{AE}{EC}$ . Then, clearly, $X,D,H$ and $X,E,I$ are collinear.

Now, note that $\angle HAB = \angle ACB = \angle HXB$ , so $(AXBH)$ is cyclic. Thus,
$Pow_{(XHI)}(D) = DX\cdot DH = DB\cdot DA = Pow_{(ABC)}(D)$ Similarly, $Pow_{(XHI)}(E) = Pow_{(ABC)}(E)$ . Thus, $DE$ is the radical axis of $(XHI)$ and $(ABC)$ , so $F,G = DE\cap (ABC)$ also lie on $(XHI)$ and we are done. $\blacksquare$ .

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#22 h Feb 28, 2022, 4:34 PM • 1 Y

Y by Johnweak

I am really surprised how no one realized that condition $BD: DA= AE: EC$ is completely redundant. The reason for this is probably that the ratio condition strongly motivates you to introduce the key point from all previous solutions. In fact, the problem is true for arbitrary points $D, E$ . Unfortunately, this makes this problem much less cute.
Let $FG \cap HI =J$ . By power of a point it suffices to prove $JI\cdot JH=JG\cdot JF$ or $JI\cdot JH=JA^2$ . This is true because from $AE||DH$ we have $\frac{JA}{JH}=\frac{JE}{JD}$ and from $IE||AD$ we have $\frac{JE}{JD}=\frac{JI}{JA}$ . Now $\frac{JA}{JH}=\frac{JI}{JA}$ yields desired conclusion.

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#23 h Jul 18, 2023, 11:05 PM

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In this picture, if we grab point $A$ that is the intersection of $IF$ and $HE$ .
Let's show that K is the tangent point of circle $FIHG$ and $BC$ . First because of the condition of the problem $BD : DA= AE : EC$ and if we use Chebas theorem at triangle $ABC$ we get $IF$ and $EH$ meet at $K$ that is on the line $BC$ . And if we look at parallel lines and the angles we can earn $IABK$ and $AHCK$ are all concyclic. Also if you use the power at point $E, F$ you will get $IFKG, HFKG$ are concyclic. So overall we get $IFKGH$ are concyclic. And because triangle $IKH$ is tangent to $BC$ (trivial) we get circle $IFKGH$ is tangent to $BC$ at point $K$ .

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#24 h Sep 22, 2023, 8:14 PM

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I hope this works, it's written up sloppily and I'm not sure if i ever used circular reasoning/forgot to say something. Note that most results are analogous if omitted.

Define J as HD\cap BC; we claim that this is the desired tangency point. Indeed, note that \frac{CE}{CA}=\frac{AD}{AB}=\frac{CJ}{CB}\Rightarrow EJ\parallel AB, so JEI is a line. Then HAB=ACB=DJB and analogous results gives HD\cdot DJ=BD\cdot DA=AE\cdot EC=IE\cdot EJ, so DE is the radax of (HIJ) and (ABC), whence if F and G lie on (ABC) they also lie on (HIJ); used in tandem with the fact that HJB=ACJ=HIJ, BC is indeed tangent to (FHIGJ) at J, as desired.

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#25 h Sep 26, 2023, 1:55 AM

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It is clear (by phantom points) that the length condition is simply equivalent to the parallel to $\overline{AC}$ through $D$ and the parallel to $\overline{AB}$ through $E$ intersecting on $\overline{BC}$ at some point $X$ . I claim that the desired tangency point is (unsurprisingly) just $X$ .

Observe that $\measuredangle HAB=\measuredangle ACB=\measuredangle DXB$ , hence $HABX$ and $IACX$ are cyclic. Furthermore, since $\measuredangle HAB=\measuredangle HIX$ , it follows that $(HIX)$ is tangent to $\overline{BC}$ .

Now, since $DH\cdot DX=DA\cdot DB$ and $EI\cdot EX=EA\cdot EC$ , $D$ and $E$ lie on the radical axis of $(HIX)$ and $(ABC)$ , hence $F$ and $G$ are the intersections of the circles, so we're done. $\blacksquare$

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#26 h Nov 17, 2023, 7:12 PM

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Very easy problem $[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.735268371031756, xmax = 15.59976627483964, ymin = -6.968258686310643, ymax = 7.423898730330016; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); draw((0.9956500476634552,2.1184228960223974)--(-1.062681436234393,-3.3604185631028978)--(7.74,-3.45)--cycle, linewidth(0) + zzttqq); draw(circle((1.0094370961590475,-1.012822014036576), 3.131275262489142), linewidth(0.4) + ccqqqq); draw(circle((5.4278305452743725,0.6180665577092871), 4.679240654777107), linewidth(0.4) + ccqqqq); /* draw figures */ draw(shift((0.9956500476634552,2.1184228960223974))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((-1.062681436234393,-3.3604185631028978))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((7.74,-3.45))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(circle((3.354041722723674,-1.893660336545912), 4.65390407701417), linewidth(0.4) + ffvvqq); draw((xmin, 0.5878222206149271*xmin + 1.5331576740495059)--(xmax, 0.5878222206149271*xmax + 1.5331576740495059), linewidth(0.4)); /* line */ draw((xmin, -0.010176607837738987*xmin-3.3712330553359005)--(xmax, -0.010176607837738987*xmax-3.3712330553359005), linewidth(0.4)); /* line */ draw(shift((3.0333502909149113,-3.402102271681033))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw((0.9956500476634552,2.1184228960223974)--(7.74,-3.45), linewidth(0.4)); draw((0.9956500476634552,2.1184228960223974)--(-1.062681436234393,-3.3604185631028978), linewidth(0.4)); draw((xmin, -0.8256426394501143*xmin-0.8976389311132736)--(xmax, -0.8256426394501143*xmax-0.8976389311132736), linewidth(0.4)); /* line */ draw((xmin, 2.661787715917385*xmin-11.476236814112772)--(xmax, 2.661787715917385*xmax-11.476236814112772), linewidth(0.4)); /* line */ draw(shift((-1.7197432167156428,0.5222543975122591))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((6.27271500785747,5.220398939252863))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((-0.10490611241868777,-0.8110239715614579))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((4.133906450997055,-0.47265540409717827))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw((xmin, 0.07982626322868441*xmin-0.8026497086172258)--(xmax, 0.07982626322868441*xmax-0.8026497086172258), linewidth(0.4)); /* line */ draw(shift((-1.1920666293411921,-0.8978079331571465))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(shift((7.684595569022911,-0.18921715991842103))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(circle((3.0832259187097506,1.4989048802186251), 4.901260927681726), linewidth(0.4) + linetype("2 2")); draw(shift((-8.201338357260022,-3.287771251129459))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw((-1.1920666293411921,-0.8978079331571465)--(3.0333502909149113,-3.402102271681033), linewidth(0.4)); draw((3.0333502909149113,-3.402102271681033)--(0.9956500476634552,2.1184228960223974), linewidth(0.4)); draw((-1.1920666293411921,-0.8978079331571465)--(-1.062681436234393,-3.3604185631028978), linewidth(0.4)); draw(shift((3.354041722723674,-1.893660336545912))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw(circle((1.9563973451805725,0.08602632717078154), 2.248037184471437), linewidth(0.4)); draw(shift((1.5065826571655647,-2.116548967617623))*scale(0.035277777777777776)*((1,0)--(0,1)--(-1,0)--(0,-1)--cycle)); /* special point */ draw((-1.7197432167156428,0.5222543975122591)--(-1.062681436234393,-3.3604185631028978), linewidth(0.4)); draw((6.27271500785747,5.220398939252863)--(7.74,-3.45), linewidth(0.4)); draw((-1.062681436234393,-3.3604185631028978)--(1.5065826571655647,-2.116548967617623), linewidth(0.4)); draw((1.5065826571655647,-2.116548967617623)--(0.9956500476634552,2.1184228960223974), linewidth(0.4)); draw((1.5065826571655647,-2.116548967617623)--(7.74,-3.45), linewidth(0.4)); draw((-0.10490611241868777,-0.8110239715614579)--(1.5065826571655647,-2.116548967617623), linewidth(0.4)); draw((1.5065826571655647,-2.116548967617623)--(4.133906450997055,-0.47265540409717827), linewidth(0.4)); draw((-8.201338357260022,-3.287771251129459)--(3.354041722723674,-1.893660336545912), linewidth(0.4)); draw((-1.1920666293411921,-0.8978079331571465)--(-1.7197432167156428,0.5222543975122591), linewidth(0.4)); draw((-1.1920666293411921,-0.8978079331571465)--(6.27271500785747,5.220398939252863), linewidth(0.4)); draw((-1.7197432167156428,0.5222543975122591)--(7.74,-3.45), linewidth(0.4)); /* dots and labels */ label("$A$", (1.0667066177563198,2.1501291594770757), NE * labelscalefactor); label("$B$", (-0.9864403144078056,-3.3249293262939164), NE * labelscalefactor); label("$C$", (7.830014159002851,-3.4054448922611367), NE * labelscalefactor); label("$D$", (3.119853549920445,-3.3651871092775263), NE * labelscalefactor); label("$E$", (-1.6305648421455705,0.5599467316244714), NE * labelscalefactor); label("$F$", (6.360605080101075,5.270107340706868), NE * labelscalefactor); label("$G$", (-0.02025352280115828,-0.7685601068346664), NE * labelscalefactor); label("$H$", (4.206813690477923,-0.42636895147397935), NE * labelscalefactor); label("$I$", (-1.1072136633586365,-0.8490756728018868), NE * labelscalefactor); label("$J$", (7.769627484527436,-0.14456447058870772), NE * labelscalefactor); label("$K$", (-8.11206790250683,-3.2444137603266956), NE * labelscalefactor); label("$O$", (3.4419158137893278,-1.8555202473921426), NE * labelscalefactor); label("$Q_A$", (1.5900577965432539,-2.076938053801999), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$ Labels in this soloution are different but it is clear from diagram.
Claim 1: $(AEBD),(AFCD)$ are cyclic
Proof
CLaim 2: $EIJD,FIJD$ is cyclic
Proof
Claim 3: $(EJIFD)$ is tangent at $D$ to $BC$ .
Proof

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#27 h Nov 29, 2023, 12:14 AM

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Let $K = \overline{HD} \cap \overline{IE}$ .
Claim: $B,K,C$ are collinear.
Proof: Let $\overline{HD}$ hit $\overline{BC}$ at $D'$ and define $E'$ similarly. It suffices to prove $D' = E'$ . Since $\overline{IE} \parallel \overline{AB}$ ,
$\frac{CE'}{CB} = \frac{CE}{CA} = \frac{1}{1+\frac{AE}{CE}} = \frac{1}{1+\frac{BD}{DA}} = \frac{DA}{AB}$ However, since $\overline{HD} \parallel \overline{AC}$ , $\frac{DA}{AB} = \frac{CD'}{CB}$ . Combining these two gives $D' = E'$ . $\square$ .

Claim: $HFKG$ and $IGKF$ are cyclic.
Proof: By symmetry, it suffices to prove $HFKG$ cyclic. Notice that
$\angle HAB = \angle ACB = \angle HKB$ so $HAKB$ cyclic. Therefore,
$HD \cdot DK = AD \cdot DB = FD \cdot DG$ which concludes. $\square$

The above claim directly implies that $FGHI$ is cyclic. Also, $(FGHI)$ must be tangent to $\overline{BC}$ at $K$ since
$\angle IHK = \angle AHK = \angle ABC = \angle IKC$

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OronSH

#28 h Feb 3, 2024, 2:24 PM • 2 Y

Y by megarnie, Spectator

In triangle $APQ,$ let $B,C$ be on sides $AP,AQ$ respectively with $PB/BA=AC/CQ.$ Let $BC$ intersect $(APQ)$ at $M,N.$ Let $X$ be on the same side of $AC$ as $B$ such that $XA\parallel BC$ and $\angle AXP=\angle BAC.$ Similarly, let $Y$ be on the same side of $AB$ as $C$ such that $YA\parallel BC,\angle AYQ=\angle BAC.$ Let $D$ be the $A$ dumpty point of $\triangle APQ.$

Claim. $AXPD,AYQD$ are cyclic.
Proof. By properties of the dumpty point we have $\angle PXA=\angle PAQ=\angle PAD+\angle DAQ=\angle PAD+\angle APD=180-\angle PDA.$

Claim. $ABDC$ is cyclic.
Proof. The spiral similarity at $D$ taking $PA$ to $AQ$ takes $B$ to $C$ by the length ratio. Thus $180-\angle ABD=\angle PBD=\angle ACD.$

Claim. $D,B,X$ and $D,C,Y$ are collinear.
Proof. The spiral similarity gives $\angle PDB=\angle ADC=\angle ABC=\angle XAP=\angle XDP,$ so $D,B,X$ are collinear. The other side is similar.

Claim. $MDNYX$ is cyclic.
Proof. By PoP $BD\cdot BX=BA\cdot BP=BM\cdot BN$ so $X$ lies on $(MDN).$ Similarly $Y$ lies on $(MDN)$ as well.

Claim. $(ABC)$ is tangent to $(XMNY).$
Proof. Since $BC\parallel XY$ there is a homothety at $D$ taking $BC$ to $XY,$ which takes $(BDC)=(ABC)$ to $(XDY)=(XMNY).$

To finish, simply invert the diagram about $A.$

inversion :love:

:love:

:love:

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#29 h Feb 5, 2024, 2:07 AM • 1 Y

Y by megarnie

Let $X$ be $HD\cap BC$ and $X'$ be $EI\cap BC$ . Note that $\angle{HAD} = \angle{C} = \angle{DXB} \implies HABX$ is cyclic. Also, $\angle{DBX} = \angle{A}$ implying that $\triangle DBX \sim \triangle DHA\sim \triangle ABC$ . We can see from symmetry that
$\triangle{DBX}\sim\triangle{EAI}\sim\triangle{EXC}\sim\triangle{DHA}\sim\triangle{ABC}$ Notice that
$BD\cdot EC = AD\cdot AE$ and that
$\frac{AE}{HD} = \frac{IE}{AD} \implies BD\cdot EC = HD\cdot IE \implies \triangle{HDB}\sim \triangle{CEI}$ Let $\angle{BDH} = \angle{IGE} = \alpha$ . Notice how, $\angle{AXC} = \angle{B}+\alpha = \angle{AX'C}$ which means that $X = X'$ . We can see that $ADXE$ is a parallelogram because $EX \parallel AD$ and $AE\parallel DX$ and that $\triangle{ABC} \sim \triangle{XHI}$ .

Consider the circumcircle of $\triangle{HIX}$ . Because $\angle{HIX} = \angle{C}$ , $(HIX)$ is tangent to $BC$ . Notice that $POW_{(ABC)}(D) = AD\cdot BD = HD\cdot DX = POW_{(HIX)}(D)$ . The same can be said for $E$ . Thus, $DE$ is the radical axis of $(HIX)$ and $(ABC)$ implying that $FG$ lies on $(HIX)$ ending the proof.

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#30 h Feb 12, 2024, 8:44 PM

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Construct point $X$ on $AB$ so that $AX$ , $BE$ and $CD$ all concur. Then note that $BD : DA = AE : EC = CX : BX$ , by Ceva. From this, we also find that $AEXD$ is a parallelogram, by angle chasing and similarity. From angle chasing, we have $\angle HIX = \angle HAD = \angle C = \angle DXB \implies BC$ is a tangent to $(IXH)$ . Note that $(AHBX)$ is cyclic, so the radical axises of $(HIX)$ , $(AHBX)$ and $(ABC)$ concur at $D$ . This implies that $F, G = (HIX) \cap (ABC)$ which finishes.

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#31 h Mar 10, 2024, 2:45 AM

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Surprisingly standard. Let $J$ be the point on $\overline{BC}$ such that $ADJE$ is a parallelogram, which exists by the ratio condition. Then, as $\angle AHJ =\angle ABJ$ , $AHBJ$ is concyclic, hence $DH \cdot DJ = DA \cdot DB = DF \cdot DG,$ i.e. $HFGJ$ is concyclic. Similarly, $IFGJ$ is concyclic, so $FHIGJ$ is concyclic and tangent to $\overline{BC}$ at $J$ as $\angle HJB = \angle C = \angle HIJ$ .

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alsk

#32 h Mar 12, 2024, 9:46 PM

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More radax usage.

Let $P = HD \cap IE$ . By the length ratios given in the problem, we must have $P \in BC$ .

Now, notice that $\angle HIP = \angle HAB = \angle C = \angle HPB$ , so $(HIP)$ is tangent to $BC$ at $P$ .

Then, Notice that we have $H, A, P, B$ cyclic by $\angle HAB = \angle HPB$ , so we have $\text{Pow}_{(HIP)}(D) = DH \cdot DP = dB \cdot DA = \text{Pow}_{(ABC)}(D)$ , so $D$ lies on the radical axis of $(HIP), (ABC)$ . Using a similar argument for $E$ , we can deduce that $DE$ is the radical axis of $(HIP), (ABC)$ . Since $(ABC)$ intersects $DE$ at $F, G$ , it follows that both $F, G$ lie on $(HIP)$ .

Thus, we have that $(HIFG)$ is tangent to $BC$ at $P$ , done.

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Markas

#33 h May 5, 2024, 12:52 PM

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Let $\angle BAC = \alpha$ and $\angle ABC = \beta$ . Let P be a point on BC, where $\frac{BD}{DA} = \frac{BP}{PC}$ $\Rightarrow$ from Tales $DP \parallel AC$ $\Rightarrow$ H, D, P lie on one line, also from $\frac{BD}{DA} = \frac{AE}{EC}$ it follows that $\frac{AE}{EC} = \frac{BP}{PC}$ $\Rightarrow$ $EP \parallel AB$ $\Rightarrow$ I, E, P lie on one line.

We will now want to prove that HFPGI is cyclic and that BC is tangent to this circle in point P.

Since AI is tangent to the circumcircle of $\triangle ABC$ , we have that $\angle IAC = \angle ABC = \beta$ . Since $HP \parallel AC$ , we have that $\angle IAC = \angle IHP = \beta$ , also since $EP \parallel AB$ , we have $\angle EPC = \angle ABC = \beta$ . Now we have that $\angle PBA = \angle PHA = \beta$ $\Rightarrow$ PBHA is cyclic $\Rightarrow$ PD.DH = AD.DB, but since AFBG is cyclic, we have that AD.DB = FD.DG $\Rightarrow$ PD.DH = FD.DG $\Rightarrow$ FPGH is cyclic.

Also $\angle CPI = \angle CAI = \beta$ $\Rightarrow$ CPAI is cyclic $\Rightarrow$ PE.EI = AE.EC, but since AFCG is cyclic, we have that AE.EC = FE.EG $\Rightarrow$ PE.EI = FE.EG $\Rightarrow$ FPGI is cyclic.

We have that FPGH is cyclic and that FPGI is cyclic $\Rightarrow$ F, P, G, H, I lie on one circle $\Rightarrow$ the only thing which is left to show is that BC is tangent to this circle in P, which is equivalent to showing that BC is tangent to the circumcircle of $\triangle HPI$ at point P. Since $\angle CPI = \angle PHI = \beta$ , than this is true $\Rightarrow$ we proved that there exists a circle passing through points F, G, H, I, and tangent to line BC $\Rightarrow$ we are ready.

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#34 h Jun 6, 2024, 5:53 PM

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The length condition forces $K := DH \cap EI$ to lie on $BC$ . Notice that $(HIK)$ is tangent to $BC$ at $K$ , as
$\measuredangle BKH = \measuredangle BCA = \measuredangle BAH = \measuredangle HIK.$
To show that $F$ , $G$ lie on this circle, first note that $BHAK$ and $CIAK$ are cyclic since
$\measuredangle ABK = \measuredangle IAC = \measuredangle AHK, \quad \measuredangle KCA = \measuredangle BAH = \measuredangle KIA.$
As a result, Power of a Point tells us line $DE$ is the radical axis of $(ABC)$ , $(HIK)$ , as desired. $\blacksquare$

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L13832

#35 h Oct 13, 2024, 6:44 AM • 1 Y

Y by alexanderhamilton124

INMO Mock and boredom :play_ball:

:play_ball:

Storage

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Eka01

#36 h Nov 13, 2024, 6:02 PM

Y by

sketch

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#37 h Dec 3, 2024, 12:48 AM

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Phantom Points :yup:

:yup:

Define $X$ to be a point on $BC$ such that: $\frac{CX}{XB} = \frac{CE}{EA}, \frac{BX}{XC} = \frac{BD}{DA}$ (which exists due to given condition).
Note that: $DX \parallel AC, EX \parallel AB$ and $(H, D, X), (X, E, I)$ are collinear. Therefore: $\triangle ABC \sim \triangle XDE$ . Thus: $DE \parallel HI$ .
Notice that: $\angle EXC = \angle ABC = \angle RDX = \angle IHX$ which implies $(HIX)$ is tangent to $BC$ at $X$ . Since $AHBX$ is cyclic, it implies $FG$ is the radical axis of $(HIX), (ABC)$ . For any two intersecting circles, the radical axis passes through the two intersection points. Thus, we must have $F, G \in (HIX)$ , and we are done.

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#38 h Dec 14, 2024, 8:08 PM

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Let the line passing through $D$ parallel to $AC$ meet $BC$ at $X$ . Since $\dfrac{BD}{DA}=\dfrac{BX}{XC}=\dfrac{AE}{EC}$ we have that $EX\parallel AB$ . Thus, we have $\overline{H-D-X}$ and $\overline{E-I-X}$ . On the other hand
$\angle XIA=\pi-\angle BAC-\angle CEI=\pi-\angle BAC-\angle CBA=\angle BCA$ implying the points $A$ , $X$ , $C$ and $I$ are concyclic. Similarly $X\in (ABH)$ . Thus a quick PoP gives
$HD\cdot DX=AD\cdot DB=FD\cdot DG$ So points $H$ , $X$ , $F$ and $G$ are concyclic. Similarly
$XE\cdot EI=AE\cdot EC=FE\cdot EG$ implies $I$ , $X$ , $F$ and $G$ are concyclic which finishes as the points $F$ , $G$ , $H$ , $I$ and $X$ all lie on a common circle as desired. Tangency is easy to obtain.

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#39 h Jan 25, 2025, 9:25 AM

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Sol sketch: show that $HD, BC, IE$ concur at a point $X$ and angle chase gives $HAXB$ and $AICX$ cyclic. Conjecture $X$ is tangency point and from here it’s just details with power of a point and angle chase.

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#40 h Feb 15, 2025, 9:21 AM

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Here is my solution

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#41 h Mar 15, 2025, 7:02 PM

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construct parallelogram $ADRE.$
Clearly, $R\in BC.$
Furthermore, $\angle HIR = \angle HAB = \angle ACB = \angle HRB,$ so $(HRI)$ is tangent to $BC$ AND $HARB$ cyclic
Thus, $HD\cdot HR = AD\cdot AB = FD\cdot DG$
Thus, $F$ lies $(HRI).$

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#42 h Apr 1, 2025, 12:11 PM

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2015/4 Canada is similar!

https://aops.com/community/p4756480
Although in this case the construction of the parallelogram is much more natural.
Claim - $H D$ , $E I$ , $B C$ concur
Let $X$ be the intersection of $H D$ and $I E$ ,
we have $D B X$ similar to $E X C$ through the ratio conditions
and
$\angle D X E = a$ ,
$\angle D X B = c$ ,
$\angle E X C = b$ ,

sums up to 180, so $B - X - C$ are collinear.
Claim - $H X B A$ and $A C X I$ are concyclic
We use alternate segment theorem,
$\angle B A H = \angle B C A = \angle H X B$
and similarly, for $A C X I$
Claim - $H F X E I$ lie on a circle, and $X$ is the point of tangency
We prove that $H F X G$ and $I E X F$ are concyclic using Power of Point.

$(A D) (B D) = (H D) (D X) = (F D) (D G)$
$(A E) (E C) = (G E) (E F) = (X E) (E I)$

Next, simply note that $X$ is the tangency point because of alternate segment theorem -
$\angle I X C = \angle I H X = \angle A B C$

$\square$
Spamming PoP is so satisfying

:starwars:

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