another geometry problem with sharky-devil point

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another geometry problem with sharky-devil point

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Source: The francophone mathematical olympiads P1

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92 posts

#1 h Jun 27, 2020, 6:59 PM • 2 Y

Y by Kanep, ItsBesi

Let $ABC$ be an acute triangle with $AC>AB$ , Let $DEF$ be the intouch triangle with $D \in (BC)$ , $E \in (AC)$ , $F \in (AB)$ ,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$ , and $X=(ABC)\cap (AEF)$ .
Prove that $B,D,G$ and $X$ are concylic

This post has been edited 1 time. Last edited by anyone__42, Jun 28, 2020, 8:35 PM

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#2 h Jun 27, 2020, 8:05 PM

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Let $T$ be the projection of $D$ on $\overline{EF}$ . Then $I,T,X$ are collinear because $T,X$ are Inverses of each other WRT $\odot(I)$ . By Reim's $GTFX$ is cyclic. Also $\Delta XFE\stackrel{+}{\sim}\Delta XBC\implies \measuredangle DBX=\measuredangle TFX=\measuredangle DGX\implies B,D,G,X$ are concyclic.

This post has been edited 1 time. Last edited by AmirKhusrau, Jun 27, 2020, 8:06 PM

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#3 h Jun 28, 2020, 12:40 PM

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Let $L$ the intersection of $(AI)$ and $(BC)$ , and $S$ the intersection of $(AI)$ and $\odot(ABC)$ ( $S$ is the South pole).
$(EA,EI)=(FA,FI)=90^\circ$ so $AFIEX$ is cyclic and $\odot(AEF)$ is tangent externally to $\odot(IBC)$ at $I$ .
Invert WRT $\odot(IBC)$ , $A \leftrightarrow L, B \leftrightarrow B, C \leftrightarrow C, I \leftrightarrow I, S \leftrightarrow \infty$ so $\odot(ABC) \leftrightarrow (BC)$ and $\odot(AEF) \leftrightarrow \odot(IDL)$ , it follows that $X \leftrightarrow D$ so $X,D,S$ are collinear.
WRT $(XD)$ and $(BG)$ , $(XB)$ and $(AI)$ are antiparallel lines and $(AI)//(DG)$ (both are perpendicular to $(EF)$ ), hence $(XB)$ and $(DG)$ are anti-parallel lines, it follows that $XBDG$ is cyclic.

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92 posts

#4 h Jun 28, 2020, 12:48 PM • 1 Y

Y by Kanep

some lemmas that kill the problem https://artofproblemsolving.com/community/c946900h1911664_properties_of_the_sharkydevil_point

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#5 h Sep 6, 2020, 5:29 PM • 1 Y

Y by Isarogu777

Here is a solution without inversion. As $AEF$ is isosceles, $AI$ is perpendicular to $EF$ . Then $AI$ and $DG$ are parallel. Then $\angle DGB=\angle \frac{A}{2}$ .As $\angle BXC=\angle A$ , we only need that $XD$ is the bisector of $\angle BXD$ to get the desired concylic. By the spiral similarity center lemma (see Yufei for example), $X$ is the spiral similarity center between $FE$ and $BC$ . Then $\Delta XFB \sim \Delta XEC$ , thus $\frac{FB}{EC}=\frac{XB}{XC}$ . The tangency with the incircle implies $BF=BD$ and $CE=CD$ and thus $\frac{XB}{XC}=\frac{DB}{DC}$ . Then the angle bisector theorem implies the desired angle bisector and we are done.

This post has been edited 2 times. Last edited by Supertinito, Oct 6, 2021, 11:33 PM
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386 posts

#6 h Jan 9, 2021, 11:31 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -100.06537040840954, xmax = 15.78830909914262, ymin = -10.713221641729424, ymax = 61.67639771070138; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((-44.34454326792231,25.867344193518576), 20.81086408567081), linewidth(0.8) + red); draw((-48.99071732558426,46.15293342869172)--(-62.50702561682165,15.707801934234288), linewidth(0.8) + blue); draw((-62.50702561682165,15.707801934234288)--(-36.09448973463682,6.761620425752361), linewidth(0.8) + blue); draw((-36.09448973463682,6.761620425752361)--(-48.99071732558426,46.15293342869172), linewidth(0.8) + blue); draw(circle((-50.2635532942482,21.0758295501552), 9.012090275992508), linewidth(0.8) + red); draw(circle((-53.711950729994,26.284530074744154), 13.755744997024323), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-58.50039721710591,24.73262778537696)--(-41.6987793032047,23.879829839916948), linewidth(0.8) + blue); draw((-51.765893027575,39.90192262405459)--(-53.15469524294922,12.540077130180736), linewidth(0.8) + blue); draw(circle((-49.62713535270761,33.614381201506156), 12.554693140778378), linewidth(0.8) + red); draw((-61.53198232877179,37.60122686388172)--(-58.50039721710591,24.73262778537696), linewidth(0.8) + blue); draw((-61.53198232877179,37.60122686388172)--(-53.15469524294922,12.540077130180736), linewidth(0.8) + blue); draw((-61.53198232877179,37.60122686388172)--(-62.50702561682165,15.707801934234288), linewidth(0.8) + blue); draw((-61.53198232877179,37.60122686388172)--(-51.765893027575,39.90192262405459), linewidth(0.8) + blue); draw((-58.50039721710591,24.73262778537696)--(-53.15469524294922,12.540077130180736), linewidth(0.8) + blue); draw((-50.2635532942482,21.0758295501552)--(-58.50039721710591,24.73262778537696), linewidth(0.8) + blue); draw((-50.2635532942482,21.0758295501552)--(-41.6987793032047,23.879829839916948), linewidth(0.8) + blue); draw(circle((-55.13314509863617,32.317275194192106), 8.298509694239405), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-61.53198232877179,37.60122686388172)--(-50.2635532942482,21.0758295501552), linewidth(0.8) + blue); draw((-48.99071732558426,46.15293342869172)--(-50.2635532942482,21.0758295501552), linewidth(0.8) + blue); draw((-50.2635532942482,21.0758295501552)--(-62.50702561682165,15.707801934234288), linewidth(0.8) + blue); draw((-50.2635532942482,21.0758295501552)--(-36.09448973463682,6.761620425752361), linewidth(0.8) + blue); draw((-61.53198232877179,37.60122686388172)--(-41.6987793032047,23.879829839916948), linewidth(0.8) + blue); draw((-61.53198232877179,37.60122686388172)--(-48.99071732558426,46.15293342869172), linewidth(0.8) + blue); /* dots and labels */ dot((-48.99071732558426,46.15293342869172),dotstyle); label("$A$", (-48.650567541986064,46.91073267542522), NE * labelscalefactor); dot((-62.50702561682165,15.707801934234288),dotstyle); label("$B$", (-62.20469083077811,16.47074629500976), NE * labelscalefactor); dot((-36.09448973463682,6.761620425752361),dotstyle); label("$C$", (-35.777936485591376,7.5356259146888025), NE * labelscalefactor); dot((-53.15469524294922,12.540077130180736),linewidth(4pt) + dotstyle); label("$D$", (-52.81524229552552,13.139006492178217), NE * labelscalefactor); dot((-41.6987793032047,23.879829839916948),linewidth(4pt) + dotstyle); label("$E$", (-41.38131706308083,24.497210365467566), NE * labelscalefactor); dot((-58.50039721710591,24.73262778537696),linewidth(4pt) + dotstyle); label("$F$", (-58.19145879554918,25.330145316175454), NE * labelscalefactor); dot((-61.53198232877179,37.60122686388172),linewidth(4pt) + dotstyle); label("$X$", (-61.2203131617597,38.20277637257005), NE * labelscalefactor); dot((-51.765893027575,39.90192262405459),linewidth(4pt) + dotstyle); label("$G$", (-51.45225783073079,40.474417147227925), NE * labelscalefactor); dot((-50.2635532942482,21.0758295501552),linewidth(4pt) + dotstyle); label("$I$", (-49.93783064762553,21.69552007672286), NE * labelscalefactor); dot((-52.5511660547788,24.430663287773335),linewidth(4pt) + dotstyle); label("$H$", (-52.28519278143868,25.027259879554403), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Would recommend

$\color{black}\rule{25cm}{1pt}$
First let's add some points and let's assume that WLOG $AB < AC$ .
Let $I$ be the incenter of $ABC$ , let $H$ be the intersection of $DG$ with $FE$ .

We start off with a lemma, which is ultra nice.
$\color{red}\rule{25cm}{0.5pt}$
Lemma: The points $I,H$ and $X$ are colinear.
Proof:
Denote with $\varphi$ the inversion around the incircle of $ABC$ . Notice that the circumcircle of $ABC$ gets sent to the nine-point circle of $DEF$ under $\varphi$ , but also notice that $\overline{FE}$ gets sent to $(FIE)$ under $\varphi$ . Since we have that $X$ lies on both $(FIE)$ and $(ABC)$ we must have that the inverse point of $X$ must lie on both the nine-point circle and on line $FE$ . Since that can't be the midpoint we have that it must be the foot of the D-altitude of $DEF$ , which is $H$ . This implies that $I,H$ and $X$ are colinear points.
$\color{red}\rule{25cm}{0.5pt}$

Now let's return to the problem at hand here.
We have that $\angle DFG=\angle DFE + \angle EFG = 90 + \frac{1}{2}\angle B$ and we have that $\angle FDG = 90 - \angle EFD = \frac{1}{2} \angle C$ .
Thus we have that $\angle FGD = \frac{1}{2} \angle A$ . But notice the following:
$\angle FXH = \angle FXI = \angle FAI = \frac{1}{2} \angle A = \angle FGD = \angle FGH$ this implies that $FXGH$ is a cyclic quadrilateral.
This implies that $\angle FXG = 90$ .

Since we have that $\angle BDG = \angle FDG + \angle FDB = 90 + \frac{1}{2}\left( \angle C - \angle B \right)$ .
To show that the problem statement is true we must have that $\angle BXG = 180-\angle BDG$ . but this implies that we should have that $\angle BXF = \frac{1}{2}\left( \angle B - \angle C \right)$ .
But notice that we have that $\angle BXF = \angle BXA - \angle FXA = 180 - \angle C - 180 + 90 - \frac{1}{2} \angle A = \frac{1}{2}\left(\angle B - \angle C \right)$ .

Thus we have that $BXGD$ is a cyclic quadrilateral.

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L567

#7 h Jan 15, 2021, 7:50 AM • 1 Y

Y by Mango247

By the properties of the sharky devil point, we know that if $M = XD \cap (ABC)$ , $M$ is the midpoint of arc $BC$ in $(ABC)$ and so $\angle BXD = \angle BXM = \angle BAM = \frac{A}{2}$ . Also, we know that $\angle AFE = 90 - \frac{A}{2}$ which gives $\angle BDG = \frac{A}{2}$ . So, $B, D, G, X$ are concyclic

This post has been edited 1 time. Last edited by L567, Jan 15, 2021, 7:51 AM

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#8 h Jun 2, 2021, 11:37 PM

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anyone__42 wrote:

Let $ABC$ be an acute triangle with $AC>AB$ , Let $DEF$ be the intouch triangle with $D \in (BC)$ , $E \in (AC)$ , $F \in (AB)$ ,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$ , and $X=(ABC)\cap (AEF)$ .
Prove that $B,D,G$ and $X$ are concylic

Sharky-Devil point :3. Clearly $X$ is the $A$ -SharkyDevil Point on $\triangle ABC$ so its known that $X-D-M_A$ where $M_A$ is the midpoint of the smaller arc $\widehat{BC}$ .
Now note that $GD \perp EF$ and $AI \perp EF$ and that means $GD \parallel EF$ so the rest is angles :3.
$\angle BXM_A=\angle BAM_A=\angle BGD \implies BDGX \; \text{cyclic}$
Thus we are done :blush:

:blush:

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338 posts

#9 h Aug 3, 2023, 3:28 AM

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anyone__42 wrote:

Let $ABC$ be an acute triangle with $AC>AB$ , Let $DEF$ be the intouch triangle with $D \in (BC)$ , $E \in (AC)$ , $F \in (AB)$ ,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$ , and $X=(ABC)\cap (AEF)$ .
Prove that $B,D,G$ and $X$ are concylic

$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
Let $P\equiv EF\cap DG$
Let $I$ be the incenter of $\triangle ABC$
Let $Y\equiv EF\cap BC$
$X$ is the A-Sharky-Devil point
$\Rightarrow I, P \text{ and }X \text{ are collinears}$ Since $IA\perp FE$ and $DG\perp FE$ :
$\Rightarrow IA//DG$ $\Rightarrow \angle FAI=\angle FGP...(I)$ Since $AEIFX$ is cyclic:
$\Rightarrow \angle FAI=\angle FXI...(II)$ By $(I)$ and $(II)$ :
$\Rightarrow FXGP \text{ is cyclic}$ Since $DG\perp EF$ and $FXGP$ is cyclic:
$\Rightarrow FX\perp XG...(III)$ Since $AXBC$ is cyclic:
$\Rightarrow 180-\angle ACB=\angle BXA=\angle BXF+\angle FXA...(IV)$ Since $AEFX$ is cyclic:
$\Rightarrow \angle FXA=180-\angle AEF...(V)$ By $(IV)$ and $(V)$ :
$\Rightarrow \angle AEF=\angle BXF+\angle ACB=\angle BXF+\angle ECB$ $\Rightarrow \angle BXF=\angle FYB...(VI)$ In $\triangle YPD$ :
$\Rightarrow \angle PYD+\angle PDY=90=\angle FYB+\angle PDY$ By $(VI)$ :
$\Rightarrow \angle BXF+\angle PDY=90$ $\Rightarrow \angle BXF=90-\angle PDY=\angle PDI$ $\Rightarrow \angle BXF+90=\angle PDI+90$ By $(III)$ and $ID\perp BC$ :
$\Rightarrow \angle BXF+\angle FXG=\angle PDI+\angle IDG$ $\Rightarrow \angle BXG=\angle PDC=\angle GDC$ $\Rightarrow XBDG \text{ is cyclic}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

This post has been edited 1 time. Last edited by hectorleo123, Aug 3, 2023, 3:29 AM

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cursed_tangent1434

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cursed_tangent1434

#10 h Mar 27, 2024, 2:12 AM

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Observe that $X$ is the $A-$ Sharkydevil Point. Since $GD \perp EF \perp AI$ , it follows that $GD \parallel AI$ . Simply note that since $XD$ bisects $\angle BXC$ (well known Sharkydevil), we have that,
$2\measuredangle DXB = \measuredangle CXB = \measuredangle CAB = 2\measuredangle IAB = 2\measuredangle DGB$ from which the desired result follows.

This post has been edited 1 time. Last edited by cursed_tangent1434, Mar 27, 2024, 2:13 AM
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#11 h Jan 11, 2025, 10:07 PM

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There is a much simpler solution by just angle chase and similar triangles.

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This post has been edited 2 times. Last edited by ItsBesi, Jan 12, 2025, 1:34 PM
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#12 h Mar 17, 2025, 12:34 AM

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$[asy] import olympiad; unitsize(15); pair A, B, C, D, E, F, P, G, X; A = (5, 12); B = origin; C = (16,0); draw(A--B--C--cycle); path w = incircle(A, B, C); D = IP(w, B..C); E = IP(w, A..C); F = IP(w, A..B); draw(D--E--F--cycle); P = foot(D, E, F); path i = circumcircle(A, B, C); path j = circumcircle(A, E, F); draw(i, dotted+red); draw(j, dotted+red); X = IP(i, j, 1); G = IP(A..B, P..P+2*(P-D)); draw(D--G); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(P); dot(X); dot(G); draw(B--X--G); draw(X--D); draw(X--F--X--C--X--E); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, NE); label("$F$", F, NW); label("$P$", P, SW); label("$X$", X, NW); label("$G$", G, NW); [/asy]$

Notice since $X,A$ lie on both circles. Thus, by spiral similarity: $\triangle XFB \sim \triangle XEC$ . So by similar triangles and incircle properties:
$\frac{XB}{XC}=\frac{BF}{CE}=\frac{BD}{CD}.$ By the angle bisector theorem, $DX$ bisects $\angle BXC$ . Now perform some angle chasing (too lazy to put) and get:
$\angle BXD = \angle BGD = \frac{\angle BAC}{2}.$ Thus proving that $B,D,G,X$ are concyclic.

This post has been edited 1 time. Last edited by zhenghua, Mar 17, 2025, 5:59 PM
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