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Source: 10th European Mathematical Cup - Problem S2

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439 posts

#1 h Dec 22, 2021, 6:35 PM • 1 Y

Y by tiendung2006

Let $ABC$ be a triangle and let $D, E$ and $F$ be the midpoints of sides $BC, CA$ and $AB$ , respectively.
Let $X\ne A$ be the intersection of $AD$ with the circumcircle of $ABC$ . Let $\Omega$ be the circle through $D$ and $X$ ,
tangent to the circumcircle of $ABC$ . Let $Y$ and $Z$ be the intersections of the tangent to $\Omega$ at $D$ with the
perpendicular bisectors of segments $DE$ and $DF$ , respectively. Let $P$ be the intersection of $YE$ and $ZF$ and
let $G$ be the centroid of $ABC$ . Show that the tangents at $B$ and $C$ to the circumcircle of $ABC$ and the line $PG$ are concurrent.

This post has been edited 4 times. Last edited by Jalil_Huseynov, Dec 23, 2021, 6:38 AM

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#2 h Dec 22, 2021, 6:35 PM

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Cute.
Solution

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439 posts

#3 h Dec 22, 2021, 6:36 PM • 1 Y

Y by ILOVEMYFAMILY

Here is quick sketch of my solution. Let tangents to $(ABC)$ from $B$ and $C$ intersect at point $Q$ .First show that $BX\cap DE\in \Omega$ and $CX\cap DF\in \Omega$ . Then easy angle chase implies that $PE$ and $PF$ are tangents to $(DEF)$ . So take homothety with center $G$ and ratio $-\frac{1}{2}$ . This homothety sends $DEF$ to $ABC$ so it sends $P$ to $Q$ . So $P-Q-G$ are collinear.

This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 22, 2021, 6:37 PM

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1184 posts

L567

#4 h Dec 23, 2021, 3:18 AM

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Let the tangents at $B,C$ to $(ABC)$ meet at $K$ . Let $N$ be the nine point center of $ABC$ and let $H$ be the orthocenter and let $A'$ be the $A$ antipode in $(ABC)$ . By homothety at $H$ , the $D$ tangent to $(DEF)$ is parallel to the tangent to $(ABC)$ at $A'$ , which is parallel to the $A$ -tangent, which is parallel to the $D$ tangent to $(DX)$ by homothety, so $(DEF)$ and $(DX)$ are tangent. Since $YE = YD$ and $ZE = ZD$ , we have that the nine point circle of $\triangle ABC$ is in fact the incircle of $\triangle PYZ$ .

Therefore, we have the angles of this triangle and so by angle chase, $PN \perp BC \perp OK$ so $PN$ is parallel to $OK$ , also $N,G,O$ are collinear with $OG = 2NG$ , so if we show that $OK = 2PN$ , then $\triangle GNP \sim \triangle GKO$ so $P,K,G$ will be collinear.

But to do this see that in $\triangle NEP$ , $\angle PNE = A, \angle PEN = 90^\circ$ and $NE = \frac{R}{2}$ , so $\cos{A} = \frac{R}{2PN}$ , so it suffices to show that $OK = \frac{R}{\cos{A}}$ , but this is true since $OK(R \cos{A}) = OK.OD = OB^2 = R^2$ , so we are done. $\blacksquare$

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Steff9

#5 h Jan 7, 2022, 11:22 AM

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Here is another way to prove that $\Omega$ is tangent to the nine-point circle of $ABC$ .
Let $\Omega$ intersect $BC$ for the second time at $K$ and let $XK$ intersect $(ABC)$ for the second time at $L$ . Let $T$ be a point on the common tangent of $(ABC)$ and $\Omega$ (such that $T$ and $K$ are on different sides of the line $AX$ ).

$\angle ALX = \angle AXT \equiv \angle DXT = \angle DKX$
$\therefore AL\parallel DK\equiv BC$
$\therefore$ $\overarc{AB}=$ $\overarc{LC}$
$\therefore \angle AXL=\angle AXC-\angle LXC=\angle ABC-\angle LBC=\angle ABC-\angle ACB=\beta-\gamma$

Also, because of the midsegments, we get $\angle EDC=\beta$ and $\angle EFD=\gamma$ .

Finally, $\angle EDY=\angle EDC-\angle YDK=\angle EDC-\angle DXK\equiv\angle EDC-\angle AXL=\beta - (\beta - \gamma) = \gamma = \angle EFD$ , so $DY$ is also tangent to $(DEF)$ .

This post has been edited 1 time. Last edited by Steff9, Jan 7, 2022, 11:22 AM

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#6 h Jun 16, 2022, 5:25 AM

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Claim $: ZD$ is tangent to $DEF$ .
Proof $:$ Let $l$ be a line tangent at $X$ to $ABC$ and $S$ an arbitrary point on it. $\angle YDX = \angle DXS = \angle AXS = \angle ABX = \angle FED + \angle XBC = \angle FED + \angle XAC = \angle FED + \angle ADF \implies \angle FED = \angle 180 - \angle YDF$
Claim $: PFE$ and $QBC$ are similar where $Q$ is intersection of tangents at $B,C$ .
Proof $:$ Note that $\angle ZDF = \angle ZFD$ and $\angle EFD = \angle EDY$ so $\angle PFE = \angle EDF = \angle 180 - \angle EDY - \angle YDX + \angle ADF = \angle 180 - \angle FYE \implies \angle PFE = \angle PEF$ . Note that $\angle PFE = \angle EDF = \angle BAC = \angle QBC = \angle QCB$ so $QBC$ and $PEF$ are similar.
Note that $EDF$ and $ABC$ are similar and $QBC$ and $PEF$ are similar and $\frac{FE}{BC} = \frac{1}{2}$ and Note that $\frac{FG}{GC} = \frac{EG}{GB} = \frac{1}{2}$ so $G$ is center of homothety which sends $ABQC$ to $DEPF$ so $P,G,Q$ are collinear.

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#7 h Dec 14, 2023, 11:11 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.20835239724974, xmax = 30.235864692064332, ymin = -12.64383299688833, ymax = 14.455129495917994; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-1.831936840255105,6.354669213014411)--(-4.42,-0.94)--(5.4966,-0.9988)--cycle, linewidth(0) + zzttqq); /* draw figures *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw(circle((0.5523629145212332,1.4023057507017074), 5.496434224429177), linewidth(0.4)); /* special point */ draw(circle((1.2374355154889833,-2.4215551559013564), 1.6116901271115784), linewidth(0.4)); /* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw((-6.742567503117793,3.9904615827866876)--(0.4768362117721101,-11.335246978583324), linewidth(0.4)); draw((0.4768362117721101,-11.335246978583324)--(12.393822444946357,13.203616111466303), linewidth(0.4)); draw((12.393822444946357,13.203616111466303)--(-6.742567503117793,3.9904615827866876), linewidth(0.4)); draw((-6.574579642600736,-4.393873449225951)--(-0.6160865260136092,7.875558095798863), linewidth(0.4)); draw((-0.6160865260136092,7.875558095798863)--(2.993615331431348,0.21270381511385997), linewidth(0.4)); draw((-6.574579642600736,-4.393873449225951)--(2.993615331431348,0.21270381511385997), linewidth(0.4)); draw(circle((-0.6538498773881692,1.5067817311563518), 2.748217112214589), linewidth(0.4)); /* dots and labels */dot((-1.831936840255105,6.354669213014411),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$A$", (-1.664917040607604,6.688708812309402), NE * labelscalefactor); dot((-4.42,-0.94),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$B$", (-4.2537239351438085,-0.6184074222685748), NE * labelscalefactor); dot((5.4966,-0.9988),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$C$", (5.683954143882266,-0.6601623721804489), NE * labelscalefactor); dot((0.5383,-0.9694),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$D$", (0.7151151043692291,-0.6184074222685748), NE * labelscalefactor); dot((1.8323315798724473,2.6779346065072054),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$E$", (2.0095185516373313,3.0142732200644766), NE * labelscalefactor); dot((-3.1259684201275526,2.7073346065072057),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$F$", (-2.959320487875706,3.0560281699763507), NE * labelscalefactor); dot((1.5216562311074888,-4.0079862608164865),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$X$", (1.6754789523423372,-3.666518765835388), NE * labelscalefactor); dot((0.4768362117721101,-11.335246978583324),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$R$", (0.6316052045454805,-11.015389950325238), NE * labelscalefactor); dot((12.393822444946357,13.203616111466303),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$S$", (12.573520879341519,13.453010698033015), NE * labelscalefactor); dot((-6.742567503117793,3.9904615827866876),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$T$", (-6.592001130208767,4.308676667332575), NE * labelscalefactor); dot((2.993615331431348,0.21270381511385997),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$Y$", (3.1786571491698106,0.5507311752639015), NE * labelscalefactor); dot((-6.574579642600736,-4.393873449225951),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$Z$", (-6.42498133056127,-4.042313315042255), NE * labelscalefactor); dot((-0.6160865260136092,7.875558095798863),linewidth(4pt) + dotstyle); label("$P$", (-0.45402349316325025,8.191887009136872), NE * labelscalefactor); dot((-0.25177894675170176,1.4719564043381368),linewidth(1pt) + dotstyle + invisible,UnFill(0)); label("$G$", (-0.0782289439563819,1.803379672620126), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Sketch of the solution: let $R,S,T$ be the intersections of tangents at $B,C; C,A; A,B$ to the circumcircle of $ABC$ respectively. Notice that the homothety centered at $R$ which sends $D$ to $A$ sends the tangent to $\Omega$ at D to the tangent to $(ABC)$ at $A$ . This means that $YZ || ST$ . Now easy angle chasing shows that $YP || TR$ and $PZ || RS$ . Since $DY=YE, DZ=ZF$ we have that $(DEF)$ is the incircle of $PYZ$ . Now consider the homothety at $G$ with scale $-2$ . It sends $\triangle{DEF}$ to $\triangle{ABC}$ . But since $(ABC)$ is the incircle of $RST$ and $YZ || ST, PY || TR, PZ || RS$ we have that this homothety sends $\triangle{PYZ}$ to $\triangle{RTS}$ . In particular it sends $P$ to $R$ and thus $P,G,R$ are collinear.

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