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Source: China TST 1 Day 1 Q3

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#1 h Jan 2, 2018, 3:19 AM • 10 Y

Y by Mathuzb, tenplusten, anantmudgal09, R8450932, nguyendangkhoa17112003, GeoMetrix, amar_04, Adventure10, Mango247, Rounak_iitr

Circle $\omega$ is tangent to sides $AB$ , $AC$ of triangle $ABC$ at $D$ , $E$ respectively, such that $D\neq B$ , $E\neq C$ and $BD+CE<BC$ . $F$ , $G$ lies on $BC$ such that $BF=BD$ , $CG=CE$ . Let $DG$ and $EF$ meet at $K$ . $L$ lies on minor arc $DE$ of $\omega$ , such that the tangent of $L$ to $\omega$ is parallel to $BC$ . Prove that the incenter of $\triangle ABC$ lies on $KL$ .

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#2 h Jan 2, 2018, 5:20 AM • 10 Y

Y by B.J.W.T, anantmudgal09, v_Enhance, igli.2001, nguyendangkhoa17112003, AlastorMoody, Arefe, IFA, AllanTian, Adventure10

Lemma: Let $ABCD$ be a cyclic quadrilateral with circumcenter $O.$ Denote $E \equiv AC \cap BD$ and $F \equiv AB \cap CD.$ Let $F'$ be the antipode of $F$ on $\odot(FBC).$ Then $O, E, F'$ are collinear.
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Proof: Denote $G \equiv AD \cap BC$ and let $M$ be the Miquel point of $ABCD.$ It is known that $O, E, M$ are collinear, and $OM \perp FG.$ On the other hand, since $M \in \odot(FBC)$ and $\overline{FF'}$ is a diameter of this circle, we have $F'M \perp FG.$ Therefore, $O, E, F'$ are collinear. $\blacksquare$
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Main Proof: Let $I$ be the incenter of $\triangle ABC.$ Note that $AI$ is the perpendicular bisector of $\overline{DE}$ ; also, $F$ is the reflection of $D$ in $IB$ and $G$ is the reflection of $E$ in $IC.$ It follows that $IF = ID = IE = IG.$ Therefore, $DEGF$ is cyclic with circumcenter $I.$
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Let $DF$ and $EG$ meet at $S.$ By the above observations, the sidelines of $\triangle SED$ are parallel to the sidelines of the intouch triangle of $\triangle ABC.$ Therefore, homothety carrying the incircle into $\omega$ carries the intouch triangle of $\triangle ABC$ into $\triangle SED.$ By homothety, we see that $S$ lies on major arc $\widehat{DE}$ of $\omega$ , and the tangent to $\omega$ at $S$ is parallel to $BC.$ Therefore, $S$ and $L$ are antipodal points on $\omega.$ Applying the lemma to cyclic quadrilateral $DEGF$ , we see that $I, K, L$ are collinear. $\square$

This post has been edited 3 times. Last edited by Dukejukem, Jan 3, 2018, 1:20 AM

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#3 h Jan 2, 2018, 11:36 AM • 4 Y

Y by Ankoganit, PHSH, Adventure10, MS_asdfgzxcvb

mofumofu wrote:

Circle $\omega$ is tangent to sides $AB$ , $AC$ of triangle $ABC$ at $D$ , $E$ respectively, such that $D\neq B$ , $E\neq C$ and $BD+CE<BC$ . $F$ , $G$ lies on $BC$ such that $BF=BD$ , $CG=CE$ . Let $DG$ and $EF$ meet at $K$ . $L$ lies on minor arc $DE$ of $\omega$ , such that the tangent of $L$ to $\omega$ is parallel to $BC$ . Prove that the incenter of $\triangle ABC$ lies on $KL$ .

Suppose $DE$ is at variable distance $\lambda$ from $A$ . Define $F, G$ as the reflections of $D, E$ in lines $BI, CI$ respectively. Let $S=DF \cap EG$ and $T=DE \cap BC$ . Let $A_1, A_2$ be the touch-points of the incircle and $A$ -excircle with $BC$ . Observe that $\angle DFB=\angle DEG=90^{\circ}-\tfrac{1}{2}\angle B$ hence $DEFG$ is cyclic. Also $I$ is the center of the circle $\odot(DEFG)$ .

Now by Brokard's Theorem, we know $IK \perp ST$ . Note that $DLES$ is cyclic and is the image of the incircle under suitable homothety at $A$ . Accordingly, we see $S \in AA_1, L \in AA_2$ and $T \in BC$ move with constant velocity. Hence we can find a spiral similarity with pivot $P$ such that $BC \mapsto AA_2$ and $T \mapsto S$ . Let $Z$ be point at infinity with $ZP \perp ST$ . Note that $L \mapsto Z$ is projective.

Thus, in order to prove $I, K, L$ collinear, it is equivalent to show $L \mapsto Z$ is a perspectivity about $I$ . Hence it is enough to verify the result for three values of $\lambda \in \mathbb{R}$ . If $DE$ is an intouch chord then we're done. Suppose $B=D$ (likewise let $C=E$ for the final case). Then $K=B$ and $\angle LDE=\tfrac{1}{2}\angle C$ hence $L$ lies on $BI$ and we're done. $\blacksquare$

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#4 h Jan 2, 2018, 1:38 PM • 6 Y

Y by tenplusten, Cristiano-Ezio, k12byda5h, Adventure10, internationalnick123456, MS_asdfgzxcvb

Too easy for China TST P3. I will provide just a sketch since I am on mobile.

By easy angle chasing, $DEFG$ is cyclic and the intersection $DF\cap EG$ lies on $\omega$ . By Reim, $LP$ is diameter of $\omega$ .

Note that $I$ is the center of $\odot(DEFG)$ . Now add Miquel point $M$ of $DEFG$ which lies on $IK$ and $\omega$ . Note that $\angle LMP=90^{\circ}$ hence $I, K,L,M$ are colinear and we are done.

This post has been edited 1 time. Last edited by MarkBcc168, Jan 2, 2018, 1:38 PM

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#6 h Jan 6, 2018, 1:28 PM • 10 Y

Y by anantmudgal09, tenplusten, Cristiano-Ezio, RAMUGAUSS, v4913, thczarif, PHSH, Adventure10, Mango247, MS_asdfgzxcvb

It is also susceptible to straight barycentric coordinates.

Substitute $a=BC=v+w$ , $b=CA=w+u$ , $c=AB=u+v$ . Let $X=(0:w:v)$ , $Y=(w:0:u)$ , $Z=(v:u:0)$ be the contact points of the incircle on $\overline{BC}$ . Then we may set $\begin{align*} D &= (v-t:u+t:0) \\ E &= (w-t:0:u+t) \\ F &= (0:w+t:v-t) \\ G &= (0:w-t:v+t) \end{align*}$ where $t=XF=XG=EY=DZ$ .

We now solve for $K = (t:y:z)$ , say. Since $K = \overline{DG} \cap \overline{EF}$ the standard determinants give $\begin{align*} t(u+t)(v+t) &= (v-t)\left[ (v+t)y-(w-t)z \right] \\ t(u+t)(w+t) &= (w-t)\left[ -(v+t)y+(w+t)z \right]. \\ \implies t(u+t)\left[ (v+t)(w-t)+(w+t)(v+t) \right] &= \left[ -(w-t)^2(v-t)+(v+t)(w-t)(w+t) \right]z \\ z &= \frac{t(u+t)(v+t)(2w)} {(w-t)\left[ (v+t)(w+t)-(w-t)(v-t) \right]} \\ &= \frac{w(u+t)(v+t)}{(w-t)(v+w)}. \end{align*}$ Thus we recover $K = \left( \frac{t(v+t)}{u+t} : \frac{v(w+t)}{v-t} : \frac{w(v+t)}{w-t} \right).$ On the other hand, by homothety we have $\begin{align*} \vec L &= \left( 1 + \frac tu \right) [2 \vec I - \vec X] - \frac tu \vec A \\ \left( 2u+2t \right) \vec I - u \vec L &= \left( u+t \right) \vec X + t \vec A \\ &= t \vec A + (u+t) \frac{w \vec B + v \vec C}{w+v} \\ \end{align*}$ Consequently the point $W = \frac{(2u+2t) \vec I - u \vec L}{u+2t} = \left( t(v+w) : w(u+t) : v(u+t) \right)$ lies on $\overline{IL}$ and is distinct from $I$ (since $u \neq 0$ ). Thus it is enough to show $W$ , $K$ , $I$ collinear: $\begin{align*} \det \begin{bmatrix} \frac{t(v+w)}{u+t} & \frac{v(w+t)}{v-t} & \frac{w(v+t)}{w-t} \\ v+w & w+u & u+v \\ t(v+w) & w(u+t) & v(u+t) \end{bmatrix} &= \det \begin{bmatrix} 0 & \frac{v(w+t)}{v-t}-w & \frac{w(v+t)}{w-t}-v \\ v+w & w+u & u+v \\ t(v+w) & w(u+t) & v(u+t) \end{bmatrix} \\ &= \det \begin{bmatrix} 0 & \frac{v(w+t)}{v-t}-w & \frac{w(v+t)}{w-t}-v \\ v+w & w+u & u+v \\ 0& u(w-t) & u(v-t) \end{bmatrix} \\ &= -(v+w) \det \begin{bmatrix} \frac{t(v+w)}{v-t} & \frac{t(v+w)}{w-t} \\ u(w-t) & u(v-t) \end{bmatrix} = 0. \end{align*}$

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#7 h Jan 7, 2018, 12:57 PM • 4 Y

Y by buratinogigle, enhanced, Adventure10, Mango247

My proof :

From simple angle chasing one can easily get that $\angle ((EDGF), (EDK)) = \angle ((EDGF), \omega)$ . So in other words $(EDK)$ is inversion image of $\omega$ wrt $(EDGF)$ , so $I$ is the inner homothety center of $\pi, (EDK)$ . Now note that the tangent to $(EDK)$ from $K$ is parallel to $BC$ , so from a homothety we get that $K, L, I$ are collinear. Done

This post has been edited 2 times. Last edited by Mosquitall, Jan 7, 2018, 1:01 PM

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#8 h Jan 9, 2018, 8:43 PM • 1 Y

Y by Adventure10

Solution.

Let $X=DF \cap EG$ . Since $DEX$ and the intouch triangle of $ABC$ are homothetic, $X \in \omega$ . Shooting lemma for $\omega$ and point $X$ implies $FGED$ is cyclic; its center is the intersection of the $\perp$ bisectors of $DF, EG$ , i.e., $I$ . Draw in the Miquel point of $FGED$ , since it is the inverse of $K$ in $FGED$ and also the foot of $I$ the polar of $K$ wrt this circle, we get that the antipode of $X$ ( $L$ ), $I$ , $K$ are collinear. Done.

This post has been edited 1 time. Last edited by WizardMath, Jan 9, 2018, 8:43 PM
Reason: Typo.

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#9 h Jan 11, 2018, 2:26 PM • 13 Y

Y by tenplusten, Pirkuliyev Rovsen, xdiegolazarox, enhanced, Cristiano-Ezio, nguyendangkhoa17112003, Sakura_jima.Mai, Avlon, k12byda5h, thczarif, PHSH, Adventure10, Mango247

Let $Y, Z$ be the second intersection of $\omega$ with $EF, DG,$ respectively. Obviously, the incenter $I$ of $\triangle ABC$ is the circumcenter of $DEFG,$ so simple angle chasing yields $\measuredangle IFG = \measuredangle LYZ,$ $\measuredangle IGF = \measuredangle LZY.$ On the other hand, by Reim's theorem we get $FG \parallel YZ,$ so $\triangle IFG$ and $\triangle LYZ$ are homothetic $\Longrightarrow I, L, K$ are collinear.

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#10 h Mar 17, 2018, 1:28 PM • 2 Y

Y by Adventure10, Mango247

You can solve this question easily if using Simson theorm and what you need to do is to find the obvious circle of DEFG and prove two right angles.Maybe a little not natural but quite simple.China`s D1 questions are often easy.Then the whole solving is your task and I just offer an idea

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#11 h Apr 1, 2018, 8:22 PM • 1 Y

Y by Adventure10

(redacted)

This post has been edited 5 times. Last edited by math163, Nov 10, 2018, 8:22 AM

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#13 h Jun 3, 2018, 1:41 PM • 2 Y

Y by Adventure10, Mango247

(redacted)

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#14 h Oct 4, 2018, 7:57 AM • 2 Y

Y by Adventure10, Mango247

It is easily to solve just by Pascal's theorem with circle DEFG. We could find that the incenter I of triangle ABC is the circumcenter of DEFG, and that make it natual to use Pascal's theorem.

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1218 posts

#15 h Oct 23, 2018, 1:50 PM • 2 Y

Y by Adventure10, Mango247

My solution: Let $I$ be the incenter of $\triangle ABC$ . Also let $DF \cap EG=T,\odot (FGT) \cap \omega=P,DE \cap BC=X$ . Note that, as $AD=AE,BD=BF,CE=CG$ , we have that $I$ lies on the perpendicular bisectors of $DE,DF,CG$ , which in turn gives $ID=IE=IF=IG$ . This means that $D,E,F,G$ lie on a circle centered at $I$ . Also, $\angle BDT=\angle BFD=\angle DET \Rightarrow T$ lies on $\omega$ , as $BD$ is tangent to $\omega$ .

Now, $DE$ is antiparallel to $FG$ as well as to the tangent to $\omega$ at $T$ , which means that the tangent to $\omega$ at $T$ is parallel to $BC$ . This also gives that $L$ is the antipode of $T$ in $\omega$ , and so $\measuredangle LPT=90^{\circ}$ . But, $P$ is the Miquel Point of the degenerate cyclic quadrilateral $DEFG$ , and so, by Brocard's Theorem, we get that $I$ is the orthocenter of $\triangle KTX$ , and that $P$ lies on $TX$ . Thus, $\measuredangle IPT=90^{\circ}=\measuredangle LPT \Rightarrow P,K,I,L$ are collinear. $\blacksquare$

This post has been edited 1 time. Last edited by math_pi_rate, Oct 23, 2018, 1:51 PM

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#16 h May 21, 2019, 3:27 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Different.

Let $I$ be the incenter of $\triangle ABC$ ; $F_1$ , $G_1$ are reflections of $F$ , $G$ in $I$ .
Clearly, $I$ lies on perpendicular bisector of $DE$ , $DF$ , $EG$ ; so $DEFGF_1G_1$ is cyclic and center is $I$ .

Claim: $D$ , $L$ , $F_1$ and $E$ , $L$ , $G_1$ are collinear.
Proof: I will prove that $\angle LDE= \angle F'DE$ , which gives the first collinearity; second one follows similarly.
$\angle LDE= \angle AEL=\angle (HE,BC)=\frac{1}2 \angle C$ $\angle F'DE=\angle F'FE=90-\angle FF'E=90-\angle EGC=\frac{1}2 \angle C$ This completes the proof of claim. $\square$

Now, by Pascal's Theorem on $FF'DGG'E$ , we get $K,I,L$ are collinear. $\square$

This post has been edited 4 times. Last edited by Jughead_0, May 25, 2019, 11:29 AM

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#17 h May 22, 2019, 6:10 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

$[asy] size(5cm); defaultpen(fontsize(10pt)); pen pri=deepblue; pen sec=deepcyan; pen tri=deepgreen; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,I,D,EE,F,G,K,X,L,M; A=dir(110); B=dir(210); C=dir(330); I=incenter(A,B,C); D=(A+4B)/5; EE=reflect(A,I)*D; F=reflect(B,I)*D; G=reflect(C,I)*EE; K=extension(D,G,EE,F); X=extension(D,F,EE,G); L=extension(K,I,X,X+(0,1)); M=foot(X,I,K); filldraw(circumcircle(D,F,G),tfil,tri); filldraw(circumcircle(D,EE,L),sfil,sec); filldraw(incircle(A,B,C),fil,pri); filldraw(D--X--EE--cycle,sfil,sec); draw(L--M--X,tri); draw(D--G,sec); draw(EE--F,sec); filldraw(A--B--C--cycle,fil,pri); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,W); dot("$I$",I,W); dot("$E$",EE,dir(30)); dot("$F$",F,SW); dot("$G$",G,SE); dot("$K$",K,dir(110)); dot("$X$",X,S); dot("$L$",L,N); dot("$M$",M,dir(240)); [/asy]$
Let $X=\overline{DF}\cap\overline{EG}$ and $M$ be the Miquel point of $DFGE$ . Note that $\triangle XED$ and the contact triangle are homothetic at $A$ , so $\overline{XL}$ is a diameter of $\omega$ .

Since $ID=IE$ , $ID=IF$ , $IE=IG$ by construction, $I$ is the circumcenter of cyclic quadrilateral $DFGE$ , thus by properties of the Miquel point, $I$ , $K$ , $M$ lie on a line perpendicular to $\overline{XM}$ . But $\angle XML=90^\circ$ , so $L$ , $I$ , $K$ , $M$ are collinear, as desired

This post has been edited 1 time. Last edited by TheUltimate123, Dec 1, 2019, 2:27 AM

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#18 h Dec 29, 2019, 2:40 AM • 2 Y

Y by Pappu123, Adventure10

We proceed with barycentric coordinates. Let $AD=AE=r$ . Then, the coordinates of $F,G$ are $(0,a-c+r,c-r)$ and $(0,b-r,a-b+r)$ respectively, and the coordinates of $D,E$ are $(c-r,r,0)$ , $(b-r,0,r)$ respectively. Computing a determinant, the equations of $EF$ and $DG$ are $(a-b+r)rx-(a-b+r)(c-r)y+(b-r)(c-r)z=0$ $(a-c+r)x+(b-r)(c-r)y-(a-c+r)(b-r)z=0$ Solving, we get that point $K$ has coordinates $((a^2+2ar-ab-ac)(b-r)(c-r),r(b-r)(a+c-b)(a-c+r),r(c-r)(a+b-c)(a-b+r))$ Also, if we consider point $X=(0,s-b,s-c)$ , the $A$ -extouch point, note that $L$ is a homothety of $X$ wrt $A$ with factor $\frac{r}{s}$ . So, it has coordinates $((s-r)a,r(s-b),r(s-c))$ . Now, if we put the coordinates of $I,K,L$ into a determinant, we get a cubic in $r$ , since it is easily verified that the quartic coefficient equals $0$ . Thus, we just need to verify that the determinant is $0$ at $4$ points. We will choose $0,b,c,s-a$ . At $0$ , we have $\begin{vmatrix}a&b&c\\(a^2-ab-ac)bc&0&0\\as&0&0\end{vmatrix}=0$ At $r=b$ we have $\begin{vmatrix}a&b&c\\0&0&ab(c-b)(a+b-c)\\a(s-b)&b(s-b)&b(s-c)\end{vmatrix}=ab(c-b)(a+b-c)(ab(s-b)-ab(s-b))=0$ and we get a similar result for $r=c$ . Finally, at $r=s-a$ , we have $\begin{vmatrix}a&b&c\\0&2(s-a)(s-c)^2(s-b)&2(s-a)(s-b)(s-c)^2\\a^2&(s-a)(s-b)&(s-a)(s-c)\end{vmatrix}\propto\begin{vmatrix}a&b&c\\0&(s-c)&(s-b)\\a^2&(s-a)(s-b)&(s-a)(s-c)\end{vmatrix}$ $=a(s-c)^2(s-a)-a(s-b)^2(s-a)+b(s-b)a^2-c(s-c)a^2=0$ as desired.

This post has been edited 1 time. Last edited by william122, Dec 29, 2019, 2:42 AM

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#19 h Jan 5, 2020, 1:29 PM • 5 Y

Y by AlastorMoody, Pappu123, amar_04, Adventure10, Rounak_iitr

Maybe similiar to the others but since i like this ill post my solution anyways.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -17.64875782250771, xmax = 12.985228170215747, ymin = -10.078567823113621, ymax = 10.151798251301829; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.43637095285195,6.498396949998834)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + sexdts); draw((-5.43637095285195,6.498396949998834)--(6.96,-2.55), linewidth(0.4) + sexdts); draw((6.96,-2.55)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + sexdts); draw(circle((-2.756624790103522,-1.8758367417264656), 5.184106943530284), linewidth(0.4) + dtsfsf); draw(circle((-3.596878261132844,0.7499638003166852), 3.55859333554701), linewidth(0.4) + dtsfsf); draw((-3.596878261132844,0.7499638003166852)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + rvwvcq); draw((-3.596878261132844,0.7499638003166852)--(6.96,-2.55), linewidth(0.4) + rvwvcq); draw((-7.676150691333448,-0.24083530163939865)--(-1.283252160384008,-2.7527894567191105), linewidth(0.4)); draw((-5.735469886770201,-2.8623169582879795)--(0.2997740533078106,2.311452466335568), linewidth(0.4) + rvwvcq); draw((-3.684395533087758,4.307480807290673)--(-3.5093609891779307,-2.8075532066573032), linewidth(0.4)); draw((-7.676150691333448,-0.24083530163939865)--(-5.735469886770201,-2.8623169582879795), linewidth(0.4) + wvvxds); draw((0.2997740533078106,2.311452466335568)--(-1.283252160384008,-2.7527894567191105), linewidth(0.4) + rvwvcq); draw((-5.43637095285195,6.498396949998834)--(-2.8841186772775,3.306702222014393), linewidth(0.4) + wrwrwr); draw(circle((-3.5968782949371954,0.7499637752913503), 4.197873997018878), linewidth(0.4) + linetype("4 4") + rvwvcq); draw((-7.676150691333448,-0.24083530163939865)--(0.2997740533078106,2.311452466335568), linewidth(0.4)); draw((-2.6291309029295515,-7.0583757054673235)--(-16.712165924866934,-3.132350942791587), linewidth(0.4)); draw((-2.8841186772775,3.306702222014393)--(-5.546903657206548,-6.2449680793904125), linewidth(0.4)); draw((-5.735469886770201,-2.8623169582879795)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4) + wvvxds); draw((-1.283252160384008,-2.7527894567191105)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4) + wvvxds); draw((-5.43637095285195,6.498396949998834)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4)); draw((-2.8841186772775,3.306702222014393)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4)); draw((-16.712165924866934,-3.132350942791587)--(-7.676150691333448,-0.24083530163939865), linewidth(0.4)); draw((-16.712165924866934,-3.132350942791587)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4)); /* dots and labels */ dot((-5.43637095285195,6.498396949998834),dotstyle); label("$A$", (-5.3553790280251805,6.690555678098006), NE * labelscalefactor); dot((-8.57057841452274,-2.9320625037227908),dotstyle); label("$B$", (-8.498346422083822,-2.738346504077927), NE * labelscalefactor); dot((6.96,-2.55),dotstyle); label("$C$", (7.037460759940218,-2.3603947288430267), NE * labelscalefactor); dot((-3.596878261132844,0.7499638003166852),linewidth(4pt) + dotstyle); label("$I$", (-3.5252967479404025,0.9019258573950597), NE * labelscalefactor); dot((-7.676150691333448,-0.24083530163939865),linewidth(4pt) + dotstyle); label("$D$", (-7.603197480738006,-0.07279187873705148), NE * labelscalefactor); dot((0.2997740533078106,2.311452466335568),linewidth(4pt) + dotstyle); label("$E$", (0.3735741965880374,2.4734095544243817), NE * labelscalefactor); dot((-3.5093609891779307,-2.8075532066573032),linewidth(4pt) + dotstyle); label("$T$", (-3.4258357544575344,-2.6388855105950584), NE * labelscalefactor); dot((-1.498834977839814,3.62429866411189),linewidth(4pt) + dotstyle); label("$M$", (-1.4167236861035932,3.786294668398246), NE * labelscalefactor); dot((-6.973851822223395,1.8722989086934119),linewidth(4pt) + dotstyle); label("$J$", (-6.887078327661354,2.0357811830997603), NE * labelscalefactor); dot((-5.735469886770201,-2.8623169582879795),linewidth(4pt) + dotstyle); label("$F$", (-5.653762008473786,-2.6985621066847796), NE * labelscalefactor); dot((-1.283252160384008,-2.7527894567191105),linewidth(4pt) + dotstyle); label("$G$", (-1.1979095004412827,-2.599101113201911), NE * labelscalefactor); dot((-3.684395533087758,4.307480807290673),linewidth(4pt) + dotstyle); label("$K$", (-3.604865542726697,4.462629424081752), NE * labelscalefactor); dot((-2.8841186772775,3.306702222014393),linewidth(4pt) + dotstyle); label("$L$", (-2.8091775948637503,3.468019489253067), NE * labelscalefactor); dot((-4.24854799053207,-1.5876392286151881),linewidth(4pt) + dotstyle); label("$K$", (-4.16184710623076,-1.425461390104063), NE * labelscalefactor); dot((-2.6291309029295515,-7.0583757054673235),linewidth(4pt) + dotstyle); label("$N$", (-2.5505790118082925,-6.8958160316618295), NE * labelscalefactor); dot((-16.712165924866934,-3.132350942791587),linewidth(4pt) + dotstyle); label("$O$", (-16.634255688982453,-2.9770528884368113), NE * labelscalefactor); dot((-5.546903657206548,-6.2449680793904125),linewidth(4pt) + dotstyle); label("$P$", (-5.474732220204623,-6.080235885102308), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Proof: Firstly we give a defination of few points. Let $DF \cap GE=N$ and $DE \cap FG=O$ . Let $J,M,T$ be the touchpoints of the incircle with $AB,AC,BC$ .
Firstly notice that $\Delta DNE$ and $\Delta JTM$ are homothetic and so taking a homothety centred at $A$ s.t. $\odot(I) \mapsto \omega \implies N \in \omega$ . Also $\{A,T,N\}$ is a collinear triplet $\implies L$ is the $N$ -Antipode w.r.t $\omega$ . Now let $P$ be the miquel point of quaderilaeral $DEGF$ . Notice that $ID=IF=IG=IE \implies I$ is the centre of $\odot(DFGE)$ . Now its well known that $K,P$ are inverses w.r.t $\odot(DEFG)\implies \{I,K,P\}$ is a collinear triplet. By brocard we have that $\angle IPN=90^\circ$ . Now notice that since $L$ is the $N$ -antipode $\implies \angle LPN=90^\circ \implies \{L,I,K,P\}$ is a collinear triplet as desired. $\blacksquare$ .

This post has been edited 1 time. Last edited by GeoMetrix, Jan 5, 2020, 1:29 PM

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#20 h Feb 21, 2020, 10:38 PM • 3 Y

Y by amar_04, Gaussian_cyber, Adventure10

Overcomplicated but different

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946 posts

Plops

#21 h Feb 22, 2020, 12:38 AM • 1 Y

Y by Adventure10

Not an answer. Just a question.

Edit: Because the problem is a cyclic quad problem.

This post has been edited 2 times. Last edited by Plops, Feb 22, 2020, 12:57 AM

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khina

#22 h Feb 25, 2020, 12:42 PM

Y by

@above the fact that u can reduce ll the incenter stuff to focus just on the quad. because we kmow so much about miquel theory, it makes sense to focus on a structure we already know a lot about

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1915 posts

#23 h Feb 26, 2020, 1:49 AM • 5 Y

Y by GeoMetrix, Hexagrammum16, Bumblebee60, Mango247, VIATON

Too easy for China TST P3. Anyways really Nice Problem.

mofumofu wrote:

Circle $\omega$ is tangent to sides $AB$ , $AC$ of triangle $ABC$ at $D$ , $E$ respectively, such that $D\neq B$ , $E\neq C$ and $BD+CE<BC$ . $F$ , $G$ lies on $BC$ such that $BF=BD$ , $CG=CE$ . Let $DG$ and $EF$ meet at $K$ . $L$ lies on minor arc $DE$ of $\omega$ , such that the tangent of $L$ to $\omega$ is parallel to $BC$ . Prove that the incenter of $\triangle ABC$ lies on $KL$ .

Note that after Simple Angle Chasing $DFGE$ we get that $DFGE$ is a cyclic quadrilateral. So, $\angle EGC=\angle GEC=\angle FDE$ , now as $AC$ is tangent to $\omega$ , hence it's obvious that $DF\cap EG=X\in\omega$ . Now it's easy to notice that $L$ is the $K-\text{Antipode}$ WRT $\omega$ . Let $LK\cap\omega=Y$ , then it's well known that $Y$ is the Miquel Point of $DEGF$ . So, the Circumcenter $(O)$ of $DEGF$ must lie on $LK$ , and if $T$ is the Circumcenter of $\omega$ then $AT$ is the perpendicular bisector of $ED$ . Hence, $ED\cap LY=O$ . Hence, $OD=OF=OE=OG$ . Hence, $BO,CO$ are the $B,C$ angle bisectors of $\triangle ABC$ . So, $O\equiv I\in LK$ . $\blacksquare$

This post has been edited 3 times. Last edited by amar_04, Feb 26, 2020, 2:11 AM

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#24 h Mar 19, 2020, 1:56 PM • 1 Y

Y by gamerrk1004

Solution

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#25 h Oct 27, 2020, 7:47 PM • 2 Y

Y by A-Thought-Of-God, VIATON

Easy for a China #3 ? Interesting problem though.

Solution

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#26 h Nov 22, 2020, 3:50 AM

Y by

Here is a moving points approach.

First, by easy angle chasing we get $DF, EG$ concur at a point $M$ on $\omega$ , and points $D,E,F,G$ all lie on a circle centered at $I$ . Let $DE,BC$ intersect at $T$ , and let $TM$ intersect $\omega$ again at $N$ . By Brokard's theorem, $KI$ passes through $N$ , so it suffices to prove that $LI$ also passes through $N$ . This can be deduced from the following claim:

Claim. Let $X = \mu D + (1-\mu)E$ , $Y = \mu F + (1-\mu) G$ for some real number $\mu$ . Denote the line through $X$ perpendicular to $DE$ by $\ell_1$ , and denote the line through $Y$ perpendicular to $FG$ by $\ell_2$ . Suppose $\ell_1 \cap \ell_2 = Z$ . Then then as $\mu$ varies, $Z$ moves on the line $NL$ linearly. In particular, taking $\mu = 0.5$ , we have $N,I,L$ collinear.

Proof. Let $\phi_1$ denote the linear map that sends $X$ to the point $Z_1 = \ell_1 \cap NL$ , and let $\phi_2$ denote the linear map that sends $Y$ to the point $Z_2 = \ell_2 \cap NL$ . It suffices to verify that $\phi_1, \phi_2$ coincide for two distinct values of $\mu$ :

Take $\mu = 1$ , then $\ell_1,\ell_2$ intersect on $NL$ because $T,D,F,N$ are concyclic.
Take $\mu = 0$ , then $\ell_1,\ell_2$ intersect on $NL$ because $T,E,G,N$ are concyclic. $\square$

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390 posts

Lcz

#27 h Nov 30, 2020, 4:58 PM

Y by

Not the best solution, but...

Bary wrt $\Delta ABC$ . Let $AD=AE=t$ so that
$\quad$ -by lengths $D=(c-t:t:0)$ , $E=(b-t:0:t)$ , $F=(0:a-c+t:c-t)$ , $G=(0:b-t:a-b+t)$ .
$\quad$ -by a homothety from $A$ with factor $\frac{t}{s}$ sending the $A$ -excircle touch point $(0:s-b:s-c)$ to $L$ we get $L=(a(s-t):t(s-b):t(s-c))$ .
Since $I=(a:b:c)$ it remains to compute $K$ , the intersection of $DG: t(a-b+t)x-(c-t)(a-b+t)y+(b-t)(c-t)z=0$ $EF: t(a-c+t)x+(b-t)(c-t)y-(b-t)(a-c+t)z=0$ and solving we get $K=(a(b-t)(c-t)(a+2t-b-c): t(a-c+t)(a-b+c)(b-t): t(a-b+t)(a+b-c)(c-t))$ Now to show the determinant $[IKL]=0$ :
-divide the first column by $a$ .
-multiply the first row by $t$ , then divide the second and third columns by $t$
-add the first row to the second, then divide the second by $s$
-subtract $t$ times the second row from the first.
Now we have $\begin{pmatrix} 0 & b-t & c-t\\ 1 & 1 & 1\\ (b-t)(c-t)(a+2t-b-c) & (a-c+t)(a-b+c)(b-t) & (a-b+t)(a+b-c)(c-t) \end{pmatrix}$
$=(b-t)(c-t)(0)=0.$ where everything cancels (yay!)

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L567

#28 h Mar 10, 2022, 8:51 AM

Y by

We have $\angle (DF, EG) = 180 - \angle DFB - \angle EGC = 90 - \frac{\angle BAC}{2} = \angle ADE$ , which is the measure of arc $DE$ , so we have $X = DF \cap EG \in \omega$ . So, $\angle BFD = \angle BDX = \angle DEG$ so $DFEG$ is cyclic.

Note that since $I$ lies on the perpendicular bisectors of $EG, DF$ , it must in fact be the circumcenter (since the quadrilateral is cyclic). Let $M$ be the miquel point of $DFEG$ , which lies on $\omega$ . It's well known that $I,K,M$ must be collinear, so it suffices to show that $MI$ goes through $L$ . Since $IK \perp XM$ , we must have that $MI \cap \omega$ is the antipode of $X$ , so it suffices to show that the tangent at $X$ to $\omega$ is parallel to $BC$ . But this is true since if $Z$ is a point on the tangent, then $\angle FXZ = \angle DXZ = \angle DEG = \angle DFB$ , so we are done. $\blacksquare$

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Om245

#29 h Apr 22, 2024, 8:56 AM • 3 Y

Y by Deadline, GeoKing, Rounak_iitr

Let $I$ be incenter of $\triangle ABC$
Claim: $D,F,G,E$ lie on circle with center as $I$ .

$\angle FDE = 180 - \angle EDA - \angle BDF = 90 - \frac{\angle C}{2} = \angle CGE$ hence $D,F,E,G$ cyclic.Note $I$ lie on perpendicular bisector of $DE$ and by condition $BD=BF$ and $CE=CG$ we have $I$ also lie on perpendicular bisector of $GF$ also from fact that $IB$ and $IC$ are angle bisector we have $I$ as center of $(DEF)$

Let $T=DF \cap GE$
Claim : $T$ lie on $(DLE)$ with $\angle TDL = \angle LET = 90$

Let perpendicular from $L$ to $BC$ meet at $H$ .If $O$ is center of $(DLE)$ then $O,E,B,H$ cyclic.
$\angle LOD = \angle CBA \implies \angle ODL = 90 - \frac{\angle B}{2} \implies \angle EDL = \frac{\angle C}{2}$ hence $\angle FDL = \angle FDE + \angle EDL = 90$

Claim : $L,I,K$ are collinear.
Let $M=(TGF)\cap (DTE)$ . Then $M$ is miquel point point of $\Box DEGF$ . hence $\overline{I-K-M}$ (well know property)
as $LT$ is diameter $\angle TML = 90$ , but it's well know that $\angle TMI = 90$ hence $\overline{L-I-M}$

Hence $L,I,K$ are collinear $\blacksquare$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(19.07651842087848cm); 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 = -4.322575073781325, xmax = 22.753943347097152, ymin = -11.155828560285837, ymax = 3.1133695277123317; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((6.993497304868754,1.7930527604426043)--(4.74121400092014,-6.783937749194878)--(15.721214000920138,-6.883937749194877)--cycle, linewidth(0.4) + ffxfqq); /* draw figures */ draw((6.993497304868754,1.7930527604426043)--(4.74121400092014,-6.783937749194878), linewidth(0.4) + ffxfqq); draw((4.74121400092014,-6.783937749194878)--(15.721214000920138,-6.883937749194877), linewidth(0.4) + ffxfqq); draw((15.721214000920138,-6.883937749194877)--(6.993497304868754,1.7930527604426043), linewidth(0.4) + ffxfqq); draw(circle((10.260220799534883,-3.6489912612960405), 6.347229752037104), linewidth(0.4)); draw((xmin, -3.673826499279709*xmin + 27.485948481710658)--(xmax, -3.673826499279709*xmax + 27.485948481710658), linewidth(0.4)); /* line */ draw(circle((8.962592300056919,-5.441060612478726), 3.7418809170599285), linewidth(0.4) + red); draw((5.343414536894738,-4.490679049058201)--(9.911189132387893,-6.831023132905878), linewidth(0.4)); draw((7.11212409920384,-6.805530737339539)--(11.600778937775456,-2.7874509836861616), linewidth(0.4)); draw((xmin, -0.009107468123861607*xmin-1.6173979879750853)--(xmax, -0.009107468123861607*xmax-1.6173979879750853), linewidth(0.4)); /* line */ draw((xmin, 4.764129681840161*xmin-44.56063720954149)--(xmax, 4.764129681840161*xmax-44.56063720954149), linewidth(0.4) + green); /* line */ draw(circle((8.538392964454843,-3.8826258523669748), 3.2523246828354737), linewidth(0.4)); draw((xmin, 0.27219576106712146*xmin-5.945133835625384)--(xmax, 0.27219576106712146*xmax-5.945133835625384), linewidth(0.4)); /* line */ draw((5.343414536894738,-4.490679049058201)--(8.996669947959758,-1.6993348727470368), linewidth(0.4)); draw((8.996669947959758,-1.6993348727470368)--(11.600778937775456,-2.7874509836861616), linewidth(0.4)); draw((7.11212409920384,-6.805530737339539)--(8.996669947959758,-1.6993348727470368), linewidth(0.4)); draw((8.996669947959758,-1.6993348727470368)--(9.911189132387893,-6.831023132905878), linewidth(0.4)); draw((5.343414536894738,-4.490679049058201)--(8.928514652154083,-9.182786352210424), linewidth(0.4)); draw((11.600778937775456,-2.7874509836861616)--(8.928514652154083,-9.182786352210424), linewidth(0.4)); draw((8.996669947959758,-1.6993348727470368)--(8.928514652154083,-9.182786352210424), linewidth(0.4)); draw(circle((8.504315316552006,-7.624351592098668), 1.6151359007370965), linewidth(0.4)); draw((4.74121400092014,-6.783937749194878)--(-2.8283481155758152,-6.714998203507301), linewidth(0.4)); draw((-2.8283481155758152,-6.714998203507301)--(8.928514652154083,-9.182786352210424), linewidth(0.4)); /* dots and labels */ dot((6.993497304868754,1.7930527604426043),dotstyle); label("$A$", (7.05464745517935,1.9534013089775137), NE * labelscalefactor); dot((4.74121400092014,-6.783937749194878),dotstyle); label("$B$", (4.115001969344539,-6.643075490139423), NE * labelscalefactor); dot((15.721214000920138,-6.883937749194877),dotstyle); label("$C$", (15.603454327498964,-6.357055929355495), NE * labelscalefactor); dot((8.962592300056919,-5.441060612478726),dotstyle); label("$O$", (9.215684136657915,-5.4354373446072834), NE * labelscalefactor); dot((5.343414536894738,-4.490679049058201),linewidth(4.pt) + dotstyle); label("$D$", (4.9253907248990005,-4.243689174674252), NE * labelscalefactor); dot((11.600778937775456,-2.7874509836861616),linewidth(4.pt) + dotstyle); label("$E$", (11.901090012907012,-2.527571809970686), NE * labelscalefactor); dot((9.911189132387893,-6.831023132905878),linewidth(4.pt) + dotstyle); label("$G$", (9.962512989815947,-6.023366441774246), NE * labelscalefactor); dot((7.11212409920384,-6.805530737339539),linewidth(4.pt) + dotstyle); label("$F$", (6.879857723589172,-6.372945904954602), NE * labelscalefactor); dot((8.996669947959758,-1.6993348727470368),linewidth(4.pt) + dotstyle); label("$L$", (9.311023990252556,-1.4152735180331897), NE * labelscalefactor); dot((8.538392964454843,-3.882625852366973),linewidth(4.pt) + dotstyle); label("$I$", (8.738984868684701,-3.957669613890324), NE * labelscalefactor); dot((8.112914677045005,-5.9096595903947025),linewidth(4.pt) + dotstyle); label("$K$", (8.500635234698095,-5.880356661382282), NE * labelscalefactor); dot((8.928514652154083,-9.182786352210424),linewidth(4.pt) + dotstyle); label("$T$", (9.199794161058806,-9.598610951573342), NE * labelscalefactor); dot((10.202416004181245,-9.995957791125791),linewidth(4.pt) + dotstyle); label("$A'$", (10.264422526198981,-9.868740536758162), NE * labelscalefactor); dot((7.489289618033217,-8.880690244372106),linewidth(4.pt) + dotstyle); label("$M$", (7.213547211170421,-8.45453270843763), NE * labelscalefactor); dot((-2.8283481155758152,-6.714998203507301),linewidth(4.pt) + dotstyle); label("$R$", (-2.76535746506883,-6.595405563342101), NE * labelscalefactor); dot((8.950012986401948,-6.822269251794895),linewidth(4.pt) + dotstyle); label("$H$", (8.389405405504345,-6.54773563654478), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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L13832

#30 h Sep 26, 2024, 12:36 PM

Y by

Finally done after :wallbash:

:wallbash:

and

:censored:

https://wiki-images.artofproblemsolving.com//7/7b/POTD_-1000.png

Let $DF\cap EG=J$ and $DE\cap FG=S$ , we have $\odot(DEGF)$ with center $I$ . Note that the $J$ is $L$ -antipode this is because $\triangle DEJ$ and contact triangle are homothetic with center $A$ , so $\overline{A-J-N}$ .
Now let the miquel point of $\odot(DEGF)$ be $M$ which is also $\odot(DJE)\cap \odot(FJG)$ , then $\overline{I-K-M}$ and finally by Brokard's theorem we have $\angle IMJ=90^{\circ}$ because $J$ is the $L$ -antipode we have $\angle LMJ=90^{\circ}$ , so $\overline{L-I-K-M}$ and we are done!
Remark

This post has been edited 2 times. Last edited by L13832, Sep 26, 2024, 12:46 PM

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#31 h Dec 22, 2024, 5:36 PM • 2 Y

Y by anantmudgal09, abeot

Here is an alternative solution using my favorite technique, the ``radical axis trick'': if you don't know how to prove three points are collinear, try to throw them on a radical axis. While there are certainly more natural solutions (see all above posts), this problem is still a nice showcase of the method.

Claim: Points $D$ , $E$ , $F$ , and $G$ lie on a circle with center $I$ .

Proof. From the given length conditions, we immediately deduce that $ID = IF$ and $IE = IG$ . But we also have $ID = IE$ from the definition of $\omega$ . $\blacksquare$

$[asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, G, I, O, K, L, X, Y, MA; real r = 0.47; A = dir(120); B = dir(210); C = dir(330); I = incenter(A, B, C); O = (1+r)*I - r*A; D = foot(O, A, B); E = foot(O, A, C); F = 2*foot(D, B, I) - D; G = 2*foot(E, C, I) - E; K = extension(D, G, E, F); L = O + abs(O-D)*dir(90); MA = dir(270); X = 2*foot(MA, D, L) - D; Y = 2*foot(MA, E, L) - E; draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(CP(O, D), heavyred); draw(D--G^^E--F, lightblue); draw(circumcircle(B, D, I), lightblue+dashed); draw(circumcircle(C, E, I), lightblue+dashed); draw(D--L--Y, heavycyan); draw(circumcircle(D, F, G), dotted+heavygreen); draw(K--L, dotted+magenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$D$", D, dir(130)); dot("$E$", E, dir(50)); dot("$I$", I, dir(120)); dot("$F$", F, dir(300)); dot("$G$", G, dir(315)); dot("$K$", K, dir(270)); dot("$L$", L, dir(90)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(60)); dot("$M_A$", MA, dir(270)); [/asy]$

Remark: If we are looking to apply the radical axis trick, we now know that $K$ automatically lies on the radical axis of any two circles $\omega_1$ and $\omega_2$ with $D, G\in \omega_1$ and $E, F\in \omega_2$ . So there are two natural choices: $\omega_1 = (DGI)$ and $\omega_2 = (EFI)$ , or $\omega_1 = (DGL)$ and $\omega_2 = (EFL)$ . Trying both briefly should convince you that the first option is much easier.

At this point, we make the following observations:

Since $DI = IG$ and $\overline{BI}$ bisects $\angle DBG$ , $B\in (DGI)$ . Similarly, $C\in (EFI)$ . (This is another of my favorite tricks.)
Straightforward angle chasing yields $\overline{DL}\parallel \overline{BI}$ and $\overline{EL}\parallel \overline{CI}$ . (For example, the homothety at $A$ sending $\omega$ to the $A$ -excircle sends $L$ , $D$ , $E$ to the excircle tangency points.)

To finish, let $X = \overline{DL}\cap (DGI)\neq D$ and $Y = \overline{EK}\cap (EFI)\neq E$ . It suffices to show that $DXEY$ is cyclic; then $K$ , $I$ , $L$ will all lie on the radical axis of $(DGI)$ and $(EFI)$ .

We will show that $D$ , $X$ , $E$ , $Y$ all lie on a circle with center $M_A$ , the midpoint of arc $BC$ . First note that $M_AD = M_AE$ , by the definition of $\omega$ . Furthermore, from the parallel lines, we see that $BDXI$ and $CYEI$ are isosceles trapezoids. Hence $DX$ and $BI$ share a perpendicular bisector, and so if $M_A$ is the midpoint of arc $BC$ , we must have $M_AD = M_AX$ . Similarly, we get $M_AE = M_AY$ , which completes the proof.

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#32 h Jan 23, 2025, 8:54 AM

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Let $I$ denote the incenter of $\triangle ABC$ .

$\angle EDF=180^\circ-\angle ADE-\angle FDB=\frac{\alpha}{2}+\frac{\beta}{2}=90^\circ-\frac{\gamma}{2}=180^\circ-\angle EGF$ . This implies $DEGF$ is cyclic. Obviously $I$ is on the perpendicular bisector of $DE$ and you can play around with lengths to easily show that the $A$ -intouch point is the midpoint of $FG$ which implies $I$ is on perpendicular bisector of $FG$ . Combining these claims imply $I$ is the center of $(DEGF)$ . From there we also get that $I$ is on $(BDG)$ and on $(CEF)$ . Let $DG$ intersect $\omega$ again in $S$ . Because the tangent in $L$ is parallel to $BC$ we know that $\angle ADL=\frac{\beta}{2}$ . So $\angle DSL=\frac{\beta}{2}$ . Also $\angle DGI=\angle DBI=\frac{\beta}{2}$ , so $\angle DSL=\angle DGI$ which implies $SL\parallel GI$ . Now with Reim we conclude that $K, I, L$ are collinear and we are done.

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