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Source: STEMS 2020 Math Category B/P4 Subjective

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15 posts

#1 h Jan 26, 2020, 12:55 PM • 4 Y

Y by Smita, mijail, Adventure10, Funcshun840

In triangle $\triangle{ABC}$ with incenter $I$ , the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$ , respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$ . The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$ . Prove that $FQ$ is parallel to $BI$ .

Proposed by Anant Mudgal

This post has been edited 2 times. Last edited by Shweta_16, Jan 26, 2020, 1:34 PM
Reason: let's chase some angels

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924 posts

#2 h Jan 26, 2020, 1:06 PM • 5 Y

Y by mueller.25, amar_04, AlastorMoody, sameer_chahar12, Adventure10

$[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 = -19.43752933627648, xmax = 32.95131572780579, ymin = -22.91737803351104, ymax = 11.67967094711988; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.88,-6.82)--(15.2,-6.7), linewidth(0.4) + rvwvcq); draw((15.2,-6.7)--(1.948187283488726,4.409082587313549), linewidth(0.4) + rvwvcq); draw(circle((5.47643671228076,-3.212986968017347), 3.574467648278019), linewidth(0.4) + dtsfsf); draw((1.88,-6.82)--(5.47643671228076,-3.212986968017347), linewidth(0.4) + wvvxds); draw(circle((6.747870334297736,-9.23935115462224), 8.825349861185424), linewidth(0.4) + wvvxds); draw(circle((8.521543675938103,-4.711348029129539), 6.968250536085029), linewidth(0.4) + sexdts); draw((2.2692084696477464,-1.63485327018204)--(1.90203496433293,-3.191281833747048), linewidth(0.4) + wrwrwr); draw((1.948187283488726,4.409082587313549)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((15.2,-6.7)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(7.772785268839476,-0.47371657695024716), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(5.5086378167960826,-6.787309569218054), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); draw((-1.7486378167960828,-6.852690430781947)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((2.2692084696477464,-1.63485327018204)--(-1.7486378167960828,-6.852690430781947), linewidth(0.4)); draw((-1.7486378167960828,-6.852690430781947)--(7.772785268839476,-0.47371657695024716), linewidth(0.4)); draw((7.772785268839476,-0.47371657695024716)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((1.948187283488726,4.409082587313549)--(1.88,-6.82), linewidth(0.4) + rvwvcq); draw((2.2692084696477464,-1.63485327018204)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((-1.7486378167960828,-6.852690430781947)--(1.88,-6.82), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((1.948187283488726,4.409082587313549),dotstyle); label("$A$", (2.0963271088950033,4.739849912657041), NE * labelscalefactor); dot((1.88,-6.82),dotstyle); label("$B$", (2.0282896477728185,-6.486331172503435), NE * labelscalefactor); dot((15.2,-6.7),dotstyle); label("$C$", (15.329613297159941,-6.350256250259066), NE * labelscalefactor); dot((5.47643671228076,-3.212986968017347),linewidth(4pt) + dotstyle); label("$I$", (5.6002563566875185,-2.9483831941498306), NE * labelscalefactor); dot((5.5086378167960826,-6.787309569218054),linewidth(4pt) + dotstyle); label("$D$", (5.634275087248612,-6.5203499030645276), NE * labelscalefactor); dot((7.772785268839476,-0.47371657695024716),linewidth(4pt) + dotstyle); label("$E$", (7.913530034841801,-0.19286601870135017), NE * labelscalefactor); dot((1.90203496433293,-3.191281833747048),linewidth(4pt) + dotstyle); label("$F$", (2.0282896477728185,-2.914364463588738), NE * labelscalefactor); dot((-1.7486378167960828,-6.852690430781947),linewidth(4pt) + dotstyle); label("$Q$", (-1.6117145222640668,-6.588387364186712), NE * labelscalefactor); dot((2.2692084696477464,-1.63485327018204),linewidth(4pt) + dotstyle); label("$K$", (2.4024956839448346,-1.34950285777849), NE * labelscalefactor); dot((11.659998503978759,-16.571322036789724),linewidth(4pt) + dotstyle); label("$I_A$", (11.791665318806333,-16.283725574098032), NE * labelscalefactor); dot((4.336270300841782,-0.7498879530434238),linewidth(4pt) + dotstyle); label("$L$", (4.47763824817147,-0.465015863190089), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
IMO the title doesnt suit the prob.
Proof: Let $I_A$ be the $A$ -excentre. Its well known(ISL 2002 G7) that $K=I_AD \cap \omega$ . Now we'll use a phantom approach . Let $\ell$ be the line through $F$ parallel to $BI$ . and let $\ell \cap BC=Q'$ We just need to show that $Q' \in \odot(KEG)$ . Firstly define $\ell \cap AI=L$ . Notice that $\angle LQ'C=\angle IBC=\angle IKC=\angle LKC \implies LQ'I_AC$ cyclic. Now notice that $\Delta FAL \cong \Delta EAL \implies \angle AFL=\angle AEL$ . But notice that $\angle AEL=\angle AFL=\angle ABI=\angle LQ'C\implies LEQ'C$ cyclic. This combined with the fact that $LQ'I_AC$ is cyclic $\implies LQ'CI_AE$ is cyclic. So now we just need to show that $K\in \odot(LQ'CI_AE)$ But notice that $\angle I_AEQ'=\angle I_ACQ'=90-\frac{\angle C}{2}$ and $\angle Q'I_AE=\angle Q'CE=\angle C \implies I_AQ=I_AE$ . FInally $\angle EKD =\angle CDE=\angle I_ACD=\angle I_AEQ'=\angle I_AQ'E\implies K \in \odot(I_AQ'E)$ as desired. $\blacksquare$

This post has been edited 9 times. Last edited by GeoMetrix, Jan 27, 2020, 6:13 AM

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1979 posts

#3 h Jan 26, 2020, 1:16 PM • 8 Y

Y by Pluto1708, biomathematics, GammaBetaAlpha, amar_04, Sumitrajput0271, DPS, Adventure10, Funcshun840

My problem.

My solution was to apply $\sqrt{DE \cdot DF}$ inversion in contact triangle $\triangle DEF$ . It is quite simple from here

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3042 posts

#4 h Jan 26, 2020, 1:44 PM • 2 Y

Y by amar_04, Adventure10

Shweta_16 wrote:

In triangle $\triangle{ABC}$ with incenter $I$ , the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$ , respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$ . The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$ . Prove that $FQ$ is parallel to $BI$ .

Proposed by Anant Mudgal

Quick sketch:

Let $I_a$ be the A-excenter. We know that $I_a,D,K$ are collinear. Angle chase to show that $I_a$ lies on $\odot(KEC)$ . This will gives us that $\angle QKD = \angle DIC$ and also we have $\angle BKD = DKC$ .

Now we'll fix $K,D$ and move $C$ linearly. Therefore $B$ and $Q$ move linearly, so just work when $C = D$ and $C$ is point of infinity to get that $BQ = BD = BF$ , so we are done.

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1566 posts

#9 h Jan 26, 2020, 5:13 PM • 4 Y

Y by GeoMetrix, amar_04, Adventure10, Mango247

Here's my solution, which is much more of a "complete the configuration" type, while ignoring $A.$ Nice problem btw.

Shweta_16 wrote:

In triangle $\triangle{ABC}$ with incenter $I$ , the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$ , respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$ . The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$ . Prove that $FQ$ is parallel to $BI$ .

Proposed by Anant Mudgal

Clearly, it suffices to show $BF=BQ.$ Let $\omega$ be tangent to $BC$ at $D.$ Since $BF=BD,$ hence it suffices to show that $BQ=BD.$ Call the circle through $B,C$ tangent to $\omega$ as $\gamma.$ We now rephrase the problem without $A:$

Rephrased Problem wrote:

Let $\gamma$ be a circle through two points $B,C,$ and let $\omega$ be a circle tangent to $BC, \gamma$ at $D,K$ respectively. Let $CE$ be the second tangent from $C$ to $\omega.$ Assume that $(KEC)$ meets $BC$ again in $Q.$ Show that $BQ=BD.$

Let $M$ be the midpoint of arc $BC$ not containing $K$ in $\gamma.$ By a well known lemma (shooting lemma), $K,D,M$ are collinear. (proof is by homothety taking $\omega$ to $\gamma$ ). Let $(M,MC)$ be the circle at $M$ with radius $MC.$

We start off by the following key lemma:

Lemma: The points $ED, CM$ meet at $X,$ where $X$ lies on $(M,MC)$
Proof: Let $O$ be the center of $\omega.$ Let $OC \cap ED=N.$ Now consider the inversion about $(M,MC).$ It is not too hard to see that $\omega$ is fixed under this inversion (since it is tangent to $BC, \gamma,$ both of which are swapped under this inversion). Hence $\omega, (M,MC)$ are orthogonal.

Thus, $ON \cdot OC=OD^2$ is the power of $O$ with respect to $(M,MC).$ Hence, $N$ lies on $(M,MC).$ Since $\angle CND=\pi/2,$ hence this implies the lemma. $\square$
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Claim: Now, we claim that $X$ also lies on $(KEC).$
Proof: Indeed,
$\measuredangle KEX=\measuredangle KED=\measuredangle KDB=\text{arc}(BK)+\text{arc}(MC)=\text{arc}(KM)$ (where the last part since $M$ is the arc midpoint.) But also $\measuredangle KCX=\measuredangle KCM=\text{arc}(KM)$ Hence $\measuredangle KEX=\measuredangle KCX$ giving that $X$ lies on $(KEC).$ $\square$

To finish, see that $\measuredangle XQC=\measuredangle XEC$ by $(KEC).$ But also $\measuredangle XEC=\measuredangle DEC=\measuredangle CDE=\measuredangle QDX$ and so $XQ=XD.$ But $XB \perp BC$ as $X \in (M,MC)$ and so $B$ is the midpoint of $QD.$ $\blacksquare$

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#10 h Jan 26, 2020, 6:29 PM • 2 Y

Y by amar_04, Adventure10

I have a solution which involve inversion about point $K$ and a little bit of lengthy angle chase.

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#11 h Jan 27, 2020, 8:54 AM • 1 Y

Y by Adventure10

Shweta_16 wrote:

In triangle $\triangle{ABC}$ with incenter $I$ , the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$ , respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$ . The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$ . Prove that $FQ$ is parallel to $BI$ .

Proposed by Anant Mudgal

Solution-
Let tangent to $\omega$ at $K$ intersect $BC$ at $T$ and $AC$ at $R$ . Let $KE$ intersect $BC$ at $S$ .
In $\Delta TRC$ , $K,E,D$ are the incircle touch points and hence $R(T,C,S,D)=-1 \implies R(D,C,S,T)=1-(-1)=2$
Hence $\frac{DS}{SC}=2\frac{DT}{TC} \implies \frac{DS}{SC}(SD-SC)=2\frac{DT}{TC}(TC-TD)\implies \frac{SD^2}{CS}-DS=2(DT-\frac{TD^2}{CT})$
But $SD^2=SE\cdot SK=SC\cdot SQ \implies \frac{SD^2}{SC}=QS$ and $TD^2=TK^2=TB\cdot TC \implies \frac {TD^2}{CT}=BT$
So we have $QS-DS=2(DT-BT) \implies QD=2BD \implies BQ=BD=BF$ and hence we have $\angle QFD=90$
But $BI\perp FD \implies BI \parallel QF$
Hence proved !!!

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282 posts

#12 h Feb 1, 2020, 7:00 PM • 2 Y

Y by GeoMetrix, Adventure10

GeoMetrix wrote:

$[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 = -19.43752933627648, xmax = 32.95131572780579, ymin = -22.91737803351104, ymax = 11.67967094711988; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.88,-6.82)--(15.2,-6.7), linewidth(0.4) + rvwvcq); draw((15.2,-6.7)--(1.948187283488726,4.409082587313549), linewidth(0.4) + rvwvcq); draw(circle((5.47643671228076,-3.212986968017347), 3.574467648278019), linewidth(0.4) + dtsfsf); draw((1.88,-6.82)--(5.47643671228076,-3.212986968017347), linewidth(0.4) + wvvxds); draw(circle((6.747870334297736,-9.23935115462224), 8.825349861185424), linewidth(0.4) + wvvxds); draw(circle((8.521543675938103,-4.711348029129539), 6.968250536085029), linewidth(0.4) + sexdts); draw((2.2692084696477464,-1.63485327018204)--(1.90203496433293,-3.191281833747048), linewidth(0.4) + wrwrwr); draw((1.948187283488726,4.409082587313549)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((15.2,-6.7)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(7.772785268839476,-0.47371657695024716), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(5.5086378167960826,-6.787309569218054), linewidth(0.4) + wvvxds); draw((7.772785268839476,-0.47371657695024716)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); draw((-1.7486378167960828,-6.852690430781947)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((2.2692084696477464,-1.63485327018204)--(-1.7486378167960828,-6.852690430781947), linewidth(0.4)); draw((-1.7486378167960828,-6.852690430781947)--(7.772785268839476,-0.47371657695024716), linewidth(0.4)); draw((7.772785268839476,-0.47371657695024716)--(11.659998503978759,-16.571322036789724), linewidth(0.4)); draw((1.948187283488726,4.409082587313549)--(1.88,-6.82), linewidth(0.4) + rvwvcq); draw((2.2692084696477464,-1.63485327018204)--(11.659998503978759,-16.571322036789724), linewidth(0.4) + wvvxds); draw((-1.7486378167960828,-6.852690430781947)--(1.88,-6.82), linewidth(0.4) + wrwrwr); draw((2.2692084696477464,-1.63485327018204)--(4.336270300841782,-0.7498879530434238), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((1.948187283488726,4.409082587313549),dotstyle); label("$A$", (2.0963271088950033,4.739849912657041), NE * labelscalefactor); dot((1.88,-6.82),dotstyle); label("$B$", (2.0282896477728185,-6.486331172503435), NE * labelscalefactor); dot((15.2,-6.7),dotstyle); label("$C$", (15.329613297159941,-6.350256250259066), NE * labelscalefactor); dot((5.47643671228076,-3.212986968017347),linewidth(4pt) + dotstyle); label("$I$", (5.6002563566875185,-2.9483831941498306), NE * labelscalefactor); dot((5.5086378167960826,-6.787309569218054),linewidth(4pt) + dotstyle); label("$D$", (5.634275087248612,-6.5203499030645276), NE * labelscalefactor); dot((7.772785268839476,-0.47371657695024716),linewidth(4pt) + dotstyle); label("$E$", (7.913530034841801,-0.19286601870135017), NE * labelscalefactor); dot((1.90203496433293,-3.191281833747048),linewidth(4pt) + dotstyle); label("$F$", (2.0282896477728185,-2.914364463588738), NE * labelscalefactor); dot((-1.7486378167960828,-6.852690430781947),linewidth(4pt) + dotstyle); label("$Q$", (-1.6117145222640668,-6.588387364186712), NE * labelscalefactor); dot((2.2692084696477464,-1.63485327018204),linewidth(4pt) + dotstyle); label("$K$", (2.4024956839448346,-1.34950285777849), NE * labelscalefactor); dot((11.659998503978759,-16.571322036789724),linewidth(4pt) + dotstyle); label("$I_A$", (11.791665318806333,-16.283725574098032), NE * labelscalefactor); dot((4.336270300841782,-0.7498879530434238),linewidth(4pt) + dotstyle); label("$L$", (4.47763824817147,-0.465015863190089), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
IMO the title doesnt suit the prob.
Proof: Let $I_A$ be the $A$ -excentre. Its well known(ISL 2002 G7) that $K=I_AD \cap \omega$ . Now we'll use a phantom approach . Let $\ell$ be the line through $F$ parallel to $BI$ . and let $\ell \cap BC=Q'$ We just need to show that $Q' \in \odot(KEG)$ . Firstly define $\ell \cap AI=L$ . Notice that $\angle LQ'C=\angle IBC=\angle IKC=\angle LKC \implies LQ'I_AC$ cyclic. Now notice that $\Delta FAL \cong \Delta EAL \implies \angle AFL=\angle AEL$ . But notice that $\angle AEL=\angle AFL=\angle ABI=\angle LQ'C\implies LEQ'C$ cyclic. This combined with the fact that $LQ'I_AC$ is cyclic $\implies LQ'CI_AE$ is cyclic. So now we just need to show that $K\in \odot(LQ'CI_AE)$ But notice that $\angle I_AEQ'=\angle I_ACQ'=90-\frac{\angle C}{2}$ and $\angle Q'I_AE=\angle Q'CE=\angle C \implies I_AQ=I_AE$ . FInally $\angle EKD =\angle CDE=\angle I_ACD=\angle I_AEQ'=\angle I_AQ'E\implies K \in \odot(I_AQ'E)$ as desired. $\blacksquare$

Well there was no need of phantom point approach
Here's how you can finish it easily:
$\triangle AEI_A\equiv \triangle AFI_A\Rightarrow I_AE=I_AF\Rightarrow I_A$ is circumcenter of $QFE$ .Since $\angle QI_AE=C$ we get $\angle QFE=180-\frac {C}{2}$ and we are done.

This post has been edited 2 times. Last edited by mmathss, Feb 1, 2020, 7:01 PM

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