A circle centered at $A$ intersects sides $AC$ and $AB$ of $\triangle ABC$ at $E$ and $F$, and the circumcircle of $\triangle ABC$ at $X$ and $Y$. Let $D$ be the point on $BC$ such that $AD$, $BE$, $CF$ concur. Let $P=XE\cap YF$ and $Q=XF\cap YE$. Prove that the foot of the perpendicular from $D$ to $EF$ lies on $PQ$.
Problem
Source: 2023 Serbia TST Problem 2
Tags: geometry, Serbia, TST, circumcircle
VZH
22.05.2023 17:10
Let $PQ \cap EF = H$, $XY \cap EF = K$. Since $((EF \cap BC), D; B, C) = -1$, all we need to prove is that $HD$ bisects $\angle BHC$, or $\frac{FH}{HE}=\frac{BF}{CE}$. Note that $\frac{FH}{HE}=\frac{FK}{KE}$ due to polar/harmonic stuff. Denote $Pow(T, \odot A) - Pow(T, \odot (ABC))$ by $f(T)$, then by Linearity of PoP we get $\frac{FK}{EK}=\frac{f(F)-f(K)}{f(E)-f(K)}=\frac{AF\cdot FB}{AE\cdot EC}=\frac {BF}{CE}$, which concludes our proof.
Paramizo_Dicrominique
22.05.2023 17:18
here is an extra property for anyone interested: $\textbf{Extra:}$ $H'$ is the orthocenter of $\triangle{DEF}$. The perpendiculars from $F,E$ to $AB,AC$ intersects at $W$. $O$ is the circumcenter of $\triangle{ABC}$. Prove that $\overline{O,W,H'}$.
gvole
22.05.2023 17:24
Here's my solution from contest. I introduced no new points so someone else can provide a suitable diagram.
First of all it is well known from the Sharky-Devil configuration that $\frac{HF}{HE}=\frac{DB}{DC}$.
$\textbf{Proof:}$ $ \mathcal{H}(EF\cap BC,D;B,C)\implies\mathcal{H}(HF,HD;HB,HC)$ which together with $HD\perp HF$ gives that $HD$ is the bisector of $\angle BHC$, the claim follows from the fact that $\triangle HFB\sim\triangle HEC$
Menelaus on $\triangle XFE$ says we want to prove: \[\frac{HF}{HE}\cdot\frac{PE}{PX}\cdot\frac{QX}{QF}=-1
\iff \frac{DB}{DC}\cdot(\frac{PY}{PX}\cdot\frac{EF}{XY})\cdot(\frac{QX}{QY}\cdot\frac{XY}{EF})=-1 \iff \frac{DB}{DC}\cdot\frac{PY}{PX}\cdot\frac{QX}{QY}=-1\]\[\iff \frac{DB}{DC}\cdot\frac{\sin{\angle \frac{YAC}2}}{{\sin{\angle \frac{XAB}2}}}\cdot\frac{\sin{\angle \frac{XAC}2}}{{\sin{\angle \frac{YAB}2}}}=-1\]
By trivial angle chasing on the two given circles. The obvious Ceva gives $\frac{DB}{DC}=-\frac{FB}{EC}$, so we just want to prove: \[\frac{FB}{EC}\cdot\frac{\sin{\angle \frac{YAC}2}}{{\sin{\angle \frac{XAB}2}}}\cdot\frac{\sin{\angle \frac{XAC}2}}{{\sin{\angle \frac{YAB}2}}}=1\].
Notice it is enough for us to prove that if we fix $A$, $X$, $Y$ and define $E$ as the intersection of ray $AC$ with the circle centered at $A$ through $X$ and $Y$, then $\frac{EC}{\sin{\frac{\angle XAC}2}\sin{\frac{\angle YAC}2}}$ is constant, since the Menelaus boils down to this ratio for $C$ being equal to the same ratio for $B$.
To prove this fact we perform inversion around $(XEY)$ and let $C'$ be the image of $C$. Using inversion distance formulas, we get that what we want to prove is $\frac{C'E}{C'A\sin{\frac{\angle XAC}2}\sin{\frac{\angle YAC}2}}$ being constant, but this is trivial using sine theorem on $\triangle C'XA$ and $\triangle C'XE$ since $C'$ is on $XY$ and trivial angle chasing (notice there are 5 points in the diagram now so this must not be hard lol).
VicKmath7
22.05.2023 18:05
Let $S$ be the Miquel point of the degenerate quadrilateral $FYEX$ (i.e. $S=(XFP) \cap (PEY)$) and let $XY \cap EF=R$. Then we know that $S=PQ \cap AR$ and $\angle ASP=90^{o}$ since $A$ is the circumcenter of this cyclic quadrilateral.
Notice that $\angle XSP=\angle PFQ=\angle PEQ=\angle PSY$, so $PS$ is interior angle bisector of $\angle XSY$, i.e. $AS$ is exterior angle bisector and since $AX=AY$, we obtain that $S \in (ABC)$. Similarly, $\angle PSF=\angle PSE$ and $AF=AE$, which combined with $\angle ASP=90^{o}$ again implies that $S \in (AFE)$.
Let $D'$ be the foot of the perpendicular from $D$ to $EF$. We will prove that $SD'$ is angle bisector in $\triangle SEF$, which will suffice as $SP$ is angle bisector of $\angle ESF$. If $EF \cap BC=V$, then $(VD;BC)=-1$, which means $D'D$ is angle bisector of $\angle BD'C$ and that $\triangle BD'F \sim \triangle CD'E$. Now, we are only left to see that by Ceva and spiral similarity using the above two facts we have $\frac{D'F} {D'E}=\frac{D'B}{D'C}=\frac{DB}{DC}=\frac {BF} {CE}=\frac {SF} {SE}$, which proves that $SD'$ is angle bisector in $\triangle SEF$, done.
Orestis_Lignos
23.05.2023 17:29
Let $K$ be the projection of $D$ on $EF$ and let $PQ$ intersect $EF$ at point $K'$. We have the following two Claims: Claim 1: $K$ divides $FE$ at ratio $BF/CE$. Proof: Let $FE$ intersect $BC$ at point $S$. Note that $(S,B,D,C)=-1$ in light of the complete quadrilateral $AFDE.BC$, and since $\angle SKD=90^\circ,$ we obtain that $\angle BKD=\angle DKC$. Therefore, $\angle FKB=90^\circ-\angle BKD=90^\circ-\angle DKC=\angle EKC$. Since $AF=AE$, we obtain that $\angle AFE=\angle AEF,$ hence triangles $FKB$ and $EKC$ are similar. Thus, $\dfrac{FK}{KE}=\dfrac{BF}{EC},$ as desired $\blacksquare$ Claim 2: $K'$ divides $FE$ at ratio $BF/CE$. Proof: Note that by repeated applications of the Ratio Lemma and the LoS, $\dfrac{FK'}{K'E}=\dfrac{FX \cdot FY}{EX \cdot EY}$ Let $XF$ and $YE$ intersect $(ABC)$ again at points $Z,W,$ respectively. Note that $\dfrac{FK'}{K'E}=\dfrac{FX \cdot FY}{EX \cdot EY}=\dfrac{\dfrac{FA \cdot FB}{FZ} \cdot FY}{\dfrac{EA \cdot EC}{EW} \cdot EX}=\dfrac{FB}{EC} \cdot \dfrac{FY/FZ}{EC/EW},$ and so in order to prove that the initial ratio is equal to $FB/EC$ we are left to prove that $\dfrac{FY}{FZ}=\dfrac{EC}{EW}$. However, this is easy since triangles $FYZ$ and $XEW$ are similar: indeed, $\angle YFZ=180^\circ-\angle XFY=180^\circ-\angle XEY=\angle XEW$ and $\angle FZY=\angle XZY=\angle XWY=\angle XWE$ $\blacksquare$ To the problem, combining both Claims we may finish.
gambi
28.08.2023 01:57
Let $\omega$ be the circle centered at $A$. Let $K$ be the second intersection point of $(ABC)$ and $(AEF)$. Let line $EF$ meet lines $BC$ and $XY$ at points $W$ and $L$, respectively. Let $N$ be the midpoint of arc $\overarc{BAC}$ in $(ABC)$, and let $M$ be the midpoint of arc $\overarc{BC}$. Let $A'$ be the antipode of $A$ in $(ABC)$. Let $J$ be the projection of $D$ onto $EF$. 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label("$L$", (-1.0525454545454453,-0.9918181818181839), NE * labelscalefactor); dot((-6.481593653497281,-2.4515228454067537),linewidth(2.pt) + dotstyle); label("$W$", (-7.072545454545445,-2.5918181818181845), NE * labelscalefactor); dot((4.069971067584266,2.720963719048588),linewidth(2.pt) + dotstyle); label("$K$", (3.567454545454554,2.8081818181818177), NE * labelscalefactor); dot((10.327432159366568,-5.014021623156887),linewidth(2.pt) + dotstyle); label("$A'$", (10.587454545454554,-5.311818181818185), NE * labelscalefactor); dot((8.209830431192039,4.750365755906017),linewidth(2.pt) + dotstyle); label("$N$", (8.387454545454553,4.948181818181818), NE * labelscalefactor); dot((8.09868362148783,-5.497370098822488),linewidth(2.pt) + dotstyle); label("$M$", (7.9274545454545535,-5.891818181818186), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The radical axes of $(ABC)$, $(AEF)$, $\omega$ concur, so $L-K-A$ are collinear. Also, since $K$ is the center of the Spiral Similarity mapping $BC$ to $FE$, then $WBFK$ is cyclic, meaning that $$ \measuredangle WKB=\measuredangle WFB=\measuredangle EFA=90^o-\frac{\measuredangle FA}{2}=90^o-\frac{\measuredangle BAC}{2}=\measuredangle ACB=180^o-\measuredangle BKA, $$so $W-K-N$ are collinear. By Brocard on $XFEY$ we get that $PQ$ is the polar of $L$ wrt $\omega$. Also, inversion around $\omega$ maps $EF$ to $(AEF)$, so $K$ also belongs to the polar of $L$ wrt $\omega$. Combining these two results, we get that $K-P-Q$ are collinear. Indeed, since $PQ$ is the polar of $L$ wrt $\omega$, line $K-P-Q$ is perpendicular to $LK$. Since $AD,BE,CF$ concur, then $$ -1=(B,C;D,W)\stackrel{(K)}=(B,C;KD\cap (ABC),N), $$meaning that $K-D-M$ concur. This way, points $D,J,K,W$ all lie on $(\overline{DW})$, and so a Spiral Similarity centered at $K$ maps $BDCN$ to $FJEA$, meaning that $$ \measuredangle DKJ=\measuredangle NKA=\measuredangle MKA'=\measuredangle DKA', $$so $K-J-A'$ are collinear, which implies that $KJ\perp KL$. Therefore points $K-P-Q-A'-J$ are all collinear in the line through $K$ perpendicular to $KL$.
SerdarBozdag
30.08.2023 23:19
Definitions: Internal and external angle bisectors of $A$ cuts $(ABC)$ at $M$ and $J$. $K=PQ \cap EF$, $H=EF \cap YX$,$X=BE \cap CF$,$G=EF \cap BC$. By radical axis theorem $H \in AI$. By Miquel Configuration $I \in P-K-Q$. Also $\frac{IB}{IC}=\frac{BF}{CE}=\frac{BD}{CE} \implies D \in IM $ and $\frac{GB}{GC}=\frac{BD}{DC}=\frac{IB}{IC} \implies I \in GJ$. We have $AH \perp PQ$ by Brocard Theorem. Thus both $\angle AIQ=90,\angle GID=90$ are true. In the end, $\angle KID=\angle AIJ=\angle B-(90^{\circ}-A/2)=\angle ABC -\angle GFB=\angle KGD$ which means $90^{\circ}=\angle GKD=\angle GID$
math_comb01
04.12.2023 17:34
too tired to write a full solution for this so here's sketch using Angle chase, harmonics and spiral [asy][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 = -10.37083874314123, xmax = 16.048545081897835, ymin = -10.470258223270344, ymax = 2.9923438436861827; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); draw((4.297950214207411,2.291977262168213)--(1.98,-6.47)--(13.6,-5.05)--cycle, linewidth(0.4)); /* draw figures */ draw((4.297950214207411,2.291977262168213)--(1.98,-6.47), linewidth(0.4)); draw((1.98,-6.47)--(13.6,-5.05), linewidth(0.4)); draw((13.6,-5.05)--(4.297950214207411,2.291977262168213), linewidth(0.4)); draw(circle((4.297950214207411,2.291977262168213), 7.413542023754758), linewidth(0.8)); 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dot((4.493230936439608,-4.177340243933741),linewidth(4pt) + dotstyle); label("$G$", (4.570314995908845,-4.030776598757901), NE * labelscalefactor); dot((-4.898737270210484,-7.310603005481832),linewidth(4pt) + dotstyle); label("$H$", (-4.826269972790616,-7.1629715883243765), NE * labelscalefactor); dot((1.2696489199740073,-1.7799626002833075),linewidth(4pt) + dotstyle); label("$R$", (1.3408468700204175,-1.6184028179737837), NE * labelscalefactor); dot((4.637288516684896,-8.949722356794856),linewidth(4pt) + dotstyle); label("$I$", (4.706497386759562,-8.797160278532973), NE * labelscalefactor); dot((7.103911380113259,-6.118903045317295),linewidth(4pt) + dotstyle); label("$J$", (7.177235049336852,-5.956784697932318), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Define $R$ as miquel point of $FPEQ$ Claim 1: $R-P-Q$ Claim 2:$R$ lies on $(ABC)$ CLaim 3: $\frac{RF}{RE}=\frac{GF}{GE}$ and then use appolonian or harmonic to finish.
shendrew7
18.06.2024 06:55
Let $H := PQ \cap EF$ lie on the polar of $R := EF \cap XY$ with respect to $(EFXY)$. This tells us \[\frac{HF}{HE} = \frac{RF}{RE}.\] Consider the function $f(X) = \operatorname{pow}(X,(ABC)) - \operatorname{pow}(X,(EFXY))$. Since $E$, $F$, $R$ collinear and $f(R)=0$, we have \[\frac{RF}{RE} = \frac{f(F)}{f(E)} = \frac{FA \cdot FB}{EA \cdot EC} = \frac{FB}{EC}.\] Combining these two ratios and $\angle BFH = \angle CEH$, we find \[\triangle BHF \sim \triangle CHE \implies HD \text{ bisects } \angle BHC,\] from which we finish by Apollonius on the harmonic bundle $(GD;BC)$, where $G = EF \cap BC$. $\blacksquare$
cursed_tangent1434
18.06.2024 14:19
Solved with Om245. Really enjoyed this problem but it felt kind of bland since it was simply a generalization of the Intouch-Miquel Configuration. In fact, in addition to the starring result, most claims of the Intouch-Miquel Configuration hold in this generalized version as well. The only block here is that the trusty inversion at $I$ which is a key feature of the traditional proof of the Intouch-Miquel Configuration is not a viable option here. We start off with defining some points. Let $I$ be the minor arc midpoint of $EF$ in circle $(AEF)$ and $M$ the minor arc midpoint of $BC$ in circle $(ABC)$. Let $S = (AEF) \cap (ABC)$, $X_D = \overline{XY} \cap \overline{EF}$ and let $R$ be the foot of the altitude from $D$ to $EF$. Finally, let $\omega$ be the circle centered at $A$ and passing through points $E$ and $F$. Now, we get rid of the annoying line $\overline{PQ}$. Claim : Points $P$ , $Q$ and $I$ are collinear. Proof : First note that, \[2\measuredangle EFI = 2\measuredangle EAI = \measuredangle EAF = 2\measuredangle EXF\]from which it is clear that $IE$ is a tangent to $\omega$. SImilarly we can show that $IF$ is a tangent to $\omega$. Thus, $I$ is the intersection of the tangents to $\omega$ at $E$ and $F$. Thus, $\overline{EF}$ is the polar of $I$ with respect to $\omega$. So, $X_D$ must lie on the polar of $I$ with respect to $\omega$ and via La Hire's it follows that $I$ lies on the polar of $X_D$ with respect to $\omega$ which due to Brokard's Theorem, is simply $\overline{PQ}$. This proves the claim. Now, note that by Radical Center on circles $(ABC)$ , $(AEF)$ and $\omega$, line $\overline{AS}, \overline{XY}$ and $\overline{EF}$ must concur, at $X_D$. Now, clearly $IS \perp AX_D$ since $AI$ is the diameter of circle $(AEF)$ (simply note that since $AE=AF$ triangles $\triangle AEI$ and $\triangle AFI$ are congruent). Further, by Brokard's Theorem on cyclic quadrilateral $XYEF$, it follows that $A$ is the orthocenter of $\triangle X_DPQ$ and thus, $PQ \perp AX_D$. Combining this with the previous claim, it follows that $S$ also lies on $\overline{PQ}$. Now, it suffices to show the following claim. Claim : Points $S$ , $R$ and $I$ are collinear. Proof : First, note that due to the given concurrency condition, applying Ceva's Theorem on $\triangle ABC$ we have \[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1\]which since $AF=AE$ implies, \[\frac{BD}{DC} = \frac{FB}{CE}\]Now, it is easy to see that $S$ is the center of spiral similarity mapping $EF \mapsto BC$, so $\triangle SFB \sim \triangle SEC$. Thus, \[\frac{BD}{DC} = \frac{FB}{CE}= \frac{SB}{SC}\]which due to the converse of the angle bisector theorem, implies that $\overline{SD}$ is the $S-$angle bisector of $\triangle SBC$. Thus, points $S$ , $D$ and $M$ must be collinear. Now, let $R' = \overline{DR} \cap \overline{SI}$. Now, it is clear that $\triangle SR'D \sim \triangle SIM$. But, this means that there exists a spiral similarity centered at $S$ mapping $R'I \mapsto DM$. We already know that there exists a spiral similarity centered at $S$ mapping $EF \mapsto BC$ and since points $D$ and $M$ lie on $\overline{BC}$ and $I$ lies on $\overline{EF}$, it follows that $R'$ must lie on $\overline{EF}$ as well, and thus $R'=R$ which proves the claim. Combining the two claims, it follows that the foot of the perpendicular from $D$ to $EF$ lies on $PQ$, as desired.
bin_sherlo
07.07.2024 02:03
Let $(ADE)\cap (ABC)=\{A,S\}, DE\cap XY=T$. Let $K$ be the altitude from $D$ to $EF$. $AB\cap XY=M,AC\cap XY=N$. We will prove that $(T,K;F,E)=-1\iff FN,EM,AK$ are concurrent. Let's prove that $FKE\sim BDC$. Let $L\in EF$ such that $FLE\sim BDC$. Since $S$ is the miquel point of $BCEF$, $S$ is the center of the spiral similarity mapping $FLE$ to $BDC$. \[\frac{BF}{CE}=\frac{DB}{DC}=\frac{LF}{LE}\]And $\angle LFB=\angle CEL$ thus $BFL\sim CEL$. \[\frac{LB}{LC}=\frac{BF}{CE}=\frac{DB}{DC}\]Hence $\angle DLB=\angle CLD$. Also $\angle BLF=\angle ELC$. These two yield $\angle DLF=\angle ELD\iff DL\perp EF\iff K=L$. We have $FKE\sim BDC$. Also $AE^2=AF^2=AY^2=AM.AB=AN.AC$ since $\angle YBA=\angle YXA=\angle AYX=\angle ACX$. Thus, $AMF\sim AFB$ and $ANE\sim AEC$. \[\frac{MA}{MF}.\frac{KF}{KE}.\frac{NE}{NA}=\frac{FA}{FB}.\frac{DB}{DC}.\frac{EA}{EC}=1\]As desired.$\blacksquare$