Spent half a hour trying to get a decent looking diagram

is a complex bash viable? i set it up but couldnt execute as i forgot general intersection formula; but it does seem feasible

GoodMorning wrote: is a complex bash viable? i set it up but couldnt execute as i forgot general intersection formula; but it does seem feasible One of my friends says that he got something with complex bashing but was unable to finish; but he's online right now and can probably give a better answer than I can

there's a nice projective solve by asdf

GoodMorning wrote: is a complex bash viable? i set it up but couldnt execute as i forgot general intersection formula; but it does seem feasible Yea I tried complex bash but it got very messy and I was running low on time. I feel like it should be possible but idk

What I got so far:

GoodMorning wrote: is a complex bash viable? i set it up but couldnt execute as i forgot general intersection formula; but it does seem feasible I was almost done with my complex bash, I never used general intersection.

Is $\angle DEF=90^\circ$ a typo?

Should be $\angle DFE = 90^\circ$

why does the problem say rays AB and AC? Shouldn't it say lines, since the lines don't intersect when ABC is acute?

If it said lines then XY may lie on the same side of extended FE and the result wouldn't hold in that case.

I hope you enjoyed this problem! My (author) solution: Let $X_1, Y_1$ be projections of $E$ onto lines $XD, YD$ correspondingly. Clearly, points $E, F, D, X_1, Y_1$ lie on a single circle with diameter $DE$. Then, $XX_1\times XD = XF \times XE = XB \times XA$, so points $A, B, D, X_1$ lie on a single circle. Similarly, points $A, C, D, Y_1$ lie on a single circle. Then, $\angle AX_1D = \angle ABD = \angle ACD = \angle AY_1D$. As $\angle EX_1D = \angle EY_1D$, it's easy to see that points $X_1, Y_1$ are symmetric with respect to the line $AD$. Therefore, $\angle XDE = \angle YDE$.

Let $D'$ be the harmonic conjugate of $D$ with respect to $BC$ and let $E'$ be the point on $XY$ such that $DE'\perp AE$. Recall that point $A$ is the harmonic conjugate of $B$ with respect to $CD$ if and only if there exists a point $E$ satisfying \[\measuredangle AEB=90^{\circ},\, \measuredangle CEB=\measuredangle BED.\]Since $DE$ bisects $\angle BEC$ we find that $D'E\perp AE$. If we project $(D',D;B,C)$ through $A$, we find that it suffices to show $A,D',E'$ are collinear. Draw $H$, the foot of the perpendicular from $E$ to $BC$. By Pascal's Theorem on $DDHFEE$ we only need to show that $A,H,F$ are collinear. But now \[\measuredangle HFE=\measuredangle HDE=\measuredangle BDE=\measuredangle ACE=\measuredangle AFE,\]done. $\blacksquare$

ihatemath123 wrote: why does the problem say rays AB and AC? Shouldn't it say lines, since the lines don't intersect when ABC is acute? this is what the test said word for word DottedCaculator wrote: Is $\angle DEF=90^\circ$ a typo? fixed, thanks for pointing it out

MS_Kekas wrote: I hope you enjoyed this problem! My (author) solution: Let $X_1, Y_1$ be projections of $E$ onto lines $XD, YD$ correspondingly. Clearly, points $E, F, D, X_1, Y_1$ lie on a single circle with diameter $DE$. Then, $XX_1\times XD = XF \times XE = XB \times XA$, so points $A, B, D, X_1$ lie on a single circle. Similarly, points $A, C, D, Y_1$ lie on a single circle. Then, $\angle AX_1D = \angle ABD = \angle ACD = \angle AY_1D$. As $\angle EX_1D = \angle EY_1D$, it's easy to see that points $X_1, Y_1$ are symmetric with respect to the line $AD$. Therefore, $\angle XDE = \angle YDE$. i feel so dumb for missing that EX1D and EY1D are right angles......................

um, this thread is a replica? the first one was here

they're two different problems

Too lazy to make proper asy... [asy][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 = -10.92, xmax = 17.88, ymin = -5.44, ymax = 8.36; /* image dimensions */ /* draw figures */ draw(circle((2.5,0.4827499410479508), 2.5461829285386783), linewidth(0.5)); draw((0,0)--(5,0), linewidth(0.5)); draw((2.5,3.028932869586629)--(2.0919201086693024,-2.0305185101671848), linewidth(0.5)); draw(circle((2.173807818499841,-1.0152592550835924), 1.0185563077484627), linewidth(0.5)); draw((2.5,3.028932869586629)--(-1.101921186355234,-1.3350581204181047), linewidth(0.5)); draw((2.5,3.028932869586629)--(7.680311326524383,-3.247393231054019), linewidth(0.5)); draw((7.680311326524383,-3.247393231054019)--(-1.101921186355234,-1.3350581204181047), linewidth(0.5)); draw((-1.101921186355234,-1.3350581204181047)--(2.2556955283303792,0), linewidth(0.5)); draw((2.2556955283303792,0)--(7.680311326524383,-3.247393231054019), linewidth(0.5)); draw(circle((1.1278477641651896,1.6152876159602325), 1.9700747857395942), linewidth(0.5) + linetype("4 4")); draw(circle((3.62784776416519,1.4136452536263968), 1.9700747857395946), linewidth(0.5) + linetype("4 4")); draw((0,3.230575231920466)--(5,2.827290507252801), linewidth(0.5)); draw((5,2.827290507252801)--(2.0919201086693024,-2.0305185101671848), linewidth(0.5)); draw((2.0919201086693024,-2.0305185101671848)--(0,3.230575231920466), linewidth(0.5)); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$C$", (0.08,0.16), NE * labelscalefactor); dot((5,0),linewidth(4pt) + dotstyle); label("$B$", (5.08,0.16), NE * labelscalefactor); dot((2.5,3.028932869586629),linewidth(4pt) + dotstyle); label("$A$", (2.58,3.18), NE * labelscalefactor); dot((2.0919201086693024,-2.0305185101671848),linewidth(4pt) + dotstyle); label("$E$", (2.18,-1.88), NE * labelscalefactor); dot((2.2556955283303792,0),linewidth(4pt) + dotstyle); label("$D$", (2.34,0.16), NE * labelscalefactor); dot((1.8261507266395756,-1.9726471067578473),linewidth(4pt) + dotstyle); label("$F$", (1.9,-1.82), NE * labelscalefactor); dot((7.680311326524383,-3.247393231054019),linewidth(4pt) + dotstyle); label("$X$", (7.76,-3.08), NE * labelscalefactor); dot((-1.101921186355234,-1.3350581204181047),linewidth(4pt) + dotstyle); label("$Y$", (-1.02,-1.18), NE * labelscalefactor); dot((1.4171238725063378,-0.3334335017346284),linewidth(4pt) + dotstyle); label("$L$", (1.5,-0.18), NE * labelscalefactor); dot((3.029987177769488,-0.46352212853255526),linewidth(4pt) + dotstyle); label("$K$", (3.1,-0.3), NE * labelscalefactor); dot((0,3.230575231920466),linewidth(4pt) + dotstyle); label("$N$", (0.08,3.4), NE * labelscalefactor); dot((5,2.827290507252801),linewidth(4pt) + dotstyle); label("$M$", (5.08,2.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $K = DX\cap (DEF)$, $L = DY\cap DEF$, $M = EK \cap (ABD)$, $N = EL \cap (ACD)$. By radical axis, $ABKD$ and $ACLD$ are cyclic. Nnote that $(DEF)$ has diameter $DE$. Then we have $$\measuredangle NAD + \measuredangle DAM = \measuredangle NLD + \measuredangle DKM = 180^{\circ},$$implying that $M$, $A$, $N$ are collinear. By LoS on $\triangle ABD$ and $\triangle ACD$, $(ABD)$ and $(ACD)$ have equal radii, and so $AD$ is the perpendicular bisector of $MN$. Thus $EM = EN$, and PoP wrt $(ABD)$ and $(ACD)$ implies that $EK = EL$, which finishes.

OMG I'M SO DUMB OK SRRYYYY DON'T MIND THAT

Perform inversion with center $D$. We get the following problem. Let $\triangle{}ABC$ be a triangle and circumcircle $\omega$ and let $D$ and $E$ be the intersections of the angle bisector of $\angle{}CAB$ with $\overline{BC}$ and $\omega$ and let $F$ be the antipode of $A$ with respect to $\omega$. Let the circumcircle of $\triangle{}DEF$ intersect the circumcircles of $\triangle{}ABD$ and $\triangle{}ACD$ at $X$ and $Y$, respectively. Prove that $\angle{}XDE=\angle{}EDY$. By radical center we see that $\overline{AB},\overline{EF},$ and $\overline{DX}$ concur at a point $P$ and that $\overline{AC},\overline{EF},$ and $\overline{DY}$ concur at a point $Q$. We see that $\triangle{}APQ$ is isosceles with axis of symmetry $\overline{AE}$ since $\overline{EF}\perp\overline{AE}$ and therefore $\angle{}XDE=\angle{}PDE=\angle{}QDE=\angle{}YDE$.

Let $\ell$ be the line through $D$ perpendicular to $\overline{DE}$. Let $Z=\ell\cap \overline{EF}$ and let $\overline{AZ}$ intersect $\omega$ and $\overline{BC}$ at $W$, $Z'$ respectively. Notice $\ell$ is tangent to $(DFE)$ at $D$ so \[ZD^2=ZF\cdot ZE=ZW\cdot ZA\]so $\triangle ZDW\sim\triangle ZAD$ so $\angle DWZ=90$. Then, \[90-\angle DWC=\angle CWA=\angle B=\angle C=180-\angle BWA=90-\angle BWD\]so $\overline{WD}$ bisects $\angle BWD$. By right angles on the bisector, \[-1=(Z'D;BC)\stackrel{A}=(ZE;XY)\]so again by right angles on the bisector we have that $\overline{DE}$ bisects $\angle XDY$. $\square$

Construct $P$ on line $FE$ such that $\angle PDA = 90^{\circ}$, and suppose $AP$ meets $\omega$ at $Q \neq A$. Let $L$ be the antipode of $A$ on $\omega$. Note that $PQ \cdot PA = PF \cdot PE = PD^2$, by Power of a Point and by the similarity of triangles $PDF$ and $PED$. Thus $\angle PQD = 90^{\circ}$, so $QD$ passes through $L$. Thus $(A, L; B, C) = (E, Q; C, B) = (E, P; Y, X) = -1$, and as $\angle PDE = 90^{\circ}$ we know $DE$ bisects $\angle XDY$, as desired.

... Clearly $(ABD)$ and $(ACD)$ are reflections across $AD$, as $2R=\tfrac{AB}{\sin\angle ADB}=\tfrac{AC}{\sin\angle ADC}$. Since $(DEF)$ is symmetric across $AD$, it suffices to show that $X$ lies on the radical axis of $(ABD)$ and $(DEF)$, and similarly for $Y$. But $X$ is in fact the radical center of $(ABC)$, $(ABD)$, and $(DEF)$, done. $\square$

imo way easier than j3 (i still cannot solve that one :sob:) prob only because im decent at radax Construct $(ADB)$ and $(ADC)$. We claim they have equal radii, hence being reflections of each other over line $\overline{AD}$. Let $R_1$ be the circumradius of $\triangle ADB$ and $R_2$ be the circumradius of $\triangle ADC$. We have \[2R_1=\frac{AB}{\sin \angle ADB} = \frac{AC}{\sin \angle ADC} = 2R_2,\] as needed. Now, notice that \[XA \cdot XB = XE \cdot XF\]\[YA \cdot YC = YE \cdot YF.\] Thus, $\overline{DX}$ is the radical axis of $(ADB)$ and $(DEF)$ while $\overline{DY}$ is the radical axis of $(ADC)$ and $(DEF)$. Thus, $\overline{DX}$ and $\overline{DY}$ are reflections with respect to $\overline{AD}$, which translates to the desired condition.

Quick projective solution: Let $P = \overline{DD} \cap \overline{EF}$ and $K = \overline{AP} \cap (ABC)$. By radical axis it follows that $\overline{PD}$ is tangent to $(AKD)$, hence $\angle AKD = 90^\circ$. So $\overline{KD}$ passes through the minor arc midpoint $M$ of $\widehat{BC}$, hence \[-1=(AM;BC) \stackrel D= (KE;BC) \stackrel A=(PE;XY).\]Bisector bundle lemma finishes. @below I had $G$ on my original diagram and forgot to change it

HamstPan38825 wrote: Quick projective solution: Let $P = \overline{DD} \cap \overline{EF}$ and $K = \overline{AG} \cap (ABC)$. By radical axis it follows that $\overline{GD}$ is tangent to $(AKD)$, hence $\angle AKD = 90^\circ$. So $\overline{KD}$ passes through the minor arc midpoint $M$ of $\widehat{BC}$, hence \[-1=(AM;BC) \stackrel D= (KE;BC) \stackrel A=(PE;XY).\]Bisector bundle lemma finishes. what is $G$?

ratio lemma lol The problem statement avoids config issues. Let $A'$ be the antipode of $E$ in $(ABC)$. This is collinear with $D,F$ from right angles. Let the parallel line to $AA'$ through $D$ intersect $EF$ at $P$. Let $Q$ be the intersection of the parallel line to $AA'$ through $E$ with $BC$. By Fact 5 and Appolonian circles, $(Q,B,D,C)=-1$. It suffices to prove that $Q,P,A$ are collinear, as then $-1=(Q,B,D,C)\overset{A}=(P,X,E,Y)$ and Appolonian circles finishes. We claim that $AQ$ is divided by $DP$ and $EF$ in the same ratio. Let $\theta=\angle BDP,\alpha=\angle ADA'$.The first ratio, by the ratio lemma on $\triangle QDA$, is \[\frac{DQ\times \sin\theta}{AD\times\sin 90^\circ}\]and the second ratio, by ratio lemma on $\triangle QEA$, is \[\frac{EQ\times \sin\alpha }{EA\times \sin{90^\circ-\alpha}}\]We equate the two, if equality holds working backwards we would be done. \begin{align*} \frac{DQ\times \sin\theta}{AD\times\sin 90^\circ}&=\frac{EQ\times \sin\alpha }{EA\times \sin{90^\circ-\alpha}}\\ \frac{DQ}{EQ}\times \frac{\sin{\theta}}{AD}&=\frac{\sin\alpha}{EA\times \cos\alpha}\\ \tan{\theta}\times\frac{EA}{AD}&=\tan{\alpha}=\frac{AA'}{AD}\\ \tan{\theta}&=\frac{AA'}{EA}=\tan{\angle AEA'} \end{align*}This is true if $\angle BDP=\theta=\angle AEA'$, taking the complement, by Thales we want \[\angle ADC=\angle EA'A=\angle ECA\]But this is true clearly, by Shooting Lemma.

This problem is so unique as symmetry plays a substantial role, but is also why this problem is so interesting Let $DX$ and $DY$ intersect $(DFE)$ at $P$ and $Q$ respectively. Since $\overline{AB}$, $\overline{PD}$, and $\overline{EF}$ are concurrent at $X$, Inverse Radical Axis implies $ADPB$ is cyclic. Similarly, $ADQC$ is as well. Now note that \[ \text{radius}[(ADPB)] = \dfrac{AD}{2 \sin(\angle ABC)} = \dfrac{AD}{2 \sin(\angle ACB)} = \text{radius}[(ADQC)]\]so $(ADPB)$ and $(ADQC)$ are symmetric about $\overline{AD}$. Since $(DFE)$ is centered at a point on this line, $P$ and $Q$ are symmetric with each other about $AD$ implying $\angle PAD = \angle QAD \implies \angle PBD = \angle QCD$. Then \[\angle XDE = \angle ABP = \angle ABC + \angle PBD = \angle ACB + \angle QCD = \angle ACQ = \angle EDY \]as desired.

Seeing the angle bisector, it is motivated to draw the intersection of the line through $D$ perpendicular to $DE$ and $EF$, we call it $T$. Now it suffices to show that $(XY;ET)=-1$. Let $AT \cap \omega = Q$ . Notice that $DT$ is tangent to the circle with diameter $DE$ so $TD^2=TE.TF=TQ.TA$. Since, $\Delta ADT$ is right angled at $D$, the length condition implies $\angle AQD=90°$ so $QD$ passes through midpoint of arc $BC$ not equal to $A$, say $M$. Now:- $$ (XY;ET) \stackrel{A}{=} (BC;QE) \stackrel{D}{=} (CB;MA)=-1$$which is what we needed to show.

Suppose the line perpendicular to $AE$ at $A$ meets $XY$ at $K$, and let $K' = AK \cap (ABC)$. Notice that Apollonian tells us it suffices to show $(XY;EK) = -1$. Projecting tells us \[(XY;KE) \overset{A}{=} (BC;EK') \overset{D}{=} (C,B;A,DK' \cap (ABC)),\] so we need $DK' \cap (ABC)$ to be the midpoint of arc $BC$, or $\angle AK'D = 90$. However, this holds true from right triangles $ADK$, $EDK$, and \[KD^2 = KE \cdot KF = KA \cdot KK'. \quad \blacksquare\]

Let $\omega$ be the circle with diameter $DE$, and denote $P, Q = DX, DY \cap \omega$. Notice $ABPD$ and $ACQD$ are cyclic, as \begin{align*} &XE \cdot XF = XA \cdot XB = XP \cdot XD \\ &YE \cdot YF = YA \cdot YC = YQ \cdot YD. \end{align*} Law of Sines using the isosceles condition tells us the two circumcircles are congruent. Thus the figure consisting of $(ABPD)$, $(ACQD)$, and $\omega$ is symmetric about $AD$, so $PD$, $QD$ are symmetric about $AD$, as desired. $\blacksquare$

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