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Source: 2021 ISL G4

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419 posts

#42 h Apr 28, 2024, 7:41 PM

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Make the problem $A-$ centric, because why not.

For clarity, we solve the following problem:

amuthup wrote:

Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $AB$ and $AC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{BD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $BC,DT,$ and $AK$ are concurrent.

All angles are directed.
Define $X=TE\cap BC$ , $Y=TF\cap BC$ and $A\ne Z=(AXB)\cap (AYC)$ .

Claim 1: $BXDE$ and $CYDF$ are cyclic.
Proof. $\angle FDC=\angle DBC=\angle DBX$ because of tangency condition. Since $ET\parallel DC$ , $\angle FDC=\angle DET=\angle DEX$ . Hence, $\angle DBX=\angle FDC=\angle DEX$ . So, $BXDE$ is cyclic. Similarly, $CYDF$ is cyclic.

Claim 2: $TZXDY$ is cyclic.
Proof. We have
$\angle YDX=180^\circ-\angle FDY-\angle XDE=\angle EBX+\angle YCF-180^\circ=\angle BAC$ Since $TE\parallel CD$ and $TF\parallel BD$ . So, $\angle XTY=\angle CDB=180^\circ-\angle BAC$ . So, $\angle XTY+\angle YDX=180^\circ$ which means $TXDY$ is cyclic.
Note that $\angle AZX=180^\circ-\angle CBA$ and $\angle YZA=180^\circ-\angle ACB$ . So,
$\angle XZY=180^\circ-\angle AZX+180^\circ-\angle YZA=\angle CBA+\angle ACB=180^\circ-\angle BAC=\angle XTY,$ and it follows that $Z$ lies on $XYT$ .

Claim 3: $TD$ is the common tangent of circumcircles $(BDE)$ and $(CDF)$ .
Proof. Note,
$\angle TDX=\angle TYX=\angle FYC=\angle FDC=\angle FEX=\angle DEX$ which means $TD$ is tangent to $(BDE)$ . Similarly, we get $TD$ is tangent to $(CDF)$ .

We now involve $K$ into the diagram (aka mess).

Claim 4: $K$ lies on circumcircle $(TZXDY)$
Proof. $TD$ is tangent to $(BDE)$ which means $\angle TDE=180^\circ-\angle EXD$ . So,
$\angle TKD=\angle KDT=\angle EXD=\angle EBD=\angle ACD=\angle TYD$ which means $TYKD$ is cyclic.

Claim 5: $A, Z, K$ are collinear.
Proof. We have,
$\angle AZX=\angle EBX=\angle KDX=180^\circ-\angle KZX$ and $\angle AZX+\angle KZX=180^\circ$ gives the result.

Claim 6: $AK, BC, TD$ are concurrent.
Proof. Let $J$ be the radical center of circles $(ABX), (BDE), (CDF)$ . Then, $J$ lies on lines $BX, TD$ . So, $J=TD\cap BC$ . Note that $BC$ is also the radical axis of $(ACY), (CYF)$ . Hence, $J=BC\cap TD$ is the radical center of $(BDE), (CDF), (ACY)$ as well. So, $J$ has equal power from all four circles. Hence, $J$ also lies on the radical axis of $(ABX), (ACY)$ which is $AK$ . So, $J$ lies on $TD$ , $BC$ and $AK$ , as desired.

The proof is complete. $\blacksquare$

Attachments:

This post has been edited 1 time. Last edited by Pyramix, Apr 28, 2024, 7:43 PM
Reason: attached diagram

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470 posts

#43 h May 1, 2024, 12:43 PM

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let point $G$ on $\Omega$ exist so that $BG$ and $EF$ are parallel, and let $BG$ and $AD$ intersect at $H$
angle chasing gives that $DBG=DGB=BDE=GDF=EAD=BCD=HAB$ , so $DTG$ are collinear with $DB=DG$ and $DT=KT$ , so triangles $BDW$ and $DTK$ are similar
more angle chasing yields that $TFK=HDE=DHB$ , so since $HBD=FKT$ , triangles $HBD$ and $FKT$ are similar
then, we can just map a homothety for our sets of similar triangles and we are done

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#44 h May 17, 2024, 3:03 AM • 1 Y

Y by JollyEggsBanana

I don't see why we can't move some points, fellas
(correct me if i'm wrong tho)

Attachments:

ISL2021G4 - Moving Points.pdf (338kb)

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896 posts

#45 h May 18, 2024, 1:26 PM

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Let $TE$ and $TF$ intersect $AC$ at $G$ and $H$ , respectively.

Claim: $EAGD$ is concyclic
$\angle GAD=\angle CAD=\angle TEF=\angle GED$ Claim: $FCHD$ is concyclic
$\angle HCD=\angle ACD=\angle TFE=\angle HFD$ Claim: $TGHD$ is concyclic
$180^{\circ}-\angle GDH=\angle GDE+\angle HDF=\angle BAC+\angle BCA=\angle ADC=\angle ETF=\angle GTH$ Claim: $K$ lies on $(TGHD)$
$\angle TKD=\angle TDK=\angle TDA+\angle ADE=\angle TGH+\angle DTH=\angle TGD$ Claim: $KGHT\sim ABCD$
$\angle GKE=\angle GHD=\angle CHD \Rightarrow GK|| BC$ $\angle HKF=\angle HGD=\angle AGD \Rightarrow HK|| BA$ As $GH$ lies on $AC$ the claim follows. Since $KGHT$ and $ABCD$ are homothetic about a $180^{\circ}$ angle lines $BK$ , $TD$ , and $AC$ concur.

Attachments:

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1481 posts

CT17

#46 h Jun 9, 2024, 2:41 PM • 1 Y

Y by MathLuis

Ratio lemma never fails! I claim that this is one of the most motivated solutions to this problem.

First, note that if $P = BD\cap (BEF)$ , then $\measuredangle PEF = \measuredangle DBC= \measuredangle FDC = \measuredangle FET$ so $T$ is the reflection of $P$ over $EF$ by symmetry. In particular, $PDTK$ is a rhombus. Now we can directly finish from here. Let $w,x,y,z = \frac{1}{2}\widehat{BA}, \frac{1}{2}\widehat{BC}, \frac{1}{2}\widehat{DA}, \frac{1}{2}\widehat{DC}$ . By angle chase $\angle FPK = w$ and $\angle EPK = x$ , so

$\frac{KE}{KF} = \frac{PE}{PF}\cdot \frac{\sin{\angle KPE}}{\sin{\angle KPF}} =\frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{x}}{\sin{w}}$
Then if $X = BK\cap AC$ , we have

$\frac{XA}{XC} =\frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{x}}{\sin{w}}\cdot \frac{BF}{BE}\cdot \frac{BA}{BC} = \frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{(w+2y)}}{\sin{(x+2z)}}.$
On the other hand, if $X' = DT\cap AC$ , we have

$\frac{X'A}{X'C} = \frac{DA}{DC}\cdot \frac{\sin{\angle TDA}}{\sin{\angle TDC}} = \frac{\sin{y}}{\sin{z}}\cdot \frac{\sin{(w+2y)}}{\sin{(x+2z)}}$
and we're done.

This post has been edited 1 time. Last edited by CT17, Jun 9, 2024, 2:42 PM

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4184 posts

#47 h Jul 12, 2024, 3:51 AM • 1 Y

Y by GoodMorning

Solved with mathhater420

Let $ET$ and $FT$ meet $AC$ at $P$ and $Q$ respectively.

Note that $\angle PED=\angle CDF=\angle CBD=\angle CAD$ , so $PAED$ is cyclic. Similarly, $QCFD$ is cyclic.

Claim 1: $TPDQ$ is cyclic. We have $\angle PDQ=180-\angle PDE-\angle QDF=180-\angle BAC-\angle BCA=\angle ABC$ $=\angle ABD+\angle CBD=\angle ACD+\angle CAD=\angle TPQ+\angle TQP.$
Claim 2: $TD$ is tangent to both $(PAED)$ and $(QCFD)$ . We have $\angle TDP=\angle TQP=\angle PAD,$ and similarly for $\angle TDQ=\angle QCD$ .

This means that the radical axis of the two circles is line $DT$ . Let $AC$ and $BK$ meet at $N$ . Thus, it suffices to show that $N$ lies on the radical axis, or in other words, $NP\cdot NA=NQ\cdot NC$ .

Claim 3: $K$ also lies on $(TPDQ)$ . We have $\angle TKD=\angle TDK=\angle EPD.$
Claim 4: $PK$ is parallel to $BC$ , and similarly $QK$ is parallel to $BA$ . We have $\angle PKD=\angle PQD=\angle CFD.$
Due to Claim 4, triangles $BAC$ and $KQP$ are homothetic. Furthermore, $N$ , which is the intersection of $BK$ and $APQC$ , is the center of (negative) homothety. Thus, $NP\cdot NA=NQ\cdot NC$ , as desired.

This post has been edited 2 times. Last edited by john0512, Aug 20, 2024, 2:15 PM

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ezpotd

#48 h Jul 31, 2024, 5:47 AM

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Let $TE \cap AC = P$ , $TF \cap AC = Q$ , let $BA \cap TF = X$ , $CA \cap TE = Y$ . Our goal will be to show the existence of a homothety from $ABCD$ to $QKPT$ , which finishes since it implies $BK$ , $DT$ , and $ACPQ$ are concurrent.

Step 1: Doxxing $TD$ .

Note that $\angle XFD = \angle XFE = \angle ADE = \angle ABD = \angle XBD$ , so $(XBFD)$ is cyclic. Now $\angle TED = \angle YED = \angle YEF = \angle CDF = \angle CBD = \angle FBD = \angle FXD = \angle TXD$ , so $(TXED)$ is cyclic.

Step 2: Figuring out $TPQ$ .

Note that $\angle APE = \angle YPC = \angle PCD = \angle ACD = \angle ADE$ , so $(APED)$ is cyclic. Symmetrically, $(CQFD)$ is cyclic. Thus $\angle PDQ = 180 - \angle DPQ - \angle DQP = \angle APD + \angle CQD - 180 = 180 - \angle AED - \angle CFD = \angle ABC$ . Now since $\angle PTQ = 180 - \angle ABC$ , we see that $(TDPQ)$ is cyclic.

Step 3: Finish.

Observe $(TXED)$ cyclic implies $\angle TDE = 180 - \angle TXE = \angle TXB = \angle DAB$ , so by the condition that $K$ is on segment $DF$ we see that $\angle TDK = \angle TDF = 180 - \angle DAB = \angle DCB$ . By isosceles properties, we see that $\angle TKD = \angle DCB$ as well.

Now notice $\angle TQD = \angle TQP + \angle PQD = \angle PQD + \angle CQF = 180 - \angle DQF = 180 - \angle DCF = \angle DCB$ , so $(TPQKD)$ is cyclic.

Now we show the final homothety exists. It is sufficient to show that the quadrilaterals are similar and that each side is parallel to the corresponding similar side. First observe we can see that $TPQ$ is similar to $DCA$ by angles we have already found, then we can see that $\angle KTP = 180 - 2\angle DCB + \angle DTP$ , we find $\angle DTP = \angle DQP = 180 - \angle DQC = \angle DFC = 180 - \angle DCF - \angle CDF = \angle DCB - \angle DBC$ . Substituting this back in, we get $\angle KTP = 180 - \angle DCB - \angle DBC = \angle BDC$ . We see that the unique point $K'$ such that $ABCD$ is similar to $QK'PT$ satisfies $\angle PTK' = \angle CDB$ , and lies on the circumcircle of $PQT$ , which is exactly $K$ . Thus the similarity is done, for the parallel sides just see that we already have two given by the problem statement and the rest follow trivially from the similarity, so the homothety exists and we are done.

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cursed_tangent1434

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cursed_tangent1434

#49 h Aug 12, 2024, 12:53 AM

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Let $P$ and $Q$ be the intersections of $\overline{ET}$ and $\overline{FT}$ with side $AC$ respectively. We start off with making some minor observations. First, since
$\measuredangle QFD = \measuredangle ADE = \measuredangle ACD = \measuredangle QCD$ we have that quadrilateral $DQCF$ is cyclic. Similarly, we can show that quadrilateral $APDE$ is also cyclic. Now, we can further note the following.

Claim : Points $P$ , $Q$ , $D$ , $T$ and $K$ are concyclic.

Proof : First, note that,
$\measuredangle TQD = \measuredangle FQD = \measuredangle FCD = \measuredangle BCD = \measuredangle EAD = \measuredangle EPD$ so quadrilateral $DPTQ$ must be cyclic. Further,
$\begin{align*} \measuredangle TKD &= \measuredangle KDT + \measuredangle FDT \\ &= \measuredangle FDQ + \measuredangle QDT \\ &= \measuredangle BCQ + \measuredangle QPT \\ &= \measuredangle BCQ + \measuredangle APE \\ &= \measuredangle BCQ + \measuredangle ADE \\ &= \measuredangle BCA + \measuredangle ACD \\ &= \measuredangle BCD\\ &= \measuredangle FCD \\ &= \measuredangle FQD \\ &= \measuredangle TQD \end{align*}$ which implies that indeed point $K$ also lies on this circle, which proves the claim.

With these observations in hand, we are now in a position to prove our key claim.

Claim : Quadrilaterals $ABCD$ and $QKPT$ are homothetic.

Proof : We are already given that $TP \parallel CD$ and $TQ \parallel AD$ , so we focus on the other two pairs of sides. Note that,
$\measuredangle FKQ = \measuredangle DPQ = \measuredangle DEA$ so we must have $KQ$ is parallel to $AB$ . Similarly, we have that $KP$ is parallel to $BC$ . Since pairs of polygons with all pairs of sides, being pairwise parallel are considered homothetic, the claim is proved.

Now, we know that lines joining corresponding vertices of homothetic polygons concur, at the center of homothety. Thus, lines $\overline{AQ} , \overline{CP},\overline{BK}$ and $\overline{DT}$ must concur. It is clear that lines $\overline{AQ}$ and $\overline{CP}$ are both just the line $\overline{AC}$ , so we have that lines $AC$ , $DT$ and $BK$ are concurrent, as desired.

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1749 posts

#50 h Aug 18, 2024, 12:10 AM

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Beautiful problem!

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-13.56,1.39); pair B = (-9.64,9.51); pair C = (-0.66,0.09); pair D = (-6.51117,-3.75196); pair E = (-16.28414,-4.25287); pair F = (2.55932,-3.28706); pair T = (-5.67034,2.71630); pair X = (-8.47125,0.87718); pair Y = (-2.29611,0.25488); pair K = (-4.17259,-3.63210); pair Z = (-5.94254,0.62234); import graph; size(10cm); pen xfqqff = rgb(0.49803,0,1); pen cczzqq = rgb(0.8,0.6,0); pen ffxfqq = rgb(1,0.49803,0); pen ffqqff = rgb(1,0,1); pen yqqqqq = rgb(0.50196,0,0); pen wwzzff = rgb(0.4,0.6,1); draw(circle((-6.86567,3.16446), 6.92551), linewidth(0.4) + linetype("2 2") + xfqqff); draw(B--A, linewidth(0.4) + red); draw(A--C, linewidth(0.4) + red); draw(C--B, linewidth(0.4) + red); draw(A--D, linewidth(0.4) + cczzqq); draw(D--C, linewidth(0.4) + cczzqq); draw(B--E, linewidth(0.4) + red); draw(B--F, linewidth(0.4) + red); draw(E--F, linewidth(0.4) + ffxfqq); draw(T--E, linewidth(0.4) + cczzqq); draw(T--F, linewidth(0.4) + cczzqq); draw(T--D, linewidth(0.4) + blue); draw(T--K, linewidth(0.4) + ffqqff); draw(B--K, linewidth(0.4) + blue); draw(B--D, linewidth(0.4) + ffqqff); draw(X--D, linewidth(0.4) + yqqqqq); draw(Y--D, linewidth(0.4) + yqqqqq); draw(circle((-11.44336,-3.11081), 4.97368), linewidth(0.4) + linetype("4 4") + green); draw(circle((-5.50064,-0.59454), 3.31519), linewidth(0.4) + wwzzff); draw(circle((-1.93351,-4.34703), 4.61617), linewidth(0.4) + linetype("4 4") + green); draw(Y--K, linewidth(0.4) + red); dot("$A$", A, dir((8.000, 20.000))); dot("$B$", B, dir((8.000, 20.000))); dot("$C$", C, dir((8.000, 20.000))); dot("$D$", D, dir((7.117, 20.196))); dot("$E$", E, dir((8.414, 16.287))); dot("$F$", F, dir((8.068, 15.706))); dot("$T$", T, dir((7.034, 15.370))); dot("$X$", X, dir((7.125, 15.282))); dot("$Y$", Y, dir((7.611, 15.512))); dot("$K$", K, dir((7.259, 16.210))); dot("$Z$", Z, dir((8.254, 16.766))); [/asy]$

Let $ET$ and $FT$ intersect $AC$ at points $X$ and $Y,$ respectively.

Claim 1: $AXDE$ and $CYDF$ are cyclic.
Proof: We have that $\measuredangle DAX = \measuredangle DAC = \measuredangle EDC = \measuredangle DET = \measuredangle DEX,$ so $AXDE$ is cyclic. We can similarly obtain that $CYDF$ is cyclic, proving our claim.

Claim 2: $TXDY$ is cyclic.
Proof: This is once again an angle chase:
$\begin{align*} \measuredangle TXD &= \measuredangle EXD = \measuredangle EAD = \measuredangle BAD \\ &= \measuredangle BCD = \measuredangle FCD = \measuredangle FYD = \measuredangle TYD. \end{align*}$
Claim 3: $TD$ is a common tangent to the circumcircles of $AXDE$ and $CYDF.$
Proof: We have that $\measuredangle TDX = \measuredangle TYX = \measuredangle TYA = \measuredangle DAY = \measuredangle DAX,$ so $TD$ is tangent to $(AXDE)$ and is similarly tangent to $(CYDF),$ as claimed.

Claim 4: $(TXDY)$ actually contains $K.$
Proof: We yet again angle chase: $\measuredangle TKD = \measuredangle KDT = \measuredangle EDT = -\measuredangle TDE = -\measuredangle DXE = \measuredangle TXD.$
Claim 5: $TK \parallel BD.$
Proof: Note that $\measuredangle TKD = \measuredangle TXD = \measuredangle EXD = \measuredangle EAD = \measuredangle BAD = \measuredangle BDE,$ implying the claim.

Now, let $Z = BK \cap TD$ be the center of a negative homothety sending $BD$ to $TK.$ Since $TY \parallel AD$ , $TK \parallel BD,$ and $YK \parallel AB$ by Reim, we see that $\triangle BAD$ and $\triangle KYT$ are homothetic, and the center of homothety is just $Z.$ Therefore, the homothety sends $A$ to $Y,$ and in particular $A,Z,Y$ are collinear, as desired.

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Eka01

#51 h Oct 13, 2024, 12:06 PM

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Sketch

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L13832

#52 h Nov 15, 2024, 9:28 AM

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Let $\{S,U\}=ET,FT\cap AC$ , we have
$\begin{align*} &\angle SED= \angle CDF = \angle SAD\implies \odot(ASDE)\\ & \angle UFD=\angle SAD=\angle UCD\implies \odot(CUDF) \end{align*}$ and
$\begin{align*} \angle STU&=180^{\circ}-(\angle TEF+\angle TFE)\\&=180^{\circ}-(\angle CAD+\angle ACD)\\&=\angle ABC\\&=180^{\circ}-(\angle ACB+\angle CAB)\\&=180^{\circ}-(\angle UDF+\angle SDE)\\&=180^{\circ}-\angle SDU \end{align*}$ So $\odot(STUD)$ is cyclic, now we show $DT$ is tangent to $\odot(CUDF)$ and $\odot(ASDE)$
$\begin{align*} &\angle SDT= \angle SUT= \angle CSF = \angle CUF = \angle CDF=\angle TED\\ &\angle UFD = \angle UCD = \angle CST = \angle UDT \end{align*}$ We angle chase more to get $K\in \odot(STUD)$
$\begin{align*} \angle DKT &=180^{\circ}-\angle TDF\\&=180^{\circ}-(180^{\circ}-\angle DCF)\\&=180^{\circ}-\angle DAE\\&=180^{\circ}-\angle DSE\\&=\angle DST \end{align*}$ If $\triangle KST$ and $\triangle BCD$ are homothetic we'll be done, so we prove
$\begin{align*} \angle TKD&=\angle TDU+\angle UDF=\angle TSU+\angle ACB=\angle ADE+\angle ADB=\angle BDE\implies KT\parallel BD\\ \angle SKE&=\angle SUD=180^{\circ}-\angle CUD =\angle CFD\implies KS\parallel BC \end{align*}$ and we are done!

$[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(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.301059888544868, xmax = 5.427247847678922, ymin = -3.9302067049047666, ymax = 7.930620330584966; /* image dimensions */ /* draw figures */ draw(circle((-1.7641834099733615,-0.43395414750665956), 2.7297207616073), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(-4,-2), linewidth(0.5)); draw((0.36842333045447384,-2.137880180828441)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((-4,-2)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((-1.8986143502387,6.4055425990452)--(-2.7477736566214856,-0.5200043091637574), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(2.0450554562137,-2.190799580619854), linewidth(0.5)); draw((0.7169072967555019,0.7042744505058862)--(-3,2), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((0.36842333045447384,-2.137880180828441)--(-3,2), linewidth(0.5)); draw((-4,-2)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw((-2.7477736566214856,-0.5200043091637574)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw(circle((-1.0709534122302529,0.24925606117104787), 1.8448543707750797), linewidth(0.5)); draw(circle((-0.33230478570055555,3.6735207021679015), 3.1491696202301815), linewidth(0.5)); draw(circle((1.2496916833029261,-0.8034932059595474), 1.599131798330576), linewidth(0.5)); draw((-0.23573917726046134,-1.3957074129133287)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-2.512254266719884,1.400837363648374)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-4,-2)--(0.7169072967555019,0.7042744505058862), linewidth(0.5)); draw((-2.512254266719884,1.400837363648374)--(0.4401007829004468,1.307652427730301), linewidth(0.5)); draw((0.4401007829004468,1.307652427730301)--(-0.23573917726046134,-1.3957074129133287), linewidth(0.5)); /* dots and labels */ dot((-3,2),linewidth(3pt) + dotstyle); label("$A$", (-3.3737381706765612,1.9371173073321764), NE * labelscalefactor); dot((-4,-2),linewidth(3pt) + dotstyle); label("$B$", (-4.335853129672412,-2.3056519380757203), NE * labelscalefactor); dot((0.36842333045447384,-2.137880180828441),linewidth(3pt) + dotstyle); label("$C$", (0.30122552516004736,-2.5737823364843977), NE * labelscalefactor); dot((0.7169072967555019,0.7042744505058862),linewidth(3pt) + dotstyle); label("$D$", (0.790169192846463,0.8803680900744474), NE * labelscalefactor); dot((-1.8986143502387,6.4055425990452),linewidth(3pt) + dotstyle); label("$E$", (-2.2381270715339183,6.432244574771769), NE * labelscalefactor); dot((2.0450554562137,-2.190799580619854),linewidth(3pt) + dotstyle); label("$F$", (2.130821184889861,-2.4949204545994923), NE * labelscalefactor); dot((-2.7477736566214856,-0.5200043091637574),linewidth(3pt) + dotstyle); label("$T$", (-3.042518266759957,-0.8545933113934656), NE * labelscalefactor); dot((-1.3505221927790012,-0.026271895669134612),linewidth(3pt) + dotstyle); label("$R$", (-1.197150230653162,-0.20792587993724349), NE * labelscalefactor); dot((0.4401007829004468,1.307652427730301),linewidth(3pt) + dotstyle); label("$K$", (0.5693559235687269,1.4008565105148214), NE * labelscalefactor); dot((-2.512254266719884,1.400837363648374),linewidth(3pt) + dotstyle); label("$S$", (-2.916339255744108,1.195815617614068), NE * labelscalefactor); dot((-0.23573917726046134,-1.3957074129133287),linewidth(3pt) + dotstyle); label("$U$", (-0.45584854093504795,-1.7693911412583652), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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142 posts

#54 h Jan 27, 2025, 10:55 PM

Y by

Beautiful problem :love:

:love:

. My solution is different from others involving $4$ extra points and using Spiral Sim.
Let $AC \cap BK=\{X\}$ , and let $AC \cap FT=\{Y\}$

Claim: Points $F$ , $C$ , $Y$ and $D$ are concyclic

Proof By angle chase we get:

$\angle DCY \equiv \angle DCA \stackrel{\Omega}{=} \angle DBA \stackrel{\Omega}{=} \angle EDA \stackrel{DA \parallel FT}{=} \angle EFG \equiv \angle DFY \implies$

$\angle DCY=\angle DFY \implies$ Points $F$ , $C$ , $Y$ and $D$ are concyclic $\square$ . Let $\odot(FCYD)=\Gamma$ .

Now note that $AY \cap BF=\{C\}$ and $\odot(CAB) \cap \odot(CYF)=\{ C, D \}$ hence we get that $D$ -is the center of Spiral Similarity $S_1$ that sends $S_1 : A \rightarrow B , S_1 : Y \rightarrow F$ so $S_1 : AY \rightarrow BF$ hence it's well known that :
$D$ is also the center of Spiral Similarity $S_2$ that sends $S_2 : AB \rightarrow YF$ .

So let $AB \cap YF=\{G\}$ so since $S_2 : AB \rightarrow YF$ and $AB \cap YF=\{G\}$ we get from Spiral Similarity properties that $G \in \odot(DAY)$ and $G \in \odot(DBF).$

Claim: Points $E$ , $D$ , $T$ and $G$ are concyclic

Proof

$\angle ETG \stackrel{\triangle ETG}{=} 180-\angle TEG-\angle TGE=-\angle TEG+180-\angle TGE \equiv -\angle TEG+180-\angle YGA \stackrel{\odot(DAGY)}{=}$
$\stackrel{\odot(DAGY)}{=} - \angle TEG+\angle YDA=-\angle TEG+180-\angle EDA-\angle TDC-\angle CDF \stackrel{\Omega}{=}-\angle TEG+180-\angle DBA-\angle TDC-\angle CDF \stackrel{\Gamma}{=}$
$\stackrel{\Gamma}{=} - \angle TEG+180-\angle DBA-\angle TFC-\angle CDF \stackrel{CD \parallel TE}{=} -\angle TEG+180-\angle DBA-\angle YFC - \angle TEF \equiv$
$\equiv -\angle TEG+180-\angle DBA-\angle GFB-\angle TED=-\angle TEG-\angle TED+180-\angle DBA-\angle GFB = -\angle GED+180-\angle DBA-\angle GFB \stackrel{\odot(BGDF)}{=}$
$\stackrel{\odot(BGDF)}{=} -\angle GED+180-\angle DBA-\angle GDB \stackrel{\triangle EDB}{=} \angle EDG \implies \angle ETG=\angle EDG$

$\implies$ Points $E$ , $D$ , $T$ and $G$ are concyclic $\square$

Now since $TY \cap EA =\{G\}$ and $\odot(GTE) \cap \odot(GYA)=\{G,D\}$ we get that $D$ -is the center of
Spiral Similarity that sends $S_3 : T \rightarrow E$ and $S_3 : Y \rightarrow A$ hence $S_3 : TY \rightarrow EA$ .

Now again as last time it's well known that since $S_3 : TY \rightarrow EA$ we have that $D$ -is also the center of Spiral Similarity that sends $S_4 : TE \rightarrow YA$ so let $TE \cap YA=\{H\}$
Hence by properties of Spiral Similarity we get that $H \in \odot(DTY)$ and $H \in \odot(DEA)$

Claim: $KY \parallel AB$

Proof:
From previous claims we found that quadrilaterals $\square EDTGH$ and $KYTD$ are cyclic and we also know that
$\overline{E-D-K}$ and $\overline{G-T-Y}$ are collinear hence from Reims's Theorem we get that $KY \parallel AB$ $\square$

Claim: $KH \parallel FB$

Proof:

$\angle DKH \stackrel{\odot(DKYTH)}{=} \angle DTH \equiv \angle DTE \stackrel{\odot(DEGE)}{=} \angle DGE = 180-\angle DGB \stackrel{\odot(DGBF)}{=} \angle DFB \implies \angle DKH=\angle DFB \implies KH \parallel FB$

Claim $TD$ -is the radical axis of $\Gamma$ and $\odot(DEAH)$

Proof:
Before we found that $S_3 : TY \rightarrow EA \implies \triangle DTY \sim \triangle DEA \implies \angle TDY =\angle EDA \stackrel{AD \parallel FT}{=} \angle EFT \equiv \angle DFY \implies \angle TDY=\angle DFY \implies$
$TD-$ tangent to $\Gamma$

Also: $\angle TDA=180-\angle ADE-\angle TDK \stackrel{\odot(HDKYT)}{=} 180-\angle ADE - 180-\angle TYK = 180-\angle ADE - \angle KYF \stackrel{KY \parallel AB}{=} 180-\angle ADE - \angle EGF \equiv 180-\angle ADE - \angle EGY \stackrel{AD \parallel GF}{=} 180-\angle ADE -\angle EAD \stackrel{\triangle EAD}{=} \angle DEA \implies \angle TDA=\angle DEA \implies$ $TD$ - is tangent to $\odot(DEAH)$

Since $TD$ is tangent to both $\Gamma$ and $\odot(DEAH)$ we get that $TD$ is the radical axis of both circles $\square$

Claim: Lines $AC,DT,$ and $BK$ are concurrent.

Proof:
Since $XY \parallel AB$ and $KH \parallel FB \equiv CB$ we get that triangles $\triangle ABC$ and $\triangle YKX$ are homothetic hence

Lines $BK$ and $AY$ meet on $X$ so $\frac{XA}{XC}=\frac{XY}{XH} \implies XA \cdot XH = XC \cdot XY \implies Pow(X, \odot(DEAH) )= Pow(X , \Gamma) \implies$

$X$ lies on the radical axis of $\Gamma$ and $\odot(DEAH)$ but before we proved that the radical axis of these circles is $TD$ hence
$X$ lies on $TD$ so lines $AC,DT,$ and $BK$ are concurrent. $\blacksquare$

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Haris1

#55 h Feb 8, 2025, 6:01 PM

Y by

Idk why i was always intimidated by this problem , but solved it in 30 mins. My solution is similar to #4.

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583 posts

#56 h Mar 13, 2025, 7:31 PM

Y by

uh.... what is angle chase...

Let $M = FT\cap BE, N = ET\cap BC.$
Claim: $BMTN$ is cyclic
Proof: $\angle BNT = \angle BCD = 180 - \angle BAD = 180 - \angle BMT.$

Let $Q = FT\cap AC, R = ET\cap AC$

Claim: $(BMDF)$ is cyclic
Proof: $\angle BMF = \angle BAD = \angle BDF,$ which finishes.

Claim: $GCDF$ is cyclic
Proof: $\angle DFG = \angle ABD = \angle DCA$ which finishes.

Claim: $TD$ is tangent to $(DCGF)$
Proof:Note that $\angle EFG = \angle DCG = \angle THG,$ so $HGFE$ is cyclic.
Now, assume that $(AHDE), (GCDF)$ aren’t tangent. Let them intersect again at $D_1.$ We get that $\angle D_1DF = \angle DCF = 180 - \angle BAD =EAD = \angle ED_1D.$
Finally, radax on $(DGCF), (AHDE), (EHFD)$ finishes.

Claim: $(TNDF)$ is cyclic
Proof:Observe that $\angle TDF = 180-\angle DCF = \angle DCN = \angle TNB$

Claim: $TK\parallel BD$
Proof: $\angle TKD = \angle TDK = 180 - \angle DCF = \angle DCN = \angle TNB = \angle BDE.$

Claim: $(THGDK)$ is cyclic
Proof:Note that $\angle HDK = 180 - \angle HDE = \angle BAC = \angle GDK,$ so $GHDK$ is cyclic. Then, observe that $\angle TKD = \angle BDE = \angle BNE = \angle BCD = 180-\angle DCF= 180 - \angle DGF = \angle TGD.$

Now, let $P = (BMTN)\cap DT.$ Observe that $\angle BPD = \angle BPT = \angle EMT = \angle EAD = \angle DCB,$ so $(BPCD)$ is cyclic.

Claim: $DG\parallel BP$
Proof:Observe that $\angle BAP = \angle BDP = \angle DTK,$ so $DG\parallel AP.$
Claim: $BP\parallel DK$
Proof:Observe that $\angle TDK = 180 - \angle TDE = 180 - \angle BMT = \angle BPT,$ so $BP\parallel DK.$
Claim: $GK\parallel AB$
Proof:Observe that $\angle GKD = \angle AHD = 180 - \angle BED = \angle EBD + \angle BFE = \angle EBF + \angle FBP = \angle PBA,$ so $GK\parallel AB.$

These 3 claims imply that $DGK\sim PBA$ , so they are homothetic, finishing the problem.
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 13, 2025, 7:40 PM
Reason: include diagram

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#57 h Apr 14, 2025, 3:03 PM

Y by

Beatiful Problem. :10:

:10:

.

Let $TE$ and $TF$ intersect $AC$ at $X , Y$ ,respectively.

And we will prove that $KXT$ and $BCD$ are homothetic i.e. $KX\parallel BC$ , $XT\parallel CD$ , $KT\parallel BD$ .The second is the condition of problem.

$\angle CDF = \angle CAD = \angle CYF$ this implies $CDFY$ is cyclic. $ADEX$ is cyclic analogously.

$\angle BDE = \angle BDC + \angle CDF = \angle BAC + \angle CYF = \angle EAX + \angle XYT = \angle EDX + \angle XDT = \angle KDT = \angle TKD = \angle TKE$ this means $KT\parallel BD$ .

There is only one left parallelity proof right now.

$\angle TYD = \angle FYD = \angle FCD = \angle BCD = \angle BDE = \angle TKE = \angle TKD$ this implies $DKTY$ is cyclic. $\angle TXD = \angle EXD = \angle EAD = \angle BCD = \angle FCD = \angle FYD = \angle TYD$ this means $DTXY$ is cyclic i.e. $D , K , T , X , Y$ are all lie on one circle.

$\angle AED = \angle AXD = \angle YXD = \angle YKD$ soo $KX\parallel BC$ .

Triangles $KXT$ and $BCD$ have pairwise parallel sides i.e. they are homothetic,soo this gives us $TD , XC$
and $KB$ are concurrent, as desired.Soo we are done. $\blacksquare$ .

.

This post has been edited 2 times. Last edited by Frd_19_Hsnzde, Apr 14, 2025, 5:10 PM

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