geometry with quadrilateral, tangent circles wanted

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trying_to_solve_br

191 posts

trying_to_solve_br

#1 h Jul 20, 2021, 8:55 PM • 5 Y

Y by centslordm, betongblander, deplasmanyollari, lian_the_noob12, Rounak_iitr

Let $ABCD$ be a convex quadrilateral with $\angle ABC>90$ , $CDA>90$ and $\angle DAB=\angle BCD$ . Denote by $E$ and $F$ the reflections of $A$ in lines $BC$ and $CD$ , respectively. Suppose that the segments $AE$ and $AF$ meet the line $BD$ at $K$ and $L$ , respectively. Prove that the circumcircles of triangles $BEK$ and $DFL$ are tangent to each other.

$\emph{Slovakia}$

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VulcanForge

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VulcanForge

#2 h Jul 20, 2021, 8:57 PM • 3 Y

Y by centslordm, Dukejukem, Chokechoke

Let $A'$ be the reflection of $A$ over $BD$ . We show the circles are tangent at $A'$ . Indeed, $BEKA'$ is cyclic because $\measuredangle BA'K = -\measuredangle BAK = \measuredangle BEA = \measuredangle BEK$ and similarly $DFLA'$ is cyclic. Hence by reflection across $BD$ it suffices to show $(ABK)$ and $(ADL)$ are tangent at $A$ : let the tangents at $A$ to these circles be $\ell_1, \ell_2$ respectively. Indeed, note
$\begin{align*} \measuredangle (\overline{AB}, \ell_1)+ \measuredangle (\ell_2, \overline{AD}) &= \measuredangle BKA + \measuredangle ALD \\ &= \measuredangle LKA + \measuredangle ALK \\ &= \measuredangle LAK = \measuredangle (\overline{CD}, \overline{CB})\\ &= \measuredangle (\overline{AB}, \overline{AD}) \end{align*}$ hence $\ell_1=\ell_2$ .

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bora_olmez

277 posts

bora_olmez

#3 h Jul 20, 2021, 9:41 PM • 1 Y

Y by centslordm

Cool problem.

$\textbf{Lemma 1:}$ The circumcircles of $\triangle ADL$ and $\triangle ABK$ are tangent at $A$
$\textbf{Proof)}$
Notice that $\angle LAK = 180^{\circ}-\angle CDB = 180^{\circ} - \angle DAB$ We consequently have that $\angle DKA + \angle ADK = 180^{\circ} - \angle DAK = \angle LAB$ Now, if we select a point $T$ inside $\triangle DAK$ such that $AT$ is tangent to the circumcircle of $\triangle DAL$ , we have that $\angle KAB = \angle DKA = \angle LAB - \angle KDA = \angle BAT + \angle LAT - \angle DLA = \angle BAT$ meaning that $AT$ is also tangent to the circumcircle of $\triangle BAK$ . $\square$

$\textbf{Lemma 2:}$ The circumcircles of $\triangle DFL$ and $\triangle BKE$ are tangent
$\textbf{Proof)}$
Notice that if we let $A'$ be the reflection of $A$ across $BD$ , then $\angle DA'L = \angle DAL = \angle DFL$ meaning that $A'$ lies on the circumcircle of $\triangle DFL$ . Analagously, $A'$ also lies on the circumcircle of $\triangle BKE$ .
Moreover, notice that the circumcircles of $\triangle A'LD$ and $\triangle A'BK$ are tangent by $\textbf{Lemma 1}$ meaning indeed that the circumcircles of $\triangle BEK$ and $\triangle DFL$ are tangent to each other. $\blacksquare$

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pad

1671 posts

pad

#4 h Jul 20, 2021, 11:26 PM

Y by

Solved with dantaxyz.

Diagram
Let $A'$ be the reflection of $A$ over $\overline{BD}$ . We claim $A'$ is the desired tangency point. First, we show it lies on each of the circles. Let $A_1$ and $E_1$ be the feet from $A$ to $\overline{BD}$ and $\overline{BC}$ respectively. Noticing that $AE_1BA_1$ is cyclic, note
$\begin{align*} \angle KEA' = \angle AEA' &= \angle AE_1A_1 = \angle ABD \\ &= 180^\circ-\angle KBA = 180^\circ-\angle KBA', \end{align*}$ so $KBA'E$ is cyclic. Hence $A' \in (BEK), (DFL)$ . To show the tangency at $A'$ , it suffices to show
$\tfrac12 \widehat{BA'} + \tfrac12 \widehat{DA'} = \angle BA'D.$ We have $\tfrac12 \widehat{BA'}=\angle BKA'=\angle BKA=90^\circ-\angle DBC$ , and similarly $\tfrac12 \widehat{DA'} = 90^\circ-\angle BDC$ , so their sum is $(90^\circ-\angle DBC)+(90^\circ-\angle BDC)=\angle BCD = \angle BAD$ , since $\angle A=\angle C$ , and we are done.

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MP8148

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MP8148

#5 h Jul 21, 2021, 2:57 AM

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$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), B = dir(195), D = dir(345), C = 2*foot(dir(80),B,D)-dir(80), P = foot(A,B,D), A1 = 2P-A, E = 2*foot(A,B,C)-A, F = 2*foot(A,C,D)-A, K = extension(B,D,A,E), L = extension(A,F,B,D), Ob = circumcenter(A1,B,E), Od = circumcenter(A1,D,F), B1 = extension(B,C,A,E), D1 = extension(C,D,A,F); draw(B--B1^^D--D1, purple+dotted); draw(A--B--C--D--A, purple); draw(B--A1--D, purple); draw(K--L^^E--A--F^^A1--A, heavycyan); draw(circumcircle(B,C,D), heavygreen); draw(circumcircle(B,K,E)^^circumcircle(D,L,F), orange); draw(Ob--Od, magenta); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(270)); dot("$D$", D, dir(315)); dot("$A'$", A1, dir(255)); dot("$P$", P, dir(45)); dot("$E$", E, dir(150)); dot("$F$", F, dir(45)); dot("$B_1$", B1, dir(135)); dot("$D_1$", D1, dir(60)); dot("$K$", K, dir(135)); dot("$L$", L, dir(60)); dot("$O_B$", Ob, dir(165)); dot("$O_D$", Od, dir(345)); [/asy]$

Let $B_1$ , $D_1$ be the midpoints of $\overline{AE}$ , $\overline{AF}$ respectively. Let $P$ be the foot from $A$ to $\overline{BC}$ and $A'$ be the reflection of $A$ over $P$ . I claim that $A'$ is the desired tangency point.

Claim: $A’BKE$ , $A’DLF$ are cyclic.

Proof. Since $\overline{B_1P}$ is the $A$ -midline in $\triangle AEA'$ , we have $\measuredangle KEA' = \measuredangle AEA' = \measuredangle AB_1P = \measuredangle ABP = \measuredangle PBA' = \measuredangle KBA',$ so $A’BKE$ is cyclic, and the other follows analogously. $\square$

To finish, let $O_B$ be the center of $(A'BKE)$ and $O_D$ be the center of $(A'DLF)$ . Note that $B$ is the center of $(AEA')$ , so $\angle O_BA'B = \frac 12 \angle A'BE = \angle A'AE.$ Thus $\begin{align*}\angle O_BA'B + \angle BA'D + \angle DA'O_D &= \angle A'AE + \angle BCD + \angle FAA' \\ &= \angle DBC + \angle BCD + \angle CDB \\ &= 180^\circ,\end{align*}$ which means $O_B$ , $A'$ , $O_D$ are collinear, and we are done.

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Reason: Typo

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DakuMangalSingh

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DakuMangalSingh

#6 h Jul 21, 2021, 3:28 AM

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Inversion

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Kagebaka

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Kagebaka

#7 h Jul 21, 2021, 4:20 AM

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The key observation is that the point of tangency is $A',$ the reflection of $A$ over $BD.$ Observe that $\measuredangle DFL = \measuredangle LAD=\measuredangle DA'L,$ so $A'\in (LDF)$ and similarly $A'\in (KBE).$ Now suppose that line $\ell$ is tangent to $(LDF)$ at $A'.$ Then we obviously have $\measuredangle (LA',\ell) = \measuredangle LDA' = \measuredangle ADB,$ but on the other hand we find that
$\begin{align*} \measuredangle LA'K &= \measuredangle KAL \\ &= \measuredangle BAD - \measuredangle BAH - \measuredangle IAD \\ &= \measuredangle BAD - (90^\circ - \measuredangle HBA) - (90^\circ - \measuredangle ADI) \\ &= \measuredangle BAD + (360^\circ - 2\measuredangle BAD) - 180^\circ \\ &= \measuredangle DBA + \measuredangle ADB, \end{align*}$ so in fact $\measuredangle (\ell, KA') = \measuredangle DBA = \measuredangle A'BK,$ meaning that $\ell$ is tangent to $(KBE)$ at $A'$ as well. $\blacksquare$

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Kamran011

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Kamran011

#8 h Jul 21, 2021, 5:16 AM

Y by

Really nice problem, though I drammatically missed it (and hence the IMO) after guessing the tangency point, more here (see Day 2).

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hukilau17

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hukilau17

#9 h Jul 21, 2021, 5:29 AM

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complex bash

side note

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JAnatolGT_00

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JAnatolGT_00

#11 h Jul 21, 2021, 6:24 PM

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Let $T$ be reflection of $A$ wrt $BD$ . Then $\measuredangle DTL=\measuredangle FAD=\measuredangle DFL\implies T\in \odot (DFL)$ . Analogously $T\in \odot (BEK)$ .
Finally $\measuredangle BTD=\measuredangle BCD=\measuredangle KAL=\measuredangle DLA+\measuredangle AKB=\measuredangle TLD+\measuredangle BKT$ , hence $T$ is desired tangency point.

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SerdarBozdag

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SerdarBozdag

#12 h Jul 21, 2021, 9:44 PM

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If $A'$ is the reflection of $A$ wrt $BD$ then $\angle BA'K=\angle BAK=\angle BEK \implies A' \in (KBE)$ . Similarly $A' \in (DLF)$ . $\angle A'KB+\angle A'DL=180-(180-\angle BCD)=\angle BA'D$ which implies $A'$ is the desired tangency point.

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franchester

1487 posts

franchester

#13 h Jul 22, 2021, 8:06 AM

Y by

turns out having a logic fault and assuming ABCD is a kite was actuallly pretty useful

Denote $A'$ as the reflection of $A$ over $BD$ . The claim is that $A'$ is the desired tangency point.

First, we show that $A'$ lies on both circles. This follows from the fact that $B$ is the circumcenter of $AEA'$ --- assume this triangle is acute (the obtuse case follows similarly): $\angle EBA'=2\angle A'AE=\angle KAA'+\angle KA'A=\angle EKA'$ which impllies the concylicity. $A'$ lying on $(DFL)$ follows similarly.

Note that in order to show the two circles are tangent, it suffices to show $\angle A'EB+\angle A'FD=\angle BA'D$ (consider constructing the tangents at $A'$ for both circles). This is, again, angle chase using the two circumcenters $B$ and $D$ : $\angle A'EB+\angle A'FD=90^{\circ}-\angle EAA'+90^{\circ}-\angle FAA'=180^{\circ}-\angle EAF$ $=90^{\circ}-\angle CAE+90^{\circ}-\angle CAF=\angle BCA+\angle DCA=\angle BCD=\angle BAD=\angle BA'D$ as desired.

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KST2003

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KST2003

#14 h Jul 22, 2021, 8:12 PM

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Let $T$ be the reflection of $A$ over $\overline{BD}$ . We claim that this is the desired tangency point. Observe that
$\measuredangle DTL = -\measuredangle DAL = \measuredangle DFL,$ and hence $T$ lies on $(DFL)$ . Similarly, $T$ also lies on $(BEK)$ . Now reflect these two circles over $\overline{BD}$ . It suffices to show that $(BKA)$ and $(DAL)$ are tangent to each other at $A$ . Just notice that
$\measuredangle ABK + \measuredangle DLA = \measuredangle BAL = \measuredangle BAK + \measuredangle KAL = \measuredangle BAK + \measuredangle BCD = \measuredangle BAK + \measuredangle DAB = \measuredangle DAK,$ and we are done.

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FAA2533

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FAA2533

#15 h Jul 23, 2021, 4:16 AM

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Solution

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Eyed

1065 posts

Eyed

#16 h Jul 23, 2021, 4:19 AM • 1 Y

Y by greenish

Define $A'$ as the reflection of $A$ over $BD$ . Observe that
$\angle KEA' = \angle ABP = 180 - \angle KBA'$ so $(EKBA')$ is cyclic. Similarly, $(FLDA')$ is cyclic. Now, if the centers of $(EKB)$ and $(FDL)$ were $O_{1}$ and $O_{2}$ respectively, then
$\angle O_{1}A'A + \angle O_{2}A'A = \angle O_{1}A'B + \angle O_{2}A'D + \angle BA'D = \angle CBD + \angle BDC + \angle BCD = 180$ Therefore, $O_{1}, A', O_{2}$ are collinear, so $(BKE)$ and $(DLF)$ are tangent at $A'$ .

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GrantStar

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GrantStar

#43 h Jan 13, 2024, 7:42 PM

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Denote the reflection of $A$ across $BC$ by $A'$ and let $AA'$ intersect $BC$ and $CD$ again at $X$ and $Y$ . The key claim is the following.

Claim: $EKBXA'$ and $FLDYA'$ are cyclic.
Proof. Note that $B$ is the orthocenter of $\triangle AKX$ . A homothety with scale factor $2$ at $A$ sends the nine point circle of $AKX$ to $(EKBXA')$ and symmetry implies the result. $\blacksquare$

Now, if $\ell$ is the line through $A'$ tangent to $(EKXBA')$ , we have $\measuredangle (\ell, \overline{AXY})=\measuredangle A'BX=\measuredangle A'BC=\measuredangle A'DC=\measuredangle A'DY$ where the fourth equality follows from $A'BCD$ being cyclic fro the given angle condition. This finishes.

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jp62

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jp62

#44 h Feb 18, 2024, 4:02 PM

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Let $X$ be the reflection of $A$ over $BC$ , then $BXCD$ is cyclic. I claim that $X$ is the tangency point (motivated by reflections about two of the sides of $\triangle BCD$ , it's worth looking at the last reflection too to get some circumcenters).
We use directed angles mod $180^\circ$ .

BKEX is cyclic
The tangents at X coincide

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Acclab

33 posts

Acclab

#45 h Mar 1, 2024, 12:00 PM

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We can draw the diagram to help visualizing. Click to reveal hidden text

Let $T$ be the reflection point of $A$ across $BD$ . Note that $BE$ and $BT$ are just reflections of $BA$ across $BC$ and $BD$ respectively, so $\angle BEK = \angle BAK = \angle BTK$ and thus $K, E, T, B$ are concyclic, so similarly $L, F, T, D$ are too.

It suffices to prove the orientation of the line tangent to $T$ from $(KEB)$ (say $l_1$ ) is the same as those from $(LFD)$ (say $l_2$ ).

Indeed, using formal sum for angle chasing we see
$KB + l_1 = KT + TB, DL + l_2 = LT+TD.$ $KB=DL$ are the same line, so $l_1 = l_2$ iff $(KT-LT) + (BT-DT) = 0$ . Indeed, this equates to $(2KB - KA + 2KB - AB) - (2KB - LA + 2KB - DA)$
$= LA + DA - KA - AB = (LA - KA) + (DA - AB) = \sphericalangle KAL + \sphericalangle BAD = \sphericalangle KAL - \sphericalangle BCD = 0,$ where the last equation follows by the fact that $A, B', C, D'$ is concyclic.

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ihatemath123

3448 posts

ihatemath123

#46 h Apr 25, 2024, 2:55 AM

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Let $T$ be the point on line $BD$ for which $\overline{TA}$ is tangent to $(ABK)$ . Then, angle chasing gives us $\angle TAD = \angle DLA$ , so $\overline{TA}$ is also tangent to $(ADL)$ . Therefore, $(ABK)$ and $(ADL)$ are tangent at $A$ .

Since $\angle KEB = 180^{\circ} - \angle KAB$ and $\angle LFD = 180^{\circ} - \angle LAD$ , it follows that $(KEB)$ and $(DFL)$ are the reflections of $(ABK)$ and $(ADL)$ over line $BD$ , respectively, so we have the desired tangency.

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john0512

4190 posts

john0512

#47 h May 25, 2024, 7:38 PM

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We claim that the tangency point is the reflection of $A$ across $BD$ . Call this point $T$ .

Claim: $T$ lies on $(KEB)$ . Note that $BE=BA$ , but $BT=BA$ as well, so $B$ is the circumcenter of $AET$ and $BE=BT$ . However, we have $\angle EKB=\angle TKB$ by symmetry, so $B$ is the arc midpoint of $ET$ . Similarly, $T$ lies on $(LFD)$ as well.

Thus, it suffices to show that they are tangent. Let $\angle BET=\angle BTE=\angle BKE=\angle BKT=\theta_b$ , and define $\theta_d$ similarly. It suffices to show that $\angle BTD=\angle BAD=\theta_b+\theta_c$ , since then the line through $T$ tangent to $(KEBT)$ would also be tangent to $(LFDT)$ as well.

This is simply a matter of angle chasing. We have $180-\theta_b-\theta_c=\angle KAL=\angle KAB+\angle BAD+\angle LAD$ $=90-(180-\angle ABT)+\angle BAD+90-(180-\angle ADT)$ $=\angle BAD+\angle ABT+\angle ADT-180=\angle BAD+(360-2\angle BAD)-180=180-\angle BAD,$ as desired.

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Shreyasharma

682 posts

Shreyasharma

#48 h Jun 28, 2024, 4:03 AM

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Reflect $A$ about $BD$ to a point $A'$ , and note that $A'BDC$ is cyclic.

Claim: $(BEK)$ passes through $A'$ .
Proof. Immediately follows from $\angle BA'K = \angle BAK = \angle BEA$ proving the claim. $\square$

Likewise $(DFL)$ passes through $A'$ . Now let $T$ on $(A'BDC)$ be chosen so that $CT \perp BD$ .

Claim: $A'T$ is the common internal tangent to $(BEK)$ and $(DFL)$ .
Proof. Proceed with complex numbers, setting $(A'BDC)$ as the unit circle. Then we compute,

$t = -bd/c$
$a = b + d - bd/a'$
$e = b + c - bc\overline{a} = b - bc/d + a'c/d$

Then to verify $\angle TA'B = \angle A'EB$ we have to check,
$\begin{align*} \frac{t - a'}{b - a'} \div \frac{a' - e}{b - e} \in \mathbb{R} \end{align*}$ However this follows as,
$\begin{align*} \frac{t - a'}{b - a'} \div \frac{a' - e}{b - e} &= \frac{-bd/c - a'}{b - a'} \div \frac{a' - b + bc/d - a'c/d}{bc/d - a'c/d}\\ &= \frac{bd + a'c}{a'c - bc} \div \frac{a'd - bd + bc - a'c}{bc - a'c}\\ &= \frac{bd + a'c}{a'c - bc} \div \frac{(b - a')(c - d)}{c(b - a')}\\ &= \frac{bd + a'c}{(a' - b)(c - d)} \end{align*}$ which is clearly self conjugate. Thus both circles are tangent at $A'$ and we're done. $\square$

This post has been edited 2 times. Last edited by Shreyasharma, Jun 28, 2024, 5:53 PM

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Eka01

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Eka01

#49 h Aug 20, 2024, 9:27 AM • 1 Y

Y by AaruPhyMath

Notice that $(ABD)$ and $(CBD)$ are reflections of each other over $BD$ , so that along with all the other reflections motivates us to reflect $A$ over $BD$ , and we call this reflection $K$ . By angle chasing, we show that $K$ lies on both the circles we are supposed to prove tangent, so it just remains to show that $K$ is the tangency point, which follows from some more angle chasing.

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SimplisticFormulas

118 posts

SimplisticFormulas

#50 h Dec 22, 2024, 2:49 PM

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All one needs to spot is a single construction and the question falls apart.

Reflect $A$ in $BD$ to get $A’$ . Let $X,Y,Z$ be feet of perpendiculars on $DB,BC,CD$ .

CLAIM 1: $A’ \in \odot(BEK)$
PROOF: Note that $\angle A’EK=\angle A’EA=\angle XYA=\angle XBA=\angle A’BX$ and $\angle A’FL=\angle A’FA=\angle XZA=\angle XDA=\angle XDA’$

CLAIM 2: $\odot(BEK)$ is tangent to $\odot(DLF)$
PROOF: Consider a line $l$ passing through $A’$ tangent to $\odot(BEK)$ which meets $B$ in $U$ . Observe that $\angle BA’U=\angle BKA’=\angle BKA$ and $\angle UA’D=\angle BA’D- \angle BA’U=\angle KLA+\angle LKA-\angle LKA=\angle KLA=\angle DLA’$ , so $l$ is also tangent to $\odot(DFL)$ and we are done. $\blacksquare$

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mariairam

8 posts

mariairam

#51 h Dec 30, 2024, 11:23 PM • 1 Y

Y by vi144

Consider $X$ the relection of $A$ across line $BD$ .

Claim 1: $X\in (BEK)\cap(DLF)$ .

Proof: Note that $\angle DFL=\angle DAL=\angle DXL$ , hence $D,X, F,L$ are concyclic. Similarly, $K,X,B,E$ are concyclic, which leads to the desired answer.

Claim 2: Point X is in fact the tangency point between the two circles.

Proof: Let $S$ be the intersection of the tangent at $X$ to $(BEKX)$ with $BD$ . All that remains is to prove that $XS$ is tangent to $(DLFX)$ . We do so by angle chase. We have that $\angle BXS=\angle XKB$ implying that $\angle BAS=\angle BKA$ . Denote $\angle BAD =\angle BCD = \alpha$ . Then, $\angle DXS= \angle DAS=\alpha -\angle BAS = \alpha -\angle BKA$ . We easily obtain that $\angle LAK = 180 - \alpha$ , thus $\angle DLA=\angle DLX= 180 - (\angle BAK + 180 - \alpha)=\alpha -\angle BAK$ . Therefore, $\angle DLX=\angle DXS$ , leading to the conclusion.

Attachments:

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Thelink_20

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Thelink_20

#52 h Jan 9, 2025, 8:15 PM

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Let $A'$ be the reflection of $A$ over $BD$ and $X,Y,Z$ be the projections on $BD, BC, CD$ , respectively.

Lemma: $(A'BKE)$ and $(A'CLF)$ .

Proof: Both claims are analogous so we'll prove $(A'BKE)$ . Notice that:
$\angle AEA' = \angle AYX = \angle ABX = \angle A'BX = 180^{\circ} - \angle A'BK \ _{_{\blacksquare}}$ Because $\angle BCD = \angle BAD = \angle BA'D$ we have $(BA'CD)$ . Let $\ \ell$ be the tangent to $(A'BKE)$ at $A'$ . We have:
$\measuredangle (BA'; \ell) = \angle BKA' = \angle BKA = 90^{\circ} - \angle KAX = 90^{\circ} - \angle DBC.$ $\measuredangle (DA'; \ell) = \angle BA'D - \measuredangle (BA'; \ell) = \angle BCD + \angle DBC -90^{\circ} = 90^{\circ} + \angle BDC = \angle DLA'.$ It implies that $\ell$ is tangent to $(A'CLF)$ and we are done. $\ _{\blacksquare}$

This post has been edited 1 time. Last edited by Thelink_20, Jan 9, 2025, 10:01 PM

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optimusprime154

23 posts

optimusprime154

#53 h Feb 8, 2025, 6:34 PM

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Let $S$ = $AE \cap BC$ and $R$ = $AF \cap CD$ .
let $T$ be the reflection of $A$ over $BD$ first i claim:
$BKTE$ , $DFLT$ cyclic.
we know $B$ is the center of $(AET$ \) and D is the center of $(AFT$ .
this means $\angle BAE = \angle BEA$ and $\angle BAK = \angle BTK$ and then we do the same for $DFKT$ which proves the claim.
now i claim $2\angle BAD$ = $180^{\circ} - \angle BAE - \angle DAF$ we know that $SARC$ is cyclic this means $\angle BAD + \angle BAE + \angle DAF =180^{\circ} - \angle BCD = 180^{\circ} - \angle BAD$
$\angle KET = 90^{\circ} - BAT$ , and $\angle DLT = 90^{\circ} - \angle TAL$ this means $\angle KET + \angle DLT$ = $180^{\circ} - \angle BAT - \angle TAL$ .
we know $\angle KTD = \angle BAD + \angle BAL$ and $\angle BAD + \angle DAF = \angle BAT + \angle TAL = \angle BAF$ .
now this along with the claim proved before gives: $\angle BAD + \angle BAE = 180^{\circ} - \angle BAT - \angle TAL$ which means $\angle BTD = \angle KET +\angle DLT$ which means these two circles are tangent

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cursed_tangent1434

635 posts

cursed_tangent1434

#54 h Feb 10, 2025, 12:51 AM

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This problem is really easy, especially with a decent diagram. However, drawing an accurate diagram is not easy. We start off with proving the following lemma, which is actually well known but we include since otherwise there is nothing at all to prove in this problem.

Lemma : Let $\triangle ABC$ be a triangle with orthocenter $H$ . Denote by $A_1$ and $A_2$ the reflections of $A$ across the $B-$ altitude and the $C-$ altitude respectively. Points $A_1$ and $A_2$ all lie on $(BHC)$ .

Proof : Note that since,
$\measuredangle BA_1C = \measuredangle CAB = \measuredangle BHC$ it follows that $A_1$ lies on $(BHC)$ . Similarly it can be show that $A_2$ also lies on $(BHC)$ , which finishes the proof.

Returning to the problem, let $A'$ denote the reflection of $A$ across line $\overline{BD}$ and let $P$ and $Q$ denote the intersections of the $A-$ altitude in $\triangle ABD$ with lines $\overline{BC}$ and $\overline{CD}$ respectively.

Claim : Points $K$ , $E$ , $B$ , $P$ and $A'$ (and similarly) are concyclic.

Proof : Note that due to the reflection, $BP \perp AK$ and by definition, $BK \perp AP$ . This implies that $B$ is the orthocenter of $\triangle ABC$ . Applying the lemma on $\triangle AKP$ , it follows that points $E$ and $A'$ lie on $(BKP)$ which proves the claim. Similarly we can show that points $F$ and $A'$ lie on $(DLQ)$ .

Now, it suffices to show the following angle condition,
$\measuredangle BKA' + \measuredangle A'LD = \measuredangle BPA + \measuredangle AQD = (90 + \measuredangle CBD) + (90 + \measuredangle BDC) = \measuredangle BCD = \measuredangle DAB = \measuredangle BA'D$ which implies that circles $(BKA')$ and $(DLA')$ are indeed tangent to each other at $A'$ .

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ihategeo_1969

235 posts

ihategeo_1969

#55 h Feb 20, 2025, 4:58 PM

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See that if $H$ is the orthocenter of $\triangle ABD$ then $C \in (BHD)$ and is in arc $\widehat{BD}$ not containing $H$ .

Now $\sqrt{bc}$ invert at $A$ in $\triangle ABD$ and see that $(BHD)$ swaps with $(BOD)$ where $O$ is circumcenter of $\triangle ABD$ . Now $C^*$ lies on that circle and on arc $\widehat{BOD}$ .

Claim: $O$ lies on $(BF^*L^*)$ and similarly on $(DE^*K^*)$ .
Proof: See that $\angle L^*OB+\angle L^*F^*B=2\angle L^*AB+\angle AF^*B=2 \angle F^*AB+\angle AF^*B=180 ^{\circ}$ and done. $\square$

To prove that $O$ is the tangency point, we just need to prove $\angle BOD=\angle BF^*O+\angle DE^*O$ . But that is true by $\angle BF^*O+\angle DE^*O=\frac{\angle AF^*B}2+\frac{\angle DE^*A}2=360 ^{\circ}- \angle AC^*B-\angle DC^*A=\angle BC^*D=\angle BOD$ And done.

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Ilikeminecraft

656 posts

Ilikeminecraft

#56 h Mar 14, 2025, 6:04 AM

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why did it actually take me an hour to do the angle chase even though the reflection was immediate

Let $G$ be the reflection of $A$ across $BD.$ Observe that $AB = BE= BG,$ so $B$ is center of circle $(AEG).$ Similarly, $D$ is center of circle $(AFG).$
Observe that $GBEK, GDLF$ are cyclic because $\angle KBG = 180 - \angle DBG = \frac12(360-\angle ABG) = \frac12(360-2\angle AEG) = 180-\angle AEG = \angle KEG.$ The other is similar.
Let $T$ be a point so that $AT$ is tangent to $(ABK).$
We will show that $AT$ is also tangent to $(ADL).$

$\angle KAL = \angle BAD + \angle LAD + \angle BAE = \angle DCB + 90 - 2\angle BCA +90- 2\angle DCA = 180 - \angle DCB.$

Hence, $\angle ALD = 180-\angle LAK - \angle AKB = \angle BAD - \angle BAC = \angle CAD,$ which finishes.

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cj13609517288

1922 posts

cj13609517288

#57 h Apr 19, 2025, 10:36 PM

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In fact, they are tangent at $A'$ , the reflection of $A$ over $BD$ .

First, to show that $A'$ lies on both circles, just note that $\angle FA'D=90^\circ-\angle FAA'=\angle ALD$ and similarly for the other circle.

Now note that
$\angle DLA'+\angle BKA'=180^\circ-\angle KA'L=180^\circ-\angle KAL=\angle DCB=\angle BA'D,$ as desired. $\blacksquare$

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