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Pascal's Theorem

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Source: EMC 2017

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157 posts

#1 h Dec 26, 2017, 9:28 PM • 3 Y

Y by tiendung2006, Adventure10, ehuseyinyigit

Let $ABC$ be a scalene triangle and let its incircle touch sides $BC$ , $CA$ and $AB$ at points $D$ , $E$ and
$F$ respectively. Let line $AD$ intersect this incircle at point $X$ . Point $M$ is chosen on the line $FX$ so that the
quadrilateral $AFEM$ is cyclic. Let lines $AM$ and $DE$ intersect at point $L$ and let $Q$ be the midpoint of segment
$AE$ . Point $T$ is given on the line $LQ$ such that the quadrilateral $ALDT$ is cyclic. Let $S$ be a point such that
the quadrilateral $TFSA$ is a parallelogram, and let $N$ be the second point of intersection of the circumcircle of
triangle $ASX$ and the line $TS$ . Prove that the circumcircles of triangles $TAN$ and $LSA$ are tangent to each
other.

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157 posts

#2 h Dec 27, 2017, 10:19 AM • 2 Y

Y by Adventure10, Mango247

Any ideas?

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#3 h Dec 27, 2017, 11:13 AM • 1 Y

Y by Adventure10

First consider $K=FE\cap AL$ , after inversion centered at $A$ with radius $AF$ it is easy to check that $K$ is the image of $M$ and that $AKDF$ is clyclic. Angle chasing to show that $\angle LAE=\angle ADL$ and $\angle LKE=\angle KDL$ , these imply that $L$ is the midpoint of $AK$ and then $L,Q,T$ are collinear and $LT\parallel FE$ .

Now, $ALDT$ is cyclic then $\angle TDA=\angle TLA$ , but $\angle TLA=\angle FKA=\angle FDA$ , and then $T,F,D$ are collinear.

$TL$ meets $AF$ at the midpoint, then $S\in TL$ , moreover $\angle ATL=\angle ADL=\angle XFE$ then $F,X,S$ are collinear.

$AS\parallel TD$ implies $\angle XAS=\angle FDA=\angle FKA=\angle SLA$ then $XA$ is tangent to the circumcircle of $\triangle ASL$ at $A$ .

$ASXN$ cyclic and $XS\parallel TA$ imply $\angle ATS=\angle XST=\angle XAN$ then $XA$ is tangent to the circumcircle of $\triangle ATN$ at $A$ and we are done.

This post has been edited 6 times. Last edited by hectorraul, Dec 27, 2017, 11:18 AM

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#4 h Dec 27, 2017, 11:24 AM • 2 Y

Y by microsoft_office_word, Adventure10

It is easy to get that $AM \mid\mid EX$ , now let $T'$ be a point of $DF$ such that $AT' \mid\mid FX$ , note that quadrilaterals $AT'DL$ and $XFDE$ are similar, thus $AT'DL$ is harmonic.Now let $P$ be the point at infinity of line $BC$ and because the tangent at $L$ to circle $ALDT'$ is parallel to $BC$ , we have that $MT'$ bisects $AE$ , because $\ -1=L(A,E;T',L)=(A,E;MT'\cap AE,P)$ . Now the tangency of the two circles can be easily established, by proving that $AD$ is tangent to both of them.

This post has been edited 2 times. Last edited by rmtf1111, Dec 27, 2017, 11:49 AM

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105 posts

#5 h Dec 27, 2017, 5:22 PM • 2 Y

Y by Adventure10, Mango247

My solution:
At first using cyclic quadrilaterals note that $\angle{XEA}=\angle{XFM}=\angle{EAM}$ so $AM \parallel XE$ .
CLAIM 1: $T,F,D$ are collinear.
Proof:Note that since $\angle{DFE}=\angle{DXE}=\angle{DAL}=\angle{LAD}=\angle{LTD}$ , it suffices to show that $TL \parallel FE$ .
Now,since $\angle{XDE}=\angle{XFE}=\angle{EAM}$ deduce that $LA$ is tangent to the circumcircle of triangle $DEA$ so $LA^2=LE\cdot LD$ (1).But since $Q$ is the midpoint of $AE$ , $QA^2=QE^2$ (2),and (1) and (2) imply that $L,Q$ have equal powers with respect to the degenerate circle with center $A$ and the incircle,so $TL$ is the radical axis of these circles and thus is perpendicular to $AI$ where $I$ is the incenter of $ABC$ .But also $AI\perp EF$ so $TL\parallel FE$ ,whih proves claim1.
CLAIM 2: $S\equiv TL\cap FX$
Proof:Let $S'\equiv TL\cap FX$ and it must be shown that $S\equiv S'$ .Let $Y\equiv TL\cap AF$ and $R\equiv XD\cap FE$ .Obviously, $Y$ is the midpoint of $AF$ and the pencil $(FA,FX,FR,FD)$ is harmonic.So since $S'\equiv TL\cap FX$ , $T\equiv TL\cap FD$ and $TL \parallel FE$ ,by a well known lemma in harmonics, $Y\equiv TL\cap FA$ must be the midpoint of $S'T$ .But it is also the midpoint of $AF$ so $ATFS'$ is a parallelogram so $S\equiv S'$ which proves claim 2.
CLAIM 3: $X,E,N$ are collinear.
Proof:Note at first that since $\angle {MAQ}=\angle {SFE}=\angle {QSF}$ and $\angle {QEM}=\angle {AFM}=\angle {ADF}=\angle {ALQ}$ ,quadrilaterals $AMQS,MQEL$ are cyclic.So by angle chasing, $\angle {AXN}=\angle {ASQ}=\angle {AMQ}=\angle {QEL}=\angle {DEC}=\angle {DXE}$ which proves claim 3
CLAIM 4:the circumcircles of triangles $TAN$ and $LSA$ are tangent to each
other.
Proof:It is easy to see that triangles $ATL,XFE$ are similar since they have parallel sides.Let $Ax$ be the tangent line to $A$ of the circumcircle of triangle $ATN$ (in fact this line is just $AX$ ,but this is not necessary to continue).Note that $\angle {SAx}=\angle {NAS}-\angle{NAX}=180-\angle{NXS}-\angle{ATN}=180-\angle {FXE}-\angle {XFE}=\angle {XEF}=\angle {ALS}$ which proves that line $Ax$ is also tangent to the circumcircle of triangle $LSA$ ,which meas that thes circles are mutually tangent,as desired $\blacksquare$

This post has been edited 1 time. Last edited by john111111, Dec 27, 2017, 5:22 PM

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#6 h Jun 16, 2019, 12:50 PM • 2 Y

Y by Adventure10, Mango247

Nice problem...
Let $V$ be the point of intersection of $FE$ and $XD$
First we prove that $AL \parallel XE$
$\angle MAE = \angle MFE = \angle XDE$
$\angle LAD = \angle LEA = \angle DEC = \angle EXD$
Now let prove that $LQ \parallel EF$ . Let $K$ be the midpoint of $XD$ . First note that triangle $ALE$ similar to triangle $DEX$ so $\angle QLE = \angle XEK = \angle FED$ since $FXED$ is harmonic.
Now consider homothety with center $D$ and which sends $XE$ to $AL$ and $LT$ to $EF$ . It is clear that it sends $(XFDE)$ to $(ATDL)$ , $XF$ to $AT$ to so $XF \parallel AT$ and $D - F - T$ are collinear.
Let prove that $S\equiv TL\cap FX$ .
Let $S'\equiv TL\cap FX$ . We prove that $S'F = AT$ and since $AT \parallel XF$ , $ATFS'$ is parallelogram.
First note that
$\angle TS'F = \angle ATS' = \angle XFE = \angle XDE$
and
$\angle LTF = \angle EFD = \angle EXD$
So triangle $TS'F$ similar to triangle $XDE$ and we have
$\frac{FS'}{DE} = \frac{TF}{XE}$

$FS' = \frac{DE*TF}{XE} = \frac{XV}{VD} *TF = \frac{AX}{AD} *TF$

Last equality holds because $(AV, XD) = -1$ .
Since $1-\frac{AX}{AD} = \frac {XD}{DA} = \frac {XF}{AT}$ It is easy to calculate $AT$ and conclude that $AT = S'F$
Now since
$\angle SAD = \angle ADT = \angle XEF = \angle ALT$
And
$\angle NAD= \angle NSF = \angle ATN$
We conclude that $AD$ is tangent to both of out circles so thet tangent to each other.
Done))

This post has been edited 3 times. Last edited by Daniil02, Jun 17, 2019, 2:18 PM

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348 posts

#7 h Sep 3, 2019, 1:35 PM • 1 Y

Y by Adventure10

This is a nice problem
* $(ABC)$ refers to the circumcircle of $\triangle ABC$

Claim 1: (ADL) is tangent to BC

Claim 2: Line LQ contains the A-midline of \triangle AEF

Claim 3: AL is parallel to XE

Claim 4: Points D,F,T are collinear

Claim 5: S is the intersection of TL and FM

Claim 6: Points N,X,E are collinear

Now, it suffices to show that $\angle ATN+\angle ALS=\angle NAS$ since it implies that we can draw a line $l$ through $A$ such that the angle formed by $l$ with $AN$ equals $\angle ATN$ and the angle formed by $l$ with $AS$ equals $\angle ALS$ . This implies that $l$ is a common tangent to $(ATN)$ and $(ALS)$ , which implies that the two circles are tangent.
Now, we have
$\angle ATN+\angle ALS=180- \angle TAL =180 -\angle SML =180-\angle SXN =\angle NAS$ Which proves the problem $\blacksquare$

Bonus property

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Reason: edit

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#8 h Mar 19, 2020, 7:31 AM • 1 Y

Y by amar_04

Let $K$ be the point where $EF$ and $AL$ meet. Note that $\angle AMX=\angle AMF=\angle AEF=\angle EDF=\angle MXE \Rightarrow AL \parallel EX$ By the converse of Reim's Theorem, we get that $BC$ is tangent to $\odot (ALD)$ at point $D$ . We have the following crucial claim:-

CLAIM $L$ is the midpoint of $AK$ .

Proof of Claim Notice that $\angle DAK=\angle DXE=\angle DFE$ which gives that $AFDK$ is cyclic. Then we have $\angle KDA=\angle KFA=\angle EFA=\angle EDF \Rightarrow \{DE,DA\} \text{ and } \{DK,DF\} \text{ are isogonal}$ This gives us $\angle KDE=\angle ADF=\angle AKF$ , and so $AK$ is tangent to $\odot (DEK)$ . Also, $\angle KAE=\angle AEX=\angle EDA$ gives that $AK$ is also tangent to $\odot (ADE)$ . Thus, $L$ lies on the radical axis as well as the common tangent of $\odot (ADE)$ and $\odot (DEK)$ , which directly gives $LA=LK$ . $\Box$

Return to the problem at hand. By our Claim, we get that $LQ$ is parallel to $EF$ ; or equivalently that it bisects $AF$ . But, as $ATFS$ is a parallelogram, so the midpoint of $AF$ must lie on line $ST$ . Thus, we have $S \in LQ$ . And, $\angle DTL=\angle DAL=\angle DXE=\angle DFE \Rightarrow \text{Using the fact that } LT \parallel EF \text{, we get } T \in DF$ Then Reim's Theorem gives $XF \parallel AT$ , which means that $S$ also lies on line $FM$ . This gives us $\angle XAN=\angle XSN=\angle ATN \Rightarrow AX \text{ is tangent to } \odot (TAN)$ Finally, since $FT \parallel AS$ , so we get $\angle XAS=\angle TDA=\angle ALS \Rightarrow AX \text{ is tangent to } \odot (LSA)$ Thus, $\odot (TAN)$ and $\odot (LSA)$ are tangent to each other at $A$ . Hence, done. $\blacksquare$

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691 posts

#9 h Dec 14, 2020, 3:18 PM

Y by

This is such a great and rich configurational geometry problem.

One of the best example to illustrate dissecting figure into pieces before assembling it into one.
Let us first delete the point $T,S,N$ for convenience.
Some Motivation before starting the problem

Claim 01. $XE \parallel AM$ .
Proof. We have $\measuredangle FXE = \measuredangle FDE = \measuredangle FEA = \measuredangle FMA$ .

Claim 02. $LQ \parallel EF$ .
Proof. Notice that $(X,D;E,F) = -1$ . Therefore,
$-1 = (X,D;E,F) \overset{E}{=} (XE \cap AM,L;A, EF \cap DL)$ Let $EF \cap DL = G$ . We then conclude that $L$ is the midpoint of $AG$ .

Claim 03. $AGDF$ is cyclic.
Proof. We have $\measuredangle DFG \equiv \measuredangle DFE = \measuredangle DXE = \measuredangle DAM = \measuredangle DAG$ .

Now, we are ready to deal with other points $T,S$ and $N$ .
Claim 04. $D,F,T$ are collinear
Proof. $\measuredangle ADT = \measuredangle ALT = \measuredangle AL'F = \measuredangle ADF$

Claim 05. $AD$ tangent $(ASL)$ .
Proof. We have
$\measuredangle ALS = \measuredangle AGF = \measuredangle ADF \equiv \measuredangle ADT \overset{AS \parallel TF}{=} \measuredangle DAS$

Claim 06. $N,X,E$ are collinear.
Proof. We have
$\measuredangle NXA = \measuredangle TSA = \measuredangle STD = \measuredangle EFD = \measuredangle EXD$

Claim 07. $AN \parallel DE$ .
Proof. From the previous claim, we have $N,X,E$ are collinear. Therefore $AL \parallel NE$ . Moreover, $Q$ is the midpoint of $AE$ and $L,Q,N$ are collinear by definition of $N$ . Then, $ANEL$ is parallelogram, forcing $AN \parallel LE \equiv DE$ .

Claim 08. $AD$ tangent $(ANT)$ .
Proof. To prove this,
$\measuredangle ATN \equiv \measuredangle ATL= \measuredangle ADL \overset{AN \parallel DE}{=} \measuredangle NAX$
To finish this, we notice that $AD$ is tangent to both $(ANT)$ and $(ASL)$ , which conclude that $(ANT)$ and $(ASL)$ are tangent to each other at $A$ .

Remark:Other Facts

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#10 h Dec 14, 2020, 3:48 PM

Y by

What is EMC?

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30629 posts

#11 h Dec 14, 2020, 4:02 PM

Y by

European Mathematical Cup

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558 posts

#12 h Mar 5, 2021, 8:09 PM • 2 Y

Y by 606234, Mango247

Nice probelm

Let $\ell$ denote the $A$ -midline of $\triangle AEF$ . Note that $\ell$ is the radical axis of $\odot (DEF)$ and $\odot (A,0)$ .

Note that we have :

$\angle FXE = \pi- \angle FDE = \pi- \angle AEF = \pi - \angle AMF = \angle LMX$
Hence $XE \parallel ML$ . We now contend that $L,T \in \ell$ .

Again we angle chase :

$\angle LAE = \angle AEX =\angle XDE$ , which implies that $LA$ is tangent to $\odot (ADE)$ . Hence $LA^2=LD\cdot LE$ . So $L$ lies on the radical axis of $\odot (DEF)$ and $\odot (A,0)$ . Hence we have $LT \parallel EF$ . Now consider the homothety at $D$ that sends $X\mapsto A$ . Note that $E\mapsto X$ and since $EF \parallel LT$ , hence $F \mapsto T$ . So $D,F,T$ are collinear. We can prove as above that $XF \parallel AT$ .

Now note that if $P$ is the midpoint of $AF$ , then $T,P,S$ are collinear, so $S \in \ell$ . Also note that $SF\parallel AT$ , so $S\in XF$ . Redefine $N$ to be the intersection of $XE$ with $\ell$ . We have $ALEN$ is a parallelogram. We show that $A,N,X,S$ lie on a circle. We have :

$\angle NAX = \angle XDE = \angle XFE = \angle NSX$ So the claim follows.

Finally we prove that $AX$ is tangent to both $\odot (TAN)$ and $\odot (LAS)$ .

Note that $\angle NAX = \angle ADL = \angle ATL = \angle ATN$ . So $AX$ is tangent to $\odot (TAN)$ . Similarly its tangent to $\odot (LAS)$ . We're done. $\square$

Motivation

PS

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#13 h Jun 18, 2022, 8:35 PM

Y by

Interesting diagram...
Claim $: AL || XE$ .
Proof $:$ $\angle MAE = \angle MFE = \angle XFE = \angle AEX$ .
Let $FE$ meet $AL$ at $P$ .
Claim $: AFDP$ is cyclic.
Proof $:$ $\angle APF = \angle XEF = \angle XDF = \angle ADF$ .
Claim $: AL = LP$ .
Proof $:$ Note that $\angle LAE = \angle AEX = \angle EDX = \angle LDA \implies LA^2 = LE.LD$ and $\angle PDA = \angle PFA \implies \angle PDL = \angle XFA = \angle XDF = \angle APF = \angle LPE \implies LP^2 = LE.LD$ .
so $LQ || EF$ .
Claim $: LMQE$ is cyclic.
Proof $:$ $\angle MEQ = \angle EPL = \angle MLQ$ .
Claim $: T,F,D$ are collinear.
Proof $:$ $\angle ADT = \angle ALT = \angle APF = \angle ADF$ .
Let $TL$ meet $MF$ at $S'$ .
Claim $: AMQS'$ is cyclic.
Proof $:$ $\angle MAQ = \angle MAE = \angle MFE = \angle MS'Q$ .
Claim $: S'$ is $S$ .
Proof $:$ $\angle AS'M = \angle AQM = \angle ALE = \angle ALD = \angle ADT \implies AS' || FT$ and $\angle TAX = \angle TAD = \angle TLD = \angle LEP = \angle FED = \angle FXD \implies AT || XF$ .
Claim $: ASL$ is tangent to $AD$ at $A$ .
Proof $:$ Note that $\angle DAM = \angle DAP = \angle DFP = \angle DFE = \angle 180 - \angle AED = \angle 180 - \angle QML = \angle 180 - \angle QSA = \angle LSA$ .
Claim $: ANT$ is tangent to $AD$ at $A$ .
Proof $:$ Note that $\angle NTA = \angle NSX = \angle NAX$ .
so $ANT$ and $ASL$ are tangent to each other at $A$ .

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#14 h Jul 23, 2022, 8:12 PM • 1 Y

Y by farhad.fritl

Wow, what a reach configuration it's!
$\angle MAE=\angle MFE=\angle XEA \implies AM||XE$ .
Let $AL\cap EF=J$ . $\angle EAJ=\angle XFE$ , $\angle AJE=\angle XEF$ , and $\angle LEJ=\angle FED=\angle FXD$ . Also $XD$ is $X-$ symmedian in triangle $XEF$ , so $EL$ is median in triangle $AEJ$ . So $AL=LJ$ , which implies $LQ||EF$ . So $LQ$ bisects $AF$ .
Let $DF\cap LQ=T'$ . $\angle ALT'=\angle AJE=\angle XEF=\angle ADT' \implies ALDT'$ is cyclic $\implies T'=T \implies T\in FD$ .
Since triangles $DEF$ and $DLT$ are homothetic, we get $(DLT)$ touches to $BC$ at $D$ .
Let $AB\cap LT=G$ and $FX\cap TL=S'$ . $\angle ATL=\angle ADL=\angle XFE=\angle TS'F\implies AT||FS'$ . Since $AG=GF$ , we get $TFS'A$ is parallelogram. So $S'=S$ .
Let $EX\cap TL=N'$ , Since $AS||DF$ , we get $\angle SAD=\angle XDF=\angle XEF=\angle SN'X \implies ASXN'$ is cyclic $\implies N'=N$ .
$\angle NAX=\angle NSX=\angle XFE=\angle XDE=\angle ATN\implies (TAN)$ touches to $AD$ at $A$ .
$\angle XAS=\angle SNX=\angle XEF=\angle AJE=\angle ALS\implies (LSA)$ touches to $AD$ at $A$ .
So $(TAN)$ and $(LSA)$ touch each other at $A$ . Done!

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1386 posts

#15 h Nov 24, 2022, 1:58 PM

Y by

Nice geo, indeed.

We start with a bit of angle chasing: $\angle MAE = \angle MFE =\angle XEA$ , so we get that $XE \parallel AL$ .

In addition, $\angle LAE= \angle MFE = \angle ADL$ , hence PoP gives that $LA^2=LE.LD$ . Using the well-known trick with point circles and midlines, it is easy to see that $L$ lies on the radical axis of $A$ and $(DEF)$ , so $LQ \parallel FE$ as a midline in $\triangle AFE$ .

Now, the homothety at $D$ taking $X$ to $A$ , $E$ to $L$ and $(XED)$ to $(ALD)$ should take $F$ to $T$ since $EF \parallel LT$ , so $XF \parallel AT$ and $D, T, F$ are collinear.

Redefine $S=XF \cap LQ$ . We proved that $AT \parallel SF$ and that $ST$ contains the midpoint of $AF$ , which is enough to imply that $FTAS$ is a parallelogram.

Now we are ready to finish: we will prove that both circles are tangent to $AD$ by angle chasing. Indeed, $\angle DAS = \angle ADT = \angle ALT$ , so $AD$ is tangent to $(ASL)$ . For the second tangency, note that $\angle NAX= \angle NSX = \angle ATN$ , so we are done.

This post has been edited 2 times. Last edited by VicKmath7, Nov 24, 2022, 3:23 PM

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#16 h Jul 11, 2024, 7:12 PM

Y by

buh nice problem ig but boring

Claim: $ALTD$ is harmonic.
Proof. It suffices to show that the $L$ tangent to $(ALD)$ is parallel to $AC$ . This is true as $\measuredangle LDA = \measuredangle EDX = \measuredangle EFX = \measuredangle EFM = \measuredangle EAM = \measuredangle EAL$
Claim: $EX \parallel AL$
Proof. We show that $\triangle EFD \sim \triangle XAM$ . To see why, $\angle AMX =\angle AEF = \angle EDF$ and $\angle MXA = \angle FXD = \angle FED$ as desired.

Thus there is a homothety at $D$ sending $F,X,E$ to $T,A,L$ . I claim that $S$ is the intersection of $XF$ and $TL$ . Note that $S$ lies on $FX$ since $FX \parallel TA$ , and similar work to the first claim gives $TL$ bisects $AF$ . Thus we showed $S$ is $MF\cap TL$ .

To finish, I claim that $AD$ is tangent to the circles we want to show are tangent. This is just angle chasing:
$\measuredangle DAN = \measuredangle XAN = \measuredangle XSN = \measuredangle XFE = \measuredangle XDE = \measuredangle ADL = \measuredangle ATN$ and similarly
$\measuredangle DAS = \measuredangle XAS = \measuredangle XNS = \measuredangle XEF = \measuredangle XDF = \measuredangle ADT = \measuredangle ALS$

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