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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
How many ordered pairs of numbers can we find?
BR1F1SZ   5
N a few seconds ago by RagvaloD
Source: 2024 Argentina TST P6
Given $2024$ non-negative real numbers $x_1, x_2, \dots, x_{2024}$ that satisfy $x_1 + x_2 + \cdots + x_{2024} = 1$, determine the maximum possible number of ordered pairs $(i, j)$ such that
\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]
5 replies
BR1F1SZ
Jan 25, 2025
RagvaloD
a few seconds ago
10^{f (n)} <10n + 1 <10^{f (n) +1}
parmenides51   4
N 4 minutes ago by AshAuktober
Source: OLCOMA Costa Rica National Olympiad, Final Round, 2018 3.4
Determine if there exists a function f: $N^*\to N^*$ that satisfies that for all $n \in N^*$, $$10^{f (n)} <10n + 1 <10^{f (n) +1}.$$Justify your answer.

Note: $N^*$ denotes the set of positive integers.
4 replies
parmenides51
Sep 20, 2021
AshAuktober
4 minutes ago
FE on Stems
mathscrazy   5
N 12 minutes ago by Levieee
Source: STEMS 2025 Category B4, C3
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$, \[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]Proposed by Aritra Mondal
5 replies
mathscrazy
Dec 29, 2024
Levieee
12 minutes ago
NEPAL TST 2019
khan.academy   10
N 14 minutes ago by MuradSafarli
Prove that there exist infinitely many pairs of different positive integers $(m, n)$ for which $m!n!$ is a square of an integer.

Proposed by Anton Trygub
10 replies
+1 w
khan.academy
May 14, 2019
MuradSafarli
14 minutes ago
No more topics!
H not needed
dchenmathcounts   44
N Mar 16, 2025 by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
Mar 16, 2025
H not needed
G H J
Source: USEMO 2019/1
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dchenmathcounts
2443 posts
#1 • 5 Y
Y by Purple_Planet, samrocksnature, HWenslawski, Rounak_iitr, Tastymooncake2
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
This post has been edited 2 times. Last edited by v_Enhance, Oct 25, 2020, 6:01 AM
Reason: backdate
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mathlogician
1051 posts
#2 • 7 Y
Y by smartninja2000, Purple_Planet, HUNTER963, AllanTian, samrocksnature, Rounak_iitr, Tastymooncake2
We define point $P$ to be the antipode of $D$ with respect to $(DBEF),$ $G$ as the intersection of lines $EF,DO,$ and $AC,$ and $K$ as the intersection of $AC$ and $BD.$

Claim: $DCGF$ is cyclic.

Proof: $\measuredangle DFG = \measuredangle DFE = \measuredangle DBE = \measuredangle DBA = \measuredangle DCA = \measuredangle DCG.$

Claim: $ADGE$ is cyclic.

Proof: $\measuredangle BAD = \measuredangle BCD = \measuredangle FCD = \measuredangle FGD = \measuredangle EGD.$

Claim: $\measuredangle DKC = 90.$

Proof: Notice that since $DP$ is a diameter of $(DBEPF),$ $\measuredangle DBP = 90.$ But also note that $\measuredangle BPD = \measuredangle BED - \measuredangle AGD \implies AG \parallel BP.$ Now $\measuredangle DKG = \measuredangle  DBP = 90,$ so $\measuredangle DKC = 90,$ as desired.


Now, we wish to prove that both $\measuredangle HFE = \measuredangle ACB$ and $\measuredangle HEF = \measuredangle CAB,$ which will prove the desired result by AA similarity. Note that $\measuredangle  HFE = \measuredangle  DFE - \measuredangle DFH = \measuredangle DFE - (90 - \measuredangle EDF) = 90 - \measuredangle FED = 90 - \measuredangle FBD = 90 - \measuredangle CBK = \measuredangle  ACB.$ Also $\measuredangle HEF=\measuredangle DEF -\measuredangle DEH = \measuredangle DEF- (90 - \measuredangle FDE) = 90 - \measuredangle EFD = \measuredangle PFD - \measuredangle EFD = \measuredangle PFE = \measuredangle PDE = \measuredangle GDE = \measuredangle GAE = \measuredangle CAB,$ so $\triangle ABC \sim \triangle EHF,$ as desired.
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jj_ca888
2726 posts
#3 • 6 Y
Y by cosmicgenius, samrocksnature, megarnie, Mango247, Mango247, Tastymooncake2
d michaelpoint
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Asuboptimal
25 posts
#4 • 5 Y
Y by TheUltimate123, zuss77, samrocksnature, PRMOisTheHardestExam, Tastymooncake2
Dumb.
  • Let concurrence point be $T$. Since $\triangle DAC\sim\triangle DEF$, cyclic quads $AETD$, $CFTD$ by spiral sim lemma.
  • If $H_D$ is reflection of $H$ over $\overline{EF}$, then $\measuredangle BDC=\measuredangle EDO=\measuredangle H_DDF$, so $\triangle DAC\cup B\sim\triangle DEF\cup H_D$ end proof.
This post has been edited 4 times. Last edited by Asuboptimal, May 24, 2020, 8:48 AM
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62861
3564 posts
#5 • 5 Y
Y by Purple_Planet, samrocksnature, aopsuser305, PRMOisTheHardestExam, Tastymooncake2
Oops, is this the official thread?

Here is an amusing solution I found when testsolving.

Solution
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dchenmathcounts
2443 posts
#6 • 2 Y
Y by Purple_Planet, samrocksnature
We claim that $AC\perp BD.$

Let $D'$ be the antipode of $D$ with respect to the circle centered at $O,$ let $P$ be the concurrence point, and let $Q$ be the intersection of $AC$ with $BD.$ It then suffices to show that $\angle BD'D=\angle QPD.$

Note that $\angle QPD=\angle CPD.$ We further claim that $\triangle BED\sim \triangle CPD.$

By the Spiral Similarity Lemma, the center of the spiral similarity that sends $B\to C$ and $E\to P$ is the intersection of $(BEF)$ and $(CPF).$ We claim this point is $D.$ Obviously $D$ lies on $(BEF),$ and also, since $\angle PCD=\angle ACD=\angle ABD=\angle EBD=\angle EFD=\angle PFD,$ $D$ lies on $(CPF)$ as well. So $\triangle BED\sim\triangle CPD.$

Since $\angle BD'D=\angle MOD$ and $\angle BD'D=\angle BED=\angle CPD=\angle QPD,$ $OM\parallel PQ,$ implying $AC\perp BD.$

Now it's well-known that
\[\angle FEH=90^{\circ}-\angle EFD\]\[\angle EHF=180^{\circ}-\angle FDE\]\[\angle HFE=90^{\circ}-\angle DEF.\]
Now notice
\[\angle BAC=90^{\circ}-\angle ABD=90^{\circ}-\angle EBD=90^{\circ}-\angle EFD=\angle FEH\]\[\angle ABC=\angle EBF=180^{\circ}-\angle EDF=\angle EHF\]\[\angle BCA=90^{\circ}-\angle DBC=90^{\circ}-\angle DBF=90^{\circ}-\angle DEF=\angle EFH,\]so $\triangle ABC\sim\triangle EHF.$
This post has been edited 1 time. Last edited by dchenmathcounts, May 23, 2020, 11:16 PM
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franchester
1487 posts
#7 • 3 Y
Y by Kagebaka, Purple_Planet, samrocksnature
oh cool, new thread
WLOG, let $C$ be inside $(DEF)$ and $A$ be on the outside.
Let the concurrency point be $T$.
Note that $D$ is the Miquel point of $BCTE$, thus $T$ lies on $(AED)$ and $(CFD)$. The problem falls to simple angle chasing from here, using the fact that $DO$ and $DH$ are isogonal (note: all angles are directed)
\[\angle CAB=\angle TAE=\angle TDE=\angle FDH=90^{\circ}-\angle EFD=\angle HEF\]\[\angle BCA=\angle FCT=\angle FDT=\angle HDE=90^{\circ}-\angle DEF=\angle EFH\]and we're done.
This post has been edited 1 time. Last edited by franchester, May 23, 2020, 11:31 PM
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Stormersyle
2785 posts
#8 • 3 Y
Y by Purple_Planet, samrocksnature, Mango247
sketch:

Angle chase to reduce to proving $AC\perp BD$. Note $D$ is the center of the spiral similarity sending $AC$ to $EF$, which also sends $N$ to $O$. So Complex bash finishes (set $d=0, n=1, e=az, f=cz, o=z$, and then prove $NO||AC$).
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nukelauncher
354 posts
#9 • 6 Y
Y by Purple_Planet, smartninja2000, crazyeyemoody907, samrocksnature, Mango247, Tastymooncake2
"Easier" way for people who lack a geometer's brain: 1/2 line of angle chasing + 4 pages of complex bash.
This post has been edited 1 time. Last edited by nukelauncher, May 23, 2020, 11:43 PM
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MP8148
888 posts
#10 • 10 Y
Y by Purple_Planet, Smkh, Aryan-23, amar_04, samrocksnature, PRMOisTheHardestExam, Mango247, Mango247, Mango247, Tastymooncake2
No one bothers to make a diagram :P
[asy]
size(8cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.35));
dotfactor *=1.5;

pair O = origin, D = dir(25), F = dir(120), B = dir(200), E = dir(340), G = extension(D,O,E,F), A = intersectionpoint(circumcircle(D,E,G),B--E+dir(E--B)*0.0069), C = extension(A,G,B,F), H = orthocenter(D,E,F);

draw(circumcircle(A,D,E)^^unitcircle, purple);
draw(A--B--C--A^^E--F--H--E, heavygreen);
draw(A--E--D--F);
draw(D--O, orange);

dot("$A$", A, dir(255));
dot("$B$", B, dir(200));
dot("$C$", C, dir(90));
dot("$D$", D, dir(45));
dot("$E$", E, dir(320));
dot("$F$", F, dir(140));
dot("$G$", G, dir(245));
dot("$H$", H, dir(30));
dot("$O$", O, dir(200));
[/asy]
Let $G = \overline{BF} \cap \overline{AC}$. It follows that $D$ is the Miquel point of $ABFG$, so $AEDG$ is cyclic. Then $$\measuredangle EHF = \measuredangle FDE = \measuredangle FBE = \measuredangle CBA$$and $$\measuredangle HEF = 90^\circ + \measuredangle DFE = \measuredangle ODE = \measuredangle GDE = \measuredangle GAB = \measuredangle CAB,$$so $\triangle ABC \overset{-}\sim \triangle EHF$ as desired. $\blacksquare$
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GeoMetrix
924 posts
#12 • 7 Y
Y by Williamgolly, Purple_Planet, amar_04, srijonrick, samrocksnature, Rounak_iitr, vangelis
Nothing but just trivial angle chase.

Let $X$ be the concurrency point. Notice that since $$\angle XFD=\angle EFD=\angle EBD=\angle ACD$$hence $(XFDC)$ is cyclic. Now just notice that $$\angle ACB=\angle XCF=\angle XDF=\angle ODF=90^\circ-\angle DBF=90^\circ-\angle DBC$$hence $\overline{AC} \perp \overline{BD}$. To finish notice that we have that $$\angle EFH=90^\circ-\angle DEF=90^\circ-\angle DBF=\angle ACB$$and also we have that $$\angle EHF=180^\circ-\angle EDF=\angle EBF=\angle ABC$$and hence the similiarity follows $\qquad \blacksquare$

[asy]
size(8cm);
pair A=(-8.5,3.69);
pair B=(-4.1004880575605736,6.553349874699234);
pair C=(4.46,-2.71);
pair D=(-9.186044509110761,-3.766637297304822);
pair O=(-5.862865536958581,1.0087848755370215);
pair E=(-11.646182892013208,1.642358691716196);
pair F=(-0.19673893348792193,2.3290820857385617);
pair H=(-9.303235260694732,-1.8127662709240977);
pair X=(-5.14334178555926,2.032391005214449);
draw(A--B--C--D--A,purple);
draw(circumcircle(A,B,C),red);
draw(A--E,purple);
draw(A--C,green);
draw(B--D,green);
draw(E--F,blue+dotted);
draw(D--X,pink);
draw(E--D,blue+white);
draw(D--F,blue+white);
draw(circumcircle(E,D,F),orange+dotted);
draw(circumcircle(D,X,F),cyan+dashed);
dot("$A$",A,N);
dot("$B$",B,N);
dot("$C$",C,SE);
dot("$D$",D,W);
dot("$E$",E,W);
dot("$F$",F,NE);
dot("$X$",X,N);
dot("$O$",O,S);
dot("$H$",H,N);
[/asy]
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wwt8167
94 posts
#13 • 3 Y
Y by samrocksnature, Mango247, Aspiring_Mathletes
ez angle chasing
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Greenleaf5002
130 posts
#14 • 3 Y
Y by Aspiring_Mathletes, samrocksnature, Mango247
Let $X=EF\cap AC\cap DO$
Claim: $AC\perp BD$
$proof:\;\angle DEF=\angle DBF=\angle DBC=\angle DAC$ which implies $AEXD$ is cyclic.
Hence, we see $\angle BDC=\angle BAC=\angle EAX=\angle EDX$ and so, $\angle XDC=\angle EDB=\angle EFB$ and hence, $DXCF$ is cyclic.

$\therefore 90^{\circ}-\angle CBD=90^{\circ}-\angle FED=\angle XDF=\angle ACB$ and so, $AC\perp BD$
It is now easy to see that $\angle FEH=90^{\circ}-\angle XFD=90^{\circ}-\angle EBD=\angle BAC$ and $EFH=90^{\circ}-\angle FED=90^{\circ}-\angle XDF=90^{\circ}-\angle BCA$ and hence $\triangle ABC \sim \triangle EHF$
[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(20cm); 
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 = -10.951719144867154, xmax = 10.205305648521268, ymin = -8.086246158049283, ymax = 4.260861279967215;  /* image dimensions */

 /* draw figures */
draw(circle((-0.8189320936592647,-0.10041731422676856), 3.59681270586667), linewidth(0.5)); 
draw((1.1309254832320157,2.9220183047606043)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); 
draw((-2.781184345331607,2.913986066525506)--(1.1309254832320157,2.9220183047606043), linewidth(0.5)); 
draw((-2.781184345331607,2.913986066525506)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); 
draw((-2.781184345331607,2.913986066525506)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); 
draw((-3.7113245978215597,-2.238436781619356)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); 
draw(circle((0.17674570293097336,-1.034699428989725), 4.070144257604352), linewidth(0.5)); 
draw((2.31409603169182,-1.8671081210299814)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); 
draw((2.8637276646324716,-4.091855331981989)--(-0.793673616147727,2.9180667698726), linewidth(0.5)); 
draw((-0.793673616147727,2.9180667698726)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); 
draw((-3.7113245978215597,-2.238436781619356)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); 
draw((-0.793673616147727,2.9180667698726)--(-1.9947619551987608,-1.3428264857492949), linewidth(0.5)); 
draw((-1.9947619551987608,-1.3428264857492949)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); 
draw((-1.9947619551987608,-1.3428264857492949)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); 
draw(circle((9.950552748009656,16.239095787241443), 3.056934751946367), linewidth(0.5)); 
draw(circle((13.43307492051343,13.410111137651443), 5.583923634788888), linewidth(0.5)); 
draw((xmin, -3.9076923076922974*xmin + 64.55197035721586)--(xmax, -3.9076923076922974*xmax + 64.55197035721586), linewidth(0.5)); /* line */
draw((xmin, 0.10780669144981424*xmin + 17.462742100974747)--(xmax, 0.10780669144981424*xmax + 17.462742100974747), linewidth(0.5)); /* line */
draw((7.68479475445666,18.291214397923607)--(7.879417498909836,13.990707729633588), linewidth(0.5)); 
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This post has been edited 3 times. Last edited by Greenleaf5002, May 24, 2020, 9:11 AM
Reason: i want to kill myself
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Greenleaf5002
130 posts
#15 • 2 Y
Y by samrocksnature, Mango247
redacted
This post has been edited 3 times. Last edited by Greenleaf5002, May 24, 2020, 9:10 AM
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Kimchiks926
256 posts
#16 • 4 Y
Y by mkomisarova, samrocksnature, Mango247, Mango247
Let $X$ be intersection of $AC$, $DO$, $EF$.

Claim: Quadrilaterals $XCFD$ and $AEXD$ are cyclic.
Proof: Note that from cyclic quadrilaterals $EBFD$ and $ABCD$ we obtain:
$$ \angle EFD = \angle EBD = \angle ABD = \angle ACD = \angle XCD$$$$ \angle DEF = \angle DBF = \angle DBC = \angle DAC = \angle DAX$$This proofs our claim.

Since $O$ is circumcenter of $\triangle DEF$, we get that $\angle ODF = 90 - \angle FED$. Since $XCFD$ is cyclic:
$$\angle ODF = \angle XDF = \angle BCX = \angle BCA = 90 - \angle FED$$But since $H$ is orthocenter of $\triangle DEF$, we get that $\angle HFE = 90 - \angle FED = \angle BCA$. Similarly we can prove that $\angle BAC = \angle HEF$, which implies that $\triangle ABC \sim \triangle EHF$$ and we are done.
This post has been edited 1 time. Last edited by Kimchiks926, May 24, 2020, 9:11 AM
Reason: typo
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Mathematicsislovely
245 posts
#18 • 1 Y
Y by samrocksnature
As first, as $H$ is the orthocentre of $DEF$ implies $D$ is the orthocentre of $HEF$.So $\angle EHF=180^{\circ}-\angle EDF$

Claim:
$\angle ABC=180^{\circ}-\angle EDF(=\angle EHF)$
proof:
$\angle DFE=\angle DBE=\angle DBA=\angle DCA$[by 2 cyclic quadrilateral].Similarly ,$\angle DEF=\angle DAC$.So
$\angle EDF=\angle ADC=180^{\circ}-\angle ABC$.
SO,$\angle ABC=180^{\circ}-\angle EDF(=\angle EHF)$$\blacksquare$

Let $AC,WE,DO$ concur at $X$.From the proof of last claim we have that
$\angle DFX=\angle DFE=\angle DBE=\angle DBA=\angle DCA=\angle DCX$.So,$XDFC$ concyclic.So $\angle XDF= \angle XCB=\angle ACB=x$.

NOW,as $O$ is the centre of $(BDF)$ So $\angle DOF=180^{\circ}-2\angle ODF=180^{\circ}-2x$ which implies$\angle DBF=90^{\circ}-x$.

Again as $DXEA$ cyclic,
So $90^{\circ}-x=\angle DBC=\angle DAX=\angle DEX$.As $D$ is orthocentre of $HEF$ so
$\angle HFE=90^{\circ}-\angle DEF=x=\angle ACB$.So $ABC$ and $EHF$ are similar.$\blacksquare$
This post has been edited 5 times. Last edited by Mathematicsislovely, May 24, 2020, 10:56 AM
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v_Enhance
6858 posts
#19 • 3 Y
Y by v4913, samrocksnature, Mango247
Also some solutions posted in https://artofproblemsolving.com/community/c5h2119379_concurrence_implies_similarity_on_cyclic_quads.
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jeteagle
480 posts
#20 • 1 Y
Y by samrocksnature
>:( forgot about directed angles so i think i only got a 6
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brianzjk
1201 posts
#21 • 4 Y
Y by Mathscienceclass, cocohearts, samrocksnature, Mango247
i was thinking during the test: "make sure to use directed angles"
me during the writeup: "normal angles go brrbrr"
This post has been edited 1 time. Last edited by brianzjk, May 24, 2020, 1:50 PM
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AopsUser101
1750 posts
#22 • 3 Y
Y by v4913, samrocksnature, Mango247
I didn’t solve this live doing math the day after you get a concussion is hard. Also not sure if this is right because I didn’t direct anything. Define $X$ to be the intersection of $EF, AC, DO$.
Claim: $AXDE$ is cyclic.
Proof. Since $ABCD$ is cyclic, $\angle DAC = \angle DBC = \angle DBF$. Since $EBFD$ is cyclic, $\angle DBF = \angle DEF$. Thus, $\angle DAC = \angle DEF$ which proves that $AXDE$ is cyclic.
Claim: $XFCD$ is cyclic.
Proof. $\angle EAD = \angle BCD$ since $ABCD$ is cyclic. Furthermore, $\angle EAD = \angle EXD  = 180 - \angle DXF$. $\angle DXF = 180 - \angle EAD$. Hence, $\angle BCD + \angle DXF = 180$, proving that $XFCD$ is cyclic.
Claim: $BD \perp AC$.
Note that $O$ is the center of the circumcircle of $BDF$. Therefore, $\angle DOF = 2 \angle BDF$, which implies that $\angle XDF = 90 - \angle DBF$. Since $XFCD$ is cyclic, $\angle XDF = \angle XCF = 90 - \angle DBF$. Hence, $BD \perp AC$.
Claim: $\triangle ABC \sim \triangle EHF$.
Let $\angle FED = a$ and $\angle ABD = b$. Then, since $EAXD$ is cyclic, $\angle DAX = a$ and since $ABCD$ is cyclic, $\angle DBC = a$ as well. Since $BD \perp AC$, $\angle XCF = 90 - a$. Similarly, since $ABCD$ is cyclic, $\angle ACD = b$ and since $BD \perp AC$, $\angle BAC = 90 - b$. As $XFCD$ is cyclic, $\angle XFD = \angle XCD = b$. Thus, $\angle ABC = b + a, \angle BCA = 90 -a, \angle BAC = 90 - b$ and $\angle DEF = a, \angle EFD = b, \angle EDF = 180 - a - b$. Since $EH \perp DF$, $\angle EHF = 90 - b$ and since $HF \perp ED$, $\angle EFH = 90 -a$, which implies that $\angle EHF = a + b$. This proves that $ABC$ and $EHF$ are similar.

EDIT: 1500th post
This post has been edited 1 time. Last edited by AopsUser101, May 24, 2020, 3:23 PM
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math_pi_rate
1218 posts
#23 • 3 Y
Y by amar_04, itslumi, samrocksnature
Solution
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mira74
1010 posts
#24 • 2 Y
Y by tworigami, Mango247
Didn't post my solution at the time but i thought it was interesting and I don't see it so

[asy]
import graph; size(10cm); 
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[/asy]
Screw directed angles.

Draw the simson line of $D$ to both $BAC$ and $BEF$ as labelled. Now, $$\angle ODS = \angle TDS = \angle TPS = \angle EPQ = \angle EDQ = 90^{\circ} - \angle DEQ = 90^{\circ} - \angle DEB = \angle ODB$$so $D$, $S$, $B$ are collinear. Thus, $DB \perp AC$, implying the result.

Edit: Here's another solution I'm surprised not to see:

Note that since $\angle FDC = \angle EDA$ by spiral similarity, $DC$ and $DE$ are isogonal in $\angle ADF$. Thus, by the second isogonality lemma, $DO$,$DB$ are isogonal in $\angle ADF$, giving $$\angle ADB = \angle FDO = 90^{\circ} - \angle DBF = 90^{\circ} - \angle DBC = 90^\circ - \angle DAC$$giving $BD \perp AC$, implying the result.
This post has been edited 2 times. Last edited by mira74, Jun 18, 2020, 1:43 AM
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poplintos
757 posts
#25
Y by
This was a very nice problem.
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crazyeyemoody907
450 posts
#26
Y by
nukelauncher wrote:
"Easier" way for people who lack a geometer's brain: 1/2 line of angle chasing + 4 pages of complex bash.

Me neither... can't launch nwooks during live olympiad
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Cindy.tw
54 posts
#27
Y by
Let $J$ be the intersection of $EF$ and $AC$, note that $D$ is the Miquel point of $\mathcal{Q}(AB, BC, AC, EF)$, hence $\measuredangle EDJ = \measuredangle BDC = \measuredangle BAC$. Since $D, O, J$ are collinear, $\measuredangle EDJ = \measuredangle HDF = \measuredangle FEH$. Combining that $\measuredangle ABC = \measuredangle EBD = \measuredangle EDF = \measuredangle FHE$, we have $\triangle ABC \sim \triangle EHF$.
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bever209
1522 posts
#28 • 2 Y
Y by centslordm, Mango247
Click to reveal hidden text
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IAmTheHazard
5000 posts
#29 • 1 Y
Y by centslordm
Let $X$ be the concurrency point. Observe that
$$\measuredangle DEX=\measuredangle DEF=\measuredangle DBF=\measuredangle DBC=\measuredangle DAC=\measuredangle DAX,$$hence $ADEX$ is cyclic. Likewise, $CDFX$ is cyclic.
Now let $M$ be the midpoint of $\overline{ED}$, and $P$ be the intersection between $\overline{AC}$ and $\overline{BD}$. Since $\overline{XD}$ and $\overline{XE}$ are radii of the same circle, we have $\overline{XM} \perp \overline{DE}$. Since we have
$$\measuredangle BAP=\measuredangle EAX=\measuredangle EDX=\measuredangle MDX,$$as well as
$$\measuredangle PBA+\measuredangle DBA=\measuredangle DCA=\measuredangle DCX=\measuredangle DFX=\measuredangle DFE=\measuredangle DXM,$$it follows that $\triangle ABP\sim \triangle DXM$. As $\angle XMD=90^\circ$, we have $\angle APB=90^\circ$, hence $\overline{AC} \perp \overline{BD}$.
Since $\overline{DF} \perp \overline{EH}$, we have $\measuredangle FEH=90^\circ-\measuredangle DFE$. Similarly, $\angle EFH=90^\circ-\angle DEF$. Now, we have
$$\measuredangle DFE=\measuredangle DBE=\measuredangle PBA=90^\circ-\measuredangle BAP,$$so $\measuredangle FEH=\measuredangle BAP=\measuredangle BAC$. Likewise, $\measuredangle EFH=\measuredangle BCA$, so $\triangle EHF$ and $\triangle ABC$ are similar, as desired. $\blacksquare$
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bora_olmez
277 posts
#30
Y by
Cool problem but somebody should get me glasses.

Let $X$ be the concurrence point.
Notice that $$\triangle DFE \sim \triangle DCA $$Now, $$\angle HFE = 90^{\circ}-\angle DEF  = 90^{\circ} - \angle DAC$$Consequently, we have to prove that $$\angle ACB = 90^{\circ} - \angle CAD$$because then $$\angle HFE = \angle ACB$$and analogously $$\angle HEF = \angle BAC$$Notice that $$\angle DFX = \angle DFE = \angle DCA$$meaning that $DCXF$ is cyclic.
Then $$\angle ACB = \angle FCX = \angle FDX = \angle FDO = 90^{\circ}-\angle DEF = 90^{\circ} - \angle DBC$$as desired. $\blacksquare$
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554183
484 posts
#31
Y by
For the given condition to hold, it is enough to show that $AC \perp BD$.
Let the concurrence point be $X$. Note that
$$\angle{XFD}=\angle{EFD}=\angle{EBD}=\angle{ABD}$$This shows that $XCFD$ is cyclic.
Now, notice
$$\angle{XDC}=\angle{XFC}=\angle{XFB}=\angle{EFB}=\angle{EDB}$$Notice
$$\angle{DOE}=2\angle{DFE}$$Therefore,
$$\angle{XDE}=90-\angle{DFE} \implies \angle{BDX}=90-\angle{DFE}-\angle{EDB} \implies \angle{BDC}=90-\angle{DFX}=90-\angle{DBA}$$As desired
This post has been edited 1 time. Last edited by 554183, Sep 22, 2021, 1:54 PM
Reason: Lol
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primesarespecial
364 posts
#32
Y by
Let $Q$ be the concurrence point.
We see that $\angle FHE=\angle ABC$ .
Thus it suffices to prove $AC \perp BD$.
Now, we see that $\angle QFD=\angle EBD=\angle ACD$.
Thus $QFCD$ is cyclic.
Which gives that $\angle BCA=\angle FCQ=\angle FDO=90- \angle FED=90-\angle DBC$
which gives the desired.
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MrOreoJuice
594 posts
#33
Y by
(Sorry if I mess up dangles) Let $X$ be the point of concurrency.
$$\measuredangle DAX = \measuredangle DBC = \measuredangle DBF = \measuredangle DEX \implies DAEX \text{ is cyclic.}$$$$\measuredangle DCX = \measuredangle DCA = \measuredangle DBE = \measuredangle DFX \implies DFCX \text{ is cyclic.}$$Note that $\{EH , EO\}$ and $\{FH,FO\}$ are isogonal wrt $\triangle EDF$.
$$\measuredangle BAC = \measuredangle EDG = \measuredangle OED = \measuredangle FEH$$$$\measuredangle BCA = \measuredangle FDG = \measuredangle OFD = \measuredangle EFH$$which finishes the problem.
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mathleticguyyy
3217 posts
#34 • 1 Y
Y by centslordm
We can see that $D$ is the miquel point of the complete quad $ABCEFX$, giving us cyclic quad $AEDX$ where $X=EF\cap AC$.

Note that $\angle XDB=\frac{180-\angle BOD}{2}=90-\angle BED=\angle AXD-90=90-\angle CXD$ (assuming that $E$ lies beyond $A$), hence $BD$ is perpendicular to $AC$. Now, the desired spiral similarity sends $B$ to a point $H'$ on $(DEF)$ such that $DH'\perp EF$, and we are done by orthocenter-reflection.
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Mogmog8
1080 posts
#35 • 5 Y
Y by centslordm, teomihai, megarnie, crazyeyemoody907, Rounak_iitr
Let $X=\overline{AC}\cap\overline{DO}\cap\overline{EF},Y=\overline{AC}\cap\overline{BD},Z=\overline{DO}\cap\overline{BC},$ and $D'$ the reflection of $D$ in $O.$ We see $\angle EHF=\angle 180-\angle EDF=\angle CBA.$ Notice $D$ is the Miquel point of $BEXZ$ so $D$ lies on $(DFX).$ Hence, $$\angle YBC=\angle DD'F=90-\angle D'DF=90-\angle YCB$$so $\overline{AC}\perp\overline{BD}.$ Therefore, $$\angle ACB=90-\angle CBD=90-\angle FED=\angle HFE.$$$\square$
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ALM_04
85 posts
#36
Y by
Since, $BEDF$ is a cyclic quadrilateral.
$$\measuredangle{FHE}=\measuredangle{EDF}=\measuredangle{EBF}=\measuredangle{ABC}$$
Let $X=\overline{EF}\cap\overline{AC}\cap\overline{DO}$.
$\measuredangle{DCX}=\measuredangle{DCA}=\measuredangle{DBA}=\measuredangle{DBE}
=\measuredangle{DFE}=\measuredangle{DFX}\implies DXFC$ is a cyclic quadrilateral.

Now,
Using the above thing and the fact that $O$ is the circumcenter of $DBF$. It can be shown that $\overline{CA}\perp \overline{BD}$.
Hence, $\measuredangle{EFH}=90^{\circ}-\measuredangle{DEF}=90^{\circ}-\measuredangle{DBF}=\measuredangle{BCA}$ which proves that $ABC$ and $EHF$ are similar. $\blacksquare$
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signifance
140 posts
#37
Y by
Let the concurrence point be P. We make a couple of observations:
\begin{itemize}
\item DAC=DBF=DEF,ACD=EBD=EFD\implies ACD\sim EFD\implies ABC=180-ADC=180-EDF=EHF
\item DFP=ACD=DCP,DEP=DAC=DAP, so we derive those two cyclic quads.
\item ACB=ODF=90-DBC\implies AC\perp BD
\item EFH=90-DEF=90-DAC=ADB=ACB
\end{itemize}
By AA similarity we get the desired conclusion.

\textbf{Remark.} It's not easy to conjecture AC\perp BD, but it would be wanted from the last item in our list (we'd want 90-DAC=90-DEF=EFH=ACB=ADB).

btw does anyone have recommendations on how to get the itemize thing to work, I've been working on overlefa with \usepackage[s3xy]{evan} which makes it work but idk how to on aops
This post has been edited 1 time. Last edited by signifance, Sep 23, 2023, 5:48 PM
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naonaoaz
329 posts
#38
Y by
Let $P = \overline{AC} \cap \overline{EF}$.
Claim: $AEPD$ and $CPDF$ are cyclic.
Proof:
\[\angle EPA = \angle CPF = 180^{\circ} - \angle PCF - \angle CFP = \angle BCA - \angle BFE\]\[ = \angle BDE + \angle EDA - \angle BFE = \angle EDA\]$CPDF$ follows similarly. $\square$

Let $\angle A$ and $\angle B$ denote the angles at $A$ and $B$ of cyclic quadrilateral $ABCD$. Notice that $\angle EDF = 180^{\circ} - \angle B$, so $\angle EHF = \angle ABC$. It suffices to show that $\angle EFH = \angle BCA$.

Now if we let $\angle BAC = \alpha$, we get $\angle FED = \angle A - \alpha$, so $\angle EFH = 90^{\circ} - \angle A + \alpha$.
Claim:
\[\alpha = \frac{90^{\circ}+\angle A - \angle B}{2}\]which finishes since then $\angle BCA = 180^{\circ} - \angle B - \alpha = \angle EFH$.
Proof:
\[\angle EFD = \angle EBD = \angle ABD = \angle ACD = 180^{\circ} - \angle CAD - \angle ADC\]\[\implies \angle EFD = 180^{\circ} - (\angle A - \alpha)- (180^{\circ} - \angle B) = \angle B - \angle A + \alpha\]\[\implies EDO = 90^{\circ} - \angle EFD = 90+\angle A-\angle B- \alpha\]Since $AEPD$ cyclic, $\alpha = \angle EAP = \angle EDP = \angle EDO$, which finishes. $\square$
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ezpotd
1247 posts
#39
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Observe that $D$ is the orthocenter of $EHF$, so it suffices to prove that $ACX$ and $EFD$ are similar where $X$ is the orthocenter of $ABC$.

Our main claim is that $D$ is the reflection of $X$ over $AC$. To see this, animate $O$ linearly along the perpendicular bisector of $BD$. Then $DEFO$ is similar to a fixed configuration, so $EF \cap DO$ moves linearly.

It is easy to see that when $O$ is the circumcenter of $ABCD$, we have $EF \cap DO$ lying on $AC$. If $BD$ is not perpendicular to $AC$, then we can choose some point $O$ such that the entire line segment $DO$ is parallel to $AC$, so $EF \cap DO$ does NOT lie on $AC$, so the line that $EF \cap DO$ moves along is never on $AC$ unless $AC = EF$, which we can discard. As a result, $BD$ must be perpendicular to $AC$ and $X$ is the desired reflection point.

Now it is easy to see the desired result, observe $ACX$ is similar to $CAD$, which is always similar to $EFD$ as we animate $O$ along the perpendicular bisector.
This post has been edited 2 times. Last edited by ezpotd, Nov 24, 2023, 6:34 PM
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ihatemath123
3430 posts
#40
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Claim: $\triangle DEF \sim \triangle DAC$.

Proof: this is true even without the concurrency information. We have
\[\angle DEA = \angle DBE = \angle DFB = \angle DFC. \]Furthermore, we have
\[\angle EDF = \angle EBF = 180^{\circ} - \angle EBC = \angle ADC, \]so $\angle EDA = \angle FDC$.

Thus, $\triangle DEA \sim \triangle DFC$, so $\triangle DEF \sim \triangle DAC$.

Let $X$ be the concurrency point in the problem, let $D'$ and $D''$ be the antipodes of $D$ WRT $(DEF)$ and $(ABCD)$, respectively, and let $Y$ be the intersection of $\overline{DG}$ with $\overline{AC}$.
Because of similarity, $\frac{DY}{DG} = \frac{DX}{DD'}$, so $\overline{GD'} \parallel \overline{YX}$.

Because $B$ lies on both $(DAC)$ and $(DEF)$, we have that $\angle DBG = \angle DBD' = 90^{\circ}$, hence $D$, $B$ and $G$ are collinear. It remains to prove that $B \neq G$; then, from there, we know that $B$ is the unique point which lies on $(ACD)$ satisfying $\overline{BG} \parallel \overline{AC}$, which means $B$ is the reflection of the orthocenter of $\triangle ACD$ across $\overline{AC}$, which means $\triangle ABC \sim \triangle EHF$ as desired.

Proof: Assume, for the sake of contradiction, that $B = G$; then, $G$ would lie on $(EDF)$. Since $\angle DCG = \angle DAG = 90^{\circ}$, line $AC$ would be the Simson line in $(DEF)$ WRT $D$ and $\triangle BEF$. So, $X$ would be the foot from $D$ to $\overline{EF}$, implying $DE = DF$, implying $DAC$ collinear, which is a contradiction.

remark: dealing with the edge case at the end was by far the most interesting part of this problem :(
This post has been edited 2 times. Last edited by ihatemath123, Jan 29, 2024, 6:34 AM
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HamstPan38825
8854 posts
#41
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The key is the following:

Claim. $\triangle DAE \sim \triangle DCF$.

Proof. $\angle EAD = \angle FCD$ is evident. Other equality comes from $BEDF$ cyclic. $\blacksquare$

So $\triangle DAC \sim \triangle DEF$ by spiral similarity, ergo for $G = \overline{AC} \cap \overline{EF}$, $GEAD$ and $GFCD$ are cyclic. So $\angle EFD = \angle ACD$ and $\angle FED = \angle CAD$. Then it suffices to show that $ABCD$ is orthodiagonal.

To see this, let $D'$ be the $D$-antipode. It suffices to show that $\angle BDD' + \angle AGD = 90^\circ$, but this follows as $\angle AGD = \angle AED$ and $\angle BDD' = \angle BED - 90^\circ$.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 10, 2024, 3:54 AM
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DroneChaudhary
4 posts
#42
Y by
Could someone check my solution...

$\text{Let the perpendicular from } D \text{ to } EF \text{ meet the circle centred at }O \text{ again at }H'$
$\text{We have that }\triangle ABC \sim \triangle EHF \cong \triangle EH'F \text{ and so,}$
$$\angle H'DF = \angle H'EF = \angle BAC = \angle BDC \text{ with } \angle CBD = \angle FBD = \angle FH'D$$$$\implies \text{ a spiral similarity centred at } D \text{ mapping } (ABCD) \text{ to }(EH'FD)$$$\text{We get that }AC \perp BD$
$\text{Now let } AC \cap EF = T;  \text{ then } \angle DCA = \angle DFE = \angle DFT \implies (DCTF) \text{ and hence we have }$
$$\angle CDT = \angle CFT = \angle BFE = \angle BDC = \angle H'DF = \angle CDO \implies D-O-T \text{ collinear}$$$\text{i.e. AC, DO, EF concurrent }$
This post has been edited 1 time. Last edited by DroneChaudhary, Jun 3, 2024, 1:50 AM
Reason: formatting
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Thapakazi
53 posts
#43 • 1 Y
Y by Rounak_iitr
sobad.

We let the concurrency point be $K$.

Claim: $DAKE$ cyclic.

By immediate angle chase, we get

\[\measuredangle DAC = \measuredangle DBC = \measuredangle DEF = \measuredangle DEK\]
implying the claim. Now notice that,

\[\measuredangle DKF = \measuredangle DKA + \measuredangle AKF =  \measuredangle DEA +  \measuredangle EDA =  \measuredangle DAB.\]
Also,

\[ \measuredangle KFD =  \measuredangle EFD =  \measuredangle EBD =  \measuredangle ABD.\]
So, $\triangle DKF \sim \triangle DAB.$ Now finally as,

\[\measuredangle EHF = - \measuredangle EDF =  -\measuredangle EBF =  \measuredangle EBC =  \measuredangle ABC\]
and,

\[ \measuredangle EFH =  \measuredangle OFD =  \measuredangle ODF =  \measuredangle KDF =  \measuredangle ADB =  \measuredangle ACB\]
we get $\triangle EHF \sim \triangle ABC$ as needed.
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shendrew7
792 posts
#44 • 1 Y
Y by Rounak_iitr
Suppose the concurrency point is $K$ and the center of $(ABCD)$ is $P$. Notice that
\[\measuredangle KFD = \measuredangle ABD = \measuredangle ACD,\]
so $CDFK$ is cyclic. Then
\[\measuredangle POD = \measuredangle CFD = \measuredangle CKD \implies AC \parallel OP \perp BD,\]
so our desired similarity is derived from
\begin{align*}
\measuredangle ABC &= \measuredangle EBF = \measuredangle EDF = \measuredangle FHE. \\
\measuredangle CAB &= 90 - \measuredangle ABD = 90 - \measuredangle EFD = \measuredangle HEF. \\
\measuredangle BCA &= 90 - \measuredangle DBC = 90 - \measuredangle DEF = \measuredangle EFH. \quad \blacksquare
\end{align*}
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L13832
250 posts
#45 • 1 Y
Y by Rounak_iitr
Solution
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Saucepan_man02
1299 posts
#46
Y by
Pure Angle-Chase:

Claim: $DEAX, DCFX$ are cyclic.
Notice that: $\angle ACD = \angle  ABD = \angle EFD = \angle XFD$ thus $CXFD$ is cyclic.
Notice that: $\angle AXD=\angle CXD=\angle CFD=\angle BFD=\angle BED = \angle AED$ thus $AEXD$ is cyclic.

Notice that: $\angle CDF = \angle CXE = \angle AXE = \angle ADE$ and thus: $\angle ADC = \angle BDF$ which implies $\angle ABC = \angle EHF$.

Notice that: $\angle ACB = \angle XCF = \angle XDF = \angle EDH = \angle EFH$. Similarly, $\angle BAC = \angle HEF$ and thus: $\triangle ABC \sim \triangle EHF$
This post has been edited 2 times. Last edited by Saucepan_man02, Feb 17, 2025, 1:34 PM
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Ilikeminecraft
298 posts
#47
Y by
nice and chill
let $X$ be the concurrency point
Note that it is given $BDEF$ is cyclic.
Thus, $\angle DEF = \angle DBC = \angle CAD,$ so $AEXD$ is cyclic.
From here, note that $\angle HEF = \angle HDF = \angle EDX = \angle BAX,$ where the second equality is from the isogonality of circumcenter and orthocenter.
Finally, observe that $\angle ABC = 180 - \angle BEF = 180 - \angle EDF = \angle EHF,$ which finishes!
This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 16, 2025, 10:26 PM
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