IMO 2016 Problem 1

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768 posts

#1 h Jul 11, 2016, 6:23 AM • 18 Y

Y by Problem_Penetrator, mathmaths, hunter117, eshan, Davi-8191, hoangbo5h5, BaoVn, Arshia.esl, BVKRB-, megarnie, Adventure10, Mango247, ohiorizzler1434, deplasmanyollari, Rounak_iitr, ItsBesi, cubres, H_Taken

Triangle $BCF$ has a right angle at $B$ . Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$ . Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$ . Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$ . Let $M$ be the midpoint of $CF$ . Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

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Juraev

61 posts

Juraev

#2 h Jul 11, 2016, 6:28 AM • 4 Y

Y by sa2001, Adventure10, Mango247, cubres

$D$ lies at $(BCF)$ , $B,E$ and $F$ are collinear, $X$ lies at $(BCF)$ , $BX$ is parallel to $FD$ . Then $ME$ passes through intersection point of diogonals of cyclic trapezoid $BXDF$ .

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AOA

1 post

AOA

#3 h Jul 11, 2016, 6:43 AM • 2 Y

Y by Adventure10, cubres

This problem was proposed by Belgium.

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rkm0959

1721 posts

rkm0959

#4 h Jul 11, 2016, 6:52 AM • 4 Y

Y by propali_mat, Adventure10, rty, cubres

Let $\omega$ be the circumcircle of $\triangle BCF$ .
First of all, we show that $BA, MX, \omega$ , and the perpendicular bisector of $AC$ meet at one point $Y$ .
First, we denote $MX \cap \omega = Y$ . Then, $\angle YMF = \angle MXD = \angle MAE = 2 \angle BAF = \angle BFC$ .
Also, if we let $BA \cap \omega = Y'$ , then $\angle Y'BF = \frac{1}{2} \angle BFC$ . This shows that $Y=Y'$ .
Finally, this gives us $\angle YCA = \angle YAC = \frac{1}{2} \angle BFC$ , so $Y$ also lies on the perpendicular bisector of $AC$ .

Now $D$ is just the reflection of $Y$ wrt $CA$ . Therefore, $D$ lies on $(BCF)$ .
Now $\angle YXD = 2 \angle YCF = \angle YCD$ , so $X$ also lies on that circle.
Also, $\angle FXD = \angle FBA = \angle FAD$ , so $FAXD$ is a parallelogram as well.
This also gives us that $MFED$ is a parallelogram since $MF=DE$ as well.
Note that $ME$ passes through the midpoint of $FD$ .
Since $BF=FA=XD$ , we get $BFDX$ as an isosceles trapezoid.

Finally, let us prove that $B, F, E$ are colinear.
It suffices to show that $\angle BFC = \angle FED$ , but we have $\angle FED = \angle FMD = 2 \angle YBF = \angle BFC$ .
Therefore, $B, F, E$ are colinear. This gives us that $BD, FX, ME$ are concurrent, as required. $\blacksquare$

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ThE-dArK-lOrD

4071 posts

ThE-dArK-lOrD

#5 h Jul 11, 2016, 6:54 AM • 2 Y

Y by Adventure10, cubres

Too easy for 1)

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manifoldaora

34 posts

manifoldaora

#6 h Jul 11, 2016, 6:56 AM • 2 Y

Y by Adventure10, cubres

Yep, too easy for a imo-1. I solved it in 30 minutes

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livetolove212

859 posts

livetolove212

#8 h Jul 11, 2016, 7:10 AM • 8 Y

Y by rkm0959, svatejas, strategos21, karitoshi, AllanTian, Adventure10, Rounak_iitr, cubres

Easy one. Just angle chasing.
Let $D'$ be a point on $(M)$ such that $D'A=D'C$ . Then $\angle D'AC=\angle D'CA=\angle D'BF$ . Moreover, $\angle FAB=\angle FBA$ then $\angle D'AB=\angle D'BA$ . We get $D'A=D'B=D'C$ . Hence $D'$ is the circumcenter of $(ABC)$ . We get $\angle D'AC=90^\circ-\frac{1}{2}\angle CD'A=\angle CBA-90^\circ=\angle FBA=\angle FAB$ . So $D'\equiv D.$
Since $\angle EDA=\angle EAD=\angle DAF$ then $DE\parallel AC$ . Let $X'$ be the intersection of $DE$ and $(M)$ . We have $\angle MFX'=\angle CBX'=\angle DBF=\angle DAF$ hence $X'F\parallel DA$ or $X'FAD$ is a parallelogram. But $MX'=MF$ and $ED=EA$ hence $FX'DA$ is a parallelogram. We get $X'\equiv X.$
Finally, since $XF=DA=DB$ then $BXDF$ is an isosceles trapezoid. We are done.

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huricane

670 posts

huricane

#9 h Jul 11, 2016, 7:12 AM • 7 Y

Y by rkm0959, TRYTOSOLVE, AnArtist, strategos21, Kanimet0, Adventure10, cubres

Note that $\angle{ADC}=180^0-2\angle{BAC}=180^0-2\angle{ABF}=2(180^0-\angle{ABC})$ and since $DA=DC$ it follows that $D$ is the circumcenter of $\triangle{ABC}$ .Hence $\angle{BDC}=2\angle{BAC}=\angle{BFC}$ ,so $BCDF$ is a cyclic quadrilateral.
Hence $\angle{FBD}=\angle{ACD}=\angle{ABF}~(\bigstar)$ and so $\angle{ABD}=2\angle{ABF}=2\angle{BAC}=2\angle{DAE}=180^0-\angle{AED}$ and so $BAED$ is a cyclic quadrilateral.
Now notice that this combined with $EA=ED$ implies that $BE$ is the angle-bisector of $\angle{ABD}$ ,so from $(\bigstar)$ we deduce that points $B,F,E$ are collinear.Hence $\angle{AEF}=\angle{AEB}=\angle{ADB}$ and from $\angle{EAF}=\angle{DAB}$ it follows that $\triangle{EAF}\sim\triangle{DAB}$ .But $AD=DB$ and so $EA=EF$ .
Therefore $EF=EA=XM$ and so $EXMF$ is an isosceles trapezoid and so $EXMF$ is cyclic.

Now in order to finish the problem we will just prove that $M\in\odot(ABDE)$ and $X\in\odot(BCDF)$ .
We have $\angle{AMB}=180^0-2\angle{BFC}=\angle{ADB}$ and so $M\in\odot(ABDE)$ and so $\angle{AME}=\angle{ABE}=\angle{BAF}=\angle{BEM}$ ,meaning that $EF=FM$ .
Therefore $XM=EA=EF=FM$ and since $M$ is the center of $\odot(BCDF)$ (because $\angle{CBF}=90^0$ ) it follows that $X\in\odot(BCDF)$ .

Now we are done,just look at circles $\odot(BMDEA),\odot(BCXDF)$ and $\odot(FMXE)$ ,their radical axis are concurrent.

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Aiscrim

409 posts

Aiscrim

#10 h Jul 11, 2016, 7:19 AM • 3 Y

Y by rkm0959, Adventure10, cubres

Note that $DA=DC$ and $\widehat{CDA}=360^\circ-2\widehat{CBA}$ so $D$ is the circumcenter of $\triangle{BCA}$ . Thus, $\widehat{DEA}=360^\circ-2\widehat{DFA}$ and $EA=EC$ , whence $E$ is the circumcenter of $\triangle{EAD}$ . As $MD=MF$ (so $ME$ is the perpendicular bisector of $DF$ ), angle chasing yields $MFED$ parallelogram, which in turn implies $FAXD$ parallelgram. It follows that $ME\cap FX$ is the midpoint of $\overarc{MF}$ in $(BMF)$ , but $BD$ is the bisector of $\widehat{MBF}$ , implying the concurrency.

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buratinogigle

2321 posts

buratinogigle

#11 h Jul 11, 2016, 7:59 AM • 9 Y

Y by baopbc, hoangA1K44PBC, kun1417, dungnguyentien, ninjalearner, Adventure10, Mango247, MS_asdfgzxcvb, cubres

My solution. From $DA=DC$ so $\angle ADC=180^\circ-2\angle FAD= 180^\circ-2\angle FBA=180^\circ-2(ABC-90^\circ)=360^\circ-2\angle ABC$ . Thus, $D$ is circumcenter of triangle $ABC$ . Now $\angle AFE=2\angle FAB=\angle FAE$ , hence $EF=EA=ED$ .

From $DA=DB$ and $FA=FB$ so $A,B$ are reflections through $FD$ . Therefore $\angle FBD=\angle FAD=\angle DCF$ so $FBCD$ is inscribed in circle $(M)$ . Circles $(BFDC)$ and $(AFD)$ have the same chord $FD$ and $\angle FBD=\angle FAD$ so its radius are equal. We get $MFED$ is rhombus. Thus, $E,M$ are reflections through $FD$ , we get $ABME$ is isocesles trapezoid, but $AMXE$ is parallelogram. So $B,X$ are reflections through $EM$ . From $MFED$ is rhombus, we get $F,D$ are reflections through $EM$ , so $FX,EM,DB$ are concurrent.

The same solution is for this extension by reflections.

Extension. Let $ABC$ be a triangle with $\angle ABC>90^\circ$ and circumcenter $D$ . Perpendicular bisector of $AB$ cuts $AC$ at $F$ . $E,M$ are circumcenter of triangles $ADF,BFC$ . Construct parallelogram $AMXE$ . Prove that $EM,FX,BD$ are concurrent.

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MathPanda1

1135 posts

MathPanda1

#12 h Jul 11, 2016, 8:01 AM • 9 Y

Y by drkim, tenplusten, ssk9208, Math_Is_Fun_101, BVKRB-, Adventure10, Mango247, cubres, ehuseyinyigit

Was this problem actually easy? I thought it was really difficult for a problem 1.

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Garfield

243 posts

Garfield

#13 h Jul 11, 2016, 8:03 AM • 5 Y

Y by Math_Is_Fun_101, Adventure10, Mango247, cubres, ehuseyinyigit

Definitely harder then IMO 2014 4th

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livetolove212

859 posts

livetolove212

#14 h Jul 11, 2016, 8:08 AM • 3 Y

Y by Adventure10, cubres, ehuseyinyigit

Got a nice and easy extension:
Given triangle $ABC$ inscribed in $(O).$ An arbitrary line through $O$ intersects the circumcircles of $AOB$ and $AOC$ at $Q$ and $P$ , respectively. $(AOB), (AOC)$ intersects $BC$ at $E, F$ , respectively. $QE$ meets $PF$ at $J$ . Then $J$ is the circumcenter of triangle $APQ.$

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Juraev

61 posts

Juraev

#15 h Jul 11, 2016, 8:22 AM • 2 Y

Y by Adventure10, cubres

What about D2 predictions?

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fprosk

681 posts

fprosk

#16 h Jul 11, 2016, 8:28 AM • 2 Y

Y by Adventure10, cubres

I'm guessing NAG

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AngeloChu

470 posts

AngeloChu

#150 h Jun 26, 2024, 9:45 PM • 1 Y

Y by cubres

almost pure angle chasing
first we just angle chase and easily prove that $D$ lies on $EX$
next, even more angle chasing yields that $D$ is the circumcenter of $ABC$ , which yields that $BFDC$ and $BAED$ are cyclic
then, $BED=BAD=BFC$ , so $EFB$ are collinear
then, still more angle chasing yields that $EM$ and $DC$ are parallel, $AD$ and $FX$ are parallel, and $EF$ and $DM$ are parallel
let $EM$ intersect $AD$ at $O$ , and let $EM$ intersect $FX$ at $P$
we easily prove $EO=MP$ and $FE=FM$ , so we get that $EFO=PFM=DXP$ and more angle chasing yields that $OF$ is parallel to $DB$
let $DB$ intersect $AC$ at $Q$ , and we do length ratio chasing to get that $ED:DX=AF:FM=FQ:QM$ , so $DPQ$ are collinear and we are done

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ezpotd

1251 posts

ezpotd

#151 h Sep 24, 2024, 4:59 AM • 1 Y

Y by cubres

Claim: $B,F,C,D,X$ are cyclic with center $M$ .
Proof: Reflect $B$ over $FC$ to get $B'$ on $AD$ . Since $\angle FB'C = 90$ , proving $AF \cdot AC = AB' \cdot AD$ will show that $D$ lies on the circle with diameter $FC$ , however this is trivial since $\triangle ADC \sim \triangle AFB \sim \triangle AFB'$ . For showing $X$ lies on the circle, note at $ED$ is parallel to $AC$ , it suffices to prove $MD$ , $MX$ , are reflections over the perpendicular from $M$ to $FC$ . Indeed, this is trivial since $\angle XMA = 180 - \angle 2 \angle CAD, \angle AMD = \angle FMD = 2 \angle FCD = 2 \angle CAD$ .

Finish: Note that arc $DX$ has measure $180 - 4 \angle CAD$ , and arc $BF$ has measure $2 \angle BCF = 180 - 4 \angle CAD$ , so $BFDX$ is an isosceles trapezoid. We then show $EM$ is the perpendicular bisector of this trapezoid, finishing. It is sufficient to prove $EF = ED$ , equivalently showing that $E$ is the center of $(ADF)$ . Indeed, we find $\angle AFD = 180 - \angle DFC = 180 - \frac 12 \angle DMC = 180 - \frac 12 (180 - \angle AMD) = 90 + \frac 12 AMD = 90 + \angle CAD$ , thus letting $Y$ be the center of $(ADF)$ we have $\angle AYD = 180 - 2 \angle CAD$ and $Y$ lies on the perpendicular bisector of $AD$ on the opposite side of $C$ , forcing $Y = E$ as desired.

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onyqz

195 posts

onyqz

#152 h Oct 5, 2024, 8:35 PM • 1 Y

Y by cubres

it was fun :-D

solution

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ItsBesi

139 posts

ItsBesi

#153 h Dec 22, 2024, 9:40 PM • 1 Y

Y by cubres

I will just write the steps since writing up the solution will take ages.

Sketch:

Step 1. $D \in \odot(BCF)$ (Do this by showing $\triangle DAF \sim \triangle CAB$ )

Step 2. $X \in \odot(BCF)$ (Do this by just angle chasing)

Step 3. $\overline{B-F-E}$ collinear (Do this by showing $\triangle EAF \thicksim \triangle MFB$ )

Step 4. $\overline{E-D-X}$ collinear (Do this by showing $\square EDMF$ is a rhombous)

Step 5. Show that qudrilaterals $\square MXEF$ and $\square EDMB$ are cyclic (Do this by showing they are isosceles trapezoid)

Step 6. Finish with the Radical Axis Theorem

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cj13609517288

1879 posts

cj13609517288

#154 h Jan 27, 2025, 4:06 PM • 1 Y

Y by cubres

Let $\alpha=\angle BAC$ and $FM=x$ . Then $AF=2x\cos 2\alpha$ , so $AC=2x(1+\cos 2\alpha)=4x\cos^2\alpha$ . So $AE=ED=x$ .

ALso, $D$ lies on $EX$ because $\angle EDA=\angle CAD$ .

Now the rest of this problem is just a lot of discoveries. In particular, eventually you get $DMC\cong EFM\cong AED\cong BMD$ and $DMX\cong BMF$ . Eventually though, we get $BF=DX$ , $EF=ED$ , $BFE$ are collinear, and $MB=MX$ . But then we can just use Ceva on triangle $BEX$ to show that $M$ lies on the $E$ -median, which is true because $EB=EX$ and $MB=MX$ . $\blacksquare$

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Tony_stark0094

44 posts

Tony_stark0094

#156 h Jan 29, 2025, 6:25 AM • 1 Y

Y by cubres

$[asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair B=D("B",origin,dir(-100)*1.5); pair C=D("C",(3,0),dir(-45)); pair F=D("F",(0,2),dir(-160)*1.5); pair A=D("A",IP(L(F,C,5,5),CP(F,B)),dir(135)); pair M=D("M",midpoint(F--C),dir(-80)*1.5); pair D=D("D",IP(CP(M,B),L(midpoint(A--C),bisectorpoint(A,C),5,5)),dir(60)); pair E=D("E",OP(L(B,F,5,5),circumcircle(A,B,M)),dir(100)*1.5); pair Y=IP(M--E,B--D); dot(Y); pair X=D("X",IP(CP(M,B),L(F,Y,5,5)),dir(-10)); D(anglemark(A,D,F,14),blue); D(anglemark(F,C,B,14),blue); D(anglemark(F,A,D,14),deepgreen); D(anglemark(B,A,F,16),deepgreen); D(anglemark(F,B,A,14),deepgreen); D(anglemark(D,C,F,16),deepgreen); D(anglemark(F,D,B,16),blue); D(B--D,dashed); D(E--M,dashed); D(F--X,dashed); D(circumcircle(A,E,D),red+dotted); D(CP(M,B),red+dotted); D(circumcircle(E,F,M),red+dotted); D(A--C--B--E); D(A--E--X--M); D(B--A--D--C); D(D--F); } b(); pathflag=false; b(); [/asy]$
claim:A,B,D,E are concyclic:
proof:
$\angle ADE =\angle EAD= \angle DAF =\angle FAB =\angle ABF$ claim: E,F,M,X are concyclic:
proof
$\angle EFM =\angle 180-BFM =180-(\angle FAB+\angle FBA )=180-\angle FAE =180 - \angle MXE$ claim: $\angle FBA=\angle FBD$ proof:
$\angle FBD=\angle DAE=\angle DAF=\angle FAB=\angle FBA$ claimB,F,D,C are concyclic points
proof:
$\angle FCD= \angle FAD =\angle FAB =\angle ABF=\angle FBD$ Hence the required result is obtained by radial axis theorem

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Reason: typo

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ljwn357

8 posts

ljwn357

#157 h Feb 8, 2025, 3:42 PM • 1 Y

Y by cubres

You can easily show the given statement by some spiral similarities. First you see $D$ is on ( $M$ ) by the similarity of first two isosceles and from that, a simple angle chase will show $X$ on the circle. Thus ( $B, D, X, C, F$ ) cyclic and $FBXD$ is a cyclic trapezoid. Some angle chasing and $AE=ED=EF$ so triangle $FED$ isosceles. Since $M$ is the center of the circle, it is obvious that ME is perpendicular to $FD$ and $BX$ so the problem ends.

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Bonime

28 posts

Bonime

#158 h Feb 10, 2025, 12:53 AM • 1 Y

Y by cubres

This problem comes down solely and exclusively to putting together a good figure, understanding the configuration and proving the facts of the same. Thus, if $\angle FAB=x$ , some lemmas follow

Lemma 1: The points $E$ , $X$ and $D$ are collinear.
From the parallelogram, we know that $EX//AC$ , however, by the hypothesis, $\angle EDA=x=\angle DAC$ , therefore $ED//AC$ , proving lemma 1.

Lemma 2:The point $E$ is the circumcenter of the triangle $AFD$ .
Define the point $T=BF\cap CD$ , see that, since $\angle DCA=x=\angle FAB \Rightarrow AB//CD \Rightarrow \angle FTD=\angle FAD$ we can conclude that the quadrilateral $AFDT$ is cyclic, but $ED=EA=ET$ , therefore its center is $E$ , proving Lemma 2.

Lemma 3: The points $B$ , $E$ and $F$ are collinear.
We know that $\angle DEF=2\angle DTF=2x=\angle TFA$ therefore, since $DE//AF$ , we conclude that $F$ , $E$ and $T$ are collinear, proving Lemma 3.

Lemma 4: The pentagon $FBCXD$ is cyclic with circumcenter $M$
Note that since $FT$ is the diameter in $(AFDT)$ $\angle FDT=90^o=\angle FDC \Rightarrow \text{D $\in$ (FBC)}$ Now, note that by parallelism the triangle $FCT$ is isosceles with base $CT$ and angle $x$ , therefore, its mean base $EM$ is such that $FE=FM$ . Finally, see that from the parallelogram $AM=AF+FM=EX=ED+DX \Rightarrow DX=AF=FB$ which tells us that $EF\cdot EB=\text{Pot}_{(BFCD)}E=ED\cdot EX \Rightarrow \text{X $\in$ (BFCD)}$ proving lemma 4.
Having finally understood the figure, just see that triangle $EBX$ is isosceles with base $BX$ , therefore $EM$ is its perpendicular bisector and since $FD//BX$ , we end by Ceva!

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Maximilian113

522 posts

Maximilian113

#159 h Feb 25, 2025, 11:49 PM • 1 Y

Y by cubres

Let $\theta = \angle EAD.$ First of all, notice that $E, D, X$ are collinear as $\angle ADE = \angle DAF \implies AF \parallel ED.$

Meanwhile, observe that from our similar triangle $\frac{AF}{AB} = \frac{AD}{AC},$ and as $\angle BAC = \angle FAD,$ it follows from SAS that $\triangle BAC \sim \triangle FAD.$ Hence $\angle AFD = 90^\circ + \theta = 180^\circ - \frac{\angle AED}{2},$ so $F$ lies on the circle centered at $E$ passing through $A, D.$ Thus, $EA=EF \implies \angle AFE = 2 \theta = \angle FAB + \angle FBA,$ so $B, F, E$ are collinear.

Now, observe that $\angle AFD = 90^\circ + \theta \implies \angle FDC=90^\circ,$ so $B, F, D, C$ are concyclic on a circle centered at $M.$ But $\angle AFD = 90^\circ + \theta \implies \angle MFD = 90^\circ - \theta \implies \angle FMD = 2\theta = \angle AFE,$ so $EF \parallel MD.$ Thus $MX = AE = EF = MD,$ so $X$ also lies on this circle. So $MB = MX.$

Meanwhile, we easily have $EF=ED,$ while $EX = AM = AF+FM=FB+MX=FB+AE=BE,$ so $FX, BD$ are symmetric with respect to $M.$ But, $E = BF \cap XD$ so the desired result follows by symmetry. QED

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Ilikeminecraft

330 posts

Ilikeminecraft

#160 h Mar 15, 2025, 4:20 AM • 1 Y

Y by cubres

why can't I do angle chase....

Let $\angle BAC = \alpha.$
Claim: $D$ lies on $(EBC)$
Proof: Let $B'$ be the reflection of $B$ across $AC, D'$ be the 2nd intersection of $AB'$ and $(FBC).$
We have that $\begin{align*}AD & = \frac{AC}2\frac1{\cos\alpha} \\ & = \frac{2(\cos2\alpha + 1)}{2}\frac1{\cos\alpha} \\ & = \frac{2\cos2\alpha(2\cos2\alpha+2)}{2\cos2\alpha\cos\alpha\cdot2} \\ & = \frac{\operatorname{Pow}_{(FBC)(A)}}{AB'} = AD'.\end{align*}$ This finishes.

From angle chase, we can get $AC\parallel ED, AMDE$ is isosceles trapezoid. From $AMDE$ isosceles trapezoid, we get $MX = ME = MD,$ so $X\in(BFC).$ Angle chase forces $FMDE$ to be a rhombus. Finally, by symmetry about $EM,$ we are done.

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bjump

995 posts

bjump

#161 h Mar 26, 2025, 3:19 PM • 2 Y

Y by cubres, imagien_bad

Note that
$\measuredangle DAE = \measuredangle MAD = \measuredangle BAM = \measuredangle DCA$ So $CD \parallel BA$ . Also note that
$2\measuredangle CBA + \measuredangle CDA = 2\measuredangle FBA+ \measuredangle CDA=2 \measuredangle BAF + \measuredangle CDA = 2\measuredangle DAC + \measuredangle CDA = 0$ Therefore $D$ is the circumcenter of $(ABC)$ , and $DF \perp CD$ , so $CDFB$ is cyclic with center $M$ . Since $MB= MD$ , and $DE=EA$ , $FB=FA$ , combined with
$\measuredangle EAM = \measuredangle DAB = 2 \measuredangle FAB = \measuredangle MFB = \measuredangle FBM$ Gives us $MEAB$ is an isosceles trapezoid by symmetry about $DF$ , with $D$ lying on the circle. By arc lengths we have $\measuredangle DEM = \measuredangle FMD$ so $XDE$ is a line. Now we have
$\measuredangle MFE = \measuredangle AFE = \measuredangle EAM = \measuredangle MXE$ So $FMXE$ is cyclic. Radical axis theorem on $(FMXE)$ , $(DFBC)$ , and $(MBADE)$ . Gives the desired concurrency.

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endless_abyss

33 posts

endless_abyss

#162 h Mar 27, 2025, 12:42 PM • 1 Y

Y by cubres

This problem was an absolute delight -

We prove that the lines are concurrent by proving that $M - P - E$ is collinear
Claim - $\angle F D A = 90 - 2 \angle C A B$
First of all, we write -
$(F A)/(A D) = (A B)/(A C)$
so, $F A D$ is similar to $B A C$ by $S-A-S$ criterion

Claim - $F$ is the incentre of $A B D$
This follows from the intersection of the angle bisectors $B F$ and $F A$
Then notice how $M P F B$ and $B F D C$ are concyclic
The rest is just angle chase, and

$\angle M P B = 2 \angle B A C$ (Through triangle sum property in $B M P$ )
$\angle B P F = \angle B M F = 180 - 4\angle B A C$ ( $B M F P$ is cyclic and using the fact that $B M = C M$ and exterior angle on $B M C$ )
$\angle F P E = 2\angle B A C$ (using angle sum property on quadrilateral $F P E A$ )

So, $M - P - E$ is collinear and our solution is complete.
$\square$
:starwars:

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This post has been edited 1 time. Last edited by endless_abyss, Mar 27, 2025, 12:44 PM
Reason: Forgor attachment cuz I'm so dumb :(

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ehuseyinyigit

799 posts

ehuseyinyigit

#163 h Mar 30, 2025, 10:30 AM

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Easy but too many Properties.

Let $\angle FBA=\alpha$ . Then $A$ is spiral center sending $BF$ to $CD$ . Thus $\angle FDA=\dfrac{\pi}{2}-2\alpha$ and $FD\perp CD$ implying $BCDF$ is cyclic. This gives $\angle DBF=\alpha$ , $BD$ bisects $\angle MBF$ and $F$ is incenter of triangle $ABD$ . On the other hand, $\alpha=\angle MBD=\angle MAD$ implies $MBDA$ is cyclic. Furthermore, $A$ is spiral center sending $BF$ to $DE$ , and $\angle ABD=2\alpha=\angle AFE$ implying $B-F-E$ collinearity and $FE=EA$ . $\angle BEA=\angle BMA$ implies $MBDEA$ is cyclic. Additionally $XM=EA=FE$ implies $XMFE$ is a cyclic-isosceles trapezoid. $M$ was center of $(BCDF)$ and since $MD=MF=FE=EA=XM$ , we have $BCDFX$ is cyclic. The concurrency of three lines $BD$ , $FX$ and $ME$ easily follows from radical axis of $(BCDFX)$ , $(XMFE)$ and $(BMDEA)$ .

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ehuseyinyigit

799 posts

ehuseyinyigit

#164 h Mar 30, 2025, 10:56 AM

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And I also want to list all the properties of the problem:

1) $FD\perp CD$ .
2) $BCDFX$ is cyclic and $M$ is center of it.
3) $F$ is the incenter of triangle $ABD$ .
4) $BD$ bisects $\angle MBF$ .
5) $CD=BD=DA$ .
6) $B-F-E$ are collinear.
7) $MBDEA$ is cyclic.
8) $XM=MF=FE=EA$ .
9) $XMFE$ is a cyclic-isosceles trapezoid.
10) $BD$ , $FX$ and $ME$ are concurrent.

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Thelink_20

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Thelink_20

#165 h Mar 30, 2025, 10:02 PM

Y by

We'll use complex coordinates, set $F=0, A=1$ .

We have $BC$ tangent to the unit circle, thus $c = \frac{2b-c}{b^2}\implies\boxed{c=\frac{2b}{b^2+1}}$ Click to reveal hidden text

$\frac{b-f}{a-f} = \frac{c-d}{a-d}\implies\boxed{d = \frac{b^2 + b}{b^2+1}}$
$\frac{b-f}{a-f} = \frac{d-e}{a-e}\implies\boxed{e = \frac{b^2}{b^2+1}}$ Click to reveal hidden text

Let $P = BD\cap ME$ , we have:

$p = \frac{(\bar{m}e - m\bar{e})(b-d) - (\bar{b}d - b\bar{d})(m-e)}{(\bar{m} - \bar{e})(b-d) - (\bar{b} - \bar{d})(m-e)} = \frac{\frac{b^3(b^2-1)(b-1)}{(b^2+1)^3}-\frac{b(b^2-1)(b-1)}{(b^2+1)^2}}{\frac{b^2(b-1)^2}{(b^2+1)^2} - \frac{(b-1)^2}{(b^2+1)^2}} =\boxed{ \frac{-b}{(b^2+1)(b-1)}}$

But $AMXE$ parallelogram $\implies \boxed{x = e+m-a = \frac{b-1}{b^2+1}}$

Now we just need to check that $\frac{p-f}{x-f} = \frac{-b}{(b-1)^2}$ is real, which is clearly true. $_{\blacksquare}$

This post has been edited 5 times. Last edited by Thelink_20, Mar 31, 2025, 12:18 PM

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