IMO Shortlist 2012, Geometry 4

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127 posts

#1 h Jul 29, 2013, 12:22 PM • 12 Y

Y by anantmudgal09, Davi-8191, Jasurbek, donotoven, geometry6, ike.chen, Adventure10, Mango247, Rounak_iitr, and 3 other users

Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$ . The bisector of $\angle BAC$ intersects $BC$ at $D$ . Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$ . The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.

This post has been edited 1 time. Last edited by Amir Hossein, Jul 29, 2013, 5:39 PM
Reason: Edited Title and Changed Format of The Problem.

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Burii

137 posts

Burii

#2 h Jul 29, 2013, 2:21 PM • 6 Y

Y by giftctb, Mobashereh, Miku3D, donotoven, Adventure10, and 1 other user

Let $H$ be the orthocenter of triangle $ABC$ . If $M$ is midpoint of $DY$ , let $P$ be the reflection of $E$ with respect point $M$ . Of course $BCPY$ is isosceles trapezoid. The bisector of $\angle BAC$ and perpendicular bisector of $BC$ intersect at $M$ , so $M$ lies on circumcircle of triangle $ABC$ . Thus we have $BD\cdot DC=AD \cdot DM$ . Points $H$ and $O$ are isogonal cojugate wrt triangle, so easy angle chasing and we have that $\angle AXD= \angle DMP$ . Thus quadrangle $AXMP$ is cyclic, so $DX\cdot DP=AD\cdot DM = BD\cdot DC$ - quadrangle $BSCP$ is cyclic, so points $B, X, C, P, T$ lie on the same circle.

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sunken rock

4379 posts

sunken rock

#3 h Jul 29, 2013, 8:32 PM • 6 Y

Y by Burii, Mobashereh, Miku3D, donotoven, Adventure10, and 1 other user

Nice problem, nice idea for solution; however it could be made shorter:

Clearly $\Delta ADX\sim\Delta AMO$ , so $\Delta ADX$ is isosceles, and so is $\Delta DPM$ , since $M$ is the center of the rectangle $DEYP$ ; as $\angle MPD=\angle MDP=\angle XDA=\angle XAD$ we infer $AXMP$ cyclic, and we are done.

Best regards,
sunken rock

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exmath89

2572 posts

exmath89

#4 h Aug 2, 2013, 5:15 AM • 7 Y

Y by sunny2000, Sx763_, Miku3D, lneis1, donotoven, Adventure10, and 1 other user

Solution

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JuanOrtiz

366 posts

JuanOrtiz

#5 h Aug 7, 2013, 8:08 PM • 4 Y

Y by AlastorMoody, donotoven, Adventure10, H_Taken

Absolutely trivial for a G4. Here is the solution. Call Y' the reflection of Y through the perpendicular bisector of BC. Clearly if we show that BXCY' is cyclic, we're done. So if we show $DX(DY)=DB(DC)$ , we're done. So if we show $DX(EY)=BD(DC)$ , we're done. Assume $\angle B \textgreater \angle C$ . Call $(A/2)+C=\delta$

Now, note that $\angle DAX = \angle DAC - \angle OAC = A/2 - 90 + B = \angle ADX = \delta$ , so $DX = \frac{AD}{2sin \delta}$ . Also, we can see that $EY=2DMtan\delta$ since $\angle EDY = \angle BDA = \delta$ . So we wish to prove $DB(DC)=\frac{DM(DA)}{cos \delta}$ . But call $Z$ the midpoint of arc $BC$ that does not contain $A$ in the circumcircle of $ABC$ . We see that $DA(DZ)=DB(DC)$ and $DZ = \frac{DM}{cos \delta}$ . So we're done.

Horrible problem, nothing interesting about it. Very poor choice for G4, in my opinion.

This post has been edited 2 times. Last edited by JuanOrtiz, Aug 8, 2013, 1:11 AM

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XmL

552 posts

XmL

#6 h Aug 7, 2013, 8:47 PM • 4 Y

Y by donotoven, Adventure10, Mango247, and 1 other user

I swear this has been posted a while ago, but I can't find it!!! my solution was the same as sunken rock's

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Math-lover123

304 posts

Math-lover123

#7 h Aug 7, 2013, 8:57 PM • 3 Y

Y by donotoven, Adventure10, Mango247

You are right XmL here it is : http://www.artofproblemsolving.com/Forum/resources.php?c=114&cid=75&year=2013&sid=7c75a2addf93750e55d6db969f0c3964

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Andrew64

33 posts

Andrew64

#8 h Aug 31, 2013, 1:51 AM • 2 Y

Y by donotoven, Adventure10

lyukhson wrote:

Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$ . The bisector of $\angle BAC$ intersects $BC$ at $D$ . Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$ . The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.

As shown in the figure below.

$H$ is the intersection of $\circ O$ and $AY$ .

$F$ is the midpoint of $AD$ .

$DG//EY$ and $DG=EY$ .

It's obvious that

$B,C,G,Y$ are concyclic. and

$HE=HD$

so $H$ is the midpoint of $YD$ .

As
$DF\times DY=\frac{AD}{2}\times DY=AD\times \frac{DY}{2}=AD\times DH=BD\times DC$

$B,F,C,Y$ are concyclic.

Further, as
$\triangle DXF \sim \triangle DYE$

we have

$DF\times DY = DX\times EY = DX\times DG$

Thus

$B,Y,G,C,F,X$ are concyclic.

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v_Enhance

6871 posts

v_Enhance

#9 h Sep 7, 2013, 4:09 AM • 18 Y

Y by TheStrangeCharm, Math-Ninja, AlastorMoody, richrow12, Miku3D, Muaaz.SY, lneis1, HamstPan38825, donotoven, Geometry285, Tellocan, Adventure10, Ritwin, shafikbara48593762, Funcshun840, NicoN9, Lcz, and 1 other user

Assume $AB < AC$ and let $L$ be the midpoint of arc $BC$ . Denote by $Y'$ the reflection of $E$ over $L$ ; evidently $BY'YC$ is a isosceles (and hence cyclic) trapezoid. Now $\triangle DLY' \sim \triangle AXD$ because they are both isosceles (since $\triangle AXD \sim \triangle AOL$ ) and share an angle, so we see that $\angle XAL = \angle XAD = \angle DY'L = \angle XY'l$ and so $AXLY'$ is cyclic; hence $XD \cdot DY' = AD \cdot DL = BD \cdot DC$ so $XBY'C$ is cyclic as well, and we're done.

$[asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(13cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (0.64, 0.96); pair B = (0.4, -1.0); pair C = (3.0, -1.0); pair D = IntersectionPoint(Line(incenter(A,B,C),A,lisf),Line(B,C,lisf)); pair E = B+C-D; pair M = midpoint(B--C); pair O = circumcenter(A,B,C); pair X = IntersectionPoint(Line(D,D+O-M,lisf),Line(A,O,lisf)); pair Y = IntersectionPoint(Line((2)*(O)-X,E,lisf),Line(A,D,lisf)); pair L = midpoint(D--Y); pair Y_prime = (2)*(L)-E; /* Draw objects */ draw(A--B, rgb(0.2,0.6,0.0)); draw(B--C, rgb(0.2,0.6,0.0)); draw(C--A, rgb(0.2,0.6,0.0)); draw(circumcircle(A,B,C), rgb(0.0,0.4,0.0) + linewidth(1.0) + linetype("4 4")); draw(circumcircle(B,C,X), rgb(0.6,0.0,0.6)); draw(A--O, rgb(0.2,0.6,0.0)); draw(E--Y, rgb(0.6,0.0,0.6)); draw(Y_prime--L, rgb(0.6,0.0,0.6) + linetype("4 4")); draw(D--X, rgb(0.0,0.6,0.6)); draw(D--Y_prime, rgb(0.6,0.0,0.6) + linetype("4 4")); draw((abs(dot(unit(L-Y_prime),unit(D-Y_prime))) < 1/2011) ? rightanglemark(L,Y_prime,D) : anglemark(L,Y_prime,D), rgb(0.6,0.0,0.6)); draw((abs(dot(unit(D-A),unit(X-A))) < 1/2011) ? rightanglemark(D,A,X) : anglemark(D,A,X), rgb(0.6,0.0,0.6)); draw(A--D, rgb(0.2,0.6,0.0)); draw(D--Y, rgb(0.6,0.0,0.6)); draw(O--M, rgb(0.2,0.6,0.0)); draw(M--L, rgb(0.6,0.0,0.6)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(E); dot(M); dot(O); dot(X); dot(Y); dot(L); dot(Y_prime); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * dir(225)); label("$C$", C, lsf * dir(-45)); label("$D$", D, lsf * dir(45)); label("$E$", E, lsf * dir(45)); label("$M$", M, lsf * dir(45)); label("$O$", O, lsf * dir(45)); label("$X$", X, lsf * dir(60)); label("$Y$", Y, lsf * dir(-90)); label("$L$", L, lsf * dir(45)); label("$Y'$", Y_prime, lsf * dir(-90)); [/asy]$

In other news, 2048th post!

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Math-lover123

304 posts

Math-lover123

#10 h Sep 7, 2013, 2:08 PM • 3 Y

Y by parola, Adventure10, Mango247

Let $F$ be a foot of perpendicular from $X$ to $AD$ .Then it it easy to prove that triangles $DEY$ and $XFD$ are similar.
So $EY/DY=EY/2DL=AD/2XD$ so $EY\cdot XD=DL\cdot AD=BD\cdot CD=CE\cdot CD$ ( we have obviously $BD=CE$ ) so triangles $CEY$ and $CDX$ are similar.Then $\angle XCY=90$ , analogously $\angle XBY=90$ so done.

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Particle

179 posts

Particle

#11 h Nov 19, 2013, 12:40 PM • 2 Y

Y by Adventure10, Mango247

Calculative proof

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AnonymousBunny

339 posts

AnonymousBunny

#12 h Jul 3, 2014, 1:45 PM • 3 Y

Y by Roct-7, Adventure10, Mango247

Sketch, will complete this soon:
Let $M$ be the midpoint of $OM.$ Extend $OM$ to meet the circumcircle of $\triangle ABC$ at $J.$ It is easy to prove that $AJ$ bisects $\angle BAC,$ and as a consequence $A,D,J$ are collinear. Now, note that $\triangle ADX \sim \triangle DJM.$ By some easy ratio computations using PoP of $D$ wrt the circumcircle of $\triangle ABC,$ it follows that $\triangle BDX \sim \triangle BEY$ and similarly $\triangle CDX \sim \triangle CEY.$ The rest is simple angle chasing.

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junioragd

314 posts

junioragd

#13 h Aug 1, 2014, 8:02 PM • 2 Y

Y by Adventure10, Mango247

Let M be the midpoint of the minor arc BC and let Y' be the reflection of Y across the perpendicular bisector of BC.Now,it is easy to see that that BCYY' is an iscocelles trapezoid and X,D and Y' are collinear and M is the midpoint of DY(simple angle chasing).Now,by angle chasing we get <Y'MA=<AXY',so AXMY' is a cyclic.Also,BMCA is a cyclic so we have DB*DC=DM*DA=DX*DY',so we are finished.

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jayme

9775 posts

jayme

#14 h Aug 2, 2014, 4:54 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=530055
Sincerely
Jean-Louis

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utkarshgupta

2280 posts

utkarshgupta

#15 h Feb 7, 2015, 5:10 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Call the midpoint of $BC$ $M$ and $AD \cap \odot ABC = P$
Obviously $OMP$ is a straight line

Consider reflection in line $OP$ as $R(OP)$ .
$R(OP) : D \to E, B \to C, O \to O, M \to M, P\to P, Y \to Y'$
$R(OP):DPY \to EPY'$
$\implies EPY'$ is a straight line.
Also easily, $XDY'$ is a straight line.
$\angle XAD = \angle \frac{A}{2}-90+ \angle B = 90-\angle C-\angle \frac{A}{2}$

$\angle PY'D = \angle EY'D = 90-\angle DEY' = 90-\angle EDP = 90 - \angle DBP - \angle DPB = 90-\angle C-\angle \frac{A}{2}$

$\implies \angle XAD=\angle PY'D$
$\implies XAY'P$ is cyclic
$\implies DX \cdot DY'=DA \cdot DP$
$\implies DX \cdot DY'=DB \cdot DC$
$\implies BXCY'$ is cyclic.
$\implies \angle BY'C = 180-\angle BXC$
But by $R(OP):\angle BY'C \to \angle CYB$
$\implies \angle CYB+\angle BXC=180$

$\implies BXCY$ is cyclic and thus we are done

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huashiliao2020

1292 posts

huashiliao2020

#76 h Aug 31, 2023, 1:27 AM • 1 Y

Y by OronSH

Define the points as shown in the diagram, with Z the intersection of perp. through D to BC with (DEY); it's evident that the circumcenter of DEYZ is the midpoint of arc BC, and that BCYY' is cyclicislscelestrapezoid. The finish is as follows: $ADX=FDZ,AO=OF\implies AX=XD,FD=FZ\implies AXD\sim DFZ\implies BD\cdot DC=AD\cdot DF=XD\cdot DZ\implies X\in(BCZ),$ and combining we have the desired. $\blacksquare$

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bobthegod78

2982 posts

bobthegod78

#77 h Sep 22, 2023, 1:54 PM • 1 Y

Y by centslordm

Let the second intersection of $XD$ with $(BCY)$ be $T$ . It is easy to see that $BTCY$ is an isosceles trapezoid, meaning it is cyclic.

Let $N$ be the arc midpoint of $BC$ . Notice that $DNT$ and $AXD$ are isosceles and have the same base angle, so they are similar. Then $\angle AXT = \angle AXD = \angle DNT = \angle ANT$ , so $AXNT$ is cyclic. Note
$XD \cdot TD = AD \cdot ND = BD \cdot CD,$ implying $BXCT$ is cyclic, which finishes.

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IAmTheHazard

5001 posts

IAmTheHazard

#78 h Sep 22, 2023, 6:19 PM • 2 Y

Y by OronSH, centslordm

Let $\overline{AD}$ intersect $(ABC)$ again at $S$ , so $S$ is the midpoint of $\overline{DY}$ . Let $N$ be the midpoint of $\overline{AD}$ . By power of a point, $BCYN$ is cyclic. Let $Y'$ be the reflection of $Y$ over the perpendicular bisector of $\overline{BC}$ , which also lies on $BCYN$ . Clearly $X,D,Y'$ are collinear, hence $\angle XY'Y=90^\circ$ . On the other hand, since $OA=OS$ , by a homothety at $A$ we also have $XA=XD$ , hence $\angle XND=\angle XNY=90^\circ$ , so $BCXYNY'$ is cyclic. $\blacksquare$

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Shreyasharma

667 posts

Shreyasharma

#79 h Sep 30, 2023, 4:31 PM

Y by

Let $P$ be the reflection of $Y$ across $M$ and let $F$ denote the midpoint of arc $BC$ . Also let $M$ be the midpoint of $BC$ . Noting that $\angle EYD = \angle 90 - \angle YDE = 90 - \angle EFD = \angle FEY$ we have that $DEPY$ is a rectangle with $F$ being the midpoint of $EP$ and $DY$ . It is also clear that $P \in (BYC)$ .

Claim: $\triangle AXD \sim \triangle EFY$
Proof: This is just angle chasing. Note that $\angle OAF = \angle OFD = \angle OFE = 90 - \angle DEF = \angle FEY$ . Also $\angle XDA = \angle YDP = \angle DYE$ . Thus proved.

Then we clearly have $XD \cdot DP = AD \cdot DF = BD \cdot BC$ so we're done.

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bjump

997 posts

bjump

#80 h Nov 19, 2023, 6:47 PM • 1 Y

Y by OronSH

Solved with emotional support from Megarnie, Tienxion, Redfiretruck, and OronSH
Define $X'$ and $Y'$ as the reflections of $X$ and $Y$ across the perpendicular bisector of $BC$ .
Note that

$XX'BC$ is a Cyclic Isosceles Trapezoid
$YY'BC$ is a Cyclic Isosceles Trapezoid
$XX'YY'$ is a Cyclic Rectangle

Now note that $XX' \parallel YY' \parallel BC$ hence the pairwise radical axes of $(XX'BC)$ , $(YY'BC)$ and $(XX'YY')$ don't concur.
Which means $(XX'BYY'C)$ cyclic which means $(BXCY)$ is cyclic $\blacksquare$

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cursed_tangent1434

569 posts

cursed_tangent1434

#81 h Dec 7, 2023, 12:55 AM

Y by

Quite boring stuff.

Let $M'$ be the arc midpoint of minor arc $BC$ . Let $H$ be the foot of the altitude from $A$ . Now, note that $AH \parallel XD \parallel MM' \parallel EY$ Notice that
$\alpha = \angle DAX = \angle DAC - \angle OAC = \angle DAB - \angle BAH=\angle HAD$ ( $\angle BAH = \angle CAO = 90 - \angle B$ ).
But, then via the parallel lines
$\alpha = \angle HAD = \angle XDA = \angle MM'D = \angle EYD$ Now, notice that $\angle DM'M=\angle EM'M$ via reflection and in turn this means
$\angle AXD = 180 - 2\alpha = 180 - 2\angle DM'M = 180 - \angle DM'M - \angle MM'E = \angle EM'Y$ Thus, we must have that $\triangle AXD \sim \triangle EM'Y$ . Now, notice the following. Since $M$ is the midpoint of $DE$ and $MM'$ is parallel to $EY$ , we have that $M'$ is the midpoint of $DY$ . By the previous similarity, we have that
$\frac{AD}{EY}=\frac{XD}{M'Y}\implies AD \cdot M'Y = XD \cdot EY \implies AD \cdot DM' = XD \cdot EY$ So,
$\begin{align*} BE \cdot BD &= BD (BM + EM)\\ &= BD (CM + DM)\\ &= BD \cdot CD \\ &= AD \cdot DM'\\ &= XD \cdot EY \text{ (via the previous similarity)} \end{align*}$ But then, this means that $\triangle XBD \sim \triangle EBY$ which in turn implies that we must have $\angle XBY = 90^\circ$ . Now, let $K$ be the intersection of $MM'$ and $XY$ . Since $M'$ is the midpoint of $DY$ and $XD \parallel MK$ , $K$ must also be the midpoint of $XY$ . Thus, $K$ is the center of $\triangle BXY$ ( $\angle XBY = 90^\circ$ ) and $BK = XK = KY$ . Further, $K$ lies on the perpendicular bisector of $BC$ ( $MM'$ ) and so it is equidistant to $B$ and $C$ . Thus, $BK=CK$ which gives us that
$XK = KY = BK = CK$ and thus $K$ must be the center of $(XBYC)$ and thus $XBYC$ is indeed cyclic.

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thdnder

194 posts

thdnder

#82 h Dec 24, 2023, 11:04 AM

Y by

Let $M = AD \cap (ABC)$ and let $Y'$ be the reflection of $Y$ wrt $OM$ . Then $BCYY'$ is isosceles trapezoid, so it suffices to prove that $BXCY'$ is cyclic. Let $A' = Y'M \cap (ABC)$ and let $N'$ be the midpoint of $AE$ . It's not hard to see that $MD = ME = MY'$ , so we have $Y'E \cdot EN' = ME \cdot A'E = AD \cdot DM = BD \cdot DC = BE \cdot EC$ , therefore $BY'CN'$ is cyclic. Thus it's enough to prove that $BCN'X$ is cyclic. Let $N$ be the midpoint of $AD$ . Then $BCNN'$ is isosceles trapezoid, so it suffices to prove that $BCXN$ is cyclic. Let $T = NX \cap BC$ . Simple angle chasing gives $AX = XD$ , thus $XN$ is the perpendicular bisector of $AD$ , hence $TA = TD$ . Therefore $AT$ is tangent of $(ABC)$ , so $TB \cdot TC = TA^2 = TN \cdot TX$ , so we're done. $\blacksquare$

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lelouchvigeo

176 posts

lelouchvigeo

#83 h Jan 5, 2024, 12:28 PM

Y by

Easy for a G4(Still spent an hour on it)
Basically we have to prove XDB is similar to BEY.
This can be done using trig observations

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john0512

4176 posts

john0512

#84 h Apr 13, 2024, 11:21 PM

Y by

We will show that $\angle XBY=90$ , which is of course sufficent by symmetry.

Let $F$ be the foot from $A$ to $BC$ , and let $L$ be the midpoint of arc $BC$ .

Note that since $BD\perp DX$ and $BE\perp EY$ , the perpendicularity we wish to show is equivalent to $BD\cdot BE=XD\cdot YE.$
We can compute three of the lengths relatively easily. By angle bisector theorem, $BD=\frac{ac}{b+c},BE=\frac{ab}{b+c},$ and $YE=2ML=a\tan\alpha/2.$ Thus, the desired conclusion is equivalent to $XD=\frac{abc}{(b+c)^2\tan\alpha/2}.$
The idea is to use similar triangles $\triangle ADX$ and $\triangle ALO$ . Due to this, $XD=R\cdot\frac{AD}{AL}=R\cdot\frac{AF}{AF+ML}=R\cdot \frac{\frac{bc}{2R}}{\frac{bc}{2R}+\frac{a}{2}\tan\alpha/2}$ $=\frac{bcR}{bc+aR\tan\alpha/2}.$
Thus, it suffices to show that $\frac{abc}{(b+c)^2\tan\alpha/2}=\frac{bcR}{bc+aR\tan\alpha/2}$ $abc+a^2R\tan\alpha/2=R(b+c)^2\tan\alpha/2$ $abc=R\tan\alpha/2((b+c)^2-a^2).$ Let $A$ be the area of the triangle. Then, $abc=4RA$ , so this becomes $4A=\tan\alpha/2(b+c-a)(b+c+a).$ Squaring and using Heron's formula, this is equivalent to $\tan^2\alpha/2=\frac{(a-b+c)(a+b-c)}{(a+b+c)(-a+b+c)}.$ However, we have $\tan^2\alpha/2=\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}=\frac{a^2-b^2-c^2+2bc}{2bc+b^2+c^2-a^2}=\frac{(a-b+c)(a+b-c)}{(a+b+c)(-a+b+c)},$ as desired.

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MagicalToaster53

159 posts

MagicalToaster53

#85 h Apr 17, 2024, 12:25 AM

Y by

Let $N$ denote the midpoint of minor arc $\hat{BC}$ , and let $Z$ denote the reflection of $Y$ over $MN$ .
Observe that $AXNZ$ is cyclic. Indeed, $\measuredangle XAD = \measuredangle OAN = \measuredangle ANO = \measuredangle DYE = \measuredangle DZE.$ Also similarly observe that $CYZB$ is trivially cyclic. Hence it now suffices to show $BZCX$ is cylic, which is trivial as $DX \cdot DZ = DA \cdot DN = DB \cdot DC,$ as desired. $\blacksquare$

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ihatemath123

3441 posts

ihatemath123

#86 h Apr 19, 2024, 3:39 AM • 1 Y

Y by OronSH

Letting $M$ be the midpoint of arc $BC$ , we have $XD \parallel OM$ , so $\triangle AXD \sim \triangle AOM$ . Therefore, $\triangle AXD$ is isosceles. Let $F$ be the midpoint of $\overline{AD}$ . By power of a point, $FBYC$ is cyclic.

Because $X$ and $Y$ are equidistant from line $OM$ , it follows that the circumcenter of right triangle $XFY$ lies on $OM$ . So, $(XFY)$ must intersect line $BC$ at two points equidistant from $O$ . For power of a point reasons, the only circle passing through $F$ and $Y$ that satisfies the aforementioned property is $(BFCY)$ ; therefore, $(BFXCY)$ is cyclic.

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L13832

256 posts

L13832

#87 h Dec 7, 2024, 3:56 PM

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Easy for a G4
$[asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; real xmin = -7, xmax = 8, ymin = -5, ymax = 9; /* image dimensions */ draw((0.9282707692837183,5.124674700939942)--(-0.2,0)--(7,0)--cycle, linewidth(0.7)); /* draw figures */ draw(circle((3.4,1.893948134588014), 4.067805248104857), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(-0.2,0), linewidth(0.7)); draw((-0.2,0)--(7,0), linewidth(0.7)); draw((7,0)--(0.9282707692837183,5.124674700939942), linewidth(0.7)); draw(circle((3.4,-0.7457494356122296), 3.676430635917937), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(4.136200965821236,0), linewidth(0.7)); draw((-5.770044925854284,0)--(0.9282707692837183,5.124674700939942), linewidth(0.7)); draw((-5.770044925854284,0)--(2.663799034178764,2.856215355809231), linewidth(0.7)); draw((-5.770044925854284,0)--(-0.2,0), linewidth(0.7)); draw((0.9282707692837183,5.124674700939942)--(4.136200965821236,-4.34771422703369), linewidth(0.7)); draw((2.663799034178764,2.856215355809231)--(2.663799034178764,-4.3477142270336895), linewidth(0.7)); draw((4.136200965821236,0)--(4.136200965821236,-4.34771422703369), linewidth(0.7)); draw((3.4,1.8939481345880143)--(3.4,-2.1738571135168425), linewidth(0.7)); dot((0.9282707692837183,5.124674700939942),linewidth(3pt) + dotstyle); label("$A$", (0.3859373501577617,5.175259523212921), NE * labelscalefactor); dot((-0.2,0),linewidth(3pt) + dotstyle); label("$B$", (-0.759590777205151,-0.68818809213788084), NE * labelscalefactor); dot((7,0),linewidth(3pt) + dotstyle); label("$C$", (7.2362103849019395,-0.30079460486524406), NE * labelscalefactor); dot((3.4,1.8939481345880143),linewidth(3pt) + dotstyle); label("$O$", (3.488340639449198,2.010391738163943), NE * labelscalefactor); dot((2.6637990341787634,0),linewidth(3pt) + dotstyle); label("$D$", (2.1765862285407382,-0.7090095907237294), NE * labelscalefactor); dot((3.4,0),linewidth(3pt) + dotstyle); label("$M$", (3.238482656419015,-0.6298310893095779), NE * labelscalefactor); dot((4.136200965821236,0),linewidth(3pt) + dotstyle); label("$E$", (4.362843580054838,-0.5298310893095779), NE * labelscalefactor); dot((2.663799034178764,2.856215355809231),linewidth(3pt) + dotstyle); label("$X$", (2.6138376988435583,3.0306451688705214), NE * labelscalefactor); dot((4.136200965821236,-4.34771422703369),linewidth(3pt) + dotstyle); label("$Y$", (4.446129574398232,-4.735773803650981), NE * labelscalefactor); dot((2.663799034178764,-4.3477142270336895),linewidth(3pt) + dotstyle); label("$Y'$", (1.9,-4.87), NE * labelscalefactor); dot((3.4,-2.1738571135168425),linewidth(3pt) + dotstyle); label("$M_{A}$", (2.678031748029836,-2.86239450661285), NE * labelscalefactor); dot((1.7960349017312414,2.56233735046997),linewidth(3pt) + dotstyle); label("$S$", (1.28126178934925,2.6350366957393994), NE * labelscalefactor); dot((-5.770044925854284,0),linewidth(3pt) + dotstyle); label("$R$", (-6.235299200142082,-0.6339385822388206), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Let $M_A$ be the arc-midpoint of minor-arc $\widehat{BC}$ and since $\overline{O-M-M_A}\parallel EY$ , due to MPT it is also the midpoint of $DY$ .
Let the $AA\cap BC=R$ so it is well-known that $TX$ is the perpendicular bisector of $AD$ , suppose that point is $S$ , by POP we have
$\begin{align*} RS\cdot RX=RA^2=RB\cdot RC\implies \odot(BSXC)\\ BD\cdot CD=AD\cdot M_AD=SD\cdot YD\implies \odot(BSCY) \end{align*}$ Thus, $\odot(BXCY)$ is cyclic.

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Aiden-1089

277 posts

Aiden-1089

#88 h Dec 9, 2024, 4:27 AM

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Let $M$ be the midpoint of $BC$ , $M_a$ be the midpoint of arc $BC$ not including $A$ on $(ABC)$ .
Since $M_a$ is the midpoint of $DE$ , we have that $DM=MY$ .
Let $Y'$ be the reflection of $Y$ across the perpendicular bisector of $BC$ . Note that $(BCYY')$ are concyclic, $DY' \perp BC$ , and $DY' = 2MM_a$ .

Note that the exists a circle centred at $X$ , passing through $A,D$ , and tangent to $(ABC)$ and $BC$ . Call this circle $\omega$ .
Let $D'$ be the antipode of $D$ in $\omega$ .
Now, $\angle DAD' = 90^\circ = \angle MM_aD$ and $\angle ADD' = \angle M_aMD$ since $DD'$ and $DM$ are both perpendicular to $BC$ .
So we have $\Delta ADD' \sim M_aMD$ , so $\frac{DD'}{MD} = \frac{AD}{MM_a} \implies DD' \cdot MM_a = DA \cdot DM$ .
By power of point, $DY' \cdot DX = 2MM_a \cdot \frac{DD'}{2} = DD' \cdot MM_a = DA \cdot DM = DB \cdot DC$ , so $(BXCY')$ concyclic.
It follows that $(BXCY)$ concyclic, so we are done.

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math-olympiad-clown

17 posts

math-olympiad-clown

#89 h Feb 24, 2025, 3:10 PM

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how can i solve this problem without thinking about making the reflection of y across line OM?
(without using some horrible coordinate bash)

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KevinYang2.71

411 posts

KevinYang2.71

#90 h Mar 15, 2025, 10:51 PM

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We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$ . Then $D=\left(0,\,\frac{b}{b+c},\,\frac{c}{b+c}\right)$ , $E=\left(0,\,\frac{c}{b+c},\,\frac{b}{b+c}\right)$ , and $O=(a^2S_A:b^2S_B:c^2S_C)$ . Let
$X=:(p:b^2S_B:c^2S_C)=\left(\frac{p}{p+b^2S_B+c^2S_C},\,\frac{b^2S_B}{p+b^2S_B+c^2S_C},\,\frac{c^2S_C}{p+b^2S_B+c^2S_C}\right)$ and let $\sigma:=p+b^2S_B+c^2S_C$ . Then $\overrightarrow{DX}=\left(\frac{p}{\sigma},\,\frac{b^2S_B}{\sigma}-\frac{b}{b+c},\,\frac{c^2S_C}{\sigma}-\frac{c}{b+c}\right)$ and $\overrightarrow{BC}=(0,\,-1,\,1)$ so
$a^2\left(\frac{b^2S_B-c^2S_C}{\sigma}-\frac{b-c}{b+c}\right)+\frac{b^2p}{\sigma}-\frac{c^2p}{\sigma}=0$ by the barycentric perpendicular formula. Multiplying by $\sigma(b+c)$ gives
$\begin{align*} a^2((b+c)(b^2S_B-c^2S_C)-(b-c)(q+b^2S_B+c^2S_C))+p(b+c)(b^2-c^2)&=0\\ \implies 2a^2(b^2cS_B-c^2bS_C)+p(b-c)((b+c)^2-a^2)&=0. \end{align*}$ Note that
$2(b^2cS_B-c^2bS_C)=b^2c(a^2+c^2-b^2)-c^2b(a^2+b^2-c^2)=bc(b-c)(a^2-(b+c)^2)$ so $p=-\frac{a^2bc(b-c)(a^2-(b+c)^2)}{(b-c)((b+c)^2-a^2)}=a^2bc$ and $\sigma=\frac{(b+c)^2(a^2-(b-c)^2)}{2}$ .

Let
$Y=:(q:b:c)=\left(\frac{q}{q+b+c},\,\frac{b}{q+b+c},\,\frac{c}{q+b+c}\right).$ Then $\overrightarrow{EY}=\left(\frac{q}{q+b+c},\,\frac{b}{q+b+c}-\frac{c}{b+c},\,\frac{c}{q+b+c}-\frac{b}{b+c}\right)$ so
$a^2\left(\frac{b-c}{q+b+c}+\frac{b-c}{b+c}\right)+(b^2-c^2)\frac{q}{q+b+c}=0$ by the barycentric perpendicular formula. Multiplying by $\frac{(b+c)(q+b+c)}{b-c}$ gives
$a^2(q+2b+2c)+q(b+c)^2=0$ so $q=-\frac{2a^2(b+c)}{a^2+(b+c)^2}$ and $q+b+c=\frac{(b+c)((b+c)^2-a^2)}{a^2+(b+c)^2}$ .

Any circle through $B$ and $C$ has the form $-a^2yz-b^2zx-c^2xy+ux(x+y+z)=0$ . It suffices to prove that $X$ and $Y$ plugged into this equation produce the same value of $u$ .

Noting that $S_C+S_B=a^2$ , the value of $u$ produced from $X$ is
$\begin{align*} u&=\frac{a^2b^2c^2S_BS_C+b^2c^2S_Ca^2bc+c^2b^2S_Ba^2bc}{a^2bc\sigma}\\ &=\frac{2bc(S_BS_C+a^2bc)}{(b+c)^2(a^2-(b-c)^2)}\\ &=\frac{bc((a^2+c^2-b^2)(a^2+b^2-c^2)+4a^2bc)}{2(b+c)^2(a^2-(b-c)^2)}\\ &=\frac{bc(a^2+(b+c)^2)(a^2-(b-c)^2)}{2(b+c)^2(a^2-(b-c)^2)}\\ &=\frac{bc(a^2+(b+c)^2)}{2(b+c)^2}. \end{align*}$ The value of $u$ produced from $Y$ is
$\begin{align*} u&=\frac{a^2bc+b^2cq+c^2bq}{q(q+b+c)}\\ &=\frac{a^2bc(a^2+(b+c)^2)^2+b^2c(-2a^2(b+c))(a^2+(b+c)^2)+c^2b(-2a^2(b+c))(a^2+(b+c)^2)}{-2a^2(b+c)^2((b+c)^2-a^2)}\\ &=-\frac{a^2bc(a^2+(b+c)^2)(a^2+(b+c)^2-2b(b+c)-2c(b+c))}{2a^2(b+c)^2((b+c)^2-a^2)}\\ &=-\frac{bc(a^2+(b+c)^2)(a^2-(b+c)^2)}{2(b+c)^2((b+c)^2-a^2)}\\ &=\frac{bc(a^2+(b+c)^2)}{2(b+c)^2}, \end{align*}$ which agrees with our other value of $u$ . $\square$

This post has been edited 1 time. Last edited by KevinYang2.71, Mar 15, 2025, 10:52 PM

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