IMO Shortlist 2013, Geometry #2

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127 posts

#1 h Jul 9, 2014, 6:24 PM • 13 Y

Y by Davi-8191, Abderrahmane_Driouch, mathroyal, A-Thought-Of-God, sotpidot, chessgocube, megarnie, Adventure10, torch, lian_the_noob12, Rounak_iitr, and 2 other users

Let $\omega$ be the circumcircle of a triangle $ABC$ . Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$ , respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$ . The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$ , respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$ . The lines $MN$ and $XY$ intersect at $K$ . Prove that $KA=KT$ .

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ThirdTimeLucky

402 posts

ThirdTimeLucky

#2 h Jul 9, 2014, 6:43 PM • 5 Y

Y by Tawan, chessgocube, Adventure10, Mango247, and 1 other user

Let $O$ be the circumcenter of $\triangle ABC$ . From Easy - Circles and Midpoints, we get $\angle NXT = 90^{\circ}$ , and $OX=OM$ . Similarly, $OY=ON$ . Thus, $MNYX$ becomes an isosceles trapezoid, and hence we have $KM=KX$ . Also, from the construction in the link, we get $XT = AM$ and $\angle KXT = \angle OXT - \angle OXY = 90^{\circ} - \angle OMN = \angle AMN$ . Therefore, $\triangle KMA \cong \triangle KXT \implies KA=KT$

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thecmd999

2860 posts

thecmd999

#3 h Jul 10, 2014, 5:32 AM • 6 Y

Y by Tawan, Davi-8191, lahmacun, chessgocube, Adventure10, Mango247

Solution

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uglysolutions

304 posts

uglysolutions

#4 h Jul 10, 2014, 6:36 PM • 6 Y

Y by Dukejukem, Tawan, thunderz28, lahmacun, Adventure10, lian_the_noob12

First we will prove that $MX \parallel AT$ . Let $X'$ be the second point where the parallel line to $AT$ passing through $M$ meets the circumcircle of $AMT$ . We will see that $X'$ lies on the perpendicular bisector of $AC$ . From this it follows that $X' = X$ and hence $MX \parallel AT$ .
We claim that $\triangle MBT$ and $\triangle X'TC$ are congruent. Since $AMX'T$ is a trapezoid inscribed in a circle, it is an isosceles trapezoid, and hence $MB = MA = X'T$ . Also, we have $BT = TC$ because $T$ is the midpoint of the arc $BC$ . Finally, $\angle X'TC = \angle X'TA + \angle ATC = \angle BAT + \angle ABC$ (using again that $AMX'T$ is an isosceles trapezoid for the first part, and inscribed angles in $\omega$ for the second part). But we also have $\angle BAT = \angle TBC$ , because the chords $BT$ and $TC$ of $\omega$ have equal lengths. Therefore, $\angle X'TC = \angle TBC + \angle ABC = \angle MBT$ . We have proven that $\triangle MBT \equiv \triangle X'TC$ . In particular $MT = X'C$ . But $MT = X'A$ , because they are the diagonals of an isosceles trapezoid. Then $X'A = X'C$ , that is, $X'$ lies on the perpendicular bisector of $AC$ , as wanted.
So far we know $MX \parallel AT$ , and moreover that $AMXT$ is an isosceles trapezoid. We can prove in an analogous way that $ANYT$ is an isosceles trapezoid. It follows that the segments $MX$ , $AT$ and $NY$ all have the same perpendicular bisector. But then $K$ , which is the intersection point of $MN$ and $XY$ , belongs to this perpendicular bisector. In particular $KA = KT$ , and we are done. $\blacksquare$

Attachments:

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fclvbfm934

759 posts

fclvbfm934

#5 h Jul 12, 2014, 6:26 AM • 3 Y

Y by Tawan, Adventure10, Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -4.300000000000004, xmax = 11.32000000000001, ymin = -3.020000000000004, ymax = 6.300000000000007; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((2.540000000000003,5.000000000000000)--(0.4000000000000005,0.8400000000000009)--(5.720000000000006,0.6800000000000008)--cycle, zzttqq); /* draw figures */ draw((2.540000000000003,5.000000000000000)--(0.4000000000000005,0.8400000000000009), zzttqq); draw((0.4000000000000005,0.8400000000000009)--(5.720000000000006,0.6800000000000008), zzttqq); draw((5.720000000000006,0.6800000000000008)--(2.540000000000003,5.000000000000000), zzttqq); draw(circle((3.099747970952588,2.081620034173432), 2.971575241504656)); draw((xmin, -12.51783780793794*xmin + 36.79530803216237)--(xmax, -12.51783780793794*xmax + 36.79530803216237)); /* line */ draw((xmin, -2.472454225932990*xmin + 6.554507712121498)--(xmax, -2.472454225932990*xmax + 6.554507712121498)); /* line */ draw((xmin, 3.330360015372642*xmin-10.91438686348901)--(xmax, 3.330360015372642*xmax-10.91438686348901)); /* line */ draw(circle((5.312765398507484,2.258409145960098), 3.899301009939554)); draw(circle((0.5621914115686578,1.878904789071374), 3.694991492297976)); draw((xmin, 0.7361111111111118*xmin-0.2001388888888885)--(xmax, 0.7361111111111118*xmax-0.2001388888888885)); /* line */ draw((xmin, -0.5144230769230775*xmin + 3.676201923076927)--(xmax, -0.5144230769230775*xmax + 3.676201923076927)); /* line */ draw((2.540000000000003,5.000000000000000)--(3.099747970952588,2.081620034173432)); draw((3.099747970952588,2.081620034173432)--(0.4000000000000005,0.8400000000000009)); draw((xmin, 0.1918009229459768*xmin + 0.6836954941066192)--(xmax, 0.1918009229459768*xmax + 0.6836954941066192)); /* line */ draw((xmin, -0.03007518796992487*xmin + 2.964210526315793)--(xmax, -0.03007518796992487*xmax + 2.964210526315793)); /* line */ draw((1.470000000000002,2.920000000000003)--(1.623769685397414,0.9951360184175412)); draw((4.130000000000004,2.840000000000003)--(4.237333239206217,1.496419920216036)); draw((0.5621914115686578,1.878904789071374)--(10.27832614694408,2.655087935430257)); /* dots and labels */ dot((2.540000000000003,5.000000000000000),dotstyle); label("$A$", (2.620000000000003,5.120000000000005), NE * labelscalefactor); dot((0.4000000000000005,0.8400000000000009),dotstyle); label("$B$", (0.4800000000000005,0.9600000000000011), NE * labelscalefactor); dot((5.720000000000006,0.6800000000000008),dotstyle); label("$C$", (5.800000000000006,0.8000000000000009), NE * labelscalefactor); dot((1.470000000000002,2.920000000000003),dotstyle); label("$M$", (1.560000000000002,3.040000000000004), NE * labelscalefactor); dot((4.130000000000004,2.840000000000003),dotstyle); label("$N$", (4.220000000000004,2.960000000000003), NE * labelscalefactor); dot((3.010417678247103,-0.8886121982839235),dotstyle); label("$T$", (3.100000000000004,-0.7600000000000009), NE * labelscalefactor); dot((3.099747970952588,2.081620034173432),dotstyle); label("$O$", (3.180000000000004,2.200000000000003), NE * labelscalefactor); dot((1.623769685397414,0.9951360184175412),dotstyle); label("$X$", (1.700000000000002,1.120000000000001), NE * labelscalefactor); dot((4.237333239206217,1.496419920216036),dotstyle); label("$Y$", (4.320000000000005,1.620000000000002), NE * labelscalefactor); dot((10.27832614694408,2.655087935430257),dotstyle); label("$K$", (10.36000000000001,2.780000000000003), NE * labelscalefactor); dot((5.312765398507484,2.258409145960098),dotstyle); label("$O_1$", (5.400000000000007,2.380000000000003), NE * labelscalefactor); dot((0.5621914115686578,1.878904789071374),dotstyle); label("$O_2$", (0.6400000000000007,2.000000000000000), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This lemma took me so long to prove I'm not even sure why D:<

Lemma: $AMXT$ is an isoceles trapezoid. By symmetry, $ANYT$ would also be an isoceles trapezoid.

Proof: Let $X'$ be on the circumcircle of $AMT$ such that $X'M||AT$ . Then, $AM = X'T$ and we already know $OT = OA$ . But we also have $\angle TON = 180^{\circ} - \angle C$ , so $X'OT \cong MOA$ since $\angle MOA = \angle C$ . So we have $MO = X'O$ . Therefore, $AOX' \cong TOM$ . Notice that $\angle TOC = \angle A$ . We also have $\angle TAB = \angle A/2$ so $\angle XMA = 180^{\circ} - A/2$ implying that $\angle X'MO = 90^{\circ} - A/2$ so $\angle XOM = A$ . Therefore, $\angle X'OC = \angle MOT$ , so $X'OC \cong MOT \cong X'OA$ , so $X'C = X'A$ , so $X'$ lies on the perpendicular bisector of $AC$ . Hence, $X' = X$ .
$\Box$

Therefore, $XMNY$ is an isoceles trapezoid; the perpendicular bisector of $XM$ and $NY$ and $AT$ are the same line. Furthermore, $MN \cap XY$ lies on the perpendicular bisector, so $KA = KT$ , as desired.

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AnonymousBunny

339 posts

AnonymousBunny

#6 h Jul 14, 2014, 5:41 PM • 3 Y

Y by Tawan, Adventure10, Mango247

Claim: $AT \parallel MX$
Proof: Let the line passing through $M$ parallel to $AT$ meet $(AMT)$ at $X'.$ Since $(AMTX')$ is cyclic and every cyclic trapezoid is isoceles, we must have that $BM = AM = TX'.$ Now, some trivial angle chasing (too lazy to post it now; will complete it later) yields $\triangle BTM \sim \triangle CTX',$ implying $X=X'.$ $\blacksquare$

Now, since $ATMX$ is cyclic, it must be isoceles. Similarly, so must be $ATNY.$ It follows that $AT, MX, NY$ share the same perpendicular bisector, so $K$ must also lie on this perpendicular bisector. $\blacksquare$

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Mathematicalx

537 posts

Mathematicalx

#7 h Jul 15, 2014, 3:26 PM • 3 Y

Y by Tawan, Adventure10, Mango247

Claim: $AT\parallel NY$
Proof:Let point $N'$ be on circumcircle of $\triangle {ANT}$ such that $AT\parallel NN'$ and let
$TN'$ intersects with $AC$ inside $\omega$ at $R$ (if outside its same), intersects with $\omega$ at $P$ .
Since $ANN'T$ is isosceles trapezoid, we have $\triangle{ART}$ , $\triangle{NN'R}$ and $\triangle{PRC}$ are isosceles which means $AN=NC=TN'=N'P$ , thus we have $N'$ is midpoint of $TP$ .
Since $AB\parallel TP$ and $ABTP$ is isosceles trapezoid, we conclude $MN'\perp TP$ $\implies$ $N'=Y$ .

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utkarshgupta

2280 posts

utkarshgupta

#8 h Jan 15, 2015, 2:02 PM • 1 Y

Y by Adventure10

AnonymousBunny wrote:

$\triangle BTM \sim \triangle CTX',$ implying $X=X'.$ $\blacksquare$

How does this imply $X=X'$ ?

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utkarshgupta

2280 posts

utkarshgupta

#9 h Jan 15, 2015, 2:18 PM • 7 Y

Y by john10, Tawan, Darghy, srijonrick, Elyson, straight, Adventure10

Claim: $AMXT$ is an isoceles trapezoid.
I will prove this indirectly.
Let $X'$ be the point on $\odot AMT$ such that $MX' ||AT$
$\implies TAMX'$ is an isosceles trapezoid.
Thus we have $OA=OT, OM=OX'$ and $AM=TX'$
$\implies \triangle OAM \cong OTX'$
$\implies \angle TOX' = \angle AOM= \angle C$

But we know $\angle TOX = 180- \angle TON = \angle C$
$\implies X=X'$
Hence the claim.

The remaining proof is trivial.

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supercomputer

491 posts

supercomputer

#10 h Apr 4, 2015, 12:58 AM • 1 Y

Y by Adventure10

Can anyone barycentric/complex bash this?

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Dukejukem

695 posts

Dukejukem

#11 h Oct 9, 2015, 7:51 PM • 3 Y

Y by Tawan, Adventure10, Mango247

Let $\tau, \ell$ be the perpendicular bisectors of $\overline{AT}, \overline{AC}$ , respectively. Let $Q$ lie on $\omega$ such that $TQ \parallel AC$ and let $X'$ be the midpoint of $\overline{TQ}.$

Since $ACTQ$ is an isoceles trapezoid, $O, N, X'$ are collinear on $\ell.$ Meanwhile, from $\widehat{AQ} = \widehat{TC} = \widehat{TB}$ , it follows that $ATBQ$ is an isoceles trapezoid as well. Then from symmetry in $\tau$ , we see that $ATX'M$ is an isoceles trapezoid too. Hence, $X'$ lies on the circumcircle of $\triangle AMT \implies X' \equiv X.$ Therefore, $\tau$ is the perpendicular bisector of $\overline{MX}$ , and analagous arguments show $\tau$ is the perpendicular bisector of $\overline{NY}$ as well. It follows that $XYNM$ is an isoceles trapezoid, so by symmetry, $K \in \tau.$ Thus, $KA = KT$ , as desired. $\square$

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EulerMacaroni

851 posts

EulerMacaroni

#12 h Nov 20, 2015, 4:42 AM • 2 Y

Y by Tawan, Adventure10

It suffices to show that $AMXT$ is an isosceles trapezoid, which implies that segments $AT, MX$ and $NY$ have the same perpendicular bisector by symmetry.

Let $X'$ be the point where the perpendicular bisector of $BC$ hits $(AMT)$ inside $\triangle ABC$ . Suppose the external angle bisector of $A$ hits $(AMT)$ at $F$ ; from $\angle TAF=90^{\circ}$ and $\angle TX'N=90^{\circ}$ , it follows that $X, N, F$ are collinear. Thus $\angle XTA=\angle XFA=\angle NFA=90^{\circ}-\angle NAF=\angle TAN=\frac{\angle A}{2}$ . But $\angle MAT=\angle BAT=\frac{\angle A}{2}$ , so we're done $.\:\blacksquare\:$

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K6160

276 posts

K6160

#13 h Jan 25, 2016, 10:42 PM • 2 Y

Y by Tawan, Adventure10

Complex bash because why not.

Let $Y'$ be the reflection of $N$ over the perpendicular bisector of $AT$ . We will show that $Y=Y'$ . We know that $Y'$ clearly lies on the circumcircle of $\triangle ANT$ . Now we can complex bash to obtain that $Y'M\perp AB$ .

Let the circumcircle of $\triangle ABC$ be the unit circle. WLOG, let $t=\bar{a}$ . This implies $y'=\bar{n}=\frac{1}{n}$ . Additionally, because $T$ is the midpoint of arc $BC$ , $\frac{1}{a}=\sqrt{bc} \Rightarrow a^2bc=1$ . Also notice that $n=\frac{a+c}{2}$ and $m=\frac{a+b}{2}$ .

We have that $\frac{y'-m}{a-b}=\frac{\overline{\frac{a+c}{2}}-\frac{a+b}{2}}{a-b}=\frac{1}{2}\cdot \frac{\frac{1}{a}+\frac{1}{c}-a-b}{a-b}=\frac{1}{2}\cdot \frac{a^2b+abc-b-a}{a-b}=\frac{1}{2}\cdot \frac{a+c-\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}-\frac{1}{b}}=-\overline{\left(\frac{y'-m}{a-b}\right )}$

So indeed, $Y'M\perp AB \implies Y'=Y$ . Similarly we can define $X'$ , and show that $X'=X$ . The result quickly follows. $\Box$

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AMN300

563 posts

AMN300

#17 h Apr 8, 2016, 3:47 AM • 3 Y

Y by Tawan, Adventure10, Mango247

Darn, this problem trolled me. D:

Let the perpendicular bisector of $AT$ be $\overline{p}$ . Observe that $\odot(AMT)$ is symmetric wrt $\overline{p}$ . Now, I claim $\overline{OM}$ and $\overline{ON}$ are symmetric with respect to $\overline{p}$ . Let the reflection of $N$ wrt $\overline{p}$ be $N'$ , and similarly for $M$ . We will show $M, O, N'$ are collinear, which proves the claim, as we can use the same argument to show that $N, O, M'$ are collinear.

To that end, we use complex numbers. Denote the complex number associated with each point by the corresponding lowercase letters. Set $\overline{p}$ as the real axis. As a result, we have $t = \frac{1}{a} = \sqrt{bc} \implies a^2bc=1$ by definition of $T$ . So we simply want
$\frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}} = \overline{\left(\frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}}\right)}$ This easily expands to $\frac{(a+b)(a+c)}{4} = \frac{(a+b)(a+c)}{4a^2bc}$ , which is true.

Therefore, $M$ and $X$ are related by reflection about $\overline{p}$ , and similarly for $N$ and $Y$ . So $K$ lies on $\overline{p}$ , and $KA=KT$ as desired.

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navi_09220114

475 posts

navi_09220114

#18 h Jul 3, 2016, 1:20 PM • 2 Y

Y by Adventure10, Mango247

I think this problem is about the same as the configuration given in IMO 2007 P4!

First let us investigate some properties of $X, Y$ . We claim that $TY \parallel AM$ as well as $TX \parallel AN$ . Let $Y'$ be a point such that $TY'\parallel AM$ and $\angle MY'T =90^{\circ}$ . If suffices to prove that $TY'=AN$ , then $ANY'T$ cyclic $\Rightarrow Y=Y'$ , because we already have $\angle NAT=\angle MAT=\angle Y'TA$ .

Let $O$ be the circumcenter of $\triangle ABC$ , then let $MO\cap AT=U$ , $NO\cap AT=V$ . Then we have $\angle AUM=90- \angle ABC=\angle AVN$ , and $\angle MAU=\angle NAV$ , thus $\triangle AMU \sim \triangle ANV$ . Then $\frac{AU}{AM}=\frac{AV}{AN}$ . But $\angle OUV=\angle OVU$ , so $AU=TV$ , so $\frac{AU}{AM}=\frac{AV}{AN}=\frac{UT}{AN} \Rightarrow \frac{AU}{UT}=\frac{AM}{AN}$ .

Back to the original problem, $TY'\parallel AM$ implies $\frac{AM}{AU}=\frac{TY'}{TU}$ , so $TY'=AM$ , so $Y=Y'$ , and thus $TY\parallel AM$ . Likewise $TX\parallel AN$ . In particular, $ANYT, AMXT$ are isosceles trapezoids, thus so as $MNXT$ . This concludes that $MN\cap XT=K$ is on the perpendicular bisector of $MX, NY, AT$ , thus $KA=KT$ .

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EpicBird08

1741 posts

EpicBird08

#71 h Mar 12, 2024, 12:57 AM

Y by

The main claim is that $TY \perp YM.$ To see why, we phantom-point $Y'$ as the intersection of the perpendicular bisector of $AB$ and the line through $T$ parallel to $AB.$ We prove that $ANY'T$ is an isosceles trapezoid, which proves our claim.

First, $\measuredangle NAT = \measuredangle CAT = \measuredangle TAB = \measuredangle ATY.$ Next reflect $T$ across point $Y'$ to get $Z,$ which lies on $(ABC)$ since $Z$ and $T$ are reflections across the perpendicular bisector of $AB.$ Thus $ABTZ$ is an isosceles trapezoid -> $AZ = BT = CT$ -> $AZCT$ is an isosceles trapezoid -> $ZT = AC$ -> $\frac{ZT}{2} = TY = \frac{AC}{2} = AN,$ proving our claim. We similarly see that $\angle TXN = 90^\circ$ and $AM = TX.$

This claim implies $3$ things:
1. $AM = TX$ and $AN = TY.$
2. $TXOY$ is cyclic, implying that $\measuredangle XTA = \measuredangle XOY = \measuredangle NOM = \measuredangle NAM.$ This implies that $\triangle AMN \cong \triangle TXY,$ so $MN = XY.$
3. $MX \parallel AT \parallel NY,$ so $MXYN$ is an isosceles trapezoid.

Therefore, the perpendicular bisector of $AT$ is just that of $MX.$ Since $MNYX$ is isosceles, $MN \cap XY = K$ lies on this perpendicular bisector, and we are done.

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Markas

105 posts

Markas

#75 h May 5, 2024, 12:36 PM

Y by

Claim: $\angle TYM = 90^{\circ}$

Let the line parallel to AB trough T be l. Let $S_{AB} \cap l = Y'$ . If we prove that $ANY'T$ is an isosceles trapezoid our claim will be shown.

Now $\angle NAT = \angle CAT = \angle TAB = \angle ATY$ . After this reflect T across point Y' to get P which lies on (ABC) since P and T are reflections across $S_{AB}$ $\Rightarrow$ ABTP is an isosceles trapezoid $\Rightarrow$ AP = BT = CT $\Rightarrow$ APCT is an isosceles trapezoid $\Rightarrow$ PT = AC, $\frac{PT}{2} = TY = \frac{AC}{2} = AN$ , which is the end of the proof for our claim. Similarly we see that $\angle TXN = 90^{\circ}$ and AM = TX.

From $\angle TYM = \angle TXN = 90^{\circ}$ , we have that AM = TX and AN = TY. Also TXOY is cyclic, from which we have that $\angle XTA = \angle XOY = \angle NOM = \angle NAM$ . This means $\triangle AMN \cong \triangle TXY$ , so MN = XY. From $TY \perp YM$ , and $TX \perp XN$ , $MX \parallel AT \parallel NY$ $\Rightarrow$ MXYN is an isosceles trapezoid. From this being true, it follows that $S_{AT} \equiv S_{MX}$ . Since MNYX is isosceles,

$MN \cap XY = K$ and $K \in S_{AT}$ , so KA = KT. We are ready.

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AngeloChu

470 posts

AngeloChu

#76 h May 13, 2024, 12:35 PM

Y by

let $O$ be the circumcenter of $ABC$ , and denote the perpendicular bisector of $AT$ as $l$
let $NO$ intersect the circumcircle of $ANT$ at $P$ , and let $MO$ intersect the circumcircle of $AMT$ at $Q$
since $T$ is the midpoint of arc $BC$ of $\omega$ , $BAT=CAT$ , and $NPT=NAT=MAT=MQT$ , we can get that $l$ is the angle bisector of $NOY$
then, $NO=YO$ , $MO=XO$ , $MN=XY$ , $MNY=XYN$ , and $NK=YK$ , so $K$ is on $l$
$T$ is the reflection of $A$ over $l$ , so $KA=KT$

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shendrew7

793 posts

shendrew7

#77 h Jun 6, 2024, 12:39 AM

Y by

Notice the perpendicular bisector of $AT$ , say $\ell$ , is both the line connecting the centers of $(AMT)$ , $(ANT)$ and the axis of symmetry between line $MOY$ and $NOX$ . Consequently, $AMXT$ and $ANYT$ are isosceles trapezoids, so segment $MN$ is simply the reflection of segment $XY$ over $\ell$ , and thus their intersection $T$ must lie on $\ell$ . $\blacksquare$

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Om245

163 posts

Om245

#78 h Jul 24, 2024, 8:46 AM

Y by

Yay!! do $\sqrt{\frac{bc}{2}}$ inversion.

If $D$ is feet of perpendicular from $A$ to $BC$ and $J=AT \cap MN$ . Then $X' = CJ \cap (ADB)$ and $Y' = BJ \cap (ACD)$ .
now using well known inversion we have $AJ^2=JX'\cdot JC=JY' \cdot JB$ So $X',Y',B,C$ cyclic.
Now back to original problem we have $M,N,X,Y$ cyclic, with $MX \parallel NY$ . As $O = MY \cap NX$ we have $J -O - K$ . $K$ will lie on perpendicular bisector of $AT$ .

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N3bula

257 posts

N3bula

#79 h Sep 18, 2024, 9:40 AM

Y by

Proving $MX \parallel AT \parallel NY$ suffices as this implies that the perpendicular bisector of $AT$ is also the perpendicular bisector of $NY$ and of $MX$ which implies the result.
To show that $MX \parallel AT$ , let $X'$ denote the second intersection of the parallel from $M$ to $AT$ with $(AMT)$ , it suffices to prove that $X'N$ is the perpendicular bisector of
$AC$ , we have that $MB=AM=TX'$ , we also have that $CT=BT$ , finally we have $CTX'=\frac{BAC}{2}+ABC$ and we have that $MBT=ABC+\frac{BAC}{2}$ which gives us that $\triangle CX'T \equiv \triangle MBT$ and thus $CX'=MT=AX'$ proving the result. A similar argument proves $NY \parallel AT$ .

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naonaoaz

329 posts

naonaoaz

#81 h Oct 22, 2024, 5:29 AM • 1 Y

Y by cursed_tangent1434

Define $X'$ as a point on $(AMT)$ such that $MX' \parallel AT$ .
Claim: $X = X'$
Proof:
$\angle TBM = \angle TBC + \angle B = \frac{\angle A}{2}+\angle B = \angle CTX'$ since $\angle CTX' = \angle ATC + \angle ATX' = \angle B + \frac{\angle A}{2}$ . By construction, $MB = MA = X'T$ and $TB = TC$ , so $\Delta TBM \cong \Delta CTX'$ . Therefore, $CX' = MT = X'A$ , so $\Delta AX'C$ is isosceles. Thus, $\overline{X'N} \perp \overline{AC}$ , so $X = X'$ . $\square$

Due to this, the perpendicular from $O$ to $\overline{AT}$ must also be perpendicular to $\overline{MX}$ and $\overline{NY}$ , which finishes as then $\Delta KAT$ is isosceles.

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onyqz

195 posts

onyqz

#82 h Oct 23, 2024, 2:46 PM

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101st post, greetings to my Taiwanese fellas :bye:

solution for storage

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fearsum_fyz

48 posts

fearsum_fyz

#83 h Oct 25, 2024, 5:40 PM

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Let $O$ denote the circumcenter of $\Delta{ABC}$ . It is obvious that lines $OM$ and $ON$ are the perpendicular bisectors of $AB$ and $AC$ .
The key claim is the following:

Claim: $\Delta{AMN}$ and $\Delta{TXY}$ are symmetric about line $OK$ .
Proof.
Let $X'$ and $Y'$ be the feet of the perpendiculars from $T$ to lines $ON$ and $OM$ respectively.
In triangles $\Delta{AMO}$ and $\Delta{TX'O}$ , we have:

$OA = OT = R$
$\measuredangle{TX'O} = 90^{\circ} = - \measuredangle{AMO}$
$\measuredangle{X'OT} = \measuredangle{NOL} = \measuredangle{NCL} = \measuredangle{ACB} = - \measuredangle{MOA}$

Hence by the SAA test, $\Delta{AMO} \cong \Delta{TX'O}$ .
Similarly, $\Delta{ANO} \cong \Delta{TY'O}$ .
This implies quadrilaterals $\square{AMON}$ and $\square{TX'OY'}$ are congruent.
Further, $\square{MX'Y'N}$ is an isosceles trapezoid and line $OK$ is the perpendicular bisector of each of $MX'$ , $NY'$ , and $AT$ .
It is not hard to see that $X = X'$ and $Y = Y'$ : Since $\square{AMX'T}$ and $\square{ANY'T}$ are both isosceles trapezoids, they must be cyclic.

Now, since $T$ is the reflection of $A$ across line $OK$ , we obviously must have $KA = KT$ .

https://i.imgur.com/qvy1BNx_d.webp?maxwidth=1520&fidelity=grand

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Saucepan_man02

1300 posts

Saucepan_man02

#84 h Dec 2, 2024, 2:15 PM

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Phantom the points:

It suffice to show that $AMXT, ANYT$ are isosceles trapezoids (we would have $MX, AT, NY$ to have common perpendicular bisector which would pass though $K$ ).

Redefine $X$ on $(AMT)$ such that $MX \parallel AT$ . We will prove that: $XA = XC$ (which implies $XN \perp AC$ ). Reflecting with respect to the perpendicular bisector of $AT$ , it suffice to show that $MT=MC'$ where $C' = YT \cap (ABC)$ (with $C' \neq T$ ) which is true as: $AB \parallel C'T$ .
Hence we are done,

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ihatemath123

3441 posts

ihatemath123

#85 h Jan 5, 2025, 1:14 PM

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Claim: We have that $AMXT$ and $ANYT$ are isosceles trapezoids.
Proof: Let $X'$ be the point on $(AMT)$ such that $AMX'T$ is an isosceles trapezoid. It suffices to show that $X'$ lies on the perpendicular bisector of $\overline{AC}$ . From $\angle X'TA = \angle MAT = \angle TAC$ , we have $\overline{X'T} \parallel \overline{AC}$ . So, if $N'$ is the foot from $X'$ to $\overline{AC}$ and $R$ is the foot from $T$ to $\overline{AC}$ , we have
$AN' = AR - N'R = AR - AM = AN,$ since $AR$ is the average of $AB$ and $AC$ by, say, Simson line. The claim is hence proven.

The claim thus follows from symmetry, as lines $MN$ and $XY$ are symmetric about the perpendicular bisector of $\overline{AT}$ .

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Ilikeminecraft

330 posts

Ilikeminecraft

#86 h Mar 15, 2025, 4:36 PM

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We introduce a different problem:
Let $M,N$ be the midpoint of $ABAC$ , define $K$ to be on $(ABC)$ so that $KB\parallel AT,$ $O$ to be the circumcenter of $(ABC),$ and $X$ is midpoint of $KT.$ Show that $X,O,N$ are collinear.
Observe that $MX$ are both half of $AB,KT,$ so $KMXB$ is an isosceles trapezoid. Let $L = KT\cap AB.$ Clearly, $OMLX$ is cyclic. Thus, $\angle MOX = \angle LXB = \angle A,$ but $\angle MON = 180 - \angle A$ since $AOMN$ is cyclic.

This finishes.

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imagien_bad

37 posts

imagien_bad

#87 h Mar 17, 2025, 5:58 PM

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Let X' and Y' be the relextions of M&N OVER PERPENDICULARBISECTOROF: AT rizzpecktiveally
Let O be the circumcenter of ABC. Note that NY'MX' is cyclic isosceles, TY' parallel to AB because we are reflecting over something perpendicular to the angle bisector of BAC . <OY' T = < OMA = 90 -> MOY' is collinear. Therefore Y'=Y. Therefore by symmetrizz X'= X. Now by symmetry we have KA=KT.

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endless_abyss

34 posts

endless_abyss

#88 h Mar 30, 2025, 6:13 PM

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Nice problem as always! It is very satisfying to proceed after the trapeziums

The key realization is that -

Claim - $A T X M$ & $A N Y T$ are isosceles trapezoids

We use phantom points, let $X'$ be a point such that $A M X' T$ is an isosceles trapezoid, it suffices to prove that the intersection of $O X'$ and $A C$ is in fact the midpoint.

$\angle A M X' = 180 - a/2$
$\angle O M X' = 90 - a/2$
since $O M X'$ is isosceles,
$\angle M O X' = a$
and $\angle M O N' = 180 - a$

so, $A O M N'$ is concyclic, forcing $O N'$ to be perpendicular to $A C$ thus $N'$ is the mid-point as required.

Claim - $O - m - K$ are collinear

Sub-claim 1 - $A M O N$ , $X O Y T$ are concyclic
First $A N$ is parallel to $X T$ because $\angle N A T = \angle A T X$
and as a result,
$\angle A N O = \angle O X T = 90 = \angle A M Y = \angle O Y T$

Sub-claim 2 - $X Y T$ and $A M N$ are congruent
$S-A-S$ criterion
next, note that $\angle X Y M = 90 - c = \angle M N X$
so, $M N X Y$ is a cyclic quad as well
Therefore $(K N)(K M) = (K Y)(K X)$
so, $K$ lies on the radical axis $O M$ of $A M O N$ , $X O Y T$
Thus, $K - O - m$ are collinear.

Fun Config Fact
:starwars:

Attachments:

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numbertheory97

40 posts

numbertheory97

#89 h Mar 31, 2025, 3:38 PM

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$[asy] size(8cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair M = (A+B)/2; pair N = (A+C)/2; pair T = dir(270); pair X = M + T + A - 2 * foot(M, A, T); pair Y = N + T + A - 2 * foot(N, A, T); pair K = extension(M, N, X, Y); pair O = (0, 0); draw(A--B--C--cycle); draw(unitcircle); draw(M--K--X, fuchsia); draw(dir(190)--K, royalblue); draw(M--Y, dotted); draw(N--X, dotted); draw(arc(circumcenter(A, M, T), A, T), dotted); draw(arc(circumcenter(A, N, T), T, A), dotted); dot("$A$", A, dir(110)); dot("$B$", B, dir(210)); dot("$C$", C, dir(310)); dot("$M$", M, dir(150)); dot("$N$", N, dir(60)); dot("$T$", T, dir(270)); dot("$O$", O, dir(270)); dot("$X$", X, dir(135)); dot("$Y$", Y, dir(315)); dot("$K$", K, dir(0)); [/asy]$
Let $O$ denote the center of $\omega$ . The key is to reflect across the perpendicular bisector of $\overline{AT}$ (call it $\ell$ ), which we claim is $\overline{OK}$ .

We can check that $\measuredangle(\ell, \overline{OM}) = \measuredangle(\overline{AT}, \overline{AB}) = \measuredangle(\overline{AC}, \overline{AT}) = \measuredangle(\overline{ON}, \ell)$ by $90^\circ$ rotations, so $\ell$ bisects $\angle MOX$ and $\angle NOY$ . Since $(AMT)$ and $(ANT)$ are fixed under the reflection, $M$ maps to $X$ and $N$ maps to $Y$ . Hence $K = \overline{MN} \cap \overline{XY}$ lies on $\ell$ , so by definition $KA = KT$ . $\square$

This post has been edited 3 times. Last edited by numbertheory97, Mar 31, 2025, 3:39 PM

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