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Source: Polish National Olympiad 2016 P2- Final Round

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54 posts

#1 h Apr 8, 2016, 7:11 PM • 1 Y

Y by Adventure10

Let $ABCD$ be a quadrilateral circumscribed on the circle $\omega$ with center $I$ . Assume $\angle BAD+ \angle ADC <\pi$ . Let $M, \ N$ be points of tangency of $\omega$ with $AB, \ CD$ respectively. Consider a point $K \in MN$ such that $AK=AM$ . Prove that $ID$ bisects the segment $KN$ .

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233 posts

#2 h Apr 8, 2016, 8:12 PM • 2 Y

Y by Adventure10, Mango247

Hello.

My solution.

$[asy]import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.435245042437634, xmax = 12.21296844940993, ymin = -6.185472488524358, ymax = 5.497069355547282; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((-0.003750461477921458,-2.8173578246173063)--(-3.7677155720410602,0.9270506794704099)--(1.6059446206206847,2.590945209683892)--(2.139151167496353,-0.6969115772172044)--cycle, aqaqaq); /* draw figures */ draw(circle((0.,0.), 2.)); draw((-0.003750461477921458,-2.8173578246173063)--(-3.7677155720410602,0.9270506794704099), uququq); draw((-3.7677155720410602,0.9270506794704099)--(1.6059446206206847,2.590945209683892), uququq); draw((1.6059446206206847,2.590945209683892)--(2.139151167496353,-0.6969115772172044), uququq); draw((2.139151167496353,-0.6969115772172044)--(-0.003750461477921458,-2.8173578246173063), uququq); draw((-1.410525261274075,-1.4178922692883624)--(3.2368374449767403,0.968526090113257)); draw((1.6059446206206847,2.590945209683892)--(3.2368374449767403,0.968526090113257)); draw((-0.5915682074235653,1.9105096325236546)--(-1.410525261274075,-1.4178922692883624)); draw((-0.5915682074235653,1.9105096325236546)--(3.2368374449767403,0.968526090113257)); draw((-0.5915682074235653,1.9105096325236546)--(1.9742071051475258,0.32016606001421505)); draw((-3.7677155720410602,0.9270506794704099)--(0.,0.)); /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle); label("$I$", (-0.1876033751113048,-0.40242021355201696), NE * labelscalefactor); dot((-0.5915682074235653,1.9105096325236546),linewidth(3.pt) + dotstyle); label("$P$", (-0.9250395712487178,1.9651381003628337), NE * labelscalefactor); dot((-1.410525261274075,-1.4178922692883624),linewidth(3.pt) + dotstyle); label("$N$", (-1.9729752183913574,-1.5473869391337889), NE * labelscalefactor); dot((1.9742071051475258,0.32016606001421505),linewidth(3.pt) + dotstyle); label("$M$", (2.218767370179201,0.16036004139495572), NE * labelscalefactor); dot((1.6059446206206847,2.590945209683892),linewidth(3.pt) + dotstyle); label("$A$", (1.6753933309200544,2.702574296500246), NE * labelscalefactor); dot((2.139151167496353,-0.6969115772172044),linewidth(3.pt) + dotstyle); label("$B$", (2.218767370179201,-0.9846066841868163), NE * labelscalefactor); dot((-0.003750461477921458,-2.8173578246173063),linewidth(3.pt) + dotstyle); label("$C$", (0.23933337002088167,-3.0804779784720937), NE * labelscalefactor); dot((-3.7677155720410602,0.9270506794704099),linewidth(3.pt) + dotstyle); label("$D$", (-4.010627865613157,0.47085949240018204), NE * labelscalefactor); dot((3.2368374449767403,0.968526090113257),linewidth(3.pt) + dotstyle); label("$K$", (3.4801713898879334,0.8395775904688882), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Let $P,$ be the tangency point of the $\omega$ with $DA$ .We have $AK=AM=AP$ ,thus $A$ is the circumcenter

of $\triangle{MPK}$ .Hence $\angle{MPK}=\frac{1}{2}\angle{MAK} \ (1)$ .

Also,the alternate chord theorem gives $\angle{MPN}=\angle{NMB}=\angle{AMK}\overset{\angle{AMK}=\angle{AKM}}\Rightarrow \angle{MPN}=90^{\circ}-\frac{1}{2}\angle{MAK} \ (2)$ .

$(1),(2)\Rightarrow PK\perp PN$ .We also have $DP=DN$ and $IP=IN$ so $ID\perp PN$ and $ID$ bisects $PN$ .

It follows that $ID\parallel PK$ and $ID$ bisects $PN$ so it passes as well through the midpoint of $KN$ ,q.e.d.

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12 posts

#3 h Apr 10, 2016, 9:30 PM • 2 Y

Y by Adventure10, Mango247

let $X$ be the tangency point of $\omega$ with $AD$
$Y$ the antipode of $N$
$K'$ = $(XY)\cap(MN)$
$U$ = $(XN)\cap(MY)$
$T$ = $(DI)\cap(MN)$
Using Pascal on $NMMYXX$ gives $K'$ , $A$ and $U$ collinear more than that simply by inversion or isogonalty $A$ is the midpoint of $UK'$ also $\angle{UMK'}=90=\angle{UXK'}$ so $KMXU$ are concyclic and $AK'=AM=AX=AU$ therfore $K'=K$
now $\angle{KXN}=90$ and $TN=TX$ gives $TN=TK$

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18 posts

#4 h Apr 12, 2016, 10:17 PM • 2 Y

Y by Adventure10, Mango247

$AB \cap DC = R$
$\omega \cap AD = Q$
$DI \cap KN = T$
$DI \cap AK = L$

Since $AK = AM$ and $MR= NR$ we have $\angle{MKA} = \angle{AMK}=\angle{RMN}=\angle{MNR}$ so
$AK \parallel DC$ and then it follows that $\angle{ADL}=\angle{TDN}=\angle{TLK}$ so $AD=AL$ .
$AK=AM=AQ$ so $KL=QD=DN$ .
Since $\angle{TLK}=\angle{TDN}$ , $\angle{TKL}=\angle{DNT}$ and $KL=DN$ triangles $KLT$ and $DTN$ are congruent so $KT=TN$ .

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49 posts

#5 h Aug 12, 2016, 2:00 AM • 2 Y

Y by Adventure10, Mango247

Let $DN \cap AM = X$ , $DI \cap AK = L$ .
Easily seen that $DN \parallel AK$ , so $A, K, L$ are collinear.
Let $X$ be the tangency point of $DA$ , then $DN = DA-DX = AL-AK =KL$ , so $DNLK$ is a parallelogram.
$DL$ and $NK$ bisect each other.

This post has been edited 1 time. Last edited by Vexation, Aug 12, 2016, 2:00 AM

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1721 posts

#6 h Feb 21, 2017, 3:21 AM • 1 Y

Y by Adventure10

Clearly it suffices to show that $ID \parallel PK$ , or just $\angle NPK = 90$ .
Note that $AP = AM = AK$ , so $A$ is the circumcenter of $PMK$ .
We have $\angle AMK = 180-\angle PMA - \angle PMN = 180- \angle PNM - \angle PMN = \angle NPM$ .
Now, $\angle NPK = \angle NPM + \angle MPK = \angle AMK + \frac{1}{2} \angle MAK = 90$ , so we are done.

This post has been edited 1 time. Last edited by rkm0959, Feb 21, 2017, 3:21 AM

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871 posts

InCtrl

#7 h Jun 21, 2018, 1:56 AM • 2 Y

Y by Adventure10, Mango247

Extend $AB$ and $CD$ past $B,C$ to intersect at $X$ . Let $DI$ intersect $MN$ at $J$ . By a well-known lemma, $\angle JIA=\frac{\pi}{2}$ .
Also notice that since $\angle XMN=\angle AMK$ and $\triangle XMN, \triangle AKM$ are both isosceles, then $AK \parallel XD$ .

Let $AJ$ intersect $XD$ at $T$ .

The problem is now equivalent to proving that $J$ is the midpoint of $AT$ , but this is immediate: $DJ$ is both altitude and angle-bisector in $\triangle ADT$ , so $\triangle ADT$ is isosceles hence $J$ is also the midpoint of $AT$ .

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1980 posts

#8 h Jun 21, 2018, 2:17 AM • 1 Y

Y by Adventure10

Pinionrzek wrote:

Let $ABCD$ be a quadrilateral circumscribed on the circle $\omega$ with center $I$ . Assume $\angle BAD+ \angle ADC <\pi$ . Let $M, \ N$ be points of tangency of $\omega$ with $AB, \ CD$ respectively. Consider a point $K \in MN$ such that $AK=AM$ . Prove that $ID$ bisects the segment $KN$ .

By right-angles on the intouch chord lemma; we know that if $L=DI \cap MN$ then $AL \perp DL \implies AL \parallel NP$ . Now let $MN \cap AD=T$ and $\infty$ be the infinity point on line $MN$ . Note that $\angle CNM=\angle BMN=\angle AKN$ so $AK \parallel DN$ . Observe that $N(P,T, A,D)=-1$ ; now observe that $AL \parallel NP, NT \parallel A\infty, NA \parallel NA, ND \parallel AK$ proving $A(\infty,L,N,K)=-1$ ; hence $L$ bisects $KN$ .

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214 posts

#9 h Aug 3, 2019, 10:39 AM • 2 Y

Y by Adventure10, Mango247

let $AB \cap CD=Z$
$X=DI \cap MN$
$DI \cap MN=X$
$AX \cap DN=Y$
and let the reflection of $N$ over $X$ is $K'$ we will prove that $K=K'$
it's well-known that $PX$ is perpendicular to $AX \implies AX=XY \implies AK'YN$ is parallelogram
we have $\angle ZMN=\angle K'MA$ and $\angle ZNM=\angle AK'M \implies AK'=AM \implies K=K'$
and we done

This post has been edited 1 time. Last edited by Ali3085, Aug 3, 2019, 10:40 AM

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251 posts

#10 h Aug 13, 2020, 7:00 AM

Y by

let $L$ be the midpoint of segment $AD$ and $S$ be the intersection of $DI$ , $MN$ . First of all notice that $AK||DC$ and if problem be true then we have $LS=LA=LD \implies$ triangle $\triangle ASD$ is right. which it easily implies from that claim:
claim: $ISMA$ cyclic.
proof: $\angle SIA + \angle SMA=180$ !
$\blacksquare$

This post has been edited 3 times. Last edited by arzhang2001, Aug 13, 2020, 7:01 AM

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1080 posts

#11 h Apr 2, 2022, 7:34 PM • 1 Y

Y by centslordm

Let $\omega$ touch $\overline{BC},\overline{AD}$ at $E,F$ and let $X=\overline{AB}\cap\overline{KN}.$ Since $A$ is the center of $(KMF),$ $\angle MKF=\tfrac{1}{2}\angle MAF=90-\angle AFM=90-\angle FNM=\angle NXI$ and $\overline{DI}\parallel\overline{KF}.$ As $\overline{DI}$ bisects $\overline{FN},$ we know $\overline{DI}$ also bisects $\overline{KN}.$ $\square$

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37 posts

Avron

#12 h Mar 14, 2025, 6:29 PM

Y by

Why is this so troll

Let $AB,CD$ intersect at $R$ , $DI, MN$ at $L$ and $\omega$ touch $AD$ at $P$ . We can now completely ignore $C$ and $D$ . It is well known that $\angle ALD = 90$ , so $AL \parallel PN$ thus $\angle MLA = \angle MNP = \angle MPA$ so $APLM$ is cyclic. We get
$\angle KPL = \angle APL - \angle APK = \angle AMK - \angle AKP = \angle AKM - \angle AKP = \angle PKL$ and we're done.

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