In an acute-angled triangle $ABC$, the altitudes intersect at point $H$, and point $K$ is the foot of the altitude drawn from the vertex $A$. Circle $c$ passing through points $A$ and $K$ intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. The line passing through point $A$ and parallel to line $BC$ intersects the circumcircles of triangles $AHM$ and $AHN$ for second time, respectively, at points $X$ and $Y$. Prove that $ | X Y | = | BC |$.
Problem
Source: Estonia IMO TST 2019 p2
Tags: geometry, circumcircle
WolfusA
24.07.2020 09:09
The same problem was Moldova JTST 2019 P7.
Let $D$ be second intersection of circle $AMN$ with segment $BC$ (it exists since $M,N$ lie on sides $AB,AC$ respectively).
$$BM\cdot AB=BK\cdot BD$$Let $F$ be the second intersection of line $BH$ and circle $AMH$. Then $$BF\cdot BH=BM\cdot AB.$$Combining these equalities
$$BF\cdot BH=BK\cdot BD\implies\text{points F,H,K,D are concyclic}.$$By Reim's theorem $AX||DK\wedge H\in AK\implies F\in DX\implies \angle DXA=\angle FXA.$
Points $A,H,K$ are collinear and $F,X,A,H$ are concyclic:
$$\angle FXA=\angle FHK.$$Points $B,K,D$ are collinear and $F,H,K,D$ are concyclic:
$$\angle FHK=\angle FDB=180^\circ-\angle XDC.$$Hence $$\angle DXA=180^\circ-\angle XDC\implies DX||AC$$which means that $$\text{quadrilateral AXDC is a parallelogram}$$so $$AX=CD.$$In a similar way we obtain $$AY=BD.$$Combining those equalities gives desired $$XY =BC.$$
EulersTurban
14.04.2021 01:40
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -11.419920180887955, xmax = 8.807612256165534, ymin = -6.5597804152189045, ymax = 6.048412449132744; /* image dimensions */ /* draw figures */ draw((-0.22,3.24)--(-4.88,-1), linewidth(0.8)); draw((-4.88,-1)--(1.34,-3), linewidth(0.8)); draw((1.34,-3)--(-0.22,3.24), linewidth(0.8)); draw(circle((-1.0561240992869259,0.6396540512176611), 2.7314652776041464), linewidth(0.8) + red); draw(circle((-2.373711511874337,2.0877571980708067), 2.4425676553532103), linewidth(0.8) + red); draw(circle((0.7362884881256596,1.0877571980708072), 2.3551298798532714), linewidth(0.8) + red); draw((-3.452248198573849,4.279308102435321)--(2.767751801426146,2.2793081024353223), linewidth(0.8) + blue); draw((-0.22,3.24)--(-1.8922481985738515,-1.960691897564678), linewidth(0.8) + linetype("2 2") + blue); draw((-4.88,-1)--(-1.2951748251748259,-0.10379370629370778), linewidth(0.8) + linetype("2 2") + blue); draw((-3.452248198573849,4.279308102435321)--(-1.8922481985738515,-1.960691897564678), linewidth(0.8) + blue); draw((-1.8922481985738515,-1.960691897564678)--(2.767751801426146,2.2793081024353223), linewidth(0.8) + blue); draw((-3.7236536174909616,0.05212632228290204)--(0.9053246700991094,-1.2612986803964374), linewidth(0.8) + linetype("2 2") + blue); draw((1.34,-3)--(-1.2951748251748259,-0.10379370629370778), linewidth(0.8) + linetype("2 2") + blue); draw(circle((-3.3861240992869255,-1.480345948782338), 1.5692028030955412), linewidth(0.8) + linetype("2 2") + red); draw(circle((-0.27612409928692516,-2.4803459487823405), 1.6976152206088715), linewidth(0.8) + linetype("2 2") + red); /* dots and labels */ dot((-0.22,3.24),dotstyle); label("$A$", (-0.12697045706212698,3.4632793798232333), NE * labelscalefactor); dot((-4.88,-1),dotstyle); label("$B$", (-4.7983512665161845,-0.7772459180792096), NE * labelscalefactor); dot((1.34,-3),dotstyle); label("$C$", (1.4377153480462708,-2.772787234739183), NE * labelscalefactor); dot((-1.2951748251748259,-0.10379370629370778),linewidth(4pt) + dotstyle); label("$H$", (-1.2154475388766646,0.0844651050239606), NE * labelscalefactor); dot((-1.8922481985738515,-1.960691897564678),linewidth(4pt) + dotstyle); label("$K$", (-1.805039291526206,-1.775016576409196), NE * labelscalefactor); dot((-3.7236536174909616,0.05212632228290204),linewidth(4pt) + dotstyle); label("$M$", (-3.641844367088238,0.24320134612191302), NE * labelscalefactor); dot((0.9053246700991094,-1.2612986803964374),linewidth(4pt) + dotstyle); label("$N$", (1.006859836494683,-1.0720417944039784), NE * labelscalefactor); dot((-3.452248198573849,4.279308102435321),linewidth(4pt) + dotstyle); label("$X$", (-3.3697250966346037,4.461050038153219), NE * labelscalefactor); dot((2.767751801426146,2.2793081024353223),linewidth(4pt) + dotstyle); label("$Y$", (2.8663415179278515,2.4655087214932467), NE * labelscalefactor); dot((-2.2940434569082533,-0.35351086422706424),linewidth(4pt) + dotstyle); label("$D$", (-2.2132181972066576,-0.16497755955853605), NE * labelscalefactor); dot((-0.6412746371126393,-0.822467026192244),linewidth(4pt) + dotstyle); label("$E$", (-0.5578259686137148,-0.6411862828523932), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Cute problem Let $D$ be the intersection point of $KX$ with $(AMH)$ and let $E$ be the intersection of $KY$ with $(AHN)$. Notice that we must have that $\angle KBM = \angle KBA = \angle XAM = \angle XDM$, this implies that $BMDK$ mus tbe a cyclic quadrilateral. Similarly we get that $AHEN$ is cyclic. Obviously we get that $KD \perp BH$. Similarily we get that $KE \perp CH$. This implies that $DHEK$ is a cylic quadrilateral. From calculating $\angle BKD$ we get that $\angle DKH = 90-\angle C$, but since we have that $90-\angle C = \angle DKH = \angle DEH$, this implies that $D,E$ and $N$ are collinear. Similarily we get that $M,D,E$ are also collinear. Now notice that this implies that $\angle C = \angle NMA = \angle XKB = \angle KXA$, thus we have that $XKCA$ is a parallelogram. Similarily we get that $AYKB$ is a parallelogram. Meaning that we must have that $KB=AY$ and $KC=XA$. Summing the last two relations we get that $BC=BK+KC=AY+AX=XY$
Steve12345
14.07.2021 13:51
@above, fakesolve. Really don't know how you can fakesolve a collinearity in geogebra. Another solution: Let $AMN$ intersect $BC$ at $D$. Let $S$ be the intersection of $DM$ with the circumcircle of $AHM$. The quadrilateral $XAHS$ is a rectangle and quadrilateral $SDCH$ is a parallelogram, done since $AX=SH=DC$ and similarly $YA=BD$.
Attachments:
EulersTurban
14.07.2021 15:56
@above, sorry I probably misread the problem, thinking that the circle was fixed
Kugelmonster
31.07.2022 22:38
Solution using Ptolemy's Sine Lemma: We are motivated to use this lemma since we only have a lot of stuff connected to $A$ and not to everything else. Also, the angles at $A$ are very convenient. Set the diameter of $(ABC)$ to $1$, so that $a = \sin \alpha, b= \sin \beta, c = \sin \gamma$. Ptolemy's Sine Lemma on $(MHAX)$: \[ XA \cdot \sin \angle MAH + HA \cdot \sin \angle XAM = MA \cdot \sin \angle XAH \Leftrightarrow XA\cos\beta + HA\sin\beta = MA \]Ptolemy's Sine Lemma on $(HNYA)$: \[ YA\cos\gamma + HA\sin\gamma = NA \]Ptolemy's Sine Lemma on $(MKNA)$: \[ AM\cos\gamma + AN\cos\beta = AK\sin\alpha = \sin\alpha\sin\beta\sin\gamma \]We need to prove \[ AX+AY = \sin\alpha \]\[\Leftrightarrow \frac{AM-AH\sin\beta}{\cos\beta} + \frac{AN-AH\sin\gamma}{\cos\gamma} = \sin\alpha\]Using $AH = cos\alpha$ we get \[ \Leftrightarrow AM\cos\gamma + AN\cos\beta - \cos\alpha (\sin\beta\cos\gamma+\sin\gamma\cos\beta) = \sin\alpha\cos\beta\cos\gamma\]Using the result from $(MKNA)$: \[ \Leftrightarrow \sin\alpha\sin\beta\sin\gamma -\sin\alpha\cos\alpha = \sin\alpha\cos\beta\cos\gamma \]which is true since $cos\alpha = -\cos(\beta+\gamma) = \sin\beta\sin\gamma - \cos\beta\cos\gamma$.