X Y = BC wanted, orthocenter, circumcircles related

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X Y = BC wanted, orthocenter, circumcircles related

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Source: Estonia IMO TST 2019 p2

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#1 h Jul 23, 2020, 12:54 PM

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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 $| X Y | = | BC |$ .

This post has been edited 1 time. Last edited by parmenides51, Jul 14, 2021, 1:43 PM
Reason: latex

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#2 h Jul 24, 2020, 6:09 AM

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The same problem was Moldova JTST 2019 P7.
my solution

This post has been edited 1 time. Last edited by WolfusA, Jul 24, 2020, 6:15 AM

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#3 h Apr 13, 2021, 10:40 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.419920180887955, xmax = 8.807612256165534, ymin = -6.5597804152189045, ymax = 6.048412449132744; /* image dimensions */ /* draw figures */ draw((-0.22,3.24)--(-4.88,-1), linewidth(0.8)); draw((-4.88,-1)--(1.34,-3), linewidth(0.8)); draw((1.34,-3)--(-0.22,3.24), linewidth(0.8)); draw(circle((-1.0561240992869259,0.6396540512176611), 2.7314652776041464), linewidth(0.8) + red); draw(circle((-2.373711511874337,2.0877571980708067), 2.4425676553532103), linewidth(0.8) + red); draw(circle((0.7362884881256596,1.0877571980708072), 2.3551298798532714), linewidth(0.8) + red); draw((-3.452248198573849,4.279308102435321)--(2.767751801426146,2.2793081024353223), linewidth(0.8) + blue); draw((-0.22,3.24)--(-1.8922481985738515,-1.960691897564678), linewidth(0.8) + linetype("2 2") + blue); draw((-4.88,-1)--(-1.2951748251748259,-0.10379370629370778), linewidth(0.8) + linetype("2 2") + blue); draw((-3.452248198573849,4.279308102435321)--(-1.8922481985738515,-1.960691897564678), linewidth(0.8) + blue); draw((-1.8922481985738515,-1.960691897564678)--(2.767751801426146,2.2793081024353223), linewidth(0.8) + blue); draw((-3.7236536174909616,0.05212632228290204)--(0.9053246700991094,-1.2612986803964374), linewidth(0.8) + linetype("2 2") + blue); draw((1.34,-3)--(-1.2951748251748259,-0.10379370629370778), linewidth(0.8) + linetype("2 2") + blue); draw(circle((-3.3861240992869255,-1.480345948782338), 1.5692028030955412), linewidth(0.8) + linetype("2 2") + red); draw(circle((-0.27612409928692516,-2.4803459487823405), 1.6976152206088715), linewidth(0.8) + linetype("2 2") + red); /* dots and labels */ dot((-0.22,3.24),dotstyle); label("$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$", (-0.5578259686137148,-0.6411862828523932), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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 $BC=BK+KC=AY+AX=XY$

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618 posts

#4 h Jul 14, 2021, 10:51 AM • 1 Y

Y by ike.chen

@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 $YA=BD$ .

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#5 h Jul 14, 2021, 12:56 PM

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@above, sorry I probably misread the problem, thinking that the circle was fixed

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#6 h Jul 31, 2022, 7:38 PM

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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 $cos\alpha = -\cos(\beta+\gamma) = \sin\beta\sin\gamma - \cos\beta\cos\gamma$ .

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