Isoceles triangle [Iran TST 2010]

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Isoceles triangle [Iran TST 2010]

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#1 h May 10, 2010, 12:02 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ an isosceles triangle and $BC>AB=AC$ . $D,M$ are respectively midpoints of $BC, AB$ . $X$ is a point such that $BX\perp AC$ and $XD||AB$ . $BX$ and $AD$ meet at $H$ . If $P$ is intersection point of $DX$ and circumcircle of $AHX$ (other than $X$ ), prove that tangent from $A$ to circumcircle of triangle $AMP$ is parallel to $BC$ .

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gilcu3

#2 h May 10, 2010, 3:53 PM • 2 Y

Y by Adventure10, Mango247

Very easy.
We have to prove that $\angle ABC =\angle APM$
$H$ is the orthocenter of triangle $ABC$ , so $\angle DHB=\angle ABC$
But $\angle ABC =\angle DHB=\angle XHA=\angle XPA=\angle PAB$ .
Then $PDBA$ is an isosceles trapezoid, hence $\angle APM=\angle BDM=\angle BCA=\angle ABC$ , as claimed.

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jgnr

#3 h May 16, 2010, 1:11 PM • 2 Y

Y by Adventure10, Mango247

Since $DP\parallel BA$ and $D$ is the midpoint of $BC$ , then $DP$ intersects $AC$ at the midpoint of $AC$ , say $N$ . We have $\angle APN=\angle AHX=\angle ACB=\angle AMN$ , so $AMPN$ is cyclic. The circumcenter of $AMP$ , say $O$ , clearly lies on $AH$ . Thus the tangent of $(AMP)$ at $A$ is perpendicular to $AO$ , which is parallel to $AH$ , which is perpendicular to $BC$ .

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#4 h May 7, 2013, 2:38 AM • 1 Y

Y by Adventure10

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sayantanchakraborty

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sayantanchakraborty

#5 h Nov 26, 2013, 11:25 AM • 1 Y

Y by Adventure10

Let $\angle{ABC}=\angle{ACB}=\theta$ .Then $\angle{BHD}=\theta=\angle{DPA}=\angle{DBA} \implies ABPD$ is cyclic.Also see that ${{\angle{BDP}=\angle{NDC}=\angle}NCD}=\theta=\angle{DBA}$ so $ABPD$ is an isosceles trapezoid.Now some trivial angle chasing gives you $\angle{PBA}=90-\theta$ so $\angle{AMP}=180-2\theta=\angle{PAX}$ where $AX$ is the tangent line from $A$ to the circumcircle of $AMP$ .Thus $\angle{DAX}=\angle{PAX}-\angle{PAD}=180-2\theta-90+2\theta=90^{\circ}=\angle{ADB} \implies AX \parallel BC$ .

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#6 h Sep 13, 2015, 3:45 AM • 2 Y

Y by Adventure10, Mango247

Let $N$ be the mid point of $AC$ . Note that it suffices to proving that if $\angle ABC=\angle ACB=\theta$ then $\angle APN=\theta$ . Now this is trivial since $\angle APN=\angle APX=\angle AHX=\theta$ . Lol. Too trivial for Iran.

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#8 h Oct 1, 2019, 3:51 AM • 1 Y

Y by Adventure10

Such a disappointment :noo:

:noo:

We want to show, $\odot (AMP)$ and $\odot (ABC)$ are internally tangent which is equivalent to show, $\odot (APM)$ bisects $AC$ . Let $E$ be midpoint of $AC$
$\angle APX=\angle AHX=\angle ACB=\angle ABC=\angle AME \qquad \blacksquare$

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Ninjasolver0201

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Ninjasolver0201

#9 h Jan 23, 2021, 2:39 PM • 1 Y

Y by teomihai

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.38667656855066, xmax = 9.682272354136112, ymin = -1.3623798338963669, ymax = 8.424293845080454; /* image dimensions */ pen ttzzqq = rgb(0.2,0.6,0); pen ccqqqq = rgb(0.8,0,0); pen qqqqcc = rgb(0,0,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqzzff = rgb(0,0.6,1); /* draw figures */ draw((0,8)--(-5,0), linewidth(2.4) + ttzzqq); draw((5,0)--(0,8), linewidth(2.4) + ttzzqq); draw(circle((0.7051282051282051,5.5625), 2.5374420260702166), linewidth(2) + ccqqqq); draw(circle((0,5.21875), 2.78125), linewidth(2) + qqqqcc); draw((0,8)--(2.1910112359550564,3.50561797752809), linewidth(2) + yqqqyq); draw((-2.5,4)--(2.1910112359550564,3.50561797752809), linewidth(1.6) + qqzzff); draw((-5,0)--(3.2051282051282053,5.128205128205129), linewidth(2.4) + red); draw((3.2051282051282053,5.128205128205129)--(0,0), linewidth(2) + red); draw((-5,0)--(0,0), linewidth(2.4) + ttzzqq); draw((0,0)--(5,0), linewidth(2.4) + ttzzqq); draw((0,8)--(0,0), linewidth(2.4) + red); /* dots and labels */ dot((0,8),dotstyle); label("$A$", (0.0522994181997128,8.139383403188845), NE * labelscalefactor); dot((5,0),dotstyle); label("$C$", (5.052477673397458,0.14764550812916896), NE * labelscalefactor); dot((0,3.125),linewidth(4pt) + dotstyle); label("$H$", (0.023808374010551855,2.8400491840048883), NE * labelscalefactor); dot((0,0),linewidth(4pt) + dotstyle); label("$D$", (0.13777255076719563,-0.2939656768028273), NE * labelscalefactor); dot((-2.5,4),linewidth(4pt) + dotstyle); label("$M$", (-2.896523655378445,3.8942178190038472), NE * labelscalefactor); dot((3.2051282051282053,5.128205128205129),linewidth(4pt) + dotstyle); label("$X$", (3.3572605441423824,4.877158843529903), NE * labelscalefactor); dot((2.1910112359550564,3.50561797752809),linewidth(4pt) + dotstyle); label("$P$", (2.3030919091434274,3.3386424573152067), NE * labelscalefactor); dot((-5,0),linewidth(4pt) + dotstyle); label("$B$", (-5.076088535849257,0.07641789765626635), NE * labelscalefactor); dot((2.191011235955056,4.49438202247191),linewidth(4pt) + dotstyle); label("$E$", (2.174882210292203,4.6777215342057765), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

To prove: Tangent from $A$ to the circumcircle of $\Delta AMP$ is parallel to $BC$ . In other words, we need to show that $\angle ABC=\angle APM$ .

Since, $H$ is the orthocentre of the triangle $ABC$ , we have $\angle DHB= \angle ABC$ . Now, we can do Angle chasing for the angles made on the two circles. $\angle DHB=\angle ABC=\angle XHA=\angle XPA$ . Now, by Angle Segment Formula, $\angle XPA=\angle PAM=\angle PAB$ . Hence, we get $PDBA$ is an isosceles trapezoid.

Using the condition $XD || AB$ we get, $\angle APM=\angle BDM=\angle BCA=\angle ABC$ . Therefore, we get the final condition $\angle APM=\angle ABC. \qquad \blacksquare$

This post has been edited 1 time. Last edited by Ninjasolver0201, Jan 23, 2021, 2:39 PM
Reason: image display

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#10 h Dec 2, 2021, 4:30 PM

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What kind of a tst problem is this

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Let $N$ be the mid point of $AC$ .Then $\angle APN =\angle AHX = \angle C= \angle B =\angle AMN \implies (AMPN)$ .
Thus the angle the said tangent makes with $AB = \angle APM = \angle ANM = \angle C= \angle B$

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