Iranian Geometry Olympiad (2)

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Source: Advanced level,P2

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237 posts

#1 h Sep 13, 2016, 3:02 AM • 3 Y

Y by Davi-8191, Adventure10, Rounak_iitr

In acute-angled triangle $ABC$ , altitude of $A$ meets $BC$ at $D$ , and $M$ is midpoint of $AC$ . Suppose that $X$ is a point such that $\measuredangle AXB = \measuredangle DXM =90^\circ$ (assume that $X$ and $C$ lie on opposite sides of the line $BM$ ). Show that $\measuredangle XMB = 2\measuredangle MBC$ .Proposed by Davood Vakili

This post has been edited 1 time. Last edited by nsato, Oct 5, 2019, 1:57 PM

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#2 h Sep 13, 2016, 4:21 AM • 2 Y

Y by Coxivai, Adventure10

Easily $\angle BAD= \angle BXD$ . And $\angle AMX+ \angle ABX= \angle MDC= \angle DMB+ \angle DBM (\star )$

Law of sines in $\triangle AXM$ :

$\frac{AX}{AM} = \frac{ \sin \angle AMX}{ \sin \angle BAD}$
Law of sines in $\triangle ABX$ :

$\frac{AB}{AX} = \frac{ \sin 90}{ \sin \angle ABX}$
Law of sines in $\triangle ABD$ :

$\frac{BD}{AB} = \frac{ \sin \angle BAD}{ \sin 90}$
Law of sines in $\triangle BDM$ :

$\frac{DM}{BD} = \frac{ \sin \angle DBM}{ \sin \angle DMB}$
Multiplying the four equations we get:

$\frac{ \sin \angle ABX}{ \sin \angle AMX} = \frac{ \sin \angle DBM}{ \sin \angle DMB}$ $(\star \star )$

By $(\star )$ and $(\star \star )$ , Two Equal Angles Lemma we get $\angle AMX= \angle DMB$ . The conclusion follows.

This post has been edited 3 times. Last edited by Math_CYCR, Sep 25, 2016, 2:24 PM

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#3 h Sep 13, 2016, 9:10 AM • 2 Y

Y by Adventure10, Mango247

My solution:Let XM cut (ABD) and BC at T and Y.
We have:TAD=TXD=180-DXM=90.
So AT parallel to BC.
So ATCY is a paralellogram.
So M is the midpoint of TY too.(1)
Because the triangle TBY=90.(2)
So MY=MB.
So XMB=2MBC.

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145 posts

#4 h Sep 13, 2016, 4:18 PM • 4 Y

Y by Reynan, Geronimo_1501, Adventure10, Mango247

Let $N$ be the midpont of $AB$ .Obviously $A,X,D,B$ are concyclic.Now,
$-\measuredangle BXM=\measuredangle DXA=\measuredangle DBA=-\measuredangle BNM$ .

So, $B,N,X,M$ are also concyclic.

Now,since $NA=NB=NX$ ,so, $NM$ bisects $\angle XMB$ .As $MN \parallel BC$ ,we are done.

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#5 h Sep 13, 2016, 4:29 PM • 3 Y

Y by I.owais, Adventure10, Mango247

Let $BM$ meets $(AB)$ at $Y$ and $MX$ meets $(AB)$ at $Z$ . Then we have $XY || BZ \implies XY\perp BC$ at $T$ and $MX=MY$ .
So we get that $\angle XMY=180^\circ-2\angle MYX=180^\circ-2\angle BNT=2\angle MBC.$

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#6 h Sep 14, 2016, 8:46 AM • 1 Y

Y by Adventure10

Let $Z$ be the midpoint of $AB$ . $BZM= \pi - DBA= DXA = BXM$ and due to the condition on the position of X, $BZXM$ is cyclic. Now since $Z$ is the center of $(AB)$ , it is the midpoint of $\overarc{BX}$ of the previously mentioned circle. Now since $MZ || BC$ we get that the bisector of $\angle XMB$ is parallel to $BC$ which completes the solution.
Click to reveal hidden text

This post has been edited 1 time. Last edited by WizardMath, Sep 15, 2016, 4:03 AM

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#7 h May 27, 2017, 2:51 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

$[asy]size(8cm); pair A=(3,6),B=(0,0),C=(10,0),D,M,Nn,X,Y; D=foot(A,B,C);M=(A+C)/2;Nn=(A+B)/2;Y=2*Nn-D;X=foot(D,M,Y); draw(circumcircle(B,D,A)^^circumcircle(D,X,M),cyan+green); D(MP("A",A,N)--MP("N",Nn,W)--MP("B",B,S)--MP("D",D,S)--MP("C",C,S)--MP("M",M,E)--A); D(M--MP("X",X,dir(275)*2)--MP("Y",Y,N),heavymagenta); draw(D--Nn--Y--A--D--M--Nn^^D--X--A^^B--X^^Y--B--M,heavymagenta); dot(A^^B^^C^^D^^X^^Y^^M^^Nn); [/asy]$
Let $MX$ meet $\odot (AB)$ again at $Y$ , and let $N$ be the midpoint of $AB$ (i.e., the center of the circle with diameter). Then $\angle YXD=\pi-\angle DXM=\pi-\frac{\pi}{2}=\frac{\pi}{2},$ so $Y$ is the antipode of $D$ in $\odot (AB)$ , so $AYBD$ is a rectangle. Since $MN$ is clearly the perpendicular bisector of $AD$ , $MY$ and $MB$ are symmetric wrt $MN$ , so $\angle XMN=\angle NMB\implies \angle XMB=\angle XMN+\angle NMB=2\angle NMB=2\angle MBC.\;\blacksquare$

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1079 posts

#8 h Oct 25, 2021, 1:49 PM

Y by

Let $N$ be the midpoint of $\overline{AB}$ and $P$ be the midpoint of $\overline{AD}$ . Observe that $P$ lies on $(XDM)$ and on $\overline{MN}$ .
Firstly, note that $BNXM$ is cyclic quadrilateral as $\begin{align*} \measuredangle NBX =\measuredangle PDX=\measuredangle PMX=\measuredangle NMX. \end{align*}$ Now, note that $\measuredangle NBX=\measuredangle BXN=\measuredangle BMN=\measuredangle MBC$ , which means that $\overline{BX},\overline{BM}$ are isogonal wrt $\angle ABC$ . Thus indeed $\measuredangle XMB=\measuredangle XMN+\measuredangle NMB=\measuredangle XBN+\measuredangle MBC=2\measuredangle MBC$ . $\blacksquare$

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#9 h Jul 30, 2023, 3:31 PM • 1 Y

Y by Halykov

Let the tangents at $A$ and $C$ to $(A B C)$ meet at point $K$ . From $(A X D B)$ cyclic, we have $180 - \angle AXM=\angle BXD=\angle BAD$ and we can easily see that $\angle AKM=\angle BAD$ , from these we get that $(A X M K)$ cyclic. Then we have $90=\angle AMK=\angle AXK$ , implying that $B$ , $X$ and $K$ are collinear. Since $BK$ is symmedian of $\angle ABC$ , we have $\angle ABX=\angle MBD$ . And we know that $90+\angle DBT=\angle BDX=90+\angle ADX=90+\angle ABX$ , then $\angle DBT=\angle ABX$ (then $\angle DBT=\angle ABX=\angle MBD$ ). From $MX//BT$ we have $\angle XMB=\angle MBT=2\angle MBD$ which completes the solution.

This post has been edited 6 times. Last edited by X.Allaberdiyev, Aug 17, 2023, 6:55 PM
Reason: Latex

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#10 h Jul 30, 2023, 6:30 PM • 1 Y

Y by Halykovmathclub

Halykov wrote:

X.Allaberdiyev wrote:

Let the tangents at $A$ and $C$ to $(ABC)$ meet at point $K$ . From $AXDB$ cyclic, we have 180 - ∠AXM=∠BXD∠=∠BAD and we can easily see that ∠AKM=∠BAD, from these we get that $AXMK$ cyclic. Then we have 90=∠AMK=∠AXK, implying that $B, X, K$ are collinear. Since $BK$ is symmedian of ∠ABC, we have ∠ABX=∠MBD. And we know that 90+∠DBT=∠BDX=90+∠ADX=90+∠ABX --> DBT=∠ABX (then ∠DBT=∠ABX=∠MBD). From $MX//BT$ we have ∠XMB=∠MBT=2∠MBD which completes the solution.

Latex form of this solution

This post has been edited 2 times. Last edited by Halykov, Jul 30, 2023, 6:49 PM

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#13 h Feb 24, 2024, 9:27 PM

Y by

$\color{red} \textbf{Geo Marabot Solve 5}$

Let, $N$ be the mid point of $AB$ and $AD\cap MN \equiv D'.$ $\angle NMB=\angle MBC.$ Hence we need to show $\angle XMN=\angle MBC$
We have, $A,B,D,X$ concyclic and $D,D',X,M$ concyclic. So,
$\angle XMN=\angle XDD'=\angle XDA=\angle XBA=\angle XBN \implies B,N,X,M$ cyclic

$\implies \angle XBN=\angle BXN=\angle BMN=\angle MBC \blacksquare$

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#14 h Jul 5, 2024, 6:08 PM

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Let $N$ be the midpoint of $AB$ . We have that $\angle AXB = 90 = \angle ADB$ , hence $A,X,D,B$ are conclycic with diameter $AB$ , so $N$ is the center of $(AXDB)$ , i.e. $NA=NX=ND=NB$ . Also, $MN \parallel BC => MN \perp AD$ and $\angle DXM = 90 => XM \perp DX$ , hence $\angle XMN = \angle XDA$ as their sides are perpendicular.
From the cyclic $AXDB$ we get that $\angle NBX = \angle ABX = \angle ADX = \angle XDA = \angle XMN => BNXM$ is cyclic and, using that $NB=NX$ , we conclude that $NM$ bisects $\angle XMB$ . Hence we get that $\angle XMB = 2 \angle NMB = 2 \angle MBC$ , where the last equalith holds because $MN \parallel BC$ .

Attachments:

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#16 h Jul 12, 2024, 3:28 PM

Y by

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.24248, xmax = 19.83192, ymin = -13.56324, ymax = 8.02316; /* image dimensions */ /* draw figures */ draw((-3.2,6.98)--(-9,-5.7), linewidth(0.8) + red); draw((-9,-5.7)--(9.86,-5.52), linewidth(0.8) + red); draw((-3.2,6.98)--(9.86,-5.52), linewidth(0.8) + red); draw((-3.2,6.98)--(-3.421008752341651,-5.646754060202625), linewidth(0.8)); draw((-3.2,6.98)--(0.574,2.066), linewidth(0.8)); draw((0.574,2.066)--(-9,-5.7), linewidth(0.8)); draw((-3.421008752341651,-5.646754060202625)--(0.574,2.066), linewidth(0.8)); draw((0.574,2.066)--(3.33,0.73), linewidth(0.8)); draw((-3.421008752341651,-5.646754060202625)--(3.33,0.73), linewidth(0.8)); draw((-6.1,0.64)--(3.33,0.73), linewidth(0.8)); draw((-6.1,0.64)--(0.574,2.066), linewidth(0.8)); draw((-9,-5.7)--(3.33,0.73), linewidth(0.8)); draw(circle((-6.271508027994867,0.7184500443509642), 6.974322147214405), linewidth(0.8) + blue); draw(circle((-1.327242851760632,-5.366665643302623), 7.671977772810744), linewidth(0.8) + blue); /* dots and labels */ dot((-3.2,6.98),dotstyle); label("$A$", (-3.10968,7.22456), NE * labelscalefactor); dot((-9,-5.7),dotstyle); label("$B$", (-9.32908,-6.15804), NE * labelscalefactor); dot((9.86,-5.52),dotstyle); label("$C$", (9.95832,-5.28684), NE * labelscalefactor); dot((-3.421008752341651,-5.646754060202625),dotstyle); label("$D$", (-3.73888,-6.23064), NE * labelscalefactor); dot((3.33,0.73),linewidth(4pt) + dotstyle); label("$M$", (3.42432,0.93256), NE * labelscalefactor); dot((0.574,2.066),dotstyle); label("$X$", (0.66552,2.31196), NE * labelscalefactor); dot((-6.1,0.64),linewidth(4pt) + dotstyle); label("$N$", (-6.64288,0.69056), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

$\color{red}\textbf{Claim:-}$ $\angle XMB=2 \angle MBC.$

$\color{blue}\textbf{Proof:-}$ Let the midpoint of $AB$ is $N.$ From this we get, $A,B,D,X$ are cyclic, Since, $AD\perp BC\implies\angle AXB=\angle ADB=90^o.$ Now Since, $MN\parallel BC\implies \angle BMN=\angle MBC$ also we get, $\angle BNM+\angle ABC=180^o\implies \boxed{\angle BNM=180^o-\angle ABC.}$
Now We claim that $MN$ bisects $\angle XMB.$

Since, $A,B,D,X$ are cyclic we get, $\angle BXD=\angle DAB=90^o-\angle ABC.$ Next we claim that the circle through $A,B,D,X$ has center at $N$ and $AB$ is diameter.
Its very obvious since, $AB$ subtends $90^o$ at $D$ and $X\implies AN=NX=BN.$
Now in triangle $BNX\implies NX=NB\implies\angle NBX=\angle NXB.$ Now we also see that $B,N,X,M$ are cyclic. Since, $\angle BNM=\angle BXM=180^o-\angle ABC.$ Now in Cyclic Quad. $B,M,X,N$ we get, $\angle NXB=\angle NMX\implies \boxed{\angle XMB=2\angle MBC.}$

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