Iranian geometry olympiad 2014(4)

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Source: Iranian geometry olympiad 2014

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237 posts

#1 h Feb 25, 2015, 7:47 PM • 5 Y

Y by HWenslawski, tiendung2006, Adventure10, Mango247, Rounak_iitr

Let $ABC$ be an acute triangle.A circle with diameter $BC$ meets $AB$ and $AC$ at $E$ and $F$ ,respectively. $M$ is midpoint of $BC$ and $P$ is point of intersection $AM$ with $EF$ . $X$ is an arbitary point on arc $EF$ and $Y$ is the second intersection of $XP$ with a circle with diameter $BC$ .Prove that $\measuredangle XAY=\measuredangle XYM$ .
Author:Ali zo'alam , Iran

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445 posts

#2 h Feb 25, 2015, 9:17 PM • 3 Y

Y by HWenslawski, Adventure10, Mango247

$EF \cap BC=Z$
$BF \cap EC=W$
$WZ \cap AM=G$
$L_1$ is the radical axis between $(EFBC)$ and $(A)$ (point $A$ )
By Brocard's theorem $M$ is orthocenter of $\triangle AZW \Rightarrow WZ \perp AM \Rightarrow GH \parallel L_1$
$WZ \cap AM=G \Rightarrow (M,P,G,A)=-1$
$L_1$ meet to $AC$ and $AM$ in $I$ and $K$ respectively.
$AI^2=IC.IF$ and $(A,H,F,C)=-1 \Rightarrow I$ is midpoint of $AH \Rightarrow K$ is midpoint of $AG$ $\Rightarrow KP.KM=KA^2=Pot_{(EFBC)}K \Rightarrow K$ and $P$ are inverses points i.e. $L_1$ is the polar of $P$
$O$ is el pole of $XY \Rightarrow O \in L_1 \Rightarrow OX=OY=OA \Rightarrow \angle XOY=2\angle XAY=2\angle XYM$

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2312 posts

#3 h Feb 25, 2015, 10:27 PM • 6 Y

Y by huricane, Not_real_name, guptaamitu1, Infinityfun, Adventure10, Mango247

My solution:

Let $H=BF \cap CE$ and $T$ be the projection of $H$ on $AM$ .

Easy to see $H$ is the orthocenter oF $\triangle ABC$ .

Since $H$ lie on the polar of $A$ WRT $\omega$ ,
so $TH$ is the polar of $A$ WRT $\omega \Longrightarrow MT \cdot MA=MX^2=MY^2$ . ... $(1)$

Since $A, E, F, H, T$ are concyclic ,
so $AP \cdot PT=EP \cdot PF=XP \cdot PY \Longrightarrow A, T, X, Y$ are concyclic . ... $(2)$

From $(1)$ and $(2)$ $\Longrightarrow MX, MY$ are the tangent of $\odot (ATXY) \Longrightarrow \angle XAY=\angle XYM$ .

Q.E.D

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#4 h Sep 5, 2016, 6:49 PM • 4 Y

Y by Systematicworker, guptaamitu1, Adventure10, Mango247

Let $S$ be the other intersection of $AM$ with $(AEF)$ . We have by power of a point, $PX\cdot PY=PE\cdot PF=PA\cdot PS$ and so $A,X,S,Y$ lie on a circle. Now, $MX=MY=ME$ and $ME^2=MS\cdot MA$ . Letting $O$ be the circumcenter of triangle $AXY$ and $r$ be its radius, we get $MO^2-r^2=MS\cdot MA=MX^2=MY^2$ yielding that $MX,MY$ are tangents to the circle $(AXY)$ . The conclusion holds.

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9787 posts

jayme

#5 h Sep 6, 2016, 8:09 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

1. we can observe that (AEF) is orthogonal to (M)
2. A, T, X, Y are concyclic where T is the second point of intersection of AM with (AEF)
and we are done...

Sincerely
Jean-Louis

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1079 posts

#7 h Oct 28, 2021, 8:20 PM

Y by

storage

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13 posts

#8 h Apr 19, 2022, 4:03 PM • 1 Y

Y by Mango247

We extend $AX$ intersect $\odot M$ at $K$ , $XM$ intersect $\odot M$ at $S$ , draw $XL$ parallel to $AY$ intersect $\odot M$ at $L$ .
We have $\angle MXY=\angle MYX$ , $\angle KXL=\angle XAY$
so if we want to prove $\angle XAY=\angle XYM$ ,we just need to prove $\angle KXL=\angle MXY$ namely $\angle KXS=\angle LXY$ .
We notice $SY \perp XY$ ,so if $KY \perp XL$ namely $KY \perp AY$ ,we have $\angle TYS=\angle TXY$ which is exactly we want.
So we turn to prove $KY \perp AY$ ,for convenience,we donate $AY$ intersect $\odot M$ at $R(R\neq Y)$ ,so we just need to prove $K,M,R$ is collinear.
Then we can put the problem into a brief way:
For a fixed circle $\odot M$ and a fixed point $A$ ,we draw any diameter $BC$ of $\odot M$ , $AB,AC$ intersect $\odot M$ at $E,F$ ,respectively, $AM$ intersect $EF$ at $P$ .Prove point $P$ is a fixed point.
You can find this problem is equivalent to the orignal problem.(Not hard,for you to think)
To solve this problem ,we need to use Menelaus in quadrilateral:
For $BEFC$ : We have $\frac{EA}{AB}\frac{BM}{MC}\frac{CA}{AF}\frac{FP}{PE}=1$ ,it can get $\frac{EP}{PF}=\frac{AC^2}{AB^2}$ .

For $PFCM$ : We have $\frac{PA}{AM}\frac{MB}{BC}\frac{CA}{AF}\frac{FE}{EP}=1$ ,

we can get $\frac{PA}{AM}=2\frac{AF}{AC}\frac{EP}{EP+PF}=2\frac{AF}{AC}\frac{AC^2}{AC^2+AB^2}=2\frac{AF AC}{AB^2+AC^2}$
Because $AFAC=AM^2-R^2$ is a constant value,and $AM^2=\frac{2AB^2+2AC^2-4R^2}{4}$ namely $AB^2+AC^2=2AM^2+2R^2$ is also a constant value.
So we can get: $\frac{PA}{AM}=\frac{AM^2-R^2}{AM^2+R^2}$ is a constant value , which indicates point $P$ is a fixed point on $AM$ .

Through the process above , we can complete this problem.The rest detailed procedure is for you.

Attachments:

This post has been edited 1 time. Last edited by Muke_Yang, Apr 19, 2022, 4:06 PM

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#9 h May 27, 2023, 9:21 AM • 1 Y

Y by preparingfor_icho

We need to prove $MY$ is tangent to $(AXY)$ .Let $AM\cap (AXY)=Z$ .Notice that $AP\cdot PZ=XP\cdot PY=EP\cdot PF \implies AEFZ$ cyclic. From 3 Tangents lemma, its easy to see that $MF$ is tangent to $(AEZF)$ , $\to MF^2=MZ \cdot MA$ and $MF=MY$ $\implies MY^2=MZ \cdot MA$ .

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154 posts

#10 h May 27, 2023, 12:36 PM

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Let $\odot(AXY)\cap AM=T$
Note that $XP\cdot PY=AP\cdot PT=EP\cdot PF$
Which means that $A,E,T,F$ is cyclic
So $\angle BTM = \angle BEM=\angle EBM$
Hence $\triangle TBM\sim \triangle BAM$
So $BT\cdot TA=BM^{2}=MY^{2}$
Which means that $\angle XAY=\angle XYM$

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lian_the_noob12

173 posts

lian_the_noob12

#11 h Feb 21, 2024, 8:03 PM

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$\color{red} \textbf{Geo Marabot Solve 3}$

Let $\odot(AXY) \cap AM \equiv T$

$XP\cdot PY=AP\cdot PT=EP\cdot PF \implies AETF$ cyclic. By Three Tangent Lemma, $ME^2 =MT \cdot MA = MY^2 \blacksquare$

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810 posts

#12 h Oct 14, 2024, 9:33 AM

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Let $AM\cap (AEHF)=T$ . Since $XP\cdot PY=EP\cdot PF=AP\cdot PT$ , the points $A, X, Y$ and $T$ are concyclic. It's well-known that $ME$ and $MF$ are tangent to $(AEHF)$ at $E$ and $F$ , respectively. On the other hand, point $M$ is on radical exis of circles $AEHF$ and $AXYT$ . Thus, $MY=ME$ and $MY$ is tangent to $(AXYT)$ implying $\angle XAY=\angle XYM$ as desired.

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