line AZ is perpendicular to BC

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line AZ is perpendicular to BC

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Source: IGO 2015 Intermediate P3

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237 posts

#1 h Jun 25, 2016, 7:48 PM • 3 Y

Y by centslordm, Adventure10, Mango247

In triangle $ABC$ , $M,N,K$ are midpoints of sides $BC,AC,AB$ ,respectively.Construct two semicircles with diameter $AB,AC$ outside of triangle $ABC$ . $MK,MN$ intersect with semicircles in $X,Y$ .The tangents to semicircles at $X,Y$ intersect at point $Z$ .Prove that $AZ \perp BC$ .(Mehdi E'tesami Fard)

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1979 posts

#2 h Jun 25, 2016, 8:08 PM • 3 Y

Y by centslordm, Adventure10, Mango247

Let $H$ be th foot of altitude from $A$ . Clearly, $H,M,K,N$ are concyclic (nine-point circle) and $\frac{XK}{YN}=\frac{AB}{AC}=\frac{HK}{HN}$ and so, $H$ is the Miquel point of the quadrilateral $XKNY$ . Now, we get points $H,M,X,Y$ concyclic and let $Z'$ be the anti-podal point of $M$ in the circle $(HMXY)$ . We get that $Z'X \perp XK$ and $Z'Y \perp YN$ and so, $Z'=Z$ and we have $ZH \perp BC$ . Thus, result follows.

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35 posts

#5 h Jun 25, 2016, 11:43 PM • 2 Y

Y by centslordm, Adventure10

Lemma: $X$ , $A$ , $Y$ are collinear. $\odot (AXB)$ $\cap$ $BC$ = $H$ , so $H$ $\in$ $\odot (AYC)$ . It´s easy to see: $\angle$ $YXM$ = $\angle$ $XYM$
( $\angle$ $AKM$ = $\angle$ $ANM$ ),so $XY$ $\perp$ $ZM$ $\longrightarrow$ $XZ$ = $YZ$ . Then $Z$ it´s in the radical axis of $\odot (AXB)$ and $\odot (AYC)$ , so $Z$ , $A$ and $H$ are collinear,done. Proof of lemma: It´s easy to see: $2$ $\angle$ $AXM$ = $\angle$ $ANM$ = $2$ $\theta$ ,so $\angle$ $YAN$ = $\theta$ .
Then $X$ , $A$ and $Y$ are collinear.

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375 posts

#6 h Jun 26, 2016, 4:18 AM • 2 Y

Y by centslordm, Adventure10

A generalizations of this problem:
Let $\triangle ABC$ . Construct two semicircles with diameter $AB,AC$ outside of $\triangle ABC$ . Let $P$ is any point on side $BC$ ( $P$ is not coincide with $B$ or $C$ ), through $P$ , construct $PX \parallel AC$ and $DF \parallel AB$ with $X$ and $Y$ on the two semi-circles respectively. The perpendicular drawn from $X$ to $PX$ and perpendicular drawn from $Y$ to $PY$ intersect at $Z$ . Prove that $AZ \perp BC$ .
Proof. Let $AH \perp BC$ at $H$ , $M$ is midpoint of $AP$ . From Cyclic Pointswe have $A, P, X, Y$ are concyclic. Let $\odot (N)\equiv (APXY)$ then $MN$ is perpendicular bisector of $HM\Rightarrow AZ\parallel MN\perp BC$ . DONE.

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851 posts

#7 h Jun 29, 2016, 2:07 AM • 2 Y

Y by centslordm, Adventure10

Note by extraversion that $X,Y$ lie on the external angle bisector of $\angle BAC$ , and that $MX=MY$ ; letting $H$ be the foot of the altitude from $A$ , we claim that $M, H, X, Y$ are concyclic. Define $L\equiv XY\cap BC$ ; it suffices to show that $MH\cdot ML=MX^2$ . WLOG $AB<AC$ ; compute
$MH\cdot ML=\bigg(\frac{a}{2}-c\cdot\frac{a^2-b^2+c^2}{2ac}\bigg)\bigg(\frac{ac}{b-c}+\frac{a}{2}\bigg)=\frac{(b+c)^2}{4}$ On the other hand, $MX^2=(MK+KX)^2=\bigg(\frac{b}{2}+\frac{c}{2}\bigg)^2$ so we get the conclusion. Then $\angle AHM=\frac{\pi}{2}=\angle ZXM=\angle ZHM$ , as desired.

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9774 posts

jayme

#8 h Jun 29, 2016, 12:15 PM • 3 Y

Y by centslordm, Adventure10, Mango247

Dear Mathlinkers,

1. according to Reim and pappus theorem, , A and Y are collinear
2. according to the pivot theorem, the circle with diameter ZM pass throug the foot of the A-altitude and we are done...

Sincerely
Jean-Louis

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1915 posts

#9 h Aug 11, 2019, 9:22 PM • 4 Y

Y by Path_to_Almighty, centslordm, Adventure10, Mango247

Let $\omega_1$ be the circle with diameter $AB$ and $\omega_2$ be the circle with diameter $AC$ . Now let $\omega_1\cap BC=K$ , hence $\angle AKB=90^\circ$ and $\omega_2\cap BC=K'\implies \angle AK'C=\angle AK'B=90^\circ\implies K\equiv K'$ . Now $AZ$ must be the radical axis of $\omega_1$ and $\omega_2$ . Hence, $AZ\perp KN\implies AZ\perp BC$ as $KN\perp BC$ .

This post has been edited 3 times. Last edited by amar_04, Oct 3, 2019, 11:55 AM

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9774 posts

jayme

#10 h May 30, 2020, 8:55 AM • 3 Y

Y by centslordm, Mango247, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%205.pdf p. 46-47.

Sincerely
Jean-Louis

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JustKeepRunning

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JustKeepRunning

#11 h Jun 11, 2021, 2:27 AM • 2 Y

Y by centslordm, Mango247

It is easy to see that $ZYMX$ is cyclic with diameter $ZM$ . Furthermore, $MX=MK+KX=MY=MN+NY=\frac{AB+AC}{2}$ . Therefore, we have that $ZX=ZY,$ or that $\text{pow}(Z, \omega_c)=\text{pow}(Z, \omega_b)$ . This means that $Z$ is on the radical axis of $\omega_c$ and $\omega_b$ . However, it is easy to see that the radical axis of $\omega_c$ and $\omega_b$ is just the $A$ -altitude(just note the right angle to the diameter), so we are done.

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111 posts

#12 h Jun 13, 2021, 7:03 AM

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$\angle X K A=2 \angle Z X A$ . Also $\angle Y N A=2 \angle Z Y A$ By mid point theorem, $AC \parallel XM$ , so $\angle X K A$ = $\angle Y N A$ . So, $\angle Z X A$ = $\angle Z Y A$ . So, $XZ=ZY$ . So, $Z$ lies on the radical axis of the two circles (can be proved by power of point). By mid point theorem, $KN \parallel BC$ . So, $AZ \perp BC$ (Hence Proved)

This post has been edited 1 time. Last edited by aansc1729, Jun 13, 2021, 7:04 AM

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#13 h Nov 30, 2021, 2:16 AM • 1 Y

Y by antimonio

Let $H$ the foot of altitude from $A$ to $BC$ . We need to prove that $Z,A,H$ are collinear.
Lemma 1. $X,A,Y$ are collinear
Proof: It´s enough to prove that: $\angle{XAK}+\angle{CAB}+\angle{YAN}=180$ $\Leftrightarrow$
$\frac{180-\angle{XKA}}{2} + \angle{CAB} + \frac{180-\angle{YNA}}{2}=180$ $\Leftrightarrow$
$180-\angle{XKA}+ 2\angle{CAB} + 180-\angle{YNA} = 360$ $\Leftrightarrow$
$\angle{XKA}+\angle{YNA}=2\angle{CAB}$
But, $\angle{XKA}=\angle{YNA}=\angle{CAB}$ . So, it´s proved.
Lemma 2. $ZX=ZY$
Proof: We know that $\angle{XKA}=\angle{ANY}$ $\Rightarrow$ $\angle{AXK}=\angle{AYN}=\frac{180-\angle{XKA}}{2}$ $\Rightarrow$
$90-\angle{AXK}=90-\angle{AYN}$ $\Rightarrow$ $\angle{ZXA}=\angle{ZYA}$ . q.e.d

Then, since $ZX=ZY$ , the power of $Z$ to both circles are the same. So, $Z$ it´s in the radical axis of $(XAB)$ and $(AYC)$ . But $H \in (XAB)$ and $H \in (AYC)$ . Then, $A,Z,H$ are colilinear.

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#14 h Dec 27, 2021, 6:08 AM

Y by

Let AH be perpendicular to BC. we will prove H,A,Z are collinear.
step1 : X,A,Y are collinear.
∠XAK = ∠XKB/2 ---> ∠BKM = ∠BAC = 180 - 2∠XAK
∠YAN = ∠YNC/2 ---> ∠CNM = ∠CAB = 180 - 2∠YAN
so ∠XAK = ∠YAN and ∠XAK + ∠KAN + ∠NAY = 180 so X,A,Y are collinear.

step2 : XZYMH is cyclic.
first note BXAH and CYAH are cyclic.
∠MXY = ∠XAK = ∠NAY = ∠CHY = ∠MHY ---> XYMH is cyclic.
∠MXZ = 90 = ∠MYZ ---> XZYM is cyclic.
so XZYMH is cyclic.

now we have ∠ZHM = ∠ZXM = 90 and we had ∠AHM = 90 so H,A,Z are collinear as wanted.
we're Done.

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672 posts

#15 h Oct 16, 2023, 7:58 PM

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Let $AB\cap YZ=K$ and $AC\cap XZ=L$
$\angle MAX=90-\ \frac{\angle A}{2}=\angle YAN \implies \angle MAX+\angle YAN+\angle CAB=180$
$\implies X,A,Y$ are collinear.
$\angle ZXY=90-\angle AXK=\frac{\angle A}{2}$ and $\angle XAL=\angle YAC=90-\frac{\angle A}{2} \implies CA\perp XZ$ and similarily $AB \perp YZ$
So $XLYC$ and $BXKY$ are cyclic.
$AL.AC=AX.AY=AK.AB$ $\implies KLCB$ are cyclic.
Also we know that $ZLKA$ are cyclic.
$\angle AZL=\angle AKL=\angle BKL=\angle BCL=\angle C$
$\angle BAZ=180-(\angle C+\frac{\angle A}{2}-90+\frac{\angle A}{2})=180-\angle C-\angle A+90=\angle B+90$ which gives the result.

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