scent of a classic geometry

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

parmenides51

30629 posts

parmenides51

#1 h Sep 4, 2018, 6:24 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a scalene acute triangle whose orthocenter is $H$ . $M$ is the midpoint of segment $BC$ . $N$ is the point where the segment $AM$ intersects the circle determined by $B, C$ , and $H$ . Show that lines $HN$ and $AM$ are perpendicular.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

L3435

104 posts

L3435

#2 h Sep 4, 2018, 6:38 PM • 2 Y

Y by Adventure10, Mango247

Let $X$ be the second intersection of the line $AN$ with the circle determined by $B,\ C$ , and $H$ . The reflection over $M$ sends $B,\ C$ and $H$ to the circumcircle of the triangle $ABC$ , so it must send $X$ to $A$ . It's well known that $AH'$ ( $H'$ being reflected $H$ ) is the diameter of the circumcircle, so $XH$ must be the diameter of the circle $BCH$ , and the result follows.

This post has been edited 1 time. Last edited by L3435, Sep 4, 2018, 6:39 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MiltonMath12

132 posts

MiltonMath12

#3 h Sep 4, 2018, 8:59 PM • 2 Y

Y by Adventure10, Mango247

I can not upload the image

Let $D$ be The intersection of $AM$ with the circumcircle of $BCH.$
Note that $ABDC$ is a prallelogram.
So $BD//AC$ therefore $\angle HBC=90$ and HD is diameter.
It follows that $\angle HND=90$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arslan

269 posts

Arslan

#4 h Aug 14, 2023, 7:53 PM

Y by

L3435 wrote:

....The reflection over $M$ sends $B,\ C$ and $H$ to the circumcircle of the triangle $ABC$ , so it must send $X$ to $A$ ....

Could you clarify this part a bit more?!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arslan

269 posts

Arslan

#5 h Aug 14, 2023, 7:54 PM

Y by

MiltonMath12 wrote:

...
Note that $ABDC$ is a prallelogram.
....

Can you explain in a more detailed way why $ABDC$ is a prallelogram?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

X.Allaberdiyev

103 posts

X.Allaberdiyev

#6 h Aug 14, 2023, 8:40 PM

Y by

Let $CH$ , $AH$ meet $AB$ , $BC$ at points $F$ and D, respectively. Let $N'$ be the feet of perpendicular from H to AM. Then $(H N' D M)$ is cyclic, then $AN'$ * $AM$ = $AH$ * $AD$ = $AF$ * $AB$ , which means that $(B F N' M)$ is cyclic. Then $\angle BN'H$ = $\angle BN'F$ - $\angle FN'H$ =180-2* $\angle B$ -(90- $\angle B$ )=90- $\angle B$ (from $(A F H N')$ cyclic and $MF$ = $MB$ (because $\angle BFC$ =90 and M is midpoint of BC, so $MF$ = $MB$ = $MC$ )) and we know that $\angle HCB$ is also equal to 90- $\angle B$ , then $(B H N' C)$ is cyclic, then $N=N'$ , so we are done.

This post has been edited 6 times. Last edited by X.Allaberdiyev, Aug 15, 2023, 4:27 PM
Reason: Corrected some mistakes

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

X.Allaberdiyev

103 posts

X.Allaberdiyev

#7 h Aug 14, 2023, 8:48 PM • 1 Y

Y by Arslan

Arslan wrote:

MiltonMath12 wrote:

...
Note that $ABDC$ is a prallelogram.
....

Can you explain in a more detailed way why $ABDC$ is a prallelogram?

Take a point $D'$ such that $ABD'C$ is parallelogram, then $A$ , $M$ , $D'$ are collinear and note that $(B H C D')$ is cyclic, then $AM$ intersects $(B H C)$ at point $D'$ , which means that $D=D'$ , then $ABDC$ is parallelogram.

This post has been edited 1 time. Last edited by X.Allaberdiyev, Aug 14, 2023, 8:50 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arslan

269 posts

Arslan

#8 h Aug 14, 2023, 9:10 PM

Y by

X.Allaberdiyev wrote:

... $(H N DM)$ is cyclic...

Shouldn't it be $HN'DM$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

X.Allaberdiyev

103 posts

X.Allaberdiyev

#9 h Aug 14, 2023, 11:21 PM

Y by

Arslan wrote:

X.Allaberdiyev wrote:

... $(H N DM)$ is cyclic...

Shouldn't it be $HN'DM$ ?

Yeah, I have corrected some mistakes, thanx.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arslan

269 posts

Arslan

#10 h Aug 15, 2023, 6:26 AM

Y by

X.Allaberdiyev wrote:

Arslan wrote:

X.Allaberdiyev wrote:

... $(H N DM)$ is cyclic...

Shouldn't it be $HN'DM$ ?

Yeah, I have corrected some mistakes, thanx.

Thank you!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arslan

269 posts

Arslan

#11 h Aug 15, 2023, 2:05 PM

Y by

X.Allaberdiyev wrote:

Let $CH$ meet $AB$ at point $F$ . Let $N'$ be the feet of perpendicular from H to AM. Then $(H N' D M)$ is cyclic, then $AN'$ * $AM$ = $AH$ * $AD$ = $AF$ * $AB$ , which means that $(B F N' M)$ is cyclic. Then $\angle BN'H$ = $\angle BN'F$ - $\angle FN'H$ =180-2* $\angle B$ -(90- $\angle B$ )=90- $\angle B$ (from $(A F H N')$ cyclic and $MF$ = $MB$ ) and we know that $\angle HCB$ is also equal to 90- $\angle B$ , then $(B H N' C)$ is cyclic, then $N=N'$ , so we are done.

It is still ambiguous. What is $D$ ?

This post has been edited 2 times. Last edited by Arslan, Aug 15, 2023, 5:08 PM
Reason: edited

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

X.Allaberdiyev

103 posts

X.Allaberdiyev

#12 h Aug 15, 2023, 3:35 PM

Y by

Arslan wrote:

X.Allaberdiyev wrote:

Let $CH$ meet $AB$ at point $F$ . Let $N'$ be the feet of perpendicular from H to AM. Then $(H N' D M)$ is cyclic, then $AN'$ * $AM$ = $AH$ * $AD$ = $AF$ * $AB$ , which means that $(B F N' M)$ is cyclic. Then $\angle BN'H$ = $\angle BN'F$ - $\angle FN'H$ =180-2* $\angle B$ -(90- $\angle B$ )=90- $\angle B$ (from $(A F H N')$ cyclic and $MF$ = $MB$ ) and we know that $\angle HCB$ is also equal to 90- $\angle B$ , then $(B H N' C)$ is cyclic, then $N=N'$ , so we are done.

It is still ambiguous. What is $D$ ?

D is the foot of perpendicular from A to BC.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ianis

402 posts

Ianis

#13 h Aug 15, 2023, 4:11 PM

Y by

Let $A'$ be the reflection of $A$ over $M$ . Since $A$ is the orthocentre of $BHC$ we get that $HBA'C$ is cyclic and $HA'$ is a diameter, so $HN\perp NA'\parallel AN$ .

Z K Y