AO and HM meet on odot(ABC)

The post below has been deleted. Click to close.

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

Potla

1886 posts

Potla

#1 h Dec 31, 2011, 12:14 PM • 2 Y

Y by Adventure10 and 1 other user

Let $ABC$ be an acute angled scalene triangle with circumcentre $O$ and orthocentre $H.$ If $M$ is the midpoint of $BC,$ then show that $AO$ and $HM$ intersect on the circumcircle of $ABC.$

This post has been edited 1 time. Last edited by Potla, Dec 31, 2011, 1:25 PM

Z K Y

The post below has been deleted. Click to close.

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

Sayan

2130 posts

Sayan

#2 h Dec 31, 2011, 1:05 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $AO$ intersects the circumcircle at $P$ . $H$ and $P$ are joined. $AD$ , $BE$ , $CF$ are altitudes. Since $AP$ is the diameter, $\angle ABP = 90^{\circ}$ and $\angle BFC = 90^{\circ}$ . Therefore, $BP$ is parallel to $CH$ . Similarly $BH$ is parallel to $PC$ , therefore, $BHCP$ is a parallelogram. $HP$ and $BC$ are diagonals, thus they bisect each other. Therefore, $M$ , the midpoint of $BC$ lies on $HP$ .

This post has been edited 1 time. Last edited by Sayan, Oct 31, 2014, 10:35 AM

Z K Y

The post below has been deleted. Click to close.

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

Virgil Nicula

7054 posts

Virgil Nicula

#3 h Dec 31, 2011, 1:13 PM • 1 Y

Y by Adventure10

Similar PP (more difficult). Let $ABC$ be a triangle with circumcircle $w$ and incentre $I$ . Let $[AA']$ be a diameter
of $w$ . Denote $D\in BC$ for which $ID\perp BC$ and $\{A,S\}=AI\cap w$ . Show that $A'I$ and $SD$ intersect on $w$

This post has been edited 1 time. Last edited by Virgil Nicula, Jan 1, 2012, 6:49 AM

Z K Y

The post below has been deleted. Click to close.

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

WakeUp

1347 posts

WakeUp

#4 h Dec 31, 2011, 2:41 PM • 1 Y

Y by Adventure10

Suppose $AO$ meets $(ABC)$ at $P$ and $PH$ crosses $BC$ at $M$ . We will show $M$ is the midpoint of $BC$ .

$\angle PBA=90^{\circ}\implies BP^2=AP^2-AB^2=4R^4-4R^2\sin^2C=4R^2\cos^2 C=CH^2$ . So $BP=CH$ and similarly $CP=BH$ . Then $BPCH$ is a parallelogram so $M$ is the midpoint of $BC$ (and $PH$ ).

Also really easy with complex numbers, letting $(ABC)$ be the unit circle in the complex plane, $P=-a$ . Now $h=a+b+c$ so the midpoint of $HP$ is $\frac{a+b+c-a}{2}=\frac{b+c}{2}$ which is also the midpoint of $BC$ , so $H,P$ and the midpoint of $BC$ are collinear.

Z K Y

The post below has been deleted. Click to close.

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

Sahil

661 posts

Sahil

#5 h Jan 1, 2012, 5:49 AM • 2 Y

Y by Adventure10, Mango247

sayandas7 wrote:

let AO intersect the circumcircle at P. HP is joined.
AD, BE, CF are altitudes

since AP is the diameter
angle ABP = 90 degree
and angle BFC = 90 degree
therfore, BP is parallel to CH
similarly BH is parallel to PC
therefore, BHCP is a parallelogram
HP and BC are diagonals, thus they bisect each other
therefore, M the midpoint of BC lies on HP

sorry, dont know how to use latex

do u also read challenges and thrills?

its the same proof as that book ( and same as mine also )

Z K Y

The post below has been deleted. Click to close.

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

Sayan

2130 posts

Sayan

#6 h Jan 1, 2012, 2:35 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

sahilsharma94 wrote:

do u also read challenges and thrills?

its the same proof as that book ( and same as mine also )

i have that book but i rarely read it. i find it similar to the Euler line's theorem, and hence proceed in this way.

Z K Y

The post below has been deleted. Click to close.

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

SCP

1502 posts

SCP

#7 h Jan 5, 2012, 9:01 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Virgil Nicula wrote:

Similar PP (more difficult). Let $ABC$ be a triangle with circumcircle $w$ and incentre $I$ . Let $[AA']$ be a diameter
of $w$ . Denote $D\in BC$ for which $ID\perp BC$ and $\{A,S\}=AI\cap w$ . Show that $A'I$ and $SD$ intersect on $w$

Someone could solve this one?
(the original question is really simple, but this one needs maybe some extra?)

trilineair: we can't define points on a circle I think.

complex: how to define $I$ if we want a calculation solution

most interested in a synthetic solution;

$S$ is midpoint of arc $BC$ but how to find the concurrence ?

Z K Y

The post below has been deleted. Click to close.

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

Virgil Nicula

7054 posts

Virgil Nicula

#8 h Jan 6, 2012, 7:15 AM • 3 Y

Y by SCP, Adventure10, and 1 other user

Similar PP (more difficult). Let $ABC$ be a triangle with circumcircle $w$ and incentre $I$ . Let $[AA']$ be a diameter

of $w$ . Denote $D\in BC$ for which $ID\perp BC$ and $\{A,S\}=AI\cap w$ . Show that $A'I$ and $SD$ intersect on $w$

Proof.

Let $L\in SD\cap A'I$ . Thus, $\left\{\begin{array}{c} AA'=2R\\\ ID=r\\\ IA\cdot IS=2Rr\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c} \frac {ID}{AI}=\frac {IS}{AA'}\\\\ \widehat{SID}\equiv\widehat{A'AI}\end{array}\right\|$ $\stackrel{(s.a.s)}{\implies}$

$\triangle ISD\sim$ $\triangle AA'I\ \implies\ \widehat{ISD}$ $\equiv\widehat{AA'I}\implies AA'SL$ is cyclically $\implies\ L\in w$ .

See PP15 from here.

Z K Y

The post below has been deleted. Click to close.

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

djmathman

7936 posts

djmathman

#9 h Nov 2, 2013, 6:00 PM • 2 Y

Y by Adventure10, Mango247

Let $H'$ be the reflection of $H$ with respect to $M$ . It's clear that $HBH'C$ is now a parallelogram, so $\angle BHC=\angle BH'C$ . Then since $180^\circ-\angle BHC=\angle HBC+\angle HCB=(90^\circ-\angle C)+(90^\circ-\angle B)=\angle A,$ we have that $ABH'C$ is cyclic.

To finish, all we need to do is to show that $H'$ and $A$ are antipodes with respect to $(ABC)$ , which is equivalent to proving that $\angle ACH'=90^\circ$ . But this is trivial, since $\angle ACH=\angle ACB+\angle BCH'=\angle ACB+\angle HBC=90^\circ.\qquad\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

jayme

9774 posts

jayme

#10 h Oct 30, 2014, 10:45 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
just a name: Carnot
Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

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

aayush-srivastava

137 posts

aayush-srivastava

#11 h Nov 23, 2015, 7:34 PM • 3 Y

Y by bond98, Adventure10, Mango247

I used the fact that $HG:GO=2:1$ and some similar triangle stuff to get that $HM=MP$ ( $P$ is the point of intersection of $H$ and $O$ .)
As $OM$ is parallel to $HA$ , By B.P.T $OA=OP$ .
$O$ being the centre and $A$ lying on the circumcirle together ensure the Result.
Q.E.D

Z K Y

The post below has been deleted. Click to close.

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

AlastorMoody

2125 posts

AlastorMoody

#12 h Feb 9, 2019, 5:43 PM • 2 Y

Y by Adventure10, Mango247

1000th post (I should have saved it for doing something more productive)

Isn't this kind'a well known?!
Anyways, Brokard gives, if $MH \cap \odot (ABC)=X$ , then, $AX \perp MX$ , Now it isn't very difficult to see that if $MH \cap \odot (ABC)=Y \cancel{=} X$ , then, infact $AO$ should pass through $Y$

This post has been edited 1 time. Last edited by AlastorMoody, Feb 9, 2019, 5:52 PM

Z K Y

The post below has been deleted. Click to close.

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

PCChess

548 posts

PCChess

#13 h Jul 17, 2020, 3:51 AM • 4 Y

Y by HoRI_DA_GRe8, Mango247, Mango247, Mango247

Denote the circumcircle of $ABC$ as $\omega$ . Clearly, when $HM$ intersects the circumcircle, the point of intersection is simply a reflection of $H$ across point $M$ . Denote this point of intersection as $P$ . Then, by Lemma 1.17 in EGMO, AP is the diameter of the circumcircle which implies that A, O, and P are collinear and implies the result.

Z K Y

The post below has been deleted. Click to close.

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

HoRI_DA_GRe8

590 posts

HoRI_DA_GRe8

#14 h Aug 11, 2021, 5:05 PM

Y by

Potla wrote:

Let $ABC$ be an acute angled scalene triangle with circumcentre $O$ and orthocentre $H.$ If $M$ is the midpoint of $BC,$ then show that $AO$ and $HM$ intersect on the circumcircle of $ABC.$

Use phantom points,assume a point $X$ where $AO$ meets the circumcircle join $H$ with $X$ and prove $BHCX$ is parallelogram, and this AVADA KEDAVRA s the problem

Z K Y

The post below has been deleted. Click to close.

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

franzliszt

23531 posts

franzliszt

#15 h Aug 11, 2021, 5:15 PM

Y by

We present two solutions.

Solution 1: Reflecting the Orthocenter.

Trivial by EGMO Lemma 1.17. $\square$

Solution 2: Complex numbers.

Set $(ABC)$ as the unit circle and let $A=1,O=0$ . Then $H=a+b+c,M=\frac{b+c}2,X=AO\cap (ABC)=-1+0i$ . It suffices to show that $X,M,H$ are colinear. To do so, we check $\frac{i}4\begin{vmatrix}-1&-1&1\\ \frac{b+c}2&\frac{c+b}{2bc}&1\\ a+b+c&\frac1a+\frac1b+\frac1c&1\end{vmatrix}=0$ which is true upon expansion. $\square$

This post has been edited 1 time. Last edited by franzliszt, Aug 11, 2021, 5:18 PM

Z K Y

The post below has been deleted. Click to close.

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

HoRI_DA_GRe8

590 posts

HoRI_DA_GRe8

#16 h Aug 11, 2021, 5:26 PM

Y by

franzliszt wrote:

We present two solutions.

Solution 1: Reflecting the Orthocenter.

Trivial by EGMO Lemma 1.17. $\square$

Solution 2: Complex numbers.

Set $(ABC)$ as the unit circle and let $A=1,O=0$ . Then $H=a+b+c,M=\frac{b+c}2,X=AO\cap (ABC)=-1+0i$ . It suffices to show that $X,M,H$ are colinear. To do so, we check $\frac{i}4\begin{vmatrix}-1&-1&1\\ \frac{b+c}2&\frac{c+b}{2bc}&1\\ a+b+c&\frac1a+\frac1b+\frac1c&1\end{vmatrix}=0$ which is true upon expansion. $\square$

Everybody doesn't have egmo so you rather provide the solution1

Z K Y

The post below has been deleted. Click to close.

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

franzliszt

23531 posts

franzliszt

#17 h Aug 11, 2021, 6:02 PM

Y by

Lemma 1.17 just states that the reflection of $H$ across $M$ is the antipode of $A$ .

Z K Y

The post below has been deleted. Click to close.

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

SilverBlaze_SY

66 posts

SilverBlaze_SY

#18 h Jan 18, 2024, 3:36 PM

Y by

Let $AD, BE, CF$ be the altitudes on sides $BC, AC$ and $AB$ respectively.
Let the reflection of $H$ about the mid-point $M$ of side $BC$ be called $H'$ .

Claim 1: The reflection of $H$ about the mid-point (say $M$ ) of any side lies on the circumcircle of $\Delta ABC$ .
Proof

Claim 2: $AH'$ is, in fact, the diameter of $(\Delta ABC)$ .
Proof: We simply angle chase:

$\angle EBC = 90^{\circ} - \angle C$ , $\angle ABE = 90^{\circ}- \angle A$ .
As $BHCH'$ is a parallogram, $\angle HCB= \angle CBH' = 90^{\circ}- \angle B$ .
$\angle ABH'= 90^{\circ}- \angle A+90^{\circ}- \angle B+90^{\circ}- \angle C= 90^{\circ}$ . (As $\angle A+ \angle B+ \angle C=180^{\circ}$ .

As the angle subtended by chord $AH'$ on the circumference is $90^{\circ}$ , $AH'$ must be the diameter of $(\Delta ABC) \Rightarrow A,H',O$ are collinear. Thus we conclude that $AO$ and $HM$ intersect on the circumcircle of $\Delta ABC$ .

Z K Y

The post below has been deleted. Click to close.

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

Jishnu4414l

154 posts

Jishnu4414l

#19 h Mar 29, 2024, 3:28 AM

Y by

It is well known that the reflection of $H$ over the midpoint of $BC$ lies on $(ABC)$ .
Let $X=HM \cap (ABC)$ .
Now let the perpendicular to $BC$ at $M$ meet $AX$ at $O'$ .
It suffices to show that $O'=O$
As $AH \parallel O'M$ , We see by intercept therom that $O'$ is the midpoint of $AX$ .
Thus $O'$ is the circumcentre of $\triangle ABC$ .
Our proof is complete.

Z K Y

The post below has been deleted. Click to close.

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

AshAuktober

953 posts

AshAuktober

#20 h Apr 6, 2024, 8:44 AM

Y by

Observe that the reflection on $A$ about $O$ and that of $H$ about $M$ both correspond to the $A$ -antipode of the circumcircle of $ABC$ , by the EGMO orthocentre reflection lemma. The result follows immediately. $\square$

Z K Y