Very simple,however beautiful geometry problem

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Very simple,however beautiful geometry problem

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Orkhan-Ashraf_2002

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Orkhan-Ashraf_2002

#1 h Nov 6, 2016, 1:56 PM • 3 Y

Y by Vanuatu, Adventure10, Mango247

Let $ABC$ be an acute triangle such that $AB\not=AC.$ Let $M$ be the midpoint $BC,H$ the orthocenter of $\triangle ABC$ $,O_1$ the midpoint of $AH$ and $O_2$ the circumcenter of $\triangle BCH$ $.$ Prove that $O_1AMO_2$ is a parallelogram.

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4071 posts

#2 h Nov 6, 2016, 4:25 PM • 3 Y

Y by Vanuatu, Math_legendno12, Adventure10

It follows from the well-known lemma that "For any $\triangle{ABC}$ with orthocenter $H$ and $M$ be the midpoint of $BC$ , if $H’$ is the reflection of $H$ over $M$ , then $AH'$ is diameter of $(ABC)$ .”
Applying the lemma for $\triangle{HBC}$ , we get that $AM\cap HO_2=T\in (HBC)$ .
So, $O_2$ is the midpoint of $HT$ and $O_1$ is the midpoint of $HA$ .
We have $O_1O_2\parallel AM$ .

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Nov 18, 2019, 12:40 PM

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#3 h Aug 31, 2018, 4:06 AM • 1 Y

Y by Adventure10

Let the circumradius of $\triangle ABC$ be $R$ .

Lemma 1: The circumradius of $\triangle BHC$ is $R$ .
Proof: Note that $\angle BHC=180^{\circ}-\angle A$ so $\sin\angle BHC=\sin\angle A$ . Hence, $\frac{BC}{\sin\angle BHC}=\frac{BC}{\sin\angle A}=2R$ , which proves the Lemma. $\Box$

Note that $\angle BO_2C=2\angle A$ . By Lemma 1, $O_2B=O_2C=R$ and $\angle O_2BC=90^{\circ}-\angle A$ . Hence, $MO_2=R\cos A$ .

Lemma 2: $AH=2R\cos A$ .
Proof: Let $B'$ and $C'$ be on $AC, AB$ respectively such that $BB'\perp AC$ and $CC'\perp AB$ . Since $BC'B'C$ is cyclic, $\angle AC'B'=\angle C$ so $\triangle AC'B'\sim \triangle ACB \implies \frac{AH}{2R}=\frac{AC'}{AC}=\cos A$ (ratios of diameters of circumcircles=ratios of sides). $\Box$

By Lemma 2, $AO_1=R\cos A$ . Hence, $AO_1=MO_2$ and $MO_2\parallel AO_1$ so $O_1AMO_2$ is a parallelogram. $\blacksquare$

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jayme

#4 h Aug 31, 2018, 8:52 AM • 3 Y

Y by Math_legendno12, Adventure10, Mango247

Dear Mathlinkers,

according to Carnot, AH = 2.MO2 and we are done...

Sincerely
Jean-Louis

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Mx.22

#5 h Nov 18, 2019, 7:09 AM • 2 Y

Y by Adventure10, Mango247

HM-point,it could be a smart and easy solution.

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#6 h Dec 11, 2019, 8:52 AM • 1 Y

Y by Adventure10

Mx.22 wrote:

HM-point,it could be a smart and easy solution.

Nice observation!

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#7 h Aug 15, 2023, 2:01 PM

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$O_2$ is the reflection of $O$ to $BC$ so $O_2M=OM=\frac{AH}{2}=AO_1$ and $AO_1 \parallel O^2M$ because $AH _\perp BC\perp O_2M$

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#9 h Dec 19, 2024, 4:53 PM

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Really easy with nine-point circle properties.

It suffices to show that $AO_2$ and $O_1M$ bisect each other but just note that $A$ is the orthocenter of $\triangle BCH$ and that $O_1M$ is the diameter of the nine-point circle of $\triangle BCH$ . It's well known that the circumcenter of the nine-point circle is the midpoint of the line that joins the circumcenter and the orthocenter, in this case $AO_2$ , and we're done.

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Ianis

#10 h Dec 19, 2024, 6:42 PM

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Observe that $A$ is the orthocentre of $BHC$ , so $AO_1=\frac{AH}{2}=\frac{2O_2M}{2}=O_2M$ and $AO_1\parallel AH\perp BC\perp O_2M$ . Hence $AO_1=O_2M$ and $AO_1\parallel O_2M$ . Done.

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Ianis

#11 h Dec 19, 2024, 6:46 PM

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Use complex numbers with $(ABC)$ as the unit circle.
We have $o_2=b+c$ because $(b+c)-b=c$ , $(b+c)-c=b$ , $(b+c)-h=-a$ and all of them have modulus equal to $1$ . Hence $a+o_2=a+b+c=\frac{a+(a+b+c)}{2}+\frac{b+c}{2}=h_1+m,$ so $O_1AMO_2$ is a parallelogram. Done.

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#12 h Dec 28, 2024, 2:51 AM

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Note that $AO_1$ and $MO_2$ are both perpendicular to $BC$ and thus parallel. Further we have $AO_1 = MO_2 = R\cos A$ , so we're done.

jayme wrote:

Dear Mathlinkers,

according to Carnot, AH = 2.MO2 and we are done...

Sincerely
Jean-Louis

What is Carnot?

This post has been edited 1 time. Last edited by AshAuktober, Dec 28, 2024, 2:58 AM

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