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AH=2*r*cos(A)=r So cos(A)=1/2 So A=60

The orthocenter $ H$ lies onto Carnot circles, i.e. symmetrical of the circle $ (ABC)$ w.r.t. its sides. The circles $ O$ and $ BHC$ being equal, there is a translation mapping them, and this translation is perpendicular onto $ BC$, i.e. $ A$ maps $ H$. Since $ AH=R$, the circumcenter $ O$ will go to the midpoint of the arc $ BC$ not containing $ A$, so $ \angle BOC=120^\circ$ and $ \measuredangle A=60^\circ$. Best regards, sunken rock

Another really cool way to do this is to put triangle ABC on the complex plane. Let O be the origin, and a,b,c lie on the unit circle. Then h = a + b + c. Thus the claim reduces to finding all <A such that |b+c| = 1. This isn't very difficult and it can be easily shown that A = 60.

pelao_malo wrote:

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I have some problems with that "well known" things. also of other posts: $AH=2Rcos(A)$ why? What is meant with the carnot circle, did he prove $BOCH$ is cyclic? How did smb get $|b+c|=1?$ tia

SCP wrote: $AH=2Rcos(A)$ why? Since $AE=c\cos A$, $2R\cos A=\frac{a}{\sin A}\cos A=\frac{c}{\sin C}\cos A=\frac{AE}{\sin A}=AH$.

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SCP wrote: What is meant with the carnot circle, did he prove $BOCH$ is cyclic? He explains what they are: the circumcircle of $ABC$ reflected in $AB,BC,CA$. These three circles coincide at $H$. It's easy to see why as $HD=DD'$ where $HD$ extended meets $ABC$'s circumcircle at $D'$. He continues by saying that since $(O)$ and $(BCH)$ are congruent, the translation - preserving size - from $(O)$ to $(BCH)$ is perpendicular to the radical axis $BC$ and so $A$ is translated onto the line $AH$. To be perfectly honest, I'm not sure he does prove $BHOC$ is cyclic, regardless here is one argument: Let $P$ be the point such that $AHPO$ is a rhombus. Then taking $O$ as the origin, $\overrightarrow{PH}=\overrightarrow{OA}=\bold{a}\implies \bold{a}=\bold{h}-\bold{p}=\bold{a}+\bold{b}+\bold{c}-\bold{p}\implies \bold{p}=\bold{b}+\bold{c}$. So $OCPB$ is a parallelogram, so $PC=BO=R=AO=PH$ so $\triangle PHC$ is isosceles, as is $PB=PH$. So $PB=PH=PC$ so $P$ is the centre of $(BCH)$. But also $AHPO$ is a rhombus so $PH=PO$ so $O$ lies on $(BHC)$. So $BHOC$ is cyclic means $2A=\angle BOC=\angle BHC=\angle FHE =180^{\circ}-\angle BAC=180^{\circ}-A$ so $A=60^{\circ}$.

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SCP wrote: How did smb get $|b+c|=1?$ The circumcircle of $\triangle ABC$ being the unit circle on the complex plane, then $|o-a|=1$. But $|o-a|=|h-a|=|a+b+c-a|=|b+c|\implies |b+c|=1$.

Let $CH$ meet $AB$ in $F$ and $N$ be the midpoint of $AC$. Now,$\angle{OAN}=\angle{HAF}=90^{\circ}-\angle{B}$ and $\angle{ANO}=\angle{AFH}=90^{\circ}$ Also $AH=AO$(given) $\Rightarrow$ the triangles $ANO$ and $AFH$ are congruent $\Rightarrow AF=AN$. But $AN=FN=NC$ (since $N$ is midpoint of hypotenuse $AC$ in right triangle $AFC$) $\Rightarrow AF=AN=FN\Rightarrow \triangle{AFN}$ is equilateral$. \Rightarrow \angle{A}=60^{\circ}$

Actually, $\angle A = 60^{\circ}$ or $\angle A = 120^{\circ}$. You need to consider acute and obtuse triangles separately.

@KrazyFK Ya ,that's a mistake. So two possible values of $\angle{A}$

orl wrote: The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$

oneplusone wrote: orl wrote: The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$ Ah, did not notice this! Thank you

Another Idea Identity: $AH+t=2R+r$ ($t$ donates the exradius opposite to vertex $A$ ) So, $t-r=R$ $Area(ABC)/(s-a) - Area(ABC)/s = abc/4Area(ABC)$ After some calculation we get ($a)^2-(b-c)^2=bc$ which implies $angle(BAC)=60$ I am sorry for bab english. and i don't know to use latex.

Solution: We see that $B'A$, where $B'$ is the foot of the altitude from $B$, is $c\cos A$, we have that $AH=\frac{c}{\sin C}\cdot \cos A$. Setting this equal to $R$, we have $\cos A=\frac{1}{2}$, so $\boxed{A=60}$. $\blacksquare$ Very silly problem for ISL...

Am I right

thegreatp.d wrote:

Am I right

No, you only proved it for one case. There are many other cases you did not consider so your proof would get a 0.

idk why aops precal is making me vector this when $AH=2R\cos A$ is literally an exercise but fine WLOG let $O$ be the origin. Then, $AO=AH$ is equivalent to $\vec{AO}\cdot\vec{AO} = \vec{AH}\cdot\vec{AH}$, or $$(\vec{O}-\vec{A})\cdot(\vec{O}-\vec{A}) = (\vec{H}-\vec{A})\cdot(\vec{H}-\vec{A}) \iff \vec{A}\cdot\vec{A} = (\vec{B}+\vec{C})\cdot(\vec{B}+\vec{C}),$$where $\vec{H} = \vec{A}+\vec{B}+\vec{C}$ is implied from $\vec{O} = 0$. Expanding, substituting $\textbf{v}\cdot\textbf{w} = \Vert v\Vert\Vert w\Vert\cos\theta$, and dividing out $\vec{A}\cdot\vec{A} = \vec{B}\cdot\vec{B} = \vec{C}\cdot\vec{C}$ yields $1 = 2+2\cos\angle BOC$, or $\cos\angle BOC = -\frac{1}{2} \iff \angle BOC = 120^\circ$ (from the acute triangle condition). Consequently, $\angle A = \boxed{60^\circ}$.