A nice one

Easy if u know ur lemmas well. Lemma 1: In a triangle $\Delta ABC$ let $l$ be the euler line and $P=l\cap AB,Q=l \cap AC$ . If $AP=AQ$ then we have that $\angle A=60^\circ$. Proof: Let $H$ be the orthocentre and $O$ be the circumcentre. Observe that $\angle HPB=180-\angle APQ$ and $\angle HQC=180-\angle AQP=180-\angle APQ$ since $\angle AQP=\angle APQ$ . So we have $\angle HPB=\angle HQC$. And also if we let $BH$ intersect $AC$ at $K$ then $\angle ABH \equiv \angle ABK=90-\angle A$ and similiarly if we let $CH$ intersect $AB$ at $T$ then $\angle HCA \equiv \angle TBA=90-\angle A\implies \angle HCA=\angle HBA $. By the two equalities we have obtained (i.e. $\angle HPB=\angle HQC$ and $\angle HBA=\angle ABH$) we have that $\Delta BPH \sim \Delta CQH$ by $AA$ similiarity.And so we have that $$\frac{PB}{CQ}=\frac{PH}{QH}...........(1)$$.Now its well known that $AH,AO$ are isogonal. Now applying theorem 4.22 egmo page 64 we have $$\frac{PH}{HQ} \times \frac{PO}{OQ}= (\frac{AP}{AQ})^2=1\implies \frac{PH}{HQ}= \frac{OQ}{PO}$$. Now substituting this in (1) we have $$\frac{OQ}{PO}=\frac{PB}{CQ}$$or that $$\frac{OQ}{PB}=\frac{PO}{CQ}..........(2)$$and now since we have already proved earlier that $\angle BPO=\angle OQC$ so by $(2)$ and $SAS$ similarity we have $\Delta BPO \sim \Delta OQC \implies \angle POB=\angle QCO=\angle HCB$ where the last step follows from a well known fact which states that $H,O$ are isogonal conjugate. So we have $BHOC$ as cyclic $\implies \angle BHC=\angle BOC$. Now $\angle PBH=90-\angle A$ (already stated earlier) and $\angle BPH=180-\angle APQ=180-(90-\frac{\angle A}{2})=90+\frac{\angle A}{2} $ and so $\angle PHB=180-(\angle PHB+\angle PBH)=180-(90-\angle A+90+\frac{\angle A}{2})=\frac{\angle A}{2}$ and now since we ad already proved $\Delta BPH \sim \Delta CQH \implies \angle PHB=\frac{\angle A}{2}=\angle QHC \implies \angle BHC=180-\angle A$. Now $\angle BOC=2\angle A$ and since $BHOC$ is cyclic $\implies \angle BOC=\angle BHC\implies 180-\angle A=2\angle A\implies \angle A=60^\circ.$ We are done with the lemma. $\blacksquare$ Now back to the main problem. We can easily draw the circumcircle by the means of a compass since we know the location of $O$ so we can draw a circle with radius $OA$ which is in fact its circumcircle. Its trivial to notice that by using a compass we can draw the foot of the altitude from $A$ onto $l$. Let this be $W$. Now by a compass just draw lines which make an angle of $30^\circ$ with $AW$ on both sides of $AW$. Let these lines intersect the circucircle at $B',C'$. By our lemma $B',C'$ are nothing but $B,C$ respectively and the triangle is thus restored. $\blacksquare$.

Nice one

I liked this problem (hope my soln is correct)

I went for my Physics lecture after posting 20 problems and the solutions to 2 of my most favorite problems. Here goes my solution for this one: We know that the Euler line ($l$) passes through the orthocenter (say $H$) and the circumcenter $O$ of a triangle (here $ABC$). We restore the triangle in a step-wise manner. $\textsf{\textbf{Step 1.}}$ Fixing the value of $\angle BAC$. Let $l$ meet $AB$ and $AC$ at $D$ and $E$. It is given in the problem that, $AD = AE$. Consider triangles $ADH$ and $AEO$ (assume, without loss of generality that $DH \leq DO$). Here, we have $\angle DAH = \angle EAO$, as $H$ and $O$ are $\emph{isogonal conjugates}$. We also have $AD = AE$, by the problem, which further leads to $\angle ADH = \angle AEO$. This implies $\triangle ADH \cong \triangle AEO$, and thus, $AH = AO$. But, as $AH = 2R \cos A$ and $AO = R$, we obtain $\boxed{\angle BAC = 60^{\circ}}$. $\square$ $\textsf{\textbf{Step 2.}}$ Restoring the triangle $ABC$. With a compass, draw a circle with center $O$ and radius $OA$. Let this circle be called $\omega$. Again, using (the same ) compass, draw the foot of the perpendicular from $A$ onto $l$. Call this line as $m$. Again, using compass, draw two lines $l_1$ and $l_2$, passing through point $A$ on both sides of $m$ which make angle of $30^\circ$ each with the line $m$. Let these meet $\omega$ at $B$ and $C$. This fixes $\triangle ABC$. $\square$

Well known TBH. Sharygin 2020 CR P9 wrote: The vertex $A$, center $O$ and Euler line $\ell$ of a triangle $ABC$ is given. It is known that $\ell$ intersects $AB,AC$ at two points equidistant from $A$. Restore the triangle. It's well known the Euler Line $\ell$ of a triangle $ABC$ intersects $AB$ and $AC$ at two equidistant points from $A$, then $\angle A=60^\circ$. Proof:- $H,O\in \ell$ where $H,O$ are the Orthocenter and the Circumcenter of $\triangle ABC$ respectively. Let $\ell\cap AB=T$ and $\ell\cap AC=K$. So, we have $$\begin{cases} AT=AK \\ \angle BAH=\angle CAO \\ \angle ATH=\angle AKO\end{cases}\implies \triangle AHT\cong\triangle AOK$$Let $R$ be the circumradius of $\triangle ABC$. Hence, we get that$$AH=AO\implies 2R\cos A=R\implies \cos A=\frac{1}{2}\implies\angle A=60^\circ$$So we must have $\angle A$ as $60^\circ$. Now coming back to the problem... We draw a line passing through $A$ (let this line be $l_1$) such that the acute angle made by the lines $l,l_1$ be $60^\circ$. Let $l\cap l_1=T$. Now we will constuct a circle $\omega$ with center at $A$ and radius $AT$. Let $l\cap\omega=K\neq T$. Extend this line $AK$ (let this line be $l_2$) So, note that after this construction we have $\angle TAK=60^\circ$. Now we will again draw a circle $\omega_1$ with center at $O$ with radius $OA$. So, $\omega_1\cap l_1=B\neq A$ and $\omega_1\cap l_2=C\neq A$. Hence, we have constructed a triangle $ABC$ with circumcenter $O$ and Euler Line intersecting $AB$ and $AC$ at two equidistant points from $A$. $\blacksquare$

Solution: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22.908cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 33.88, ymin = -11.68, ymax = 6.3; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen zzttqq = rgb(0.6,0.2,0.); draw((11.9,3.38)--(7.567481331734039,-4.587379591259801)--(15.281777636651984,-2.1464714569931074)--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw(shift((10.72,-1.14)) * scale(4.671487985642261, 4.671487985642261)*unitcircle, linewidth(2.)); draw((xmin, 0.025547445255474453*xmin-1.4138686131386862)--(xmax, 0.025547445255474453*xmax-1.4138686131386862), linewidth(2.)); /* line */ draw((10.72,-1.14)--(12.129258968386024,-5.593851048252911), linewidth(2.)); draw((xmin, -3.1604205814306483*xmin + 40.989004919024715)--(xmax, -3.1604205814306483*xmax + 40.989004919024715), linewidth(2.) + green); /* line */ draw((xmin, 0.31641358301347455*xmin-6.981833473821349)--(xmax, 0.31641358301347455*xmax-6.981833473821349), linewidth(2.)); /* line */ draw((11.9,3.38)--(7.567481331734039,-4.587379591259801), linewidth(2.) + zzttqq); draw((7.567481331734039,-4.587379591259801)--(15.281777636651984,-2.1464714569931074), linewidth(2.) + zzttqq); draw((15.281777636651984,-2.1464714569931074)--(11.9,3.38), linewidth(2.) + zzttqq); draw((xmin, -39.142857142857096*xmin + 469.17999999999944)--(xmax, -39.142857142857096*xmax + 469.17999999999944), linewidth(2.)); /* line */ /* dots and labels */ dot((11.9,3.38),ududff); label("$A$", (11.98,3.58), NE * labelscalefactor,ududff); dot((10.72,-1.14),ududff); label("$O$", (10.8,-0.94), NE * labelscalefactor,ududff); dot((12.129258968386024,-5.593851048252911),linewidth(4.pt) + uuuuuu); label("$L$", (12.2,-5.44), NE * labelscalefactor,uuuuuu); dot((13.309258968386024,-1.0738510482529118),linewidth(4.pt) + uuuuuu); label("$H$", (13.38,-0.92), NE * labelscalefactor,uuuuuu); dot((14.285291777473697,-4.158525226245165),linewidth(4.pt) + uuuuuu); label("$H'$", (14.56,-4.28), NE * labelscalefactor,uuuuuu); dot((13.79727537292986,-2.616188137249038),linewidth(4.pt) + uuuuuu); label("$D$", (13.88,-2.46), NE * labelscalefactor,uuuuuu); dot((7.567481331734039,-4.587379591259801),linewidth(4.pt) + uuuuuu); label("$B$", (7.28,-5.02), NE * labelscalefactor,uuuuuu); dot((15.281777636651984,-2.1464714569931074),linewidth(4.pt) + uuuuuu); label("$C$", (15.32,-2.6), NE * labelscalefactor,uuuuuu); dot((9.4241018250048,-1.173106887682359),linewidth(4.pt) + uuuuuu); label("$E$", (9.14,-1.02), NE * labelscalefactor,uuuuuu); dot((14.605157143381224,-1.0407441605705525),linewidth(4.pt) + uuuuuu); label("$F$", (14.68,-0.88), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.6666666666666667) * currentpicture; /* end of picture */ [/asy][/asy] As the circumcentre $O$ and $A$ are given, we can draw the circumcircle $\omega$ of $\triangle ABC$ . Let $E$ and $F$ be the touchpoints of the Euler Line with $AB$ and $AC$ respectively(currently not known). Now drop a perpendicular line $m$ from $A$ to the euler line . Let $m$ intersect $\omega$ for the second time at $L$. Claim 1: $LO\perp BC$ . Proof: As $AE=AF ; \triangle AEF$ is isosceles triangle. So the altitude from $A$ is also the angle bisector of $\angle EAF=\angle BAC$. So, by Incentre-Excentre Lemma , $L$ is the midpoint of arc $BC$ not containing $A$. So $LO\perp BC$. Now we know that $A$-altitude in $\triangle ABC$ is parallel to $LO$. So , we draw a line parallel to $LO$ through $A$. Let this line intersect the euler line and $\omega$ at $H$ and $H'$ respectively. Let $D$ be the midpoint of $HH'$. As $AH\perp BC$ and $H$ lies on the Euler Line , $H$ is the orthocentre of $\triangle ABC$. By 'Reflecting The Orthocentre Lemma'(EGMO lesson number 1) , we know that $D$ lies on $BC$. So, $BC$ is a line segment perpendicular to $AH$ and it also passes through $D$ ; and $B$ and $C$ lies on the circumcircle $\omega$. So,we draw a line perpendicular to $AH$ through $D$. Let this line intersect $\omega$ at $B$ and $C$ . Thus, $\triangle ABC$ is the required triangle. $\blacksquare$

Problem 9. The vertex $A$, the circumcenter $O$, and the Euler line $\ell$ of triangle $ABC$ are given. It is known that $\ell$ meets $AB$ and $AC$ at two points equidistant from $A$. Restore the triangle. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.914390434851895, xmax = 9.45521115587534, ymin = -8.564720046557866, ymax = 5.605292389658885; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-2.53,4.41),0.4676571761127645,-78.26142879327767,-48.26142879327767)--(-2.53,4.41)--cycle, linewidth(2) + qqwuqq); draw(arc((-2.53,4.41),0.4676571761127645,-108.26142879327767,-78.26142879327767)--(-2.53,4.41)--cycle, linewidth(2) + qqwuqq); /* draw figures */ draw(circle((-2.09,-1.81), 6.235543280260351), linewidth(0.8)); draw((xmin, 0.20779220779220783*xmin-1.3757142857142856)--(xmax, 0.20779220779220783*xmax-1.3757142857142856), linewidth(0.8) + red); /* line */ draw((-2.53,4.41)--(-6.145384714830859,-6.546650168074084), linewidth(0.8)); draw((4.039751731807417,-2.953741101125596)--(-2.53,4.41), linewidth(0.8)); draw((-1.2728164915117215,-1.6401956345998383)--(-2.53,4.41), linewidth(0.8) + linetype("4 4") + red); draw((-2.53,4.41)--(-2.09,-1.81), linewidth(0.8) + blue); draw((-6.145384714830859,-6.546650168074084)--(4.039751731807417,-2.953741101125596), linewidth(0.8)); draw((-2.53,4.41)--(0.5072204730689345,-4.199876943811298), linewidth(0.8) + blue); /* dots and labels */ dot((-2.09,-1.81),dotstyle); label("$O$", (-2.2829839645550493,-2.1734053063500913), NE * labelscalefactor); dot((-2.53,4.41),dotstyle); label("$A$", (-2.3921039723146946,4.6232123198220805), NE * labelscalefactor); dot((-1.2728164915117215,-1.6401956345998383),linewidth(4pt) + dotstyle); label("$Z$", (-1.2073724594956911,-1.5186852597922213), NE * labelscalefactor); label("$\alpha = 30^\circ$", (-1.7685610708310084,4.030846563412579), NE * labelscalefactor,qqwuqq); label("$\beta = 30^\circ$", (-3.545658340059514,3.781429402819105), NE * labelscalefactor,qqwuqq); dot((-6.145384714830859,-6.546650168074084),linewidth(4pt) + dotstyle); label("$X$", (-6.4295442594215615,-6.927919930163193), NE * labelscalefactor); dot((4.039751731807417,-2.953741101125596),linewidth(4pt) + dotstyle); label("$Y$", (4.201862210875285,-3.171073948723988), NE * labelscalefactor); dot((-0.45563298302344346,-1.4703912691996765),linewidth(4pt) + dotstyle); label("$H$", (-0.3967666875668993,-1.3472109618842078), NE * labelscalefactor); dot((0.5072204730689345,-4.199876943811298),linewidth(4pt) + dotstyle); label("$H_1$", (0.569724809732814,-4.075211155875332), NE * labelscalefactor); dot((-4.765898569797838,-2.3660308716463043),linewidth(4pt) + dotstyle); label("$I$", (-4.699212707804333,-2.23575959649846), NE * labelscalefactor); dot((2.2202655867743934,-0.9143603975533724),linewidth(4pt) + dotstyle); label("$J$", (2.2844677888129507,-0.786022350548891), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Since we are given the point $A$ and $O$, we can easily construct the circumcircle of triangle $ABC$ by drawing a circle with center $O$ which passes through $A$. Now, we next we draw a line through $A$ perpendicular to $\ell$ and let it meet $\ell$ at some point $Z$. Now, draw a lines $\ell_1,\ell_2$ passing through $A$ which cuts the circle with center $O$ for the second time at $X$ and $Y$ respectively, such that $\angle ZAX=\angle ZAY=30^\circ$. We claim that $\triangle AXY$ is nothing but $\triangle ABC$. More precisely, $X=B,Y=C$. If the euler line intersects sides $AB,AC$ at $I,J$ respecively, then $\angle AZI=90\implies \angle AIZ=90-\angle IAH=90-30=60$ and similarly, $\angle AJZ=90-\angle ZAJ=90-30=60\implies IAJ$ is an equilateral triangle. Now, obviously $O$ lies on $\ell$ and so, if we can show that the orthocenter of $\triangle AXY$ lies on $\ell$ as well, then we are done as in that case $\ell$ will be the euler line of $\triangle AXY$ which clearly cuts the side $AX,AY$ at points equidistant from $A$ as $\angle AIJ=\angle AJI=60$ giving us the desired triangle. So, it is enough to show that the orthocenter of $\triangle AXY$ lies on $\ell$. For that, let feet of perpendicular from $A$ to $XY$ meet $\ell$ at some point $H$. Clearly, the orthocenter of $\triangle AXY$ lies on $AH$ and let us denote it by $H'$. We also know that if $R$ denotes the circumradius of $\triangle AXY$, then $AO=R$ and $AH'=2R\cos(\angle XAY)=2R\cos(60)=R\implies AO=AH'$. Also, as $H',O$ are isogonal conjugates of $\triangle AXY$, we must have $\angle H'AJ=\angle OAI$ and since, $AI=AJ$ because $AIJ$ is equilateral, we get $\triangle AIO$ is congruent to $\triangle AJH'\implies \angle AIO=60=\angle AJH'$, but $\angle AJH=60\implies H=H'\implies H'\in\ell$ completing the proof. Feedback: A simple observation that $\angle A=60$, kills the problem .

Ohh man the official solution is so easy and it completely avoided the fact that $\angle A=60^\circ$. Here it goes. Proof: We have that the Euler line is parallel to the exterior angle bisector at $A$. Since $AO$ and $AH$ are isogonal rays with respect to $\angle A$, it follows that $AO = AH$. Thus we can recover $H$ as the second point where the circle with center $A$ and radius $AO$ meets the Euler line. Furthermore let line $AH$ meet the circumcircle (which we can recover because we know its center $O$ and one point on it, namely $A$) again at $D$. Then $B$ and $C$ are the points where the perpendicular bisector of segment $HD$ meets the circumcircle. Done $\blacksquare$.

Thats why we are not them

GeoMetrix wrote: Ohh man the official solution is so easy and it completely avoided the fact that $\angle A=60^\circ$. Here it goes. Proof: We have that the Euler line is parallel to the exterior angle bisector at $A$. Since $AO$ and $AH$ are isogonal rays with respect to $\angle A$, it follows that $AO = AH$. Thus we can recover $H$ as the second point where the circle with center $A$ and radius $AO$ meets the Euler line. Furthermore let line $AH$ meet the circumcircle (which we can recover because we know its center $O$ and one point on it, namely $A$) again at $D$. Then $B$ and $C$ are the points where the perpendicular bisector of segment $HD$ meets the circumcircle. Done $\blacksquare$. Wait , I just realised that even I didn't use it at all

@above you are the next Mr. Akopyan

GeoMetrix wrote: Ohh man the official solution is so easy and it completely avoided the fact that $\angle A=60^\circ$. Here it goes. Proof: We have that the Euler line is parallel to the exterior angle bisector at $A$. Since $AO$ and $AH$ are isogonal rays with respect to $\angle A$, it follows that $AO = AH$. Thus we can recover $H$ as the second point where the circle with center $A$ and radius $AO$ meets the Euler line. Furthermore let line $AH$ meet the circumcircle (which we can recover because we know its center $O$ and one point on it, namely $A$) again at $D$. Then $B$ and $C$ are the points where the perpendicular bisector of segment $HD$ meets the circumcircle. Done $\blacksquare$. Wheres the official solution?

Draw the circle $\omega$ with center $O$ and radius $OA$. Note that since $O, H$ are isogonal conjugates wrt $\Delta ABC$, we have $AO = AH$. Thus we can construct $H$ by drawing a circle with center at $A$ and with radius $AO$, and intersecting it with $l$ again. Let $H_A$ be the second intersection of $AH$ and $\omega$. It is well known that $BC$ is the perpendicular bisector of $HH_A$; thus by drawing this perpendicular bisector and intersecting it with $\omega$, we have reconstructed $\Delta ABC$ as desired.

Solution without using $\angle A=60^\circ$ or $AO=AH$: Let $A'$ be the reflection of $A$ across $l$. Since $l \perp AI$, we have that $A'$ is the midpoint of arc $BC$ not containing $A$ on $(ABC)$. Draw the line $l'$ parallel to $l$ such that $d(A,l)=d(l',l)$ and $A,l'$ lie on opposite sides of $l$. Draw $M=OA' \cap l'$, then $M$ is the midpoint of $BC$. Now draw the line through $M$ perpendicular to $OM$, which intersects $(ABC)$ at $B$ and $C$.