parmenides51 wrote: Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $ABC$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $ABC$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . Sam Bealing, United Kingdom On the second and the third sentence, should $ABC$ be fixed to $\odot(ABC)$ ?

indeed, both have been changed, thanks

Claim 01. Let $H_M$ be the $A$-Humpty point of $\triangle ABC$ and let $M$ be the midpoint of $BC$. Let $H_M'$ is the reflection of $H_M$ over $BC$. Let $H$ be the orthocenter of $\triangle ABC$ and let $H'$ be the reflection of $H$ over $BC$. Then, $D$, the intersection of $(HH'H_MH_M')$ of $AM$ lies on the line through $H'$ parallel to $BC$ and $H_M'H'$ is tangent to $(AH'D)$. Proof. It is well-known that $H_M'$ lies on $(ABC)$ and lies on the $A$-symmedian of $\triangle ABC$. Also, we have $H,B,H_M,C$ concyclic. Firstly we show that $H'D\parallel BC$. Indeed, \begin{align*} \measuredangle H'DH_M&=\measuredangle H'H_M'H_M=\measuredangle HH_MH_M'\\&=\measuredangle HH_MC+\measuredangle CH_MH_M'\\&=\measuredangle HBC+90^\circ+\measuredangle H_MCB\\&=\measuredangle ACB+\measuredangle MAC\\&=\measuredangle AMB. \end{align*}Now, $\measuredangle H_M'H'D=\measuredangle H_M'H_MD=\measuredangle H_M'H_MM=\measuredangle H'AD$, which means that $H'H_M'$ is tangent to $(AH'D)$. [asy][asy] import geometry; defaultpen(fontsize(10pt)); size(10cm); pair A,B,C,M,HM,H,H1,HM1,D; A = dir(155); B = dir(228); C = dir(312); M=midpoint(B--C); H=orthocenter(A,B,C); H1=2*foot(A,B,C)-H; HM=intersectionpoints(circumcircle(C,B,H),A--M)[0]; HM1=2foot(HM,B,C)-HM; D=intersectionpoints(circumcircle(HM,H1,H),A--M)[1]; draw(circumcircle(A,B,C),royalblue);draw(circumcircle(H,H1,HM),royalblue);draw(A--B--C--cycle,red);draw(A--M,red); draw(A--H,red);draw(H--foot(B,A,C),red);draw(B--foot(B,A,H),red);draw(H1--D,orange);draw(HM--HM1,orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(90)); dot("$M$", M, dir(M)); dot("$H$", H , dir(H)); dot("$H_M$", HM, dir(HM)); dot("$H_M' $", HM1, dir(HM1)); dot("$H'$", H1, dir(H1)); [/asy][/asy] Now, in our original problem, as $AG=GH$, we get that $L$ is the midpoint of $AH$ and as $BC$ bisects $AL$, by the intercept theorem, we get that $LG\parallel BC$. Now, this is the exact set-up of the claim, hence we are done.

Let $D,E$ be the feet of the altitudes from $A,C$ to the opposite sides, respectively. Let $L' = AH \cap (ABC)$ , $J = AG \cap BC$ and $O$ be the circumcenter of $\triangle ABC$. We claim that $AL'JO$ is a parallelogram. Note that $OJ \perp BC$ and $AL' = AH \perp BC$, hence $OJ \parallel AL'$. By Euler line, $H,G,O$ are collinear and $2GJ = AG = HG = 2GO \implies HJ = AO$. Since $OJ \parallel AH$ and $HJ = AO$, this means $HJOA$ is an isosceles trapezoid, and hence cyclic. Also by a well-known property we know that $L'$ is the reflection of $H$ across $BC$. Now note that $\angle AL'J = 180 - \angle HL'J = 180 - \angle L'HJ = 180 - \angle AHJ = \angle AOJ$, combining this angle equality with $OJ \parallel AL'$ gives that $AL'JO$ is a parallelogram. Since $2AL' = 2OJ = AH$, $L$ is the midpoint of $AH$. So, $2 = \frac{AG}{GJ} = \frac{AL'}{L'E} \implies GL' \parallel CE \implies \angle AL'G = 90 \implies L' = L$. Let $N$ be the midpoint of $AG$. Since $N$ is the center of $\omega$, we have $90 = \angle NLK$. Also because $NL$ is midline, $NL \parallel OH$. So this implies $OH \perp KL$ but since $OK = OL$, this means $GH$ is the perpendicular bisector of $LK$, giving that $90 = \angle HKG$. Done. $\square$

Attachments:

hakN wrote: Let $D,E$ be the feet of the altitudes from $A,C$ to the opposite sides, respectively. Let $L' = AH \cap (ABC)$ , $J = AG \cap BC$ and $O$ be the circumcenter of $\triangle ABC$. We claim that $AL'JO$ is a parallelogram. Note that $OJ \perp BC$ and $AL' = AH \perp BC$, hence $OJ \parallel AL'$. By Euler line, $H,G,O$ are collinear and $2GJ = AG = HG = 2GO \implies HJ = AO$. Since $OJ \parallel AH$ and $HJ = AO$, this means $HJOA$ is an isosceles trapezoid, and hence cyclic. Also by a well-known property we know that $L'$ is the reflection of $H$ across $BC$. Now note that $\angle AL'J = 180 - \angle HL'J = 180 - \angle L'HJ = 180 - \angle AHJ = \angle AOJ$, combining this angle equality with $OJ \parallel AL'$ gives that $AL'JO$ is a parallelogram. Since $2AL' = 2OJ = AH$, $L$ is the midpoint of $AH$. So, $2 = \frac{AG}{GJ} = \frac{AL'}{L'E} \implies GL' \parallel CE \implies \angle AL'G = 90 \implies L' = L$. This part can be shown easly like this : Let $L'=\omega\cap HA, F=\text{midpoint of BC}$ and $J=BC\cap HA$ $\implies \angle GL'A=90 \implies CJ||GL' \implies \frac{AL'}{L'J}=\frac{AG}{GF}=\frac{2}{1}$. Also $GH=GA$ $\implies HL'=AL' =2JL' \implies HJ=L'J \implies \angle BCA=\angle BHL'=\angle BL'H$. So $ABCL'$ is cyclic $\implies L'\equiv L$ which we want.

btw, degrees should be written as $90^\circ$ instead of $90^o$

john0512 wrote: btw, degrees should be written as $90^\circ$ instead of $90^o$ Yeah you are right. It kills the problem.

I always write the degrees as $90^o$ , as the reader understands exactly the same thing and this can be written faster

Let $AH \cap (\triangle ABC) = L'$, $AH\cap BC =D$, $M$ be the midpoint of $BC$ and $O$ be the circumcenter of $\triangle ABC$. Since $AG=GH$ and $GM=OG$, so $AO=MH$ and $AH\parallel OM$, so $AHMO$ is a isosceles trapezoid. We have $AO = MH =ML'$ and $AL' \parallel OM$, so since $AHMO$ is a isosceles trapezoid, $AL'MO$ is a parallelogram. We have $AH=2OM$, therefore $L'$ is the midpoint of $AH$. Since, $AL'=L'H = 2L'D$ and $AG=2GM$, so $L'G\parallel DM$ or $AL'\perp L'G$, which implies $L=L'$ or points $A,L,H$ are collinear. Let $LK\cap GH=T$, $N$ be the center of $\omega $, we have $LN\parallel$ and $KL \perp NL$, so $KL\perp GH$. Since $O,G,H$ are collinear, so $T$ is the midpoint of $LK$. Now $GH\cap BC = X$. Since $D$ is the midpoint of $LH$, so $X$ is the midpoint of $GH$, $NX \perp BC$ and $NX$ bisects $LG$. Assume $NX\cap LG=P$, then we have points $L,D,X,T,P$ are concyclic. So, $\angle DTP = 90^\circ$. Now taking homothety centered at $L$, with factor $2$, sends $D$ to $H$, $P$ to $G$ and $T$ to $K$, which solves the problem.

Let $SH$ meet $BC$ at $S$. Claim1 : $A,L,H$ are collinear. Proof : Let $AH$ meet $\omega$ at $L'$ we will show $L'$ is $L$. we have $\angle AL'G = \angle 90 = \angle ASB$ so $AL' = 2L'S$ so $HS = SL'$ so $\angle HL'B = \angle L'HB = \angle BCA $ so $AL'BC$ is cyclic so $L'$ is $L$. Let $\omega$ meet $AC$ at $P$. Note that $LH = AL = LP$ so $\angle LKH = \angle LAC = \angle LPA = \angle LGA = \angle LGH$ so $LGKH$ is cyclic and we know $\angle GLH = \angle 90$ so $HKG = 90$ as wanted. we're Done.

Let $O$ be center of $\odot(ABC)$ ; $M,N,T$ be the midpoints of segments $BC,OM,AG$, respectively ; $A'$ antipode of $A$ wrt $\odot(ABC)$ ; $U = \overline{LT} \cap \odot(ABC) \ne L$. Observe $L-G-A'$ and $U-O-K$. Since $GO = GM$, so $\overline{GN} \perp \overline{OM}$. Now, $$ \overline{GN} \parallel \overline{TO} \parallel \overline{GA'} \implies \text{points } A',N,G,L \text{ are collinear} $$ So now $H \in \overline{AL}$. Now $\angle TLG = \angle TGL = \angle OGN$, so $\overline{UL} \parallel \overline{OG}$. By Reims theorem points $A',O,G,K$ are concyclic. Since $OA' = OK$, so $\overline{OG}$ bisects $\angle A'GK$ (externally). Hence, $$ \angle KGH = \angle OGA' = \angle TLA' = 90^\circ - \angle KLG = \angle KLH $$So points $K,H,L,G$ are concyclic. Since $\angle HLG = 90^\circ$, hence $\angle HKG = 90^\circ$, as desired. $\blacksquare$

Let $D = AH \cap BC$, the midpoint of $BC$ be $M$, and $X = GH \cap KL$. In addition, redefine $L$ as the projection of $G$ onto $AH$. Clearly, $$\frac{AH}{AD} = \frac{2 \cdot AL}{AD} = 2 \cdot \frac{AG}{AM} = \frac{4}{3}$$so $$DH = \frac{AD}{3} = DL.$$This means that $L$ is the reflection of $H$ over $D$, so it lies on $(ABC)$. Thus, we have recovered the original definition of $L$, as $\angle ALG = 90^{\circ}$ still holds. It's well-known that $H, G, O$ are collinear along the Euler Line of $ABC$. Thus, $$\angle OXL = \angle GXL = 180^{\circ} - \angle XLG - \angle XGL = 180^{\circ} - \angle LAG - \angle HGL$$$$= 180^{\circ} - \angle LAG - \angle AGL = 90^{\circ}$$so $X$ is the midpoint of $KL$. Hence, we know that $K$ and $L$ are symmetric about $OX \equiv GH$, which implies $$\angle HKG = \angle HLG = 90^{\circ}$$as desired. $\blacksquare$ Remark: This length condition is too nice! Also, there is no need to explicitly state $AH > AD$, as the length condition $GA = GH$ fixes $\frac{AH}{AD}$.

Claim: $L$ is the midpoint of $AH$. Proof: Let $M$ be the midpoint of $BC$ and let $D$ be the foot of the altitude from $A$ to $BC$. In addition, define "phantom point" $L'$ as the midpoint of $AH$. The length condition implies that $\angle ALG = 90^\circ$. But then $\frac{AL'}{DL'} = \frac{AG}{MG} = 2$, so $H$ is the reflection of $L'$ over $BC$. It follows immediately that $L'$ lies on $(ABC)$, and the claim is proved. Next, we have $$\angle KLG + \angle LGH = \angle LAG + \angle LGA = 90^\circ$$ so $LK\perp HG$. But $HG$ is the Euler line of $\triangle ABC$, so it passes through the circumcenter $O$ of $\triangle ABC$. It follows that $K$ is the reflection of $L$ over $OH$, and we're done.

This is a UK proposal by Sam Bealing (me). Let $L'$ be the midpoint of $AH$. Then we claim $L'$ lies on $\odot ABC$. Indeed, let $D$ be the foot of the $A$-altitude on $BC$ then: $$AG=GH \Rightarrow \angle GL'A=90^{\circ} \Rightarrow GL' \parallel BC \Rightarrow DL'=\frac{AL'}{2}=\frac{HL'}{2} \Rightarrow DL'=HD$$where in the last step we have used that if $M$ is the midpoint of $BC$, $AG:GM=2:1$ and that $\angle B$ is obtuse so $H,A$ lie on opposite sides of line $BC$. This means that $L'$ is the reflection of $H$ in $BC$ which is well-known to lie on $\odot ABC$. Also $AG=GH \Rightarrow \angle GL'A=90^{\circ}$ so $L'$ lies on $\omega$ and hence in fact $L \equiv L'$. Let $O$ be the midpoint of $AG$ then $OL \bot LK$. Homothety of factor $2$ at $A$ takes $OL \rightarrow HG$ so $HG \parallel OL$ and hence $LK \bot HG$. But the centre of $\odot ABC$ lies on $HG$ so this means $K$ is the reflection of $L$ across line $HG$ and hence as $\angle HLG=90^{\circ}$ it follows $\angle HKG=90^{\circ}$.