Let $ABC$ be an acute triangle and $H$ be its orthocenter. Let $E$ be the foot of the altitude from $C$ to $AB$, $F$ be the foot of the altitude from $B$ to $AC$. Let $G \neq H$ be the intersection of the circles $(AEF)$ and $(BHC)$. Prove that $AG$ bisects $BC$. Proposed by Kang Taeyoung, South Korea
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Tags: geometry
dkshield
17.03.2024 17:37
$G$ is the A-humpty point of the triangle $ABC$, because $\angle AGH = 90^\circ$, and $G\in \odot BHC$, so G lies on the A-median. So $AG$ bisects $BC$
vsamc
17.03.2024 18:22
Let $X = AG\cap BC$. Note that $HGXD$ is cyclic as $\angle HGX = \angle HGA = 90^{\circ}$ and $\angle HDX = 90^{\circ}$, so $AH\cdot AD = AG\cdot AX$ by power of a point. Now, let $H' = AD\cap (HBC)$ and let $X' = AX\cap (HBC)$. Then, by power of a point again, $AH\cdot AH' = AG\cdot AX'$, so dividing these two relations, $$\frac{AH'}{AD} = \frac{AX'}{AX}.$$Therefore, by SAS similarity, $\triangle DAX \sim \triangle H'AX'$, so $\angle AH'X' = \angle ADX = 90^{\circ}$. However, this means $\angle HH'X' = 90^{\circ}$, so $X'$ is the antipode of $H$ on $(BHC)$. Therefore, if we reflect the orthocenter of $(BHC)$ about the midpoint $M$ of $BC$, we get $X'$. However, the orthocenter of $(BHC)$ is $A$, so if we reflect $A$ about the midpoint $M$ of $BC$, we get $X'$. Thus, $A, M, X'$ are collinear, but $A, X, X'$ are collinear and $X$ and $M$ lie on BC, therefore $X = M$, as desired. $\blacksquare$
DroneChaudhary
17.03.2024 20:41
Lemma: If $(AH)$ intersects $(ABC)$ at $P$ then $PH$ bisects $BC$.
Proof:
Let $E$ and $F$ be the feet of the perpendiculars from $B$ and $C$ to $AC$ and $AB$ respectively.
Construct $A'$, the $A$-antipode on $(ABC)$, We have $P-H-A'$ as $\angle APH = 90^{\circ} = \angle APA'$
Notice that $BHCA'$ is a parallelogram as
$$\angle BCA' = \angle BAA' = \angle BAO = \angle HAC = \angle HEF = \angle CBH \iff BH//A'C$$Because $O$ and $H$ are isogonal conjugates in a triangle, $BEFC$ cyclic; and a similar angle chase shows $CH//A'B$
We get that if $M$ is the midpoint of BC then $H-M-A'$ which combined with first collinearity implies $P-H-M$.
Now applying the lemma in triangle $BHC$ the result follows as $A$ is its orthocenter, $(HA)$ meets $(BHC)$ at $G$ and so if $M$ is the midpoint of $BC$
then we have $A-G-M$ collinear $\square$
$G$ is the $A$ Humpty point and $P$ is the $A$ Queue Point
meth4life2020
17.03.2024 21:40
Surprised nobody has found this very nice elementary solution yet.
Let $H'$ be the $H$-antipode with respect to circle $(BHC)$, and let $M$ be the midpoint of $BC$. Then, $H'$ is also the reflection of the orthocenter of $BHC$ across $M$, but this orthocenter is just $A$. Therefore, $A$, $M$, and $H'$ are collinear. We also have $AGH = 90^{\circ}$ because $AH$ is a diameter of $AEF$ due to $AEH = AFH = 90^{\circ}$, and $HGH' = 90^{\circ}$. Therefore, $AGH' = 180^{\circ}$, so $A$, $G$, $M$, and $H'$ are collinear, done.
EpicBird08
17.03.2024 21:50
This is a classic Humpty-point configuration, but I present a solution without it for completeness. First note that the center $O$ of $(AEF)$ is the midpoint of $AH,$ and if $A'$ is the reflection of $A$ over the midpoint $M$ of $BC,$ then $BHCA'$ is cyclic with $A'$ being the $H$-antipode, and so the center $P$ of $(BHC)$ is the midpoint of $AA'.$ Thus by homothety, we have that $OP \parallel AA' \parallel AM,$ so $HG$ is perpendicular to $AM.$ To finish, we note that if $G'$ is the foot of the altitude from $H$ to $AM,$ then $G'$ lies on $(AEF)$ by definition and so $G = G',$ done.
breloje17fr
18.03.2024 00:30
Another solution without Humpty point : Let Bx and Cy be the parallels, respectively, to AC from B and to AB from C, and let K be their intersection. So ABKC is a parallelogram since it has two pairs of parallel sides. The altitudes BF and CE of ABC thus are respectively perpendicular to Bx and Cy, so the quadrilateral BHCK, comprised of both right triangles HBK and HCK, is inscribed in the circle (BHC), with HK being a diameter of this circle and thus having the center J as its middle. Similarly, the circle (AEF) has AH as its diameter. In the triangle HAK, the respective middles of HA and HK are the centers O and J, and the line OJ thus is parallel to the third side AK. However, this line OJ, simultaneously, is the centerline of the circles (AEF) and (BHC), and thus the perpendicular bissector of the segment HG. G thus belongs to the line AK. And since AK, as a diagonal of the parallelogram ABKC, bissects the other diagonal BC, we are done ...
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eibc
18.03.2024 00:41
humpty lol Let $M$ be the midpoint of $\overline{BC}$, and $G'$ be the foot from $H$ onto $\overline{AM}$. I claim that $G' = G$, which will solve the problem. Reflect $A$ over $M$ to $A'$. Since $(ABC)$ and $(BHC)$ are reflections across $M$, we see by orthocenter reflections that $\overline{A'H}$ is a diameter of $(BHC)$. Thus, because $\overline{HG'} \perp \overline{AG'}$, $G'$ lies on $(BHC)$. Additionally, since $\overline{AH}$ is a diameter of $(AEF)$ and $\overline{AG'} \perp \overline{HG'}$, we also find that $G'$ lies on $(AEF)$, which finishes.
Leo.Euler
18.03.2024 06:10
Fairly easy for a P3. Here is a somewhat bashy computational solution. 1. Let $G'$ denote the intersection of the radical axis of $(AEF)$ and $(BHC)$ with $AM$. 2. Let $f(\bullet) = \text{Pow}(\bullet, (AEF)) - \text{Pow}(\bullet, (BHC))$. Then for some $t$, $f(At+M(1-t))=0$, where $G'=At+M(1-t)$. 3. Solve for $t$ in terms of the side lengths of $\triangle ABC$. 4. Verify that $\angle AG'H = 90^{\circ}$ by showing that $\overrightarrow{G'A} \cdot \overrightarrow{G'H} = 0$.
khan.academy
18.03.2024 06:22
Leo.Euler wrote: Fairly easy for a P3. Here is a somewhat bashy computational solution. They were not in any particular order of difficulty. I think P2 > P1 > P3 > P4
p.lazarov06
18.03.2024 11:15
Perform an inversion with center $A$ and radius $\sqrt{AE*AC}=\sqrt{AF*AB}$. Then $B\leftrightarrow F$ and $C\leftrightarrow E$. So $(AEHF)\leftrightarrow BC$ And $H$ goes to $D$ - the foot from $A$ to $BC$ . Then the circles $(BHC)\leftrightarrow(EFD)$. So $G’=BC\cap (EFD)$ which is the midpoint of $BC$ because $(EFD)$ is the nine-point circle. But $G\in AG’$ so we are done.
Bata325
21.08.2024 21:08
For the sake of completeness, proof that A-HM(T) point passes through $(HBC)$ in triangle $ABC$ with orthocenter $H$. Let $M$ be the midpoint of the side $BC$ and let $A'$ be a point in line $AM$ s.t. $AM= A'M$ . Then, since the diagonals bisect each other $ABA'C$ is a parallelogram. Now, $ \angle BA'C = \angle BAC $. We know that $BHC = 180^{\circ} - \angle ABC$. Therefore, $ \angle BA'C + \angle BHC = 180^{\circ}$. HBA'C is a cylic quadrilateral. Now, $ \angle HBA' = 90^{\circ}$ as $BH \perp AC$ and $AC \parallel BA'$. Now, from definition, $T$ is A-hm point. So, $ \angle HBA' + \angle HTA' = 180^{\circ}$. So, $T$ lies on $(BHC)$. Let $X$ be the set of Points such that $Ax$ bisects BC for $x \in X$. A-HM point is one of them. Let it be $T$. We know, $T \in (BCH)$. Now, $ HT \perp AT$, $\angle HTA = 90^{\circ}$ and obviously $\angle HFA = 90^{\circ}$. SO, $AFHT$ is cyclic which means $T \in (AEHF)$. So, $T$ is the second interaction point which implies $T = G$. Therefore, $AG$ bisects $BC$.