NMO (Nepal) Problem 3

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#1 h Mar 17, 2024, 2:22 PM

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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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#2 h Mar 17, 2024, 2:37 PM • 1 Y

Y by khan.academy

$G$ 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 $BC$

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vsamc

#3 h Mar 17, 2024, 3:22 PM • 1 Y

Y by centslordm

Solution

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#4 h Mar 17, 2024, 5:41 PM

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Solution
Remark

This post has been edited 2 times. Last edited by DroneChaudhary, Mar 17, 2024, 5:44 PM
Reason: latex

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#5 h Mar 17, 2024, 6:40 PM • 1 Y

Y by Bata325

Surprised nobody has found this very nice elementary solution yet.

sol

This post has been edited 1 time. Last edited by meth4life2020, Mar 17, 2024, 6:43 PM

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#6 h Mar 17, 2024, 6:50 PM

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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.

This post has been edited 1 time. Last edited by EpicBird08, Mar 17, 2024, 6:53 PM

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#7 h Mar 17, 2024, 9:30 PM

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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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#8 h Mar 17, 2024, 9:41 PM

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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.

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#9 h Mar 18, 2024, 3:10 AM

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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 $\overrightarrow{G'A} \cdot \overrightarrow{G'H} = 0$ .

This post has been edited 1 time. Last edited by Leo.Euler, Mar 18, 2024, 3:10 AM

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#10 h Mar 18, 2024, 3:22 AM

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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

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#11 h Mar 18, 2024, 8:15 AM • 2 Y

Y by DroneChaudhary, Sudip_sah

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.

This post has been edited 1 time. Last edited by p.lazarov06, Mar 18, 2024, 8:16 AM

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#12 h Aug 21, 2024, 6:08 PM

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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 $BC$ .

This post has been edited 2 times. Last edited by Bata325, Aug 21, 2024, 6:12 PM
Reason: Latex mistake again

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#13 h Apr 1, 2025, 4:39 AM

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let $G'$ be defined such that $AG'$ bisects $BC$ and $G'$ lies on $\odot \ (BHC)$ now it is enough to prove that $G' \in \ \odot \ (AEHF)$
consider inversion with centre $A$ and radius $\sqrt{AH*AD}$
This inversion maps $\odot \ (BHC)$ to $\odot \ (FDE)$
which means $\odot \ (BHC)$ is mapped to nine point circle of $\odot \ (ABC)$ which also passes through $I(midpoint \ of \ BC)$
and since $A-G'-I$ and $G' \in \ \odot \ (BHC)$
$G'$ is the inverse of $I$
$\implies AG'*AI=AH*AD$ $\implies$ $H,G',I,D$ are concyclic points
$\implies \angle HG'A=\angle HDI=90$ $\implies G' \in \odot (AEHF)$
$\implies G' \equiv G$

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