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Orthocenter madness once again!

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Source: USEMO 2023 Problem 4

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1471 posts

#1 h Oct 22, 2023, 10:58 PM • 7 Y

Y by ihatemath123, mathmax12, crazyeyemoody907, starchan, Shreyasharma, LoloChen, Rounak_iitr

Let $ABC$ be an acute triangle with orthocenter $H$ . Points $A_1$ , $B_1$ , $C_1$ are chosen in the interiors of sides $BC$ , $CA$ , $AB$ , respectively, such that $\triangle A_1B_1C_1$ has orthocenter $H$ . Define $A_2 = \overline{AH} \cap \overline{B_1C_1}$ , $B_2 = \overline{BH} \cap \overline{C_1A_1}$ , and $C_2 = \overline{CH} \cap \overline{A_1B_1}$ .

Prove that triangle $A_2B_2C_2$ has orthocenter $H$ .

Ankan Bhattacharya

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815 posts

#2 h Oct 22, 2023, 10:59 PM • 5 Y

Y by OronSH, ihatemath123, mathmax12, megarnie, Aryan-23

Haha Benny L

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3441 posts

#3 h Oct 22, 2023, 10:59 PM • 6 Y

Y by crazyeyemoody907, OronSH, centslordm, mathmax12, Math4Life7, Rounak_iitr

$[asy] unitsize(1.7cm); pair A = (-1.6, 5.84); pair A1 = (-0.12,0); pair A2 = (-1.6, 2.87); pair B = (-3.5, 0); pair B1 = (-0.28, 4.17); pair B2 = (-2.19, 1.03); pair C= (3,0); pair C1 = (-3.03, 1.44); pair C2 = (-0.16, 1.03); pair D = (-1.6,0); pair D1 = (-2.28, 2.19); pair E = (0.51, 3.16); pair E1 = (-1.9, 0.88); pair F = (-2.88, 1.91); pair F1 = (-0.18, 1.55); pair H = (-1.6, 1.5); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); draw(A1--B1--C1--cycle, royalblue); draw(A1--D1, royalblue); draw(B1--E1, royalblue); draw(C1--F1, royalblue); dot("$A$", A, dir(90), black+0); dot("$B$", B, dir(225), black+0); dot("$C$", C, dir(-45), black+0); dot("$D$", D, dir(270), black+4); dot("$E$", E, dir(45), black+4); dot("$F$", F, dir(160), black+4); dot("$A_1$", A1, dir(270), royalblue+4); dot("$B_1$", B1, dir(45), royalblue+4); dot("$C_1$", C1, dir(160), royalblue+4); dot("$D_1$", D1, dir(135), royalblue+4); dot("$E_1$", E1, dir(240), royalblue+4); dot("$F_1$", F1, dir(0), royalblue+4); dot("$A_2$", A2, dir(160), heavyred+4); dot("$B_2$", B2, 2.5*dir(180), heavyred+4); dot("$C_2$", C2, dir(30), heavyred+4); dot("$H$", H, 2*dir(-60), black+4); [/asy]$
Let $D$ , $E$ and $F$ be the feet of $\triangle ABC$ ; let $D_1, E_1$ and $F_1$ be the feet of $\triangle A_1 B_1 C_1$ .

The condition in the problem is equivalent to proving that $\overline{B_2 C_2} \parallel \overline{BC}$ and likewise conditions on the other two sides.

To prove this we use power of a point: We have that $\angle B_2 E_1 B_1 = \angle B_2 E B_1 = 90^{\circ}$ , so $B_2 E_1 EB_1$ is cyclic. The diagonals intersect at $H$ , so
$E_1 H \cdot HB_1 = B_2H \cdot HE. \qquad (\heartsuit)$ Similarly, $C_2 F_1 FC_1$ is cyclic, so
$F_1 H \cdot HC_1 = C_2 H \cdot HF. \qquad (\diamondsuit)$
We also have $\angle C_1E_1 B_1 = \angle C_1 F_1 B_1 = 90^{\circ}$ , so $C_1E_1F_1B_1$ is cyclic. Then,
$F_1 H \cdot HC_1 = E_1H \cdot HB_1.$ Substituting the LHS with $(\diamondsuit)$ and the RHS with $(\heartsuit)$ , we have
$C_2H \cdot HF = B_2 H \cdot HE.$ Dividing both sides by $CH \cdot HF=BH \cdot HE$ (because $BFEC$ is cyclic), we have
$\frac{C_2H}{CH} = \frac{B_2H}{BH} \implies \overline{B_2C_2} \parallel \overline{BC},$ as desired.

This post has been edited 4 times. Last edited by ihatemath123, Mar 3, 2024, 3:40 PM

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#4 h Oct 22, 2023, 11:00 PM • 2 Y

Y by crazyeyemoody907, mathmax12

We claim that $\tfrac{HA_2}{HA}=\tfrac{HB_2}{HB}=\tfrac{HC_2}{HC}$ , after which we can conclude by noticing that $ABCH$ and $A_2B_2C_2H$ are homothetic. Let $D$ and $D_1$ be the feet of the altitudes from $A$ to $\overline{BC}$ and from $A_1$ to $\overline{B_1C_1}$ , respectively. Since $\angle A_1DA_2=\angle A_1D_1A_2=90^\circ$ , we know that $A_1DD_1A_2$ is cyclic, so power of a point at $H$ gives
$HA_2 \cdot HD=HA_1 \cdot HD_1 \Longrightarrow \frac{HA_2}{HA}=\frac{HA_1 \cdot HD_1}{HA \cdot HD}.$ This is equal to $\tfrac{\operatorname{Pow}_{(A_1B_1C_1)}(H)/2}{\operatorname{Pow}_{(ABC)}(H)/2}$ by orthocenter reflections. Since this is symmetric with respect to $A$ , $B$ , and $C$ , we have $\tfrac{HA_2}{HA}=\tfrac{HB_2}{HB}=\tfrac{HC_2}{HC}$ , as desired. $\square$

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crazyeyemoody907

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crazyeyemoody907

#5 h Oct 22, 2023, 11:00 PM • 4 Y

Y by mathmax12, OronSH, sixoneeight, v4913

wow projective actually worked

Working backwards: suffices to prove $\overline{AA_2H}\perp\overline{B_2C_2}$ , or in other words, $\overline{BC}\parallel\overline{B_2C_2}$ .
$\iff HB_2/HB=HC_2/HC$ .
$\iff$ There exists a point $A_3\in\overline{HA_1}$ with $\overline{BA_3}\parallel\overline{A_1C_1}$ and $\overline{CA_3}\parallel\overline{A_1B_1}$ . Indeed, this point would be chosen so that
$\frac{HA_3}{HA}=\frac{HB_2}{HB}=\frac{HC_2}{HC},$ lengths directed. In still other words, we want $\overline{HA_1}$ , $\overline{B\infty_{A_1C_1}}$ , $\overline{C\infty_{A_1B_1}}$ concurrent.

For this we employ a massive cross-ratio chase:
$\begin{align*} (\infty_{A_1C_1}\infty_{A_1B_1}; \infty_{\perp B_1C_1}\infty_{BC}) &\overset{\text{rotate 90}^\circ}= (\infty_{HB_1}\infty_{HC_1};\infty_{B_1C_1}\infty_{HA})\\ &\overset H= (B_1C_1;\infty_{B_1C_1}A_2)\\ &\overset A= (\overline{AC},\overline{AB};\overline{B_1C_1},\overline{AH})\\ &\overset{\text{rotate 90}^\circ}= (\overline{HB},\overline{HC};\overline{HA_1},\overline{BC})\\ &\overset H= (BC;A_1\infty_{BC}) \end{align*}$ and the concurrence follows by prism lemma.

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297 posts

#6 h Oct 22, 2023, 11:01 PM • 1 Y

Y by mathmax12

i think the closest comparison to this problem i can make is IMO 2005/1, although they're not very similar

In fact, I claim that triangles $ABC$ and $A_2B_2C_2$ are homothetic with center $H$ , which will immediately finish since if $\overline{AH} \perp \overline{BC}$ and $\overline{B_2C_2} \parallel \overline{BC}$ , then $\overline{A_2H} \perp \overline{BC}$ and similar.

First we define the following points:

$D = \overline{AH} \cap \overline{BC}$ , $E = \overline{BH} \cap \overline{AC}$ , $F = \overline{CH} \cap \overline{AB}$ and
$D_1 = \overline{A_1H} \cap \overline{B_1C_1}$ , $E_1 = \overline{B_1H} \cap \overline{A_1C_1}$ , $F = \overline{C_1H} \cap \overline{A_1B_1}$ .

Claim: $EA_2B_2D$ , $FC_2B_2E$ , $DA_2C_2F$ are all cyclic.

Proof. Note that since $\measuredangle A_2D_1H = \measuredangle A_2D_1A_2 = 90 = \measuredangle ADA_1 = \measuredangle A_2D_1A_1$ , $A_2D_1DA_1$ is cyclic - similarly, $B_2E_1EB_1$ , $C_2F_1FC_1$ are cyclic. Also, since $H$ is the orthocenter, $B_1C_1F_1E_1$ , $C_1A_1E_1D_1$ , $A_1B_1D_1F_1$ are cyclic. Hence
$A_2H \cdot HD = A_1H \cdot HD_1 = B_1H \cdot HF_1 = EH \cdot HB_2$ so $EA_2B_2D$ is cyclic as desired. The other concyclities follow similarly. $\square$

To finish, we have
$\measuredangle HAB = \measuredangle DAB = \measuredangle DEB = \measuredangle DEB_2 = \measuredangle DA_2B_2 = \measuredangle HA_2B_2$ so $\overline{A_2B_2} \parallel \overline{AB}$ . Similarly, $\overline{B_2C_2} \parallel \overline{BC}$ and $\overline{C_2A_2} \parallel \overline{CA}$ , hence triangles $ABC$ and $A_2B_2C_2$ are homothetic as desired. $\blacksquare$

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DottedCaculator

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DottedCaculator

#7 h Oct 22, 2023, 11:02 PM • 2 Y

Y by centslordm, mathmax12

$\frac{A_2H}{AH}=\frac{B_2H}{BH}=\frac{C_2H}{CH}=\frac{\operatorname{Pow}_{(A_1B_1C_1)}(H)}{\operatorname{Pow}_{(ABC)}(H)}$

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5000 posts

#8 h Oct 22, 2023, 11:02 PM • 3 Y

Y by centslordm, mathmax12, Assassino9931

A somewhat innovative coordinate-based approach...

Use coordinates. Let $H=(0,0)$ and WLOG set $\overline{BC}$ as $y=-1$ , so $B=(b,-1)$ and $C=(c,-1)$ . Then $A=(0,a)$ lies on the $y$ -axis. Since $\overline{CH} \perp \overline{AB}$ we find $a=-bc-1$ .

Let $A_1=(t,-1) \in \overline{BC}$ . $\overline{AB}$ has equation $y=cx-bc-1$ and $\overline{AC}$ has equation $y=bx-bc-1$ . Thus let $B_1=(p,bp-bc-1)$ and $C_1=(q,cq-bc-1)$ . Then $\overline{B_1H}$ has slope $\tfrac{bp-bc-1}{p}$ and $\overline{C_1H}$ has slope $\tfrac{cq-bc-1}{q}$ .

Since $\overline{B_1C_1}$ has slope $\tfrac{bp-cq}{p-q}$ , we can compute $\overline{B_1C_1} \cap \overline{AH}=(0,-\tfrac{(b-c)pq}{p-q}-bc-1)$ .

We now compute the $y$ -coordinate of $\overline{A_1B_1} \cap \overline{CH}$ : here, the key idea is to not use the slope of $\overline{A_1B_1}$ as obtained from $A_1$ and $B_1$ directly, but rather to take the negative reciprocal of the slope of $\overline{C_1H}$ . This $y$ -coordinate ends up being $-\tfrac{qt-cq}{bc+1}-1$ and likewise the $y$ -coordinate of $\overline{A_1C_1} \cap \overline{BH}$ as $-\tfrac{pt-bp}{bc+1}-1$ . Thus we can calculate
$\frac{HA_2}{HA}=\frac{(b-c)pq}{(bc+1)(p-q)}+1\qquad \frac{HB_2}{HB}=\frac{qt-cq}{bc+1}+1 \qquad \frac{HC_2}{HC}=\frac{pt-bp}{bc+1}+1.$
I claim that these are equal. Indeed,
$\frac{HB_2}{HB}=\frac{HC_2}{HC} \iff qt-cq=pt-bp \iff \frac{bp-cq}{p-q}=t,$ which is the equation from $\overline{B_1C_1} \perp \overline{A_1H}$ , and
$\frac{HB_2}{HB}=\frac{HA_2}{HA} \iff qt-cq=\frac{(b-c)pq}{p-q} \iff (p-q)(t-c)=(b-c)p \iff (p-q)t=bp-cq,$ which also follows from $\overline{B_1C_1} \perp \overline{A_1H}$ . Thus $\triangle A_2B_2C_2$ and $\triangle ABC$ are homothetic with center $H$ , so we're done. $\blacksquare$

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#10 h Oct 22, 2023, 11:05 PM • 2 Y

Y by centslordm, mathmax12

redacted

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DottedCaculator

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DottedCaculator

#11 h Oct 22, 2023, 11:06 PM • 7 Y

Y by centslordm, GrantStar, ihatemath123, EpicBird08, mathmax12, mistakesinsolutions, crazyeyemoody907

signifance wrote:

I've never bashed ever before, and am against bashing, but I didn't want to synthetic at the time. For reference, this is my first oly problem ever bashed and is incredibly inefficient, and omitted details. How many points is this worth

zero

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140 posts

#12 h Oct 22, 2023, 11:07 PM

Y by

redacted

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DottedCaculator

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DottedCaculator

#13 h Oct 22, 2023, 11:15 PM • 2 Y

Y by GrantStar, mathmax12

It's a zero because there's no synthetic progress and the bash is clearly incomplete.

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577 posts

#14 h Oct 22, 2023, 11:19 PM

Y by

imagine complex bashing and not writing up
imagine not solving p1 or p4
:skull:

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496 posts

#15 h Oct 22, 2023, 11:22 PM • 1 Y

Y by crazyeyemoody907

We have $\triangle CB_1C_2\sim \triangle BHC_1$ . Therefore, $\frac{CC_2}{CH}=\frac{BC_1\cdot CB_1}{BH\cdot CH}=\frac{BB_2}{BH}$ . So,
$\frac{CC_2}{CH}=\frac{BB_2}{BH}=\frac{AA_2}{AH}\Rightarrow \frac{C_2H}{CH}=\frac{B_2H}{BH}=\frac{A_2H}{AH}.$ So, a homothety centered at $H$ sends $\triangle A B C$ to $\triangle A_2B_2C_2$ . So, $H$ is the orthocenter of $\triangle A_2B_2C_2$ .

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1728 posts

OronSH

#16 h Oct 22, 2023, 11:40 PM • 1 Y

Y by mathmax12

i trigbashed this :sob: (im sorry) but it was somewhat clean ig

It suffices to show that $\frac{AA_2}{HA_2}=\frac{BB_2}{HB_2}=\frac{CC_2}{HC_2}.$ Law of Sines on $\triangle AC_1A_2$ and $\triangle HC_1A_2$ give $\frac{AA_2}{HA_2}=\frac{\sin{AC_1B_1} \sin{AHC_1}}{\sin{BAH} \sin{B_1C_1H}}=\frac{\sin{AC_1B_1} \sin{CA_1B_1}}{\cos{ABC} \cos{A_1B_1C_1}}.$ This last expression is symmetric w.r.t. $B,$ so $\frac{AA_2}{HA_2}=\frac{CC_2}{HC_2}$ and similarly for $B.$

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#17 h Oct 22, 2023, 11:50 PM • 1 Y

Y by Rounak_iitr

In-contest Sol

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#18 h Oct 23, 2023, 12:04 AM • 1 Y

Y by LoloChen

We present two solutions.

Power of a point solution, by Nikolai Beluhov In this solution, all lengths are signed. Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ , and $\triangle D_1 E_1 F_1$ be the orthic triangle of $\triangle A_1 B_1 C_1$ . We define two common quantities, through power of a point: $\begin{align*} k &\coloneqq HA \cdot HD = HB \cdot HE = HC \cdot HF. \\ k_1 &\coloneqq HA_1 \cdot HD_1 = HB_1 \cdot HE_1 = HC_1 \cdot HF_1. \\ \end{align*}$ $[asy] size(8cm); pair A = dir(110), B = dir(210), C = dir(330); pair H = orthocenter(A, B, C); pair B1 = A + 0.35*(C-A), C1 = A + 0.651*(B-A); pair A1 = orthocenter(H, B1, C1); pair A2 = extension(A, H, B1, C1), B2 = extension(B, H, C1, A1), C2 = extension(C, H, A1, B1); filldraw(A--B--C--cycle, invisible, red); filldraw(A1--B1--C1--cycle, invisible, blue); /* filldraw(A2--B2--C2--cycle, red+opacity(0.1), red); */ pair D = extension(A, H, B, C); pair D1 = extension(A1, H, B1, C1); draw(A--D, deepgreen); draw(A1--D1, deepgreen); draw(circumcircle(D, A1, A2), deepgreen+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A_1$", A1, dir(H-A)); dot("$B_1$", B1, dir(H-B)); dot("$C_1$", C1, dir(H-C)); dot("$A_2$", A2, dir(dir(C1-A2)+dir(A-A2))); dot("$H$", H, dir(dir(A2-H)+dir(C2-H))); dot("$D$", D, dir(-90)); dot("$D_1$", D1, dir(H-A1)); [/asy]$
Because quadrilateral $A_2D_1DA_1$ is concyclic (with circumdiameter $\overline{A_1A_2}$ ), by power of a point, we get $\begin{align*} HA_2 \cdot HD &= HD_1 \cdot HA_1 = k_1 \\ \implies HA_2 &= \frac{k_1}{HD} = \frac{k_1}{k} \cdot HA. \end{align*}$ Since $k_1/k$ is fixed, a symmetric argument now gives $\frac{HA_2}{HA} = \frac{HB_2}{HB} = \frac{HC_2}{HC} = \frac{k_1}{k}.$ Therefore, $H$ is the center of a homothety mapping $\triangle A_2 B_2 C_2$ to $\triangle ABC$ . In particular, it is also the orthocenter of $\triangle A_2 B_2 C_2$ .

Author's ratio-based solution We are going to prove:
Claim: We have $\overline{B_2 C_2} \parallel \overline{BC}$ .
Proof. Refer to the diagram below.
$[asy] size(8cm); pair A = dir(110), B = dir(210), C = dir(330); pair H = orthocenter(A, B, C); pair B1 = A + 0.35*(C-A), C1 = A + 0.651*(B-A); pair A1 = orthocenter(H, B1, C1); pair A2 = extension(A, H, B1, C1), B2 = extension(B, H, C1, A1), C2 = extension(C, H, A1, B1); draw(A--H^^B--H^^C--H); filldraw(A--B--C--cycle, invisible, red); filldraw(A1--B1--C1--cycle, invisible, blue); filldraw(A2--B2--C2--cycle, invisible, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A_1$", A1, dir(H-A)); dot("$B_1$", B1, dir(H-B)); dot("$C_1$", C1, dir(H-C)); dot("$A_2$", A2, dir(dir(C1-A2)+dir(A-A2))); dot("$B_2$", B2, dir(dir(A1-B2)+dir(B-B2))); dot("$C_2$", C2, dir(dir(B1-C2)+dir(C-C2))); dot("$H$", H, dir(dir(A2-H)+dir(C2-H))); [/asy]$
Note that $\begin{align*} \frac{C_1A_2}{A_2B_1} & = \frac{[AC_1H]}{[AB_1H]} = \frac{AC_1 \cdot d(H, \overline{AB})}{AB_1 \cdot d(H, \overline{AC})}\\ &= \frac{AC_1 / HC}{AB_1 / HB} = \frac{HB}{HC} \cdot \frac{\sin \angle AB_1C_1}{\sin \angle AC_1B_1}\\ &= \frac{HB}{HC} \cdot \frac{\sin \angle BHA_1}{\sin \angle CHA_1} = \frac{[HBA_1]}{[HCA_1]} = \frac{BA_1}{A_1C}. \end{align*}$ Similarly, $\tfrac{A_1B_2}{B_2C_1} = \tfrac{CB_1}{B_1A}$ and $\tfrac{B_1C_2}{C_2A_1} = \tfrac{AC_1}{C_1B}$ . Hence, $[BB_2C] = [BC_1C] \cdot \frac{B_2A_1}{C_1A_1} = [BAC] \cdot \frac{B_2A_1}{C_1A_1} \cdot \frac{C_1B}{AB} = [ABC] \cdot \frac{B_1C}{AC} \cdot \frac{C_1B}{AB}.$ Similarly, $[BC_2C]$ also equals this quantity, so $\overline{B_2C_2} \parallel \overline{BC}$ and $\overline{A_2H} \perp \overline{B_2C_2}$ . $\blacksquare$
Repeating this we see that $H$ is the orthocenter of $\triangle A_2B_2C_2$ , as wanted.

Remark: In the first equality chain, we obtained $[AC_1H] \cdot [CA_1H] = [AB_1H] \cdot [BA_1H].$ Similarly, $[BC_1H] \cdot [CB_1H]$ also equals this quantity, and so we see that $\frac{\sin\angle BHC_1 \cdot \sin\angle CHB_1}{AH \cdot A_1H} = \frac{\sin\angle CHA_1 \cdot \sin\angle AHC_1}{BH \cdot B_1H} = \frac{\sin\angle AHB_1 \cdot \sin\angle BHA_1}{CH \cdot C_1H}.$ Intuitively, this result is symmetric under swapping $\triangle ABC$ and $\triangle A_1B_1C_1$ , and doesn't depend upon $\triangle A_1B_1C_1$ being inscribed in $\triangle ABC$ , in the sense that scaling $\triangle ABC$ or $\triangle A_1B_1C_1$ by any factor (with center $H$ ) preserves this property. Thus, this offers an intuitive explanation for why ``swapping'' the triangles preserves the common orthocenter.
It might be possible to adapt this into a phantom-point approach to directly settle the problem, but I don't see how to do that.

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eibc

#19 h Oct 23, 2023, 12:14 AM

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In some terrible notation, let $B_3, C_3$ be the feet of the $B$ and $C$ altitudes in $\triangle ABC$ , and let $B_4, C_4$ be the feet of the $B_1$ and $C_1$ altitudes in $\triangle A_1B_1C_1$ .

Because
$90^{\circ} = \measuredangle B_2B_3B_1 = \measuredangle B_2B_4B_1 = \measuredangle C_1B_4B_1 = \measuredangle C_1C_4B_1 = \measuredangle C_1C_4C_2 = \measuredangle C_1C_3C_2,$ we find that $B_1B_2B_3B_4$ , $C_1C_2C_3C_4$ , and $B_1B_4C_1C_4$ are all concyclic. Thus,
$HB_2 \cdot HB_3 = HB_1 \cdot HB_4 = HC_1 \cdot HC_4 = HC_2 \cdot HC_3,$ so $B_2C_2B_3C_3$ is cyclic. But $BCB_3C_3$ is cyclic too, so $\overline{B_2C_2} \parallel \overline{BC}$ b y Reimand hence $\overline{A_2H} \perp \overline{B_2C_2}$ . Similarly $\overline{B_2H} \perp \overline{C_2A_2}$ and $\overline{C_2H} \perp \overline{A_2B_2}$ , so we are done.

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1137 posts

#20 h Oct 23, 2023, 12:34 AM • 1 Y

Y by Leo.Euler

I WROTE THE POINT NAMES WRONG (in the length calculations). I spent 5 minutes on this (I only had 15 minutes left at the end) then gave up because what I got didn't seem true...

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#21 h Oct 23, 2023, 1:44 AM

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We simply need to prove that $A_2B_2C_2$ and $ABC$ are homothetic about $H$ . Let the foot of the altitude from $A$ be $D$ , the foot of the altitude from $B$ be $E$ , and the foot of the altitude from $C$ be $F$ . Similarly denote $D_1, E_1, F_1$ the same way for $A_1, B_1, C_1$ . It is clear that $D_1, D \in \left( A_1A_2 \right)$ , etc. It is easy to prove that $AH \cdot HD = BH \cdot HE = CH \cdot HF$ . Applying this to $A_1B_1C_1$ , we get the same result. Now simply using this, and Power of a Point we get the desired result.

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#22 h Oct 23, 2023, 2:39 AM • 1 Y

Y by YIYI-JP

Let $D_1$ be the intersection of $A_1H$ and $B_1C_1$ , ie: the foot of the altitude, and define $E_1$ and $F_1$ similarly.

Claim: $A_1B_2D_1E$ is cyclic.

First note that $\angle A_1F_1H = \angle A_1E_1H = \angle A_1DH = 90^\circ \implies A_1DF_1HE_1$ is cyclic. Likewise, $B_1ED_1HF_1$ and $C_1FE_1HD_1$ are cyclic. Thus, $\angle B_2A_1D_1 = \angle E_1A_1H = 90^\circ - \angle E_1HA_1 = 90^\circ - \angle D_1HC_1 = \angle HC_1D_1 = \angle D_1EB_2$ Thus, $A_1B_2D_1E$ is cyclic as desired. $\square$

Likewise, $A_1C_2D_1F$ is cyclic. Note that the intersection of $B_2E$ and $C_2F$ is $H$ which lies on the radical axis. So, $EFB_2C_2$ is cyclic. Clearly, $EFBC$ is cyclic so $B_2C_2 \parallel BC$ by Reim's theorem. From here just apply the same logic to get $C_2A_2 \parallel CA$ and $A_2B_2 \parallel AB$ to get $ABC$ and $A_2B_2C_2$ are homothetic with homothety ah $H$ .

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#23 h Oct 23, 2023, 3:41 AM • 2 Y

Y by OronSH, ihatemath123

Let $DEF$ and $D_1E_1F_1$ be the orthic triangles of $ABC$ and $A_1B_1C_1$ .

Claim: $B_2A_2 \parallel BA$ and cyclic relations
Proof. It suffices to show $\frac{HA_2}{HB_2}=\frac{HA}{HB}$ . But notice that as $A_1DD_1A_2$ is cyclic and $B_1EE_1B_2$ is cyclic by right angles, we have $HD\cdot HA_2=HD_1\cdot HA_1$ and $HE\cdot HB_2=HE_1\cdot HB_1$ . But as $H$ is the orthocenter of $A_1B_1C_1$ , $HA_1\cdot HD_1=HB_1\cdot HE_1$ so $HD\cdot HA_2=HE\cdot HB_2$ or $\frac{HA_2}{HB_2}=\frac{HE}{HD}$ . Then, it's obvious that $HD\cdot HA=HE\cdot HB$ from $ADEB$ cyclic so $\frac{HA_2}{HB_2}=\frac{HE}{HD}=\frac{HA}{HB}$ . $\blacksquare$

Then, as $H,A_2,A,D$ are collinear and $AH\perp BC$ , we have $AH\perp B_2C_2$ so $A_2H\perp B_2C_2$ . Cyclic relations hold, thus we are done.

This post has been edited 1 time. Last edited by GrantStar, Oct 23, 2023, 4:06 AM

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#24 h Oct 23, 2023, 3:53 AM • 2 Y

Y by GrantStar, OronSH

Haha Grant L

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#26 h Oct 23, 2023, 4:43 AM • 3 Y

Y by Assassino9931, p.lazarov06, mxlcv

this is a good problem
(quick?) solution

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#27 h Oct 23, 2023, 8:38 AM

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Solution

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CT17

#28 h Oct 23, 2023, 1:05 PM • 2 Y

Y by blackbluecar, crazyeyemoody907

Revenge or something. I have no idea how I missed this in contest but whatever.

Let $D,E,F$ be the feet from $A,B,C$ in $\triangle ABC$ and let $D_1,E_1,F_1$ be the feet from $A_1,B_1,C_1$ in $\triangle A_1B_1C_1$ . As $C_1C_2F_1F$ and $B_1B_2E_1E$ are cyclic with diameters $C_1C_2$ and $B_1B_2$ respectively, we have

$HC_2\cdot HF = HC_1\cdot HF_1 = HB_1\cdot HE_1 = HB_2\cdot HE$
so $B_2C_2EF$ is cyclic. Hence, $B_2C_2\parallel BC$ by Reim, so $AH\perp B_2C_2$ as desired.

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#29 h Oct 23, 2023, 8:58 PM

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Let $\omega$ be the circle with diameter $AA_1$ and $\Gamma$ be the circle with diameter $A_1A_2$ . Let $A_0$ be the foot of the altitude from $A$ to $\overline{BC}$ , and similarly define $A_0'$ in $\triangle A_1B_1C_1$ . Then by power of a point on $H$ with respect to each of the two circles, we have $\text{Pow}(H, \omega) = HA \cdot HA_0$ (cyclic variants hold) and $\text{Pow}(H, \Gamma) = HA_0' \cdot HA_1 = HA_0 \cdot HA_2$ (cyclic variants hold). Thus, $\frac{\text{Pow}(H, \omega)}{\text{Pow}(H, \Gamma)} = \frac{HA}{HA_2}.$ Since cyclic variants hold for all prior calculations, it is clear that $\frac{HX}{HX_2}$ is constant for $X \in \{A, B, C\}$ . Thus, $\triangle A_2B_2C_2 \sim \triangle ABC$ , and since $\angle B_2HC_2 = \angle BHC = 180^{\circ} - \angle A = 180^{\circ} - \angle A_2,$ $H$ is the orthocenter of $\triangle A_2B_2C_2$ , as desired.

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#30 h Oct 26, 2023, 10:00 AM • 9 Y

Y by YIYI-JP, LoloChen, eibc, CyclicISLscelesTrapezoid, starchan, David-Vieta, crazyeyemoody907, TestX01, Aryan-23

Vectors.
$\begin{aligned} \overrightarrow{AH}\cdot \overrightarrow{B_2C_2} &= \overrightarrow{AH}\cdot (\overrightarrow{B_2A_1} + \overrightarrow{A_1C_2}) \\ &= \overrightarrow{AB_1}\cdot \overrightarrow{B_2A_1} + \overrightarrow{AC_1}\cdot \overrightarrow{A_1C_2} \\ &= \overrightarrow{AB_1}\cdot \overrightarrow{HA_1} + \overrightarrow{AC_1}\cdot \overrightarrow{A_1H} \\ &= \overrightarrow{A_1H}\cdot \overrightarrow{B_1C_1} \\ &= 0\end{aligned}$

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#31 h Oct 26, 2023, 1:28 PM • 6 Y

Y by ihatemath123, David-Vieta, IAmTheHazard, The_Great_Learner, CyclicISLscelesTrapezoid, TestX01

So here is my unique cute bash

:
Set $O(0, 0, 0), A(a, 0, 0), B(0, b, 0), C(0, 0, c)$ in the $x-y-z$ coordinate, so $OH \perp$ face $ABC$ .
Let $C_1(p_1a, p_2b, 0), A_1(0, q_1b, q_2c), B_1(r_2a, 0, r_1c)$ , where $p_1+p_2=q_1+q_2=r_1+r_2=1$ .
We still have $A_1B_1 \perp OC$ in 3D space, so $\overrightarrow{A_1B_1}\cdot \overrightarrow{OC}=0$ , which is $p_1ar_2a-p_2bq_1b=0$ , so $p_1r_2a^2=q_1p_2b^2=r_1q_2c^2=k$ .
Note that the legal vector of face $ABC$ is $i=(\frac{1}{a},\frac{1}{b},\frac{1}{c})$ , let $B_3=(0, b'b, 0)$ on line $OB$ such that $B_2B_3 \parallel i$ , which is face $A_1C_1B_3 \parallel i$ .
This means that ${i}$ is linearly related with $\overrightarrow{A_1B_3}=(0, (q_1-b')b, q_2c)$ and $\overrightarrow{C_1B_3}=(p_1a, (p_2-b')b, 0)$ , which indicates:
$\frac{(q_1-b')b}{q_2c^2}+\frac{(p_2-b')b}{p_1a^2}=\frac{1}{b^2}$ Combining $\frac{1}{p_1a^2}+\frac{1}{q_2c^2}=\frac{r_2}{k}+\frac{r_1}{k}=\frac{1}{k}$ , $b'=\frac{-\frac{1}{b^2}+\frac{p_2}{p_1a^2}+\frac{q_1}{q_2c^2}}{\frac{1}{p_1a^2}+\frac{1}{q_2c^2}}=1-\frac{k}{a^2}-\frac{k}{b^2}-\frac{k}{c^2}$ .
If we define $A_3, C_3, a', c'$ similarly, we will find $a'=b'=c'=1-\frac{k}{a^2}-\frac{k}{b^2}-\frac{k}{c^2}$ , so $\triangle A_3B_3C_3$ is homothetic to $\triangle {ABC}$ , and so is $\triangle A_2B_2C_2$ , thus it has orthocenter ${H}$ .

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Reason: Add a graph

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#32 h Nov 5, 2023, 7:31 PM

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Storage

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#33 h Apr 3, 2024, 12:01 AM • 3 Y

Y by CyclicISLscelesTrapezoid, MarkBcc168, crazyeyemoody907

Quicker presentation of #30. Solved with mira74.

Let $H=0$ . Note that $A\cdot B_2=B_1\cdot B_2=B_1\cdot C_1=C_1\cdot C_2=A\cdot C_2,$ as desired.

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#34 h Aug 28, 2024, 10:27 PM

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Subjective Rating (MOHs) $\text{[math]}$

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#36 h Mar 31, 2025, 12:53 AM

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Let $D, E, F,$ be the feet of the altitudes from $A, B, C,$ wrt. $\triangle ABC$ and let $X, Y, Z,$ be the feet of the altitudes from $A_1, A_2, A_3,$ wrt. $\triangle A_1 A_2 A_3$ .

Clearly $\angle HXA_1 = 90^\circ = \angle HDA_1$ so $A_2XDA_1$ is cyclic so PoP yields
$HA_2 \cdot HD = HX \cdot HA_1 \implies \dfrac{HA_2}{HA} = \dfrac{HA_1 \cdot HX}{HA \cdot HD}$ but by PoP we obviously have
$HA \cdot HD = HB \cdot HE = HC \cdot HF \qquad HA_1 \cdot HX = HB_1 \cdot HY = HC_1 \cdot HZ$ thus $\dfrac{HA_2}{HA} = \dfrac{HB_2}{HB} = \dfrac{HC_2}{HC}$ . So $H$ is the homothetic center sending $\triangle ABC \to \triangle A_2 B_2 C_2$ so it is also the orthocenter of $\triangle A_2 B_2 C_2$ as needed.

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