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Another perpendicular to the Euler line

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Source: German TST 2022, exam 2, problem 3

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#1 h Mar 11, 2022, 1:01 PM • 11 Y

Y by megarnie, BVKRB-, buratinogigle, Sprites, TheCollatzConjecture, Hedra, HrishiP, HWenslawski, livetolove212, amar_04, RedFlame2112

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Let $P$ be a point in the plane such that $AP \perp BC$ . Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$ , respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$ . Let $Z$ be the orthogonal projection of $Q$ onto $AB$ . Assume that $H \neq O$ and $Y \neq Z$ . Prove that $YZ \perp HO$ .

$[asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]$

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9801 posts

jayme

#2 h Mar 11, 2022, 1:47 PM • 3 Y

Y by buratinogigle, amar_04, RedFlame2112

Dear Darij,
very nice to hear You again on Mathlinks (AoPS)...

Sincerely
Jean-Louis

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6555 posts

#3 h Mar 11, 2022, 1:54 PM • 2 Y

Y by amar_04, RedFlame2112

Nice to see you again, Jean-Louis. I checked with your collection before posing this problem

This post has been edited 2 times. Last edited by darij grinberg, Mar 11, 2022, 1:55 PM

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799786

#5 h Mar 11, 2022, 2:52 PM • 1 Y

Y by RedFlame2112

Let $P$ varies on the plane. We only need to consider when segment $AP$ cuts segment $BC$ .
Claim. (main claim) $YZ$ is perpendicular to a fixed line.
Proof
Then the rest is to move $P$ to a special point, then we are done!

(I don't think this is easy)

This post has been edited 2 times. Last edited by 799786, Mar 11, 2022, 2:58 PM

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#6 h Mar 11, 2022, 2:58 PM • 1 Y

Y by RedFlame2112

wardtnt1234 wrote:

Then the rest is to move $P$ to a special point, then we are done!

This is easier said than done!

Checking that $YZ$ is parallel for all points $P$ satisfying $AP \perp BC$ is actually pretty easy without any trigonometry; just observe that if $P$ undergoes a homothety with center $A$ , then $Q$ , $R$ , $Y$ and $Z$ undergo the same homothety. But I don't know of any choice of $P$ (other than $P = A$ , which is too degenerate to be useful) that makes the problem significantly simpler.

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1052 posts

799786

#7 h Mar 11, 2022, 3:00 PM • 1 Y

Y by RedFlame2112

@above I have the same thought. I'm trying to find a good choice of position $P$ .
(The best position so far I found is $P \equiv H$ )

This post has been edited 1 time. Last edited by 799786, Mar 11, 2022, 3:01 PM

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#8 h Mar 11, 2022, 3:15 PM • 1 Y

Y by RedFlame2112

By the way, it seems that this can be complex bashed (picking $o=0$ , $h=a+b+c$ ), though I will try this later.

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3 posts

Kvon

#9 h Mar 11, 2022, 3:58 PM • 2 Y

Y by chystudent1-_-, RedFlame2112

First of all we define S= RY \cap QZ and X is the the projection of P onto AC (or the midpoint of PQ) and W is the projection of P onto AB. Now notice that PRSQ is a parallelogram and AXPW is a cyclic quad which also means that AXW is similar to ABC.

Let O' and H' be the cirumcenter and orthocenter of AXW. Now notice that O' is the midpoint of AP and H' is the midpoint of PS. To see the latter note that RS is parallel to WH' and PR is parallel to XH'.

Now we know that AS is parallel to O'H'. Now by an easy angle chase, we get:

90° - \angle AYZ=\angle ZYS = \angle ZAS=\angle BAS = \angle (O'H';AB)=\angle (OH;AC)

This means YZ \perp OH and we are finished.

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6555 posts

#10 h Mar 11, 2022, 6:16 PM • 1 Y

Y by RedFlame2112

Nicely done, Kvon! This is the second of my proposed solutions.

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1052 posts

799786

#11 h Mar 12, 2022, 4:04 AM • 1 Y

Y by RedFlame2112

Continue from #7, let $P \equiv H$ then do like #9
Or try this: (also let $P \equiv H$ )
Let $AH$ cuts $BC$ at $D$ . $DM$ cuts $AC$ at $C'$ , $DN$ cuts $AB$ at $B'$ then use Thales.

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jayme

#12 h Mar 12, 2022, 6:52 AM • 1 Y

Y by RedFlame2112

Dear Mathlinkers,

1. J, K the midpoints of AC, AB
2. V, W the feet of the perpendiculars to AC, AB though H.

If we prove that AY/AZ = KW/JV we are done...

Sincerely
Jean-Louis

This post has been edited 1 time. Last edited by jayme, Mar 12, 2022, 7:09 AM
Reason: typo

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jayme

#13 h Mar 12, 2022, 9:50 AM • 1 Y

Y by RedFlame2112

Dear Mathlinkers,

1. (QR) has a fix direction
2. we chioce for P the foot A' of the A-altitude. or H
3. J, K the midpoints of AC, AB
4. V, W the feet of the perpendiculars to AC, AB though A' or H;

one of these two points lead to a not so heavy calculation...

If we prove that AY/AZ = KW/JV we are done...

Sincerely
Jean-Louis

This post has been edited 1 time. Last edited by jayme, Mar 12, 2022, 10:00 AM
Reason: typos

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khina

#14 h Mar 13, 2022, 6:05 PM • 1 Y

Y by RedFlame2112

nice

motivation

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#15 h Mar 13, 2022, 10:09 PM • 1 Y

Y by RedFlame2112

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885; /* image dimensions */ /* draw figures */ draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); /* dots and labels */ dot((-4.356336480424067,10.601149161007438),dotstyle); label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); dot((-11.291822257619955,-6.524184426443613),dotstyle); label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); dot((9.757778495559513,-6.875988622852955),dotstyle); label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$ .

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$ .
From here we have that $p=-bc$ .

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$ , respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$ , where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$ .

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$ , or we could say $q=\overline{r}$ , which is going to help us in our calculations since from here we have that:
$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$
From here we see that:
$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$ implying that $YZ \perp OH$ , hence we are done

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8082 posts

jmiao

#16 h Mar 13, 2022, 10:15 PM • 1 Y

Y by RedFlame2112

jayme wrote:

Dear Mathlinkers,

Sorry if this is taken out of context, but what do you mean by Mathlinkers?

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i3435

#17 h Mar 13, 2022, 10:51 PM • 1 Y

Y by RedFlame2112

Let $O_A$ be the reflection of $O$ over $\overline{BC}$ . Let $E,F$ be the feet from $P$ to $\overline{AC},\overline{AB}$ respectively. $EF=FY=EZ$ . Let $P_{1}=\overline{CO}\cap\overline{BO_A}$ (a point at infinity), and let $P_2=\overline{BO}\cap\overline{CO_A}$ . Then $(\overline{HO},\overline{HO_A});(\overline{HB},\overline{HC});(\overline{HP_1},\overline{HP_2})$ is an involution by DDIT. Let $m$ be the line at infinity. Then projecting from $H$ to $m$ , $(\overline{HO}\cap m,\overline{HO_A}\cap m);(\overline{HB}\cap m,\overline{HC}\cap m);(P_1,P_2)$ is an involution. $(\overline{YZ}\cap m,\overline{FE}\cap m);(\overline{EY}\cap m,\overline{FZ}\cap m);(\overline{FY}\cap m,\overline{EZ}\cap m)$ is also an involution by DIT. However $\overline{FE}\perp\overline{HO_A},\overline{EY}\perp\overline{HB},\overline{FZ}\perp\overline{HC},\overline{FY}\perp\overline{CO},\overline{EZ}\perp\overline{BO}$ by angle chasing, so $\overline{YZ}\perp\overline{HO}$ as desired.

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859 posts

#18 h Mar 14, 2022, 4:35 AM • 2 Y

Y by amar_04, RedFlame2112

Nice problem!
I checked other special cases when $P$ moves on fixed line through $A$ and get 2 similar problems:
1) If $AP\perp OH$ then $YZ\perp BC.$
2) If $AP$ passes through nine-point circle's center then $YZ\parallel BC.$

Attachments:

This post has been edited 1 time. Last edited by livetolove212, Mar 14, 2022, 4:35 AM

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jayme

#19 h Mar 14, 2022, 3:55 PM • 1 Y

Y by RedFlame2112

Dear Darij,
has this nice problem an author?

Can you send me a scan of this problem if you have some time at <(jeanlouisayme@yahoo.fr>...

Very sincerely
Jean-Louis

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9801 posts

jayme

#20 h Mar 21, 2022, 10:02 AM • 2 Y

Y by darij grinberg, RedFlame2112

Dear Mathlinkers,

here

then

La perpendiculaire de Darij Grinberg

Sincerely
Jean-Louis

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527 posts

#21 h Apr 5, 2022, 9:08 PM • 1 Y

Y by RedFlame2112

Dear Darij, J-L, and MLs

Construct two perp lines at each vertex of ABC to the sides at that vertex.
We get two triangles with sides orthogonal to sides of ABC.
Thus their Euler Lines are perp to EL of ABC.
Now when a side of a triangle moves parallel to itself EL of this triangle also
moves parallel to itself.

In this problem we move sides of the two triangles orthogonal to ABC.

Friendly,
M.T.

Darij, is there a connect to your paper with Nikos' solution and JPE's
note about directrix of hyperbola? BTW, where are Nikos and JPE
these days?

This post has been edited 1 time. Last edited by armpist, Apr 5, 2022, 9:19 PM

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#22 h Apr 5, 2022, 9:19 PM • 1 Y

Y by RedFlame2112

@armpist: What is the exact connection between the two orthogonal triangles and the line $YZ$ ? How do you make $YZ$ the Euler line of any triangle? (Incidentally, when I was thinking up problems for this exam, one of the things I looked are were these two orthogonal triangles... but I didn't see any relation.)

PS. I just recalled that you asked me a while ago about Thebault's isogonal triangles theorem, but when I wanted to respond, your email wasn't active any more. Here is a recent reference with a very nice proof (using transformations, but it shouldn't be hard to rewrite it in purely Euclidean terms): Waldemar Pompe, Three reflections, 2021-12-19. This was, of course, not the proof I've had in mind, but I doubt mine was any simpler.

This post has been edited 1 time. Last edited by darij grinberg, Apr 5, 2022, 9:20 PM

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Imposter-xDDDDD

3 posts

Imposter-xDDDDD

#23 h Dec 8, 2022, 6:58 PM • 1 Y

Y by Mango247

EulersTurban wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885; /* image dimensions */ /* draw figures */ draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); /* dots and labels */ dot((-4.356336480424067,10.601149161007438),dotstyle); label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); dot((-11.291822257619955,-6.524184426443613),dotstyle); label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); dot((9.757778495559513,-6.875988622852955),dotstyle); label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$ .

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$ .
From here we have that $p=-bc$ .

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$ , respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$ , where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$ .

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$ , or we could say $q=\overline{r}$ , which is going to help us in our calculations since from here we have that:
$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$
From here we see that:
$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$ implying that $YZ \perp OH$ , hence we are done

why did u define P? If u define P,your solution is only the case of P to be on $(ABC)$ .Can u explain clearly,pls?

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#24 h Mar 20, 2023, 2:25 PM

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Imposter-xDDDDD wrote:

EulersTurban wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885; /* image dimensions */ /* draw figures */ draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); /* dots and labels */ dot((-4.356336480424067,10.601149161007438),dotstyle); label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); dot((-11.291822257619955,-6.524184426443613),dotstyle); label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); dot((9.757778495559513,-6.875988622852955),dotstyle); label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$ .

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$ .
From here we have that $p=-bc$ .

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$ , respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$ , where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$ .

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$ , or we could say $q=\overline{r}$ , which is going to help us in our calculations since from here we have that:
$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$
From here we see that:
$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$ implying that $YZ \perp OH$ , hence we are done

why did u define P? If u define P,your solution is only the case of P to be on $(ABC)$ .Can u explain clearly,pls?

Intuitively, Y and Z move linearly at the same velocity as a function of P, and moreover they coincide at A when P=A. This means that, for any choice of P, the line YZ will always have the same slope.

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#25 h Sep 30, 2023, 3:26 PM

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My student's solution:
https://youtu.be/d4B2KZXWV2w

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jkmmm3

#26 h Oct 3, 2023, 9:51 PM

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It is clear that we just need to solve this problem when $P = H$ . Then by homothety, $\angle AZY$ will be constant so $ZY \perp OH$ no matter our choice of $P$ .
Since $Q$ and $R$ are the reflections of $H$ across lines $AB$ and $AC$ respectively, they're both on the circumcircle of $\triangle ABC$ . We now define points $D$ and $E$ to be the foot of the perpendicular from $C$ to $AB$ and $B$ to $AC$ respectively. Now, let $D'$ the foot of the perpendicular from $D$ to $AC$ and $E'$ be the foot of the perpendicular from $E$ to $AB$ . Because $HD = DR$ , It follows that $ED' = D'Y$ because $RY \parallel D'D \parallel EH$ .

We utilize cartesian coordinates where $A = (0, 0), B = (b,d), C = (c,0)$ . The orthocenter is just $H = (b, \frac{bc - b^2}{d})$ by power of point. $OH$ is just the Euler line, so we will use the more convenient centroid $G = (\frac{b+c}{3}, \frac{d}{3})$ .

We first calculate $Y$ . The line $CD$ has the equation $y = -\frac{b}{d} (x-c)$ , while the line $AB$ has the equation $y = \frac{c-b}{d}x$ . Solving this system of equations, we get that $D = (\frac{cb^2}{b^2+d^2}, \frac{cbd}{b^2+d^2})$ . Now, $AC$ is just the x-axis, so $D' = (\frac{cb^2}{b^2+d^2})$ . Reflecting $E = (b,0)$ across $D'$ , we get that $Y = (\frac{2cb^2 - b^3 - bd^2}{b^2+d^2}, 0)$ .

We now calculate $Z$ . Because $EE' \parallel CD$ , $\frac{AE'}{AD} = \frac{AE}{AC} = \frac{b}{c}$ . So, $\vec{E}' = \frac{b}{c} \vec{D} = (\frac{b^3}{b^2+d^2}, \frac{b^2d}{b^2+d^2})$ . Reflecting $D$ across $E'$ , we can get that $Z = (\frac{(2b-c)b^2}{b^2+d^2}, \frac{(2b-c)bd}{b^2+d^2})$ .

The slope of the line $XY$ can then be calculated to be $\frac{(2b-c)d}{3b^2 - 3bc + d^2}$ . We now calculate the slope of the Euler line $GH$ , which is just $\frac{\frac{3bc - 3b^2 - d^2}{3d}}{\frac{2b - c}{3}} = \frac{3bc - 3b^2 - d^2}{(2b-c)d}$ . Multiplying these slopes results in $-1$ , which finishes.

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1556 posts

#27 h Apr 30, 2025, 2:42 AM

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Absolutely beautiful problem.
Let $PQ,PR$ meet $AC,AB$ at $V,W$ respectively then let $RY \cap QZ=S$ , notice $SRPQ$ is a parallelogram, but also that $BWVC$ is cyxlic from antiparallels or some $\sqrt{bc}$ +homothety, let $M,N$ midpoints of $AP,SP$ then $M,N$ are in fact circumcenter, orthocenter of $\triangle AVW$ respectively so now notice internal angle bisector of the angle $\angle (MN,OH)$ is parallel to the external angle bisector of $\angle BAC$ , this is due to anti-parallels transform and so if you consider $\infty_{\perp YZ}$ then $A\infty_{\perp YZ}$ is symetric to $AS$ on the external angle bisector of $\angle BAC$ but as a result it is parallel to $OH$ thus $YZ \perp OH$ as desired thus we are done :cool:

:cool:

.

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