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][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][/asy]
Problem
Source: German TST 2022, exam 2, problem 3
Tags: geometry, Euler Line, Triangle Geometry, circumcircle, geometric transformation, reflection
11.03.2022 16:47
Dear Darij, very nice to hear You again on Mathlinks (AoPS)... Sincerely Jean-Louis
11.03.2022 16:54
Nice to see you again, Jean-Louis. I checked with your collection before posing this problem
11.03.2022 17:52
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.
Then the rest is to move $P$ to a special point, then we are done! (I don't think this is easy)
11.03.2022 17:58
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.
11.03.2022 18:00
@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 $P \equiv H$)
11.03.2022 18:15
By the way, it seems that this can be complex bashed (picking $o=0$, $h=a+b+c$), though I will try this later.
11.03.2022 18:58
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.
11.03.2022 21:16
Nicely done, Kvon! This is the second of my proposed solutions.
12.03.2022 07:04
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.
12.03.2022 09:52
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
12.03.2022 12:50
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
13.03.2022 21:05
14.03.2022 01:09
[asy][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][/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
14.03.2022 01:15
jayme wrote: Dear Mathlinkers, Sorry if this is taken out of context, but what do you mean by Mathlinkers?
14.03.2022 01:51
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.
14.03.2022 07:35
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:


14.03.2022 18:55
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
21.03.2022 13:02
Dear Mathlinkers, here then La perpendiculaire de Darij Grinberg Sincerely Jean-Louis
06.04.2022 00:08
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?
06.04.2022 00:19
@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.
08.12.2022 21:58
EulersTurban wrote: [asy][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][/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?
20.03.2023 17:25
Imposter-xDDDDD wrote: EulersTurban wrote: [asy][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][/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.
30.09.2023 18:26
My student's solution: https://youtu.be/d4B2KZXWV2w
04.10.2023 00:51
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.