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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Many Equal Sides
mathisreal   3
N 2 minutes ago by QueenArwen
Source: Brazil EGMO TST 2023 #1
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$. Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$. Prove that $DX=YD$.
3 replies
mathisreal
Nov 10, 2022
QueenArwen
2 minutes ago
LOTS of recurrence!
SatisfiedMagma   4
N 5 minutes ago by Reacheddreams
Source: Indian Statistical Institute Entrance UGB 2023/5
There is a rectangular plot of size $1 \times n$. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size $1 \times 1$, the blue tiles are of size $1 \times 1$ and the black tiles are of size $1 \times 2$. Let $t_n$ denote the number of ways this can be done. For example, clearly $t_1 = 2$ because we can have either a red or a blue tile. Also $t_2 = 5$ since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

[list=a]
[*]Prove that $t_{2n+1} = t_n(t_{n-1} + t_{n+1})$ for all $n > 1$.

[*]Prove that $t_n = \sum_{d \ge 0} \binom{n-d}{d}2^{n-2d}$ for all $n >0$.
[/list]
Here,
\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]for integers $m,r$.
4 replies
SatisfiedMagma
May 14, 2023
Reacheddreams
5 minutes ago
combi/nt
blug   1
N 6 minutes ago by blug
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
1 reply
blug
Yesterday at 3:37 PM
blug
6 minutes ago
Inequality, inequality, inequality...
Assassino9931   9
N 14 minutes ago by ZeroHero
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
9 replies
Assassino9931
Today at 9:38 AM
ZeroHero
14 minutes ago
No more topics!
Another perpendicular to the Euler line
darij grinberg   25
N Apr 30, 2025 by MathLuis
Source: German TST 2022, exam 2, problem 3
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 $YZ \perp HO$.

IMAGE
25 replies
darij grinberg
Mar 11, 2022
MathLuis
Apr 30, 2025
Another perpendicular to the Euler line
G H J
Source: German TST 2022, exam 2, problem 3
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darij grinberg
6555 posts
#1 • 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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jayme
9792 posts
#2 • 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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darij grinberg
6555 posts
#3 • 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
1052 posts
#5 • 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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darij grinberg
6555 posts
#6 • 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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799786
1052 posts
#7 • 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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VicKmath7
1389 posts
#8 • 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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Kvon
3 posts
#9 • 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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darij grinberg
6555 posts
#10 • 1 Y
Y by RedFlame2112
Nicely done, Kvon! This is the second of my proposed solutions.
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799786
1052 posts
#11 • 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.
This post has been edited 1 time. Last edited by 799786, Mar 12, 2022, 4:08 AM
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jayme
9792 posts
#12 • 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
9792 posts
#13 • 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
994 posts
#14 • 1 Y
Y by RedFlame2112
nice

motivation
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EulersTurban
386 posts
#15 • 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 :D
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jmiao
8083 posts
#16 • 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
1350 posts
#17 • 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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livetolove212
859 posts
#18 • 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
9792 posts
#19 • 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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jayme
9792 posts
#20 • 2 Y
Y by darij grinberg, RedFlame2112
Dear Mathlinkers,

here

then

La perpendiculaire de Darij Grinberg

Sincerely
Jean-Louis
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armpist
527 posts
#21 • 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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darij grinberg
6555 posts
#22 • 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.
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Imposter-xDDDDD
3 posts
#23 • 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 */

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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)); 
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draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
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label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); 
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label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); 
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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); 
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[/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 :D
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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mathleticguyyy
3217 posts
#24
Y by
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)); 
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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); 
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label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); 
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label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); 
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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); 
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[/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 :D
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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mikestro
74 posts
#25
Y by
My student's solution:
https://youtu.be/d4B2KZXWV2w
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jkmmm3
56 posts
#26
Y by
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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MathLuis
1524 posts
#27
Y by
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:.
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