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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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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Both a and a+1997 are roots of P, Q(P(x))=1 has no solutions
WakeUp   2
N 35 minutes ago by Rohit-2006
Source: Baltic Way 1997
Let $P$ and $Q$ be polynomials with integer coefficients. Suppose that the integers $a$ and $a+1997$ are roots of $P$, and that $Q(1998)=2000$. Prove that the equation $Q(P(x))=1$ has no integer solutions.
2 replies
WakeUp
Jan 28, 2011
Rohit-2006
35 minutes ago
greatest volume
hzbrl   1
N 38 minutes ago by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
1 reply
hzbrl
Yesterday at 9:56 AM
hzbrl
38 minutes ago
Gheorghe Țițeica 2025 Grade 9 P2
AndreiVila   5
N 40 minutes ago by sqing
Source: Gheorghe Țițeica 2025
Let $a,b,c$ be three positive real numbers with $ab+bc+ca=4$. Find the minimum value of the expression $$E(a,b,c)=\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}-(a-b)^2.$$
5 replies
+1 w
AndreiVila
Mar 28, 2025
sqing
40 minutes ago
Geometry Parallel Proof Problem
CatalanThinker   4
N an hour ago by CatalanThinker
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
4 replies
CatalanThinker
2 hours ago
CatalanThinker
an hour ago
Circle and square
Marrelia   1
N Apr 10, 2025 by sunken rock
Given a circle with center $O$, and square $ABCD$. Point $A$ and $B$ are on the circle, and $CD$ is tangent to the circle at point $E$. Let $M$ represent the midpoint of $AD$ and $F$ represent the intersection between $AD$ and circle. Prove that $MF = FD$.
1 reply
Marrelia
Apr 10, 2025
sunken rock
Apr 10, 2025
Two geometry and algebra problems
Kempu33334   5
N Apr 9, 2025 by joeym2011
Here are two problems I made...

1) Let there be a triangle $ABC$ such that $AB = 5$, $AC = 6$, and $BC = 7$. Then, let the circumcenter of $\triangle ABC$ be $O$. Furthermore, let the reflections of $A$, $B$, and $C$ across $O$, be $A'$, $B'$, and $C'$ respectively. Find $[AB'CA'BC']$.

2) If \[\underbrace{\dfrac{v_2(4!)}{v_2(8!)}\cdot\dfrac{v_2(16!)}{v_2(32!)}\dots}_\text{50 terms} = \dfrac{1}{2^n}\]find $\left\lceil \dfrac{n}{3} \right\rceil$.

@2below Yes, that’s the intended product.
5 replies
Kempu33334
Apr 8, 2025
joeym2011
Apr 9, 2025
Simple Combinatorics on seating in a circle
duttaditya18   6
N Mar 6, 2025 by S_14159
Five persons wearing badges with numbers $1, 2, 3, 4, 5$ are seated on $5$ chairs around a circular table. In how many ways can they be seated so that no two persons whose badges have consecutive numbers are seated next to each other? (Two arrangements obtained by rotation around the table are considered different)
6 replies
duttaditya18
Aug 11, 2019
S_14159
Mar 6, 2025
Ellipse not parallel to x- or y-axis
chess64   9
N Mar 4, 2025 by daijobu
An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis?
9 replies
chess64
Jan 1, 2006
daijobu
Mar 4, 2025
[PMO25 Qualifying I.14] Grid Division
kae_3   0
Feb 23, 2025
How many ways are there to divide a $5\times5$ square into three rectangles, all of whose sides are integers? Assume that two configurations which are obtained by either a rotation and/or a reflection are considered the same.

$\text{(a) }10\qquad\text{(b) }12\qquad\text{(c) }14\qquad\text{(d) }16$

Answer Confirmation
0 replies
kae_3
Feb 23, 2025
0 replies
Reflection Geometry
Kempu33334   1
N Feb 17, 2025 by Kempu33334
Let there be a triangle $ABC$, and a point $P$. In addition, let the reflections of $P$ across the sides of $\triangle ABC$ be $X,Y,Z$.

1) Prove that $P$ is in the interior of $\triangle XYZ$ if and only if it is in the interior of $\triangle ABC$.

2.1) Prove that it is impossible for $X$, $Y$, $Z$ to all lie inside (not on) $(ABC)$ or provide a counterexample.

2.2) Prove that it is impossible for $A$, $B$, $C$ to all lie inside (not on) $(XYZ)$ or provide a counterexample.

3) Let all points $P$ such that they lie on a singular circle be the set $\mathcal{P}$. Prove that for every point in $\mathcal{P}$, the locus of the corresponding $X$, $Y$, $Z$ are circles.

4) Let all points $P$ such that they lie on a singular line be the set $\mathcal{L}$. Prove that for every point in $\mathcal{L}$, the locus of the corresponding $X$, $Y$, $Z$ is a line.

5.1) Let all points $P$ such that they lie on a singular ellipse be the set $\mathcal{E}$. Prove that for every point in $\mathcal{E}$, the locus of the corresponding $X$, $Y$, $Z$ is an ellipse.

5.2) Let all points $P$ such that they lie on a singular ellipse be the set $\mathcal{E}$. Prove that for every point in $\mathcal{E}$, the locus of the corresponding $X$, $Y$, $Z$ is an ellipse congruent to $\mathcal{E}$.

Note, 3) and 4) are corollaries of 5), so proving that suffices.

6) Generalize.
1 reply
Kempu33334
Feb 16, 2025
Kempu33334
Feb 17, 2025
Geometry
hn111009   0
Feb 17, 2025
Let triangle $ABC.$ $K$ and $H$ be the reflection of $B$ through $CA$ and $C$ through $BA$. Let $E$ be the center of Euler circle of triangle $ABC.$ Prove that $AE$ passed through the center of $\odot(AKH).$
0 replies
hn111009
Feb 17, 2025
0 replies
intermediate algebra
ANGEW   2
N Jan 28, 2025 by ANGEW
Find all pairs of real numbers (a, b) such that (x - a)^2 + (2x - b)^2 = (x - 3)^2 + (2x)^2 for all x.
The answer is (a,b)=(3,0) and (-9/5, 12/5)
By graphing y=2x and plotting these two points, we can see that the two points are reflections upon the line y=2x. Why?
2 replies
ANGEW
Jan 22, 2025
ANGEW
Jan 28, 2025
Questions about angle notation
tsun26   2
N Jan 13, 2025 by AbhayAttarde01
What's the difference between writing $\angle BAL \cong \angle DAC$ and $\angle BAL = \angle DAC$?
2 replies
tsun26
Jan 13, 2025
AbhayAttarde01
Jan 13, 2025
Properties of Reflections
Dachel2018   3
N Dec 28, 2024 by Mathzeus1024
In Cartesian coordinates, prove that it is impossible to map the point $(0,0)$ to the point $(0,1)$ only by using reflections through lines passing through at least 2 lattice points. Clarifications
3 replies
Dachel2018
Dec 27, 2024
Mathzeus1024
Dec 28, 2024
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]
Z K Y
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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
Z K Y
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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.
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
#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)); 
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)); 
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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((-11.291822257619955,-6.524184426443613),dotstyle); 
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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); 
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
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); 
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label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); 
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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
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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N Quick Reply
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