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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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Angle Relationships in Triangles
steven_zhang123   1
N 6 minutes ago by Double07
In $\triangle ABC$, $AB > AC$. The internal angle bisector of $\angle BAC$ and the external angle bisector of $\angle BAC$ intersect the ray $BC$ at points $D$ and $E$, respectively. Given that $CE - CD = 2AC$, prove that $\angle ACB = 2\angle ABC$.
1 reply
steven_zhang123
Yesterday at 11:09 PM
Double07
6 minutes ago
IMO 2010 Problem 1
canada   121
N 12 minutes ago by maromex
Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$

Proposed by Pierre Bornsztein, France
121 replies
canada
Jul 7, 2010
maromex
12 minutes ago
Nice original fe
Rayanelba   7
N 21 minutes ago by Rayanelba
Source: Original
Find all functions $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ that verfy the following equation :
$P(x,y):f(x+yf(x))+f(f(x))=f(xy)+2x$
7 replies
Rayanelba
4 hours ago
Rayanelba
21 minutes ago
Israeli Mathematical Olympiad 1995
YanYau   26
N 21 minutes ago by zuat.e
Source: Israeli Mathematical Olympiad 1995
Let $PQ$ be the diameter of semicircle $H$. Circle $O$ is internally tangent to $H$ and tangent to $PQ$ at $C$. Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB\perp PQ$ and is tangent to $O$. Prove that $AC$ bisects $\angle PAB$
26 replies
YanYau
Apr 8, 2016
zuat.e
21 minutes ago
4-vars inequality
xytunghoanh   1
N 28 minutes ago by xytunghoanh
For $a,b,c,d \ge 0$ and $a\ge c$, $b \ge d$. Prove that
$$a+b+c+d+ac+bd+8 \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})$$.
1 reply
xytunghoanh
2 hours ago
xytunghoanh
28 minutes ago
ARO 2011 11-8
sartt   19
N 31 minutes ago by Double07
Let $N$ be the midpoint of arc $ABC$ of the circumcircle of triangle $ABC$, let $M$ be the midpoint of $AC$ and let $I_1, I_2$ be the incentres of triangles $ABM$ and $CBM$. Prove that points $I_1, I_2, B, N$ lie on a circle.

M. Kungojin
19 replies
sartt
May 3, 2011
Double07
31 minutes ago
APMO 2017: (ADZ) passes through M
BartSimpsons   78
N an hour ago by Ihatecombin
Source: APMO 2017, problem 2
Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.

Olimpiada de Matemáticas, Nicaragua
78 replies
BartSimpsons
May 14, 2017
Ihatecombin
an hour ago
Inspired by Baltic Way 2005
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 , a+b+(a+b)^2=6$. Prove that
$$ \frac {a}{b+2}+\frac {b}{a+2}+\frac {1}{ab+2}\leq \frac{3}{2} $$Let $ a,b,c>0 , a+b+(a-b)^2=2$. Prove that
$$ \frac {a}{b+2}+\frac {b}{a+2}+\frac {1}{ab+2}\leq 1 $$Let $ a,b,c>0 , a+b+a^2+b^2=4$. Prove that
$$ \frac {a}{b+2}+\frac {b}{a+2}+\frac {1}{ab+2}\leq \frac{1+\sqrt{17}}{4} $$Let $ a,b,c>0 , a+b+a^2+b^2+ab=5$. Prove that
$$ \frac {a}{b+2}+\frac {b}{a+2}+\frac {1}{ab+2}\leq \frac{1+\sqrt{21}}{4} $$
5 replies
sqing
4 hours ago
sqing
an hour ago
JBMO TST Bosnia and Herzegovina 2022 P3
Motion   8
N an hour ago by AylyGayypow009
Source: JBMO TST Bosnia and Herzegovina 2022
Let $ABC$ be an acute triangle. Tangents on the circumscribed circle of triangle $ABC$ at points $B$ and $C$ intersect at point $T$. Let $D$ and $E$ be a foot of the altitudes from $T$ onto $AB$ and $AC$ and let $M$ be the midpoint of $BC$. Prove:
A) Prove that $M$ is the orthocenter of the triangle $ADE$.
B) Prove that $TM$ cuts $DE$ in half.
8 replies
Motion
May 21, 2022
AylyGayypow009
an hour ago
A sharp one with 3 var
mihaig   2
N an hour ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
2 replies
mihaig
Tuesday at 7:20 PM
mihaig
an hour ago
IMO Shortlist 2014 N1
hajimbrak   46
N an hour ago by cursed_tangent1434
Let $n \ge 2$ be an integer, and let $A_n$ be the set \[A_n = \{2^n  - 2^k\mid k \in \mathbb{Z},\, 0 \le k < n\}.\] Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_n$ .

Proposed by Serbia
46 replies
hajimbrak
Jul 11, 2015
cursed_tangent1434
an hour ago
Find all p(x) such that p(p) is a power of 2
truongphatt2668   2
N an hour ago by truongphatt2668
Source: ???
Find all polynomial $P(x) \in \mathbb{R}[x]$ such that:
$$P(p_i) = 2^{a_i}$$with $p_i$ is an $i$ th prime and $a_i$ is an arbitrary positive integer.
2 replies
truongphatt2668
3 hours ago
truongphatt2668
an hour ago
Interesting inequalities
sqing   0
an hour ago
Source: Own
Let $a,b,c \geq 0 $ and $ abc+2(ab+bc+ca) =32.$ Show that
$$ka+b+c\geq 8\sqrt k-2k$$Where $0<k\leq 4. $
$$ka+b+c\geq 8 $$Where $ k\geq 4. $
$$a+b+c\geq 6$$$$2a+b+c\geq 8\sqrt 2-4$$
0 replies
sqing
an hour ago
0 replies
x+yz+zx=n where n is a postive integer
Jackson0423   0
an hour ago
Source: Own
Let \( f(n) \) denote the number of ordered triples of positive integers \( (x, y, z) \) satisfying
\[
x + yz + zx = n.
\]
(1) Find \( f(10) \) and \( f(2025) \).
(2) Let \( d(n) \) denote the number of positive divisors of \( n \). Express \( f(n) \) in terms of \( d(n) \).
0 replies
Jackson0423
an hour ago
0 replies
I need some pure geometry :))
grobber   26
N Jan 11, 2025 by Double07
Source: IMO Shortlist 1996 problem G3
Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.
26 replies
grobber
Oct 4, 2003
Double07
Jan 11, 2025
I need some pure geometry :))
G H J
Source: IMO Shortlist 1996 problem G3
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grobber
7849 posts
#1 • 5 Y
Y by nguyendangkhoa17112003, Adventure10, mathematicsy, Mango247, ehuseyinyigit
Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.
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ya
4 posts
#2 • 3 Y
Y by sayantanchakraborty, Adventure10, Mango247
Let E be the midpoint of AC, G of OP.
angle OFP = angle OEP = 90
OFPG- inscribed in circle with center G
Let K be midpiont of OH.
It is obvious that K is the center ot the Euler's (also known as nine-point) circle for the triangle ABC.
Than K, G lie on the perpendicular bisector of the common chord FE.
angle FHP = 90- angle EFH
and angle EFH = angle EFC = angle ECF = 90- angle A
angle FHP = angle A
There you go!
Irina
P.S. I think I've seen this somewhere before... Was it an IMO problem? I'm too lazy to check...
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grobber
7849 posts
#3 • 6 Y
Y by Sx763_, Illuzion, Adventure10, Yunis019, Mango247, and 1 other user
It's from a shortlist (can't remember which one); I don't know if it was actually given at an IMO, but I doubt it. Nice soln! The problem I was referring to (the one I said you could use in order to prove this) is the Butterfly problem.

Here's my soln:

Let T be the intersection between the altitude CH and the circumcircle of ABC. Let the chord FP (a chord in the circumcircle of ABC) cut the chord BT at Q. OF perpendicular to chord PF and O is the center of the circumcircle => F is the midpt of the chord PF and, because of the butterfly property, F must be the midpt of PQ (*). It's well-known that F is the midpt of HT (**). From (*) and (**) we get triangles FHP and FTQ equal, so HP || TQ=TB, so angle FHP=angle FTB=angle BAC Q.E.D.
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sam-n
793 posts
#4 • 2 Y
Y by Adventure10, Mango247
u find it in our olympiad (14-th Iranian Mathematical Olympiad 1996/1997 (1375)september).
it's beatifuly solved by batterfly theorem.
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darij grinberg
6555 posts
#5 • 4 Y
Y by Amir Hossein, Adventure10, Mango247, and 1 other user
If somebody is still interested, I have another solution:

I use the orthologic triangles theorem, which states that if ABC and A'B'C' are two non-degenerate triangles, then the lines $A\;\overline{B^{\prime }C^{\prime }}$, $B\;\overline{C^{\prime }A^{\prime }}$, $C\;\overline{A^{\prime }B^{\prime }}$ concur if and only if the lines $A^{\prime }\;\overline{BC}$, $B^{\prime }\;\overline{CA}$, $C^{\prime }\;\overline{AB}$ concur. Hereby, for any point P and any line g, the notion $P\;\overline{g}$ means the perpendicular from the point P to the line g.

For your problem, I will work with directed angles modulo 180°, and I will prove that < FHP = < CAB.

Let C' be the reflection of the point C in the line AB, or, equivalently, the reflection of the point C in the point F. Let also Z be the reflection of the point C in the point O. Then, the segment CZ is a diameter of the circumcircle of triangle ABC; hence, < CAZ = 90°, and thus $ZA \perp AC$. Similarly, $ZB \perp BC$.

Since the points O and F are the midpoints of the segments CZ and CC', we have OF || C'Z.

Now, since the point C' is the reflection of the point C in the line AB, we have < CAB = < BAC'. Thus, instead of proving < FHP = < CAB, it will be enough to show < FHP = < BAC'. But < FHP = < (FH; HP) = < (FH; AB) + < (AB; HP) = 90° + < (AB; HP), and < BAC' = < (AB; AC'). So we have to prove that 90° + < (AB; HP) = < (AB; AC'). This is equivalent to 90° = < (AB; AC') - < (AB; HP), what is obviously equivalent to 90° = < (HP; AC'). Thus, we must show that 90° = < (HP; AC'), i. e. we must show that $HP \perp AC^{\prime}$. In other words, we must show that the point P lies on the line $H\;\overline{AC^{\prime }}$.

Now, the point P is defined as the point of intersection of the lines $F\;\overline{OF}$ and AC. Since OF || C'Z, we can rewrite $F\;\overline{OF}$ as $F\;\overline{C^{\prime }Z}$, and since $ZA \perp AC$, we can rewrite AC as $A\;\overline{ZA}$. Thus, we must prove that the point P, defined as the point of intersection of the lines $F\;\overline{C^{\prime }Z}$ and $A\;\overline{ZA}$, lies on the line $H\;\overline{AC^{\prime }}$. Or, simply, we have to prove that the lines $F\;\overline{C^{\prime }Z}$, $A\;\overline{ZA}$, $H\;\overline{AC^{\prime }}$ concur. By the orthologic triangles theorem, applied to the triangles FAH and AC'Z, this is equivalent to proving that the lines $A\;\overline{AH}$, $C^{\prime }\;\overline{HF}$, $Z\;\overline{FA}$ concur. In order to prove this, we denote by S the point of intersection of the lines $A\;\overline{AH}$ and $C^{\prime }\;\overline{HF}$, and try to show that this point S lies on the line $Z\;\overline{FA}$, i. e. that we have $ZS\perp FA$.

Well, since the point S lies on the line $A\;\overline{AH}$, we have $AS \perp AH$, and together with $AH \perp BC$, this gives AS || BC. Since the point S lies on the line $C^{\prime }\;\overline{HF}$, we have $C^{\prime } S \perp HF$, and since $HF \perp AB$, this yields C'S || AB. If the lines CS and AB meet at K, then from C'S || AB, we have CK : KS = CF : FC', and since CF : FC' = 1 (the point C' is the reflection of the point C in the point F), we have CK : KS = 1, too, so that the point K is the midpoint of the segment CS. On the other hand, AS || BC yields BK : KA = CK : KS, what now shows us that BK : KA = 1, and the point K is the midpoint of the segment AB. Thus, the segments AB and CS have the point K as their common midpoint, i. e. these segments bisect each other, and it follows that the quadrilateral ACBS is a parallelogram. Hence, not only AS || BC, but also BS || AC. Now, BS || AC together with $ZA \perp AC$ yields $ZA \perp BS$, while AS || BC together with $ZB \perp BC$ yields $ZB \perp AS$. Hence, the point Z lies on two of the three altitudes of the triangle ABS; this means that the point Z is the orthocenter of this triangle, and hence also lies on the third altitude. And this yields $ZS \perp AB$, or, in other words, $ZS \perp FA$. Proof complete.

Well, this is a really monstrous solution, but it doesn't use the butterfly theorem, does it?

Darij
This post has been edited 1 time. Last edited by darij grinberg, Mar 5, 2006, 10:26 AM
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orl
3647 posts
#6 • 2 Y
Y by Adventure10, Mango247
Have a look at page 27/52.
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Virgil Nicula
7054 posts
#7 • 2 Y
Y by Adventure10, Mango247
See the problem $P3$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=46146
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yetti
2643 posts
#8 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Let the altitude CF meet the circumcircle (O) of the triangle $\triangle ABC$ again at a point D and consider the cyclic quadrilateral ADBC with the diagonal intersection F. Let the perpendicular to OF at F meet AC at P, DB at P', the circumcircle arc DC opposite to the vertex B at X, and the circumcircle arc DC opposite to the vertex A at X'. Since $XX' \perp OF$, FX = FX'. By the butterfly theorem, FP = FP' as well, i.e., P' is a reflection of P in the line OF. Reflect the cyclic quadrilateral ADBC in the line OF into a cyclic quadrilateral A'D'B'C' with the same circumcircle (O) and the same diagonal intersection F. Then D'B' meets AC at P and A'C' meets DB at P'. (This is true for any cyclic quadrilateral ADBC, not necessarily with perpendicular diagonals $AB \perp CD$.)

D is a reflection of the orthocenter H of the triangle $\triangle ABC$ in the line AB, FH = FD. By symmetry, FD' = FD, hence FH = FD = FD' and the triangle $\triangle DD'H$ has right angle $\angle DD'H = 90^\circ$. But D' is a reflection of D in OF, hence $DD' \perp OF$, so that $OF \perp FP$ are midlines of this right angle triangle, i.e., $HD' \perp FP$. Consequently, the quadrilateral FHPD' is a kite, which means that the triangles $\triangle FHP \cong \triangle FD'P$ are (oppositely) congruent and $\angle FHP = \angle FD'P$. But obviously, $\angle FD'P \equiv \angle C'D'B' = \angle CDB = \angle CAB$, which is what we were supposed to prove.

Butterfly theorem can be proved in various ways, synthetically or by trigonometry. For example, see http://www.cut-the-knot.org/pythagoras/Butterfly.shtml.
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xeroxia
1140 posts
#9 • 2 Y
Y by Adventure10, Mango247
Unfortunately, there is a trigonometric solution. And I will write it in $\LaTeX$, tenaciously.
Let $FP$ intersects the circumcircle of $\triangle AFC$ at $J$. We should show $H,P,J,C$ are concyclic because $\angle FAC = \angle FJC$ and we are asked to show $\angle FHP = \angle FAP$.
This yields $FP \cdot FJ = FH \cdot FC = AF \cdot FB$.
Let $\angle FCA = \alpha$, $\angle FCB = \beta$, and $\angle AFP = \theta$.

$\frac {AF}{FP} = \frac {\sin (90^{\circ} + \alpha - \theta)} {\sin (90^{\circ} - \alpha)}$

$\frac {FC}{FJ} = \frac {\sin (90^{\circ}  - \alpha )} {\sin (\alpha+ \theta)}$

$AF \cdot FC = FP \cdot FJ \cdot \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)}$

We will show $\frac {FC}{BF} = \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \beta} {\sin \beta}$.

Let $R=1$. Thus $AC = 2\cos \beta, BC=2\cos \alpha, BF=2\cos \alpha \sin \beta$, $AF = 2\cos\beta \sin \alpha, AB= 2\sin(\alpha + \beta)$.

So $MF = \sin(\beta - \alpha)$ and $OM = \cos (\alpha+\beta)$. And $\angle FOM = \angle AFP = \theta$. Then $\tan \theta = \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}$.

$\frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \alpha \cos \theta + \sin \alpha \sin \theta}{\sin \alpha \cos \theta + \cos \alpha \sin \theta} = \frac {\cos \alpha + \sin \alpha \tan \theta}{\sin \alpha  + \cos \alpha \tan \theta}$.

$\Rightarrow \frac {\cos \alpha + \sin \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}}{\sin \alpha  + \cos \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}} = \frac {\cos \alpha \cos (\alpha + \beta) + \sin \alpha \sin(\beta - \alpha)}{\sin \alpha \cos (\alpha + \beta)  + \cos \alpha \sin(\beta - \alpha)}$ $\Rightarrow \frac {\cos \beta (\cos^2 \alpha - \sin^2\alpha)}{\sin \beta (\cos^2 \alpha - \sin^2\alpha)} = \frac {\cos \beta} {\sin \beta}$ $Q.E.D$
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Luis González
4149 posts
#10 • 3 Y
Y by wiseman, Adventure10, and 1 other user
Since $\angle HFA=\angle OFP=90^{\circ}$ and $\angle HAF=\angle OAP,$ it follows that $O,H$ are isogonal conjugates with respect to $\triangle APF.$ Consequently, if $M,N$ denote the midpoints of $AB,AC,$ then $\triangle FNM$ is the pedal triangle of $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$ $\Longrightarrow$ $\angle FHP=\angle NFA.$ Since $\triangle ANF$ is N-isosceles, then $\angle FHP=\angle BAC.$
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mathreyes
109 posts
#11 • 2 Y
Y by Adventure10, Mango247
Luis González wrote:
...$\triangle FNM$ is the pedal triangle of $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$...

why? I think this is not a useful reason to ensure that perpendicularity.

The real reason (for me, at least) is:

$\measuredangle ONP=\measuredangle OFP=90\Longrightarrow NPFO$ is cyclic, so $\measuredangle FNP=\measuredangle FOP$ but $\measuredangle NPH=\measuredangle OPF$.
Finally $\measuredangle FNP+\measuredangle NPH=\measuredangle FOP+\measuredangle OPF=90$, so $HP \perp FN$.

(note that in my argument, there was no need to construct either point $M$ nor pedal triangle of $O$.)
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Mosquitall
571 posts
#12 • 4 Y
Y by AlastorMoody, Adventure10, Mango247, and 1 other user
Generalization:
Triangle $ABC$, and point $F$, such that $\angle BFC=\angle CFA=\gamma$, $\angle FAC=\beta$, point $H$ is on $CF$ and $\angle FHB=\beta$, point $P$ is on $AC$ and $\angle PHF=\beta$, point $O$ with $\angle CBO=\angle OCB= \alpha$, $\alpha+\beta+\gamma=180$. Then prove that $\angle PFA=\angle OFC$.
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duanby
76 posts
#13 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
MY SOLUTION:
Let P' be the reflection of P wrt CF then P' is the isogonal conjugate point of O wrt ACF
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vslmat
154 posts
#14 • 3 Y
Y by minlat7, Adventure10, Mango247
To avoid using the Butterfly theorem as well as advanced geometry, we can do this way:
Let $CF$ cuts the circumcircle at $D$. On $AC$ let’s choose point $P'$ so that $\angle FAP' = \angle BAC$, $P'F$ cuts $BD$ at $Q$ and cuts the circumcirle at $M$ and $N$. Easy to see that $HP'\parallel BD$. As $DF = FH$ is a well known property, $QF = FP'$.
If we can prove that $P'M = NQ$ then $F$ is the midpoint of $MN$ and $OF\perp MN$, thus $P'\equiv P$ and we are done.
Now using sinus theorem we have
$\frac{P'C}{sinF_{1}} = \frac{FP'}{sinC_{1}} $ and $\frac{QD}{sinF_{1}} = \frac{QF}{sinD_{1}}$, thus $\frac{P'C}{QD} = \frac{sinD_{1}}{sinC_{1}}$. Similarly, we get $\frac{BQ}{AP'} = \frac{sinA_{1}}{sinB_{1}}$
Therefore, $\frac{BQ}{AP'} = \frac{P'C}{QD}$, or $BQ.QD = AP'. P'C$
But notice that $QD. BQ = NQ. QM$ and $AP'. P'C = P'M. P'N$, it follows that $NQ = P'M$
$F$ is indeed the midpoint of $MN$ and we are done.
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XmL
552 posts
#15 • 2 Y
Y by Adventure10, Mango247
Let the line through $O$ perpendicular to $AC$ meet $AC,CF,AB$ at $M,L,K$ respectively, thus $\angle OKA=90-\angle A=\angle ACF$ and $M$ is the midpoint of $AC$. From the first result we deduce that $\triangle ACF\sim \triangle LKF$. Since $\angle PFO=\angle CFA=90$, thus $\angle AFP=\angle LFO$, which means that $O,P$ are corresponding points concerning similar triangles $LKF$ and $ACF$. Now let $H'$ be the point that corresponds to $H$, thus $H'$ is on $FK$ and $\triangle FPH\sim \triangle FOH'$ $\Rightarrow$ we now just need to prove $\angle OH'F=\angle A=\angle MFA$ $\iff$ $OH'\parallel MF$ $\iff$ $\frac {MO}{OK}=\frac {FH'}{KH'}=\frac {FH}{CH}(*)$.
Since $\triangle AMK\sim \triangle AFC$ and $\angle HAF=\angle MAO$, thus $O,H$ are two corresponding points concerning those two triangles, which means that (*) is true. Q.E.D
This post has been edited 1 time. Last edited by XmL, Jul 12, 2013, 2:26 AM
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IDMasterz
1412 posts
#16 • 2 Y
Y by Adventure10, Mango247
Let $DEF$ be the orthic triangle of $\triangle ABC$.Since $\angle (OF, AF) = \angle FPH$ and we already have that $O, H$ as isogonal wrt $\angle FAP$, we get $H, O$ are isogonal conjugates wrt $\triangle AFP$. If we let $M$ be the midpoint of $AC$, then note that $AH \perp FM$ (since they are the feet of the pedals from $O$). Now, $M$ is the centre of $DFAC$ so $\angle MFC = 90 - A$ so $\angle FHP = \angle BAC$ as desired.
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IDMasterz
1412 posts
#17 • 2 Y
Y by Adventure10, Mango247
@mathreyes

It is well-known that for two isogonal conjugates $X, Y$, we have $AX, BX, CX$ is perpendicular to the sides of the pedal triangles.

I realise now that my solution is basically the same as Luis's, sorry I posted on impulse hehe
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jred
290 posts
#18 • 2 Y
Y by AlastorMoody, Adventure10
duanby wrote:
MY SOLUTION:
Let P' be the reflection of P wrt CF then P' is the isogonal conjugate point of O wrt ACF
there's a typo, it should be $\triangle BCF$ instead of $\triangle ACF$.
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jayme
9795 posts
#19 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

see

http://www.artofproblemsolving.com/community/c6t48f6h1167200_angle_equal

Sincerely
Jean-Louis
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Blast_S1
358 posts
#20 • 1 Y
Y by Adventure10
[asy]
size(8cm);
pair A=(0,0), B=(120,0), C=(34.5,140.9), F=(34.5,0), O=(60,60), P=(3.3,13.3), Q=(136,97.7), H=(34.5,20.9), X=(19.4,-14.5), Y=(34.4,-20.9);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NW);
label("$Q$", Q, NE);
label("$H$", H, NW);
label("$O$", O, NW);
label("$P$", P, NW);
label("$H_1$", X, SW);
label("$H_2$", Y, S);
draw(A--B--C--cycle, linewidth(0.5)+blue);
draw(circle(O, 60sqrt(2)), linewidth(0.4)+blue);
draw(C--F, linewidth(0.5)+blue);
draw(F--P, linewidth(0.5)+blue);
draw(X--O--F--cycle, linewidth(0.5)+red);
draw(F--Y--O, linewidth(0.5)+red);
draw(F--Q--C, linewidth(0.4)+dashed+blue);
draw(circle((17.2,2.8), 17.42), linewidth(0.4)+dashed+grey);
dot(A);
dot(B);
dot(C);
dot(F);
dot(O);
dot(P);
dot(Q);
dot(H);
dot(X);
dot(Y);
[/asy]
Let $H_1$ and $H_2$ be the reflections of $H$ over $\overline{PF}$ and $\overline{AB}$ respectively, and let $\theta=\angle BFO$. Clearly $\angle PFC=\angle PFH_1=\theta$ too.
Lemma: $H_1\in (ABC)$

Proof: It is well-known that $H_2\in(ABC)$, so it suffices to prove that $OH_1=OH_2$. Clearly, $H_1F=H_2F$, $OF=OF$, and
$$\angle OH_1F=90^\circ+\theta=\angle HFO,$$so $\triangle OFH_1\cong\triangle OFH_2\implies OH_1=OH_2$, as desired.
Now, let $Q$ be the second intersection of $\overline{H_1F}$ and $(ABC)$. Since $H_1F=H_2F$, we must have that $CF=QF$ as well. This yields that
$$\angle FCQ=\frac{180^\circ-\angle CFQ}{2}=\frac{180^\circ-(180^\circ-2\theta)}{2}=\theta=\angle AFP,$$or that $\overline{CQ}\parallel\overline{PF}$. Finally, this must mean that
$$\angle PAH_1=180^\circ-\angle CQF=180^\circ-\angle PFH_1,$$so $PAH_1F$ is cyclic and
$$\angle BAC=\angle FH_1P=\angle FHP.$$Yep...
This post has been edited 2 times. Last edited by Blast_S1, Jan 14, 2020, 10:16 PM
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jayme
9795 posts
#21 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%200.pdf p. 51...

Sincerely
Jean-Louis
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Mercury_is_small
15 posts
#22 • 1 Y
Y by Adventure10
IMOSL 1996 G3 wrote:
Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.

We use phantom points :)
Notations:
Let $P'$ be a point on $AC$ such that $\angle FHP'=\angle A$. Now, let $H'$ be the reflection of $H$ on $AB$. Let $\angle AFP'=x\implies \angle P'FH=90-x$. Let $X=H'B\cap P'F$.

Two line proof: :D
Clearly, $H'B\| P'H$ and so, $\triangle PHF\equiv \triangle XH'F$ and so, $F$ is the midpoint of $XP'$. Now, obviously sine rule in triangles $XH'F,XFB,AFP',FP'C$ gives power of $X=$ power of $P'$ and so, $XOP'$ is isosceles and thus, $\angle OFP'=90\implies P=P'$.
QED
This post has been edited 1 time. Last edited by Mercury_is_small, Dec 17, 2019, 1:04 PM
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Bunrong123
79 posts
#23
Y by
Let $\Gamma$ be the circle with center $O$ of triangle $ABC$, Extend $CF$ intersect $\Gamma$ at $G$, $FP$ intersect $GB$ at $I$, and The point $F$ be a point on chord $DE$ of $\Gamma$ such that $OF \perp DE$.
Then We have $F$ is midpoint of $DE$
Since $AB$ intersect $GC$ at F
By Butterfly's Theorem,
We get $IF=IP$
Since $GBCA$ is cyclic quadrilateral of $\Gamma$
Implies $\angle{GBA}=\angle{GCA}$
$\angle{BGC}=\angle{BAC}$.
and $\angle{GCA}=\angle{FCA}=90^\circ -\angle{BAC}=\angle{ABH}=\angle{FBH} =\angle{GBA} = \angle{GBF}$
Since $GH \perp BF$
We deduce $GF=FH$
Then $\triangle{GFI} \cong \triangle{HFP}$ $(S.A.S)$
We obtained $\angle{BAC}=\angle{FGI}=\angle{FHP}$
The result Follows. $\blacksquare$
This post has been edited 1 time. Last edited by Bunrong123, May 18, 2020, 2:38 PM
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Jishnu4414l
155 posts
#24 • 1 Y
Y by ehuseyinyigit
Reflect $H$ over $AB$ to $H_C$. It is a very well known fact that $H_C$ lies on $(ABC)$.
Now let $PF$ meet $BH_C$ at $Q$. By Butterfly theorem, we have $PF$=$FQ$.
Now notice that $\triangle PFH \cong \triangle QFH_C$ by SAS congruency.
Thus $\angle CAB=\angle CH_CB=\angle FHP$. Our proof is thus complete.
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Assassino9931
1353 posts
#25
Y by
Observe that $\angle BAH = \angle BCH = \angle CAO = \angle BAO = 90^{\circ} - \angle ABC$. Now let $K$ be a point on $AH$ such that $\angle KCH = \angle BCH$. Then $O$ and $K$ are isogonal conjugates in triangle $ACD$, thus $\angle ADO = \angle CDK$. On the other hand, $\angle ADO = 180^{\circ} - \angle EDO - \angle BDE = 90^{\circ} - \angle BDE = \angle CDE$. Hence $\angle CDK = \angle CDE$ and together with $\angle KCD = \angle ECD$ it follows that $\triangle KCD \cong \triangle ECD$ and that $CD$ is the perpendicular bisector of $KE$. But then $\angle DHE = \angle DHK = 90^{\circ} - \angle BAH = \angle ABC$ and we are done.
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LeYohan
46 posts
#26
Y by
This is direct use of Butterfly Thorem.

Let $H'$ be the reflection of $H$ over $AB$ which we know lies on $(ABC)$. Let the line $FP$ intersect $(ABC)$ at $X, Y$. Because $O \perp FP$ it's clear that $F$ is the midpoint of $XY$. Let $Z$ be the intersection of $H'$ and $B$, then using Butterfly Theorem we know that $FP=FZ$. Now because $H'$ is the reflection of $H$ over $AB$, we know $H'F=FH$ so $H'BHP$ is a parallelogram $\implies H'B \parallel HP \implies \angle BAC = \angle CH'B = \angle FHP$ and we're done. $\square$
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Double07
90 posts
#27 • 2 Y
Y by HotSinglesInYourArea, Calamarul
Nice bashing here we goooooooooooo:

Take $(ABC)$ the unit circle. Then
$h=a+b+c$ and $f=\dfrac{ab+bc+c^2-ab}{2c}$

The equation of line $OF$ is $\overline{z}=\dfrac{\overline{f}}{f}\cdot z=\dfrac{ab+bc+ca-c^2}{ab(ac+bc+c^2-ab)}\cdot z$.

Consider $X$ and $Y$ the two intersections of line $OF$ with the unit circle.
Then $x$ and $y$ are the solutions of the equation $\dfrac{1}{z}=\dfrac{ab+bc+ca-c^2}{ab(ac+bc+c^2-ab)}\cdot z\iff z^2=\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}$, which by Viete's implies that $x+y=0$ and $x\cdot y=-\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}$.

$P\in AC\iff P$ is its own projection on chord $AC\iff p=a+c-ac\overline{p}$.
$OF\perp FP\iff F$ is the projection of $P$ on chord $XY\iff f=\dfrac{1}{2}(p+x+y-xy\overline{p})\iff$
$\iff\dfrac{ac+bc+c^2-ab}{c}=a+c-(ac+xy)\overline{p}\iff \dfrac{bc-ab}{c}=-(ac+xy)\overline{p}\iff$
$\iff \left(ac-\dfrac{ab(ac+bc+c^2-ab)}{ab+bc+ca-c^2}\right)\overline{p}=\dfrac{b(a-c)}{c}\iff \dfrac{a(a-c)(b^2+c^2)}{ab+bc+ca-c^2}\cdot \overline{p}=\dfrac{b(a-c)}{c}\iff$
$\iff \overline{p}=\dfrac{b(ab+bc+ca-c^2)}{ac(b^2+c^2)}\iff p=\dfrac{c(ac+bc+c^2-ab}{b^2+c^2}$.

Ok, now we just need to prove the angle condition, which is equivallent to proving that $\widehat{BAC}+\widehat{CHP}=180^\circ\iff \dfrac{b-a}{c-a}\cdot\dfrac{c-h}{p-h}\in \mathbb{R}$
But $c-h=-a-b$ and $p-h=\dfrac{ac^2+bc^2+c^3-abc}{b^2+c^2}-a-b-c=\dfrac{-abc-ab^2-b^2c-b^3}{b^2+c^2}=(-b)\cdot\dfrac{(a+b)(b+c)}{b^2+c^2}$, so

$\dfrac{b-a}{c-a}\cdot\dfrac{c-h}{p-h}=\dfrac{b-a}{c-a}\cdot\dfrac{b^2+c^2}{b(b+c)}$, which is clearly real after conjugating, so we're done.
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