Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
2024 Japan Mathematical Olympiad Preliminary
parkjungmin   0
2 minutes ago
2024 Japan Mathematical Olympiad Preliminary
0 replies
parkjungmin
2 minutes ago
0 replies
ISI UGB 2025 P2
SomeonecoolLovesMaths   8
N 12 minutes ago by lakshya2009
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
8 replies
SomeonecoolLovesMaths
Yesterday at 11:16 AM
lakshya2009
12 minutes ago
2025 Japanese Mathematics Olympiad
parkjungmin   0
16 minutes ago
2025 Japanese Mathematics Olympiad
0 replies
+1 w
parkjungmin
16 minutes ago
0 replies
Points on the sides of cyclic quadrilateral satisfy the angle conditions
AlperenINAN   3
N 25 minutes ago by Primeniyazidayi
Source: Turkey JBMO TST 2025 P1
Let $ABCD$ be a cyclic quadrilateral and let the intersection point of lines $AB$ and $CD$ be $E$. Let the points $K$ and $L$ be arbitrary points on sides $CD$ and $AB$ respectively, which satisfy the conditions
$$\angle KAD = \angle KBC \quad \text{and} \quad \angle LDA = \angle LCB.$$Prove that $EK = EL$.
3 replies
AlperenINAN
Yesterday at 7:15 PM
Primeniyazidayi
25 minutes ago
Calculating sum of the numbers
Sadigly   4
N 25 minutes ago by MITDragon
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
4 replies
Sadigly
May 9, 2025
MITDragon
25 minutes ago
Inequality
Sadigly   8
N an hour ago by Sadigly
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
8 replies
Sadigly
May 9, 2025
Sadigly
an hour ago
Minimum value of a 3 variable expression
bin_sherlo   5
N an hour ago by Primeniyazidayi
Source: Türkiye 2025 JBMO TST P6
Find the minimum value of
\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]where $x,y,z>1$ are reals.
5 replies
+1 w
bin_sherlo
Yesterday at 7:16 PM
Primeniyazidayi
an hour ago
Nice R+ FE
math_comb01   5
N an hour ago by jasperE3
Source: XOOK 2025 P3
Find all functions $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ so that for any $x, y \in \mathbb{R}_{>0}$, we have \[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]
Proposed by Anmol Tiwari and MathLuis
5 replies
+1 w
math_comb01
Feb 9, 2025
jasperE3
an hour ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   3
N an hour ago by Primeniyazidayi
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
3 replies
bin_sherlo
Yesterday at 7:13 PM
Primeniyazidayi
an hour ago
Help me solve this problem please. Thank you so much!
illybest   1
N an hour ago by GreekIdiot
Give two fixed points B and C, and point A moving on the circle (O). Let D be a point on (O) such that AD is perpendicular to BC. Let O' be the point symmetric to O with respect to BC, M be the midpoint of BC, and N ( dinstinct from D) be the intersection of MD with the circumcircle of triangle AOD. Suppose DO' intersects the circle (O) again at S.
a) Prove that the circle (OMN) is tangent to the circle (DNS)
b) Let d be the line tangent to (DNS) at N. Prove that d always passes through a fixed point when A moves along the arc BC of (O)
1 reply
illybest
Sep 8, 2024
GreekIdiot
an hour ago
Looks like power mean, but it is not
Nuran2010   6
N an hour ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
6 replies
Nuran2010
Yesterday at 11:51 AM
sqing
an hour ago
Expressing polynomial as product of two polynomials
Sadigly   5
N 2 hours ago by Sadigly
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
5 replies
Sadigly
Yesterday at 9:10 PM
Sadigly
2 hours ago
Simple inequality
sqing   19
N 2 hours ago by sqing
Source: Old? Where?
Let $a,b,c$ be positive real numbers .Prove that
$$(a+b)^2+(a+b+4c)^2\geq \frac{100 abc}{a+b+c}$$
19 replies
sqing
Jan 16, 2021
sqing
2 hours ago
Interesting inequalities
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b>0  $ . Prove that
$$ \frac{a^2+b^2}{ab+1}+ \frac{4}{ (\sqrt{a}+\sqrt{b})^2} \geq 2$$$$ \frac{a^2+b^2}{ab+1}+ \frac{3}{a+\sqrt{ab}+b} \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{4}{(a+b)^2}  \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{3}{a^2+ab+b^2}  \geq 2$$$$\frac{a^2+b^2}{ab+2}+ \frac{1}{2\sqrt{ab}}  \geq \frac{2+3\sqrt{2}-2\sqrt{2(\sqrt{2}-1)}}{4} $$
2 replies
sqing
Today at 4:34 AM
sqing
2 hours ago
Midpoint of base in isosceles triangle [<APM + <BPC = 180°]
warut_suk   22
N Jul 26, 2024 by Grasshopper-
Source: Poland National Olympiad 2000, Day 1, Problem 2
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
22 replies
warut_suk
Jan 25, 2005
Grasshopper-
Jul 26, 2024
Midpoint of base in isosceles triangle [<APM + <BPC = 180°]
G H J
Source: Poland National Olympiad 2000, Day 1, Problem 2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
warut_suk
803 posts
#1 • 3 Y
Y by Adventure10, jhu08, Mango247
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grobber
7849 posts
#2 • 5 Y
Y by Adventure10, jhu08, Mango247, and 2 other users
By an angle chase, we can see that $P$ moves on a circle tangent in $A,B$ to $CA,CB$ respectively. Let $Q$ be the second intersection of $CP$ with this circle. $PAQB$ is a harmonic cyclic quadrilateral, so, if $C'$ is the intersection between the tangent at $P$ to the circle and $AB$, we find $(PA,PB;PQ,PC')=-1$, meaning that $PQ$ is the symmedian of $PAB$, which is precisely what we want.

I was in a hurry while writing this, but I'll clarify if needed.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#3 • 3 Y
Y by Adventure10, jhu08, Mango247
Rewriting your problem with different notations:

Let ABC be an isosceles triangle with AB = AC. Let P be a point inside the triangle ABC such that < PBC = < PCA. Let M be the midpoint of the segment BC. Prove that < BPM + < APC = 180°.

Now, this problem was solved in posts #8 and #9 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (see the Lemma in post #8), with the only difference being that an additional assumption was used, namely the assumption that < PCB = < PBA; but this assumption is actually unnecessary, since it follows from the other assumptions of the problem (namely, since the triangle ABC is isosceles with AB = AC, we have < ACB = < ABC, and since < PBC = < PCA, we have < PCB = < ACB - < PCA = < ABC - < PBC = < PBA, so that < PCB = < PBA comes out as a consequence of the other assumptions of the problem).

Darij
This post has been edited 2 times. Last edited by darij grinberg, Mar 15, 2007, 9:35 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Arne
3660 posts
#4 • 7 Y
Y by LoveMath4ever, r31415, Adventure10, jhu08, Mango247, and 2 other users
warut_suk wrote:
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.

Take point Q such that APBQ is a parallellogram.

Take point R on AQ such that ACB, APR are similar. (I assume R lies between A and Q, the other case is similar) Then BPRQ is cyclic, and ACP, ABR are similar.

Hence BPQ = BRQ = 180 - ARB = 180 - APC. Hence result.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
#5 • 7 Y
Y by LoveMath4ever, Abdollahpour, Adventure10, jhu08, Mango247, and 2 other users
$S\in AB\cap CP$, $CA=CB=a$, $AB=b$, $m(\widehat {PCA})=u$, $m(\widehat {PCB})=v$,
$m(\widehat {PAB})=m(\widehat {PBC})=x$, $m(\widehat {PAC})=m(\widehat {PBA})=y$.

$\blacktriangle$ The theorem of Sinus in the triangle $APB\Longrightarrow \frac{PA}{PB}=\frac{\sin y}{\sin x}\ \ (1)$.
$\blacktriangle$ The trigonometrical form of the Menelaus' theorem in
the triangle $ABC$ for the inner point $P\Longrightarrow$
$\sin u\sin^2x=\sin v\sin^2y$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\left(\frac{\sin y}{\sin x}\right)^2\ \ (2)$.
$\blacktriangle\ \frac{SA}{SB}=\frac{CA}{CB}\cdot \frac{\sin \widehat {ACS}}{\sin \widehat {BCS}}$ $\Longrightarrow$ $\frac{SB}{SC}=\frac{\sin u}{\sin v}\ \ (3).$

From the relations $(1)$, $(2)$, $(3)$ results: $\frac{SA}{SB}=\left(\frac{PA}{PB}\right)^2$, i.e. the ray $[PS$ is
the symmedian from the vertex $P$ in the triangle $APB$, meaning

$\widehat {APM}\equiv \widehat {BPS}\Longrightarrow$ $\boxed {\ m(\widehat {APM})+m(\widehat {BPC})=180^{\circ}\ }.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
treegoner
637 posts
#6 • 2 Y
Y by Adventure10, jhu08
The problem can be generalized as follow (which is well-known)
Let $PAB$ be a triangle and $M$ be the midpoint of $AB$. Suppose $C$ is the intersection of the tangents of the circumcircle at $A$ and $B$. Then $\measuredangle{BPM}= \measuredangle{CPB}$.
[Moderator edit: This is basically the assertion of http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .]
Back to the problem , it is easy to see that $CA$ and $CB$ are the tangents of $(PAB)$ at $A$ and $B$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
#7 • 3 Y
Y by Adventure10, jhu08, Mango247
See http://www.mathlinks.ro/Forum/viewtopic.php?t=55581
http://www.mathlinks.ro/Forum/viewtopic.php?t=43202
where are mentioned the properties of the harmonical quadrilateral.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zbghj
1 post
#8 • 3 Y
Y by Adventure10, jhu08, Mango247
Let us see the circumcircle of APB. Let v=<ABC.
Use the center of the circumcircle as origin of coordinate and give the coordinates as:

C (0,1)
M (0, (cos v)^2)
P (cos v*cos u, cos v*sin u)

It is easy to obtain: PM/PC=cos v

So, we can get: BC/PC=AM/PM.
So, <BPC=<APM or <BPC+<APM=180.

When <BPC=<APM, then <AMP=<BCP. So, <BPC=<BMC=90. As a result, <BPC+<APM=180 too.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
yetti
2643 posts
#9 • 3 Y
Y by Adventure10, jhu08, Mango247
P is on circle (U) tangent to BC at B and passing through A(then $ \angle PAB = \angle PBC$). $ \triangle ABC$ is isosceles $ \Longrightarrow$ (U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in $ CM \perp AB,$ $ P' \in (U)$ and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, $ M \equiv PQ' \cap P'Q \Longrightarrow$ $ \angle APM \equiv \angle APQ' = \angle BP'Q = \angle BPQ = 180^\circ - \angle BPC.$
This post has been edited 1 time. Last edited by yetti, Dec 9, 2007, 3:02 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Leonhard Euler
247 posts
#10 • 3 Y
Y by Adventure10, jhu08, Mango247
We have $ \angle PBA=\angle PAC$. Let $ X$ be on $ PM$ such $ PAXB$ is parallogram. Then $ \angle XAB=\angle PAC,\angle XBA=\angle PBC$. Hence $ C,X$ are isogonal points wrt $ \triangle PAB$. Hence $ PC,PX$ is isogonal line $ \triangle PAB$. This imply $ \angle APM+\angle BPC=180$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Learner94
634 posts
#11 • 4 Y
Y by samirka259, jhu08, Adventure10, Mango247
Lemma: Let $ABC$ be a triangle and $\Gamma $ its circumcircle. Let the tangent to $\Gamma $ at $B$ and $C$ intersect at $D$. Then $AD$ coincides with the $A$- symmedian of $\triangle ABC$. The proof can be found here.

By angle chasing we can show that $AC$ and $BC$ are tangents to the circumcircle of $\triangle PAB$ at points $A$ and $B$ respectively. The tangents $AC$ and $BC$ intersect at $C$. So $PC$ coincides with the $C$- symmedian of $\triangle ABC$. Since $PM$ is the median of $\triangle ABC$, $\angle APM = \angle BPE$. or $\angle APM = 180^{\circ} - \angle BPC$. So $\angle APM +  \angle BPC = 180^{\circ} $.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DaChickenInc
418 posts
#12 • 3 Y
Y by jhu08, Adventure10, Mango247
Translation of Official Solution
Official Solutions
Problem Significance
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Shauryajain123
152 posts
#13 • 2 Y
Y by AllanTian, jhu08
A nice problem for symmedians :)
We see that AC and AB are tangents to the circumcircle of $\triangle{APB} \implies CP $ is the symmedian.
Then the rest easily follows from how symmedians are defined
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PCChess
548 posts
#14 • 1 Y
Y by jhu08
Since $\triangle ABC$ is isosceles, we know that $\angle PAC=\angle PBA$. Consider $(APB)$. Since $\angle PAB=\angle PBC$ and $\angle PAC=\angle PBA$, lines $AC$ and $BC$ are tangent to $(APB)$. This implies that $PC$ lies on the $P$-symmedian. Let $CP$ intersect $AB$ at point $E$. By the definition of the symmedian, $\angle APM=\angle BPE$. Clearly $\angle BPE+\angle BPC=180^{\circ}$, so $\angle APM + \angle BPC = 180^{\circ}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JustKeepRunning
2958 posts
#15 • 1 Y
Y by jhu08
Let $CP\cap AB=G$. Then the problem is equivalent to $\angle APM=\angle BPG$, aka. $CG$ is a symmedian in $\triangle APG$. Notice that $CB$ and $CA$ are the tangents to $(APB),$ so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#16 • 2 Y
Y by centslordm, jhu08
Since $\angle CBP=\angle BAP$ and $\angle PAC=\angle PBA,$ $C$ is the intersection of the tangents to $(PAB)$ at $A$ and $B.$ Hence, $\overline{PC}$ is the $P$-symmedian point of triangle $PAB.$ Notice that $$\measuredangle BPC=-\measuredangle(\overline{CP},\overline{PB})=-\angle APM.$$$\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JustKeepRunning
2958 posts
#17 • 1 Y
Y by jhu08
Nice problem!

It is well known that $P$ is the $A$ humpty point, and so $M_1:=AP\cap BC$ is the midpoint of $BC$. Furthermore, we have that $\angle PCB=\angle PAC=\angle PBA$ and $\angle PAB=\angle PBC$. Hence, we get that $\angle APM+\angle BPC=\angle APM+\angle CPM_1+\angle M_1PB=\angle APM+\angle MPB+\angle BPM_1=180^{\circ},$ where this follows from similar triangles $\triangle PMB$ and $\triangle PM_1C$.

EDIT: oops just realized I already solved this before. At least this is a slightly different/more motivated solution.
This post has been edited 1 time. Last edited by JustKeepRunning, Dec 11, 2021, 1:48 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Albert123
204 posts
#18 • 1 Y
Y by jhu08
Let $(APB) \cap CP=L$
As well: $CA$ and $CB$ are tangent of $(APB)$
Then: $PQ$ is symedian of $(APB)$
$\implies \angle APM=\angle LPB \implies \angle APM + \angle BPC=180$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahdi_Mashayekhi
695 posts
#19
Y by
Let $CP$ meet circumcircle of $APB$ at S. Since $\angle PAB = \angle PBC$ then $CB$ is tangent to $APB$ and since $CB = CA$ then $SP$ is symmedian of $ASB$ so $\angle APM = \angle BPS$ and so $\angle APM + \angle BPC = \angle 180$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EulersTurban
386 posts
#20
Y by
Let $CP \cap AB = D$. The angle condition implies that $AC$ and $AB$ are tangent to the circumcircle of $APB$, then we know that $PD$ is the $P$-symmedian of $APB$, which implies that $\angle TPC = \angle APD = \angle MPB$ thus we have that $\angle BPC = \angle MPT$, which clearly gives us that $\angle APM + \angle BPC = 180$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8864 posts
#21
Y by
Let $\overline{PM'}$ be a symmedian in triangle $BPC$, so it suffices to show that $A$ lies along the symmedian. But this is evident by the tangent intersection definition.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ianis
411 posts
#22
Y by
Complex


Synthetic
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Grasshopper-
10 posts
#23
Y by
Let CP intersect AB at N, the circumradius of tri. ACP(center O) and tri. BCP(center O') be r, r' respectively. We have to prove that CP is P -symmedian in tri. APB and to show that we need to prove AN/BN=AP^2/BP^2 ,
We have [tri. APN]/[tri. BPN] = AN/BN = [tri. APC]/[tri. BPC]=(AP × sin<CAP)/(BP × sin<CBP)
by sine rule,
sin<CAP/sin<CBP=r'/r, by angle chasing we will get that tri. OO'P and tri. APB are similar so, r'/r = AP/BP then, we get that AN/BN=AP^2/BP^2
so, PN is P-symmedian in tri. APB. This implies <APM = <BPN; <BPN + <CPB=180°=> <APM + <BPC=180° .
Z K Y
N Quick Reply
G
H
=
a