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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.

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)!

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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
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
Hardest in ARO 2008
discredit   29
N a few seconds ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
29 replies
discredit
Jun 11, 2008
JARP091
a few seconds ago
Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N 4 minutes ago by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
4 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   9
N 5 minutes ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
9 replies
+1 w
OgnjenTesic
May 22, 2025
JARP091
5 minutes ago
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N 5 minutes ago by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
5 minutes ago
JBMO Shortlist 2023 N3
Orestis_Lignos   9
N 30 minutes ago by Just1
Source: JBMO Shortlist 2023, N3
Let $A$ be a subset of $\{2,3, \ldots, 28 \}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by $29$ also belongs to $A$.

Find the minimum possible value of $|A|$.
9 replies
Orestis_Lignos
Jun 28, 2024
Just1
30 minutes ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N 43 minutes ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
43 minutes ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N an hour ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
1 viewing
v_Enhance
Dec 31, 2012
cursed_tangent1434
an hour ago
IMO Shortlist 2008, Geometry problem 2
April   42
N an hour ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
an hour ago
3^n + 61 is a square
VideoCake   25
N an hour ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
an hour ago
Problem 5
blug   3
N an hour ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
an hour ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N 2 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
2 hours ago
Problem 7
SlovEcience   7
N 2 hours ago by GreekIdiot
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
7 replies
SlovEcience
May 14, 2025
GreekIdiot
2 hours ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N 2 hours ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
2 hours ago
Inequality
lgx57   1
N 2 hours ago by sqing
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
1 reply
lgx57
3 hours ago
sqing
2 hours ago
Find all possible values of BT/BM
va2010   53
N Apr 25, 2025 by ja.
Source: 2015 ISL G4
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
53 replies
va2010
Jul 7, 2016
ja.
Apr 25, 2025
Find all possible values of BT/BM
G H J
G H BBookmark kLocked kLocked NReply
Source: 2015 ISL G4
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bjump
1035 posts
#48
Y by
DottedCaculator wrote:
[asy]
unitsize(1cm);
pair A, B, C, M, P, Q, T;
A=(2,2.5sqrt(5));
B=(0,0);
C=(8,0);
M=(B+C)/2;
pair X;
X=(1,0);
P=intersectionpoints(A--B,circumcircle(A,M,X))[1];
Q=intersectionpoints(A--C,circumcircle(A,M,X))[1];
T=P+Q-A;
draw(A--B--C--A--P--T--Q--A);
draw(circumcircle(A,M,X));
draw(circumcircle(A,B,C));
label("$B$", A, N);
label("$A$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);
label("$P$", P, W);
label("$Q$", Q, E);
label("$T$", T, S);
label("$X$", X, NE);
[/asy]

bashy bary
less bashy
best solution

How do you learn to make diagrams like this also this got liked by tapir orz orz orz
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fuzimiao2013
3311 posts
#50 • 2 Y
Y by Mango247, Mango247
oh, great, here we go again. i am now thoroughly convinced that in this white wasteland of geometry, bary is just problem suicide - your last resort, unless it's not.

We use barycentric coordinates.

Preliminaries:
Let $ABC$ be the reference triangle, and $P = (p, 1-p, 0), Q = (0, 1-q, q)$. The circle through $B, M$ is characterized by \begin{align*}-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz) &= 0,\\ u+w &= \frac{b^2}{2}.\end{align*}So after plugging in $P$ and $Q$ and heavily simplifying, \begin{align*}c^2p &= c^2-u \\ a^2q &= a^2-w.\end{align*}
Actual Problem:
By distance formula, we can now proceed:
\begin{align*}\frac{BT^2}{BM^2} &= \frac{a^2p-wp+a^2q-wq+c^2p-up+c^2q-uq-b^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{(a^2-w+c^2-u)(p+q)+a^2q-a^2pq-a^2q^2+c^2p-c^2p^2-c^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+(a^2q+c^2p)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+\left(a^2+c^2-\frac{b^2}{2}\right)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)}{\frac 12 \left(a^2+c^2-\frac{b^2}{2}\right)} = 2.
\end{align*}Since length is positive, the only possible value is $\sqrt 2$, done. $\square$
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Knty2006
50 posts
#51
Y by
G #1/100

Let the midpoint of $BT$ be $N$, Let $PQ$ intersect $AC$ at $D$ and the Miquel point of $APQC$ be $R$

Claim: $DRNM$ cyclic

Proof: Due to the definition of a Miquel point,
$APRD$ cyclic, $QCDR$ cyclic
So, $R$ is the center of the spiral similarity sending $QC$ to $PA$, which means that it is the center of the spiral similarity sending $NM$ to $PA$

Hence, our claim must be true.

Claim: $MB$ tangent to $MNRT$

Proof: $$\angle BMR=\angle BPR$$$$=180-\angle APR$$$$=\angle MDR$$
Hence, our claim must be true.

Using the tangency, we have
$MB^2=(NB)(BT)=\frac{1}{2} BT^2$

So, $$\frac{BT}{MB}=\sqrt{2}$$
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Taco12
1757 posts
#52 • 1 Y
Y by Mango247
Solved with a hint from Evan Chen's solution.

Let $X$ be the second intersect of $\omega$. Set $T=(p,q,r)$. The parallelogram condition yields $P=(p:q+r:0)$ and $Q=(0:p+q:r)$. Then, we have $$b^2=qc^2+rc^2+a^2p+a^2q.$$By distance formula, we get $BT^2=a^2r+c^2p$, which gives us $BT^2=2BM^2$, so $\sqrt2$ is the only value.
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john0512
4191 posts
#53
Y by
We will use barycentric coordinates. Let $$T=(r,s,t)$$for which $r+s+t=1.$ Then, from the parallelogram, $$P=(r,s+t,0),Q=(0,r+s,t).$$Consider the equation of $(BQMP)$. By putting in $B$, $v=0$. By putting in $P$, we have $u=c^2(1-r)$. By plugging in $Q$, we have $w=a^2(1-t).$ Therefore, the equation of the circle is $$(x+y+z)(c^2(1-r)x+a^2(1-t)z)=a^2yz+b^2xz+c^2xy.$$Plugging in $(1/2,0,1/2)$ for $M$ reveals that $$b^2=2(c^2(1-r)+a^2(1-t)) (*).$$We will return to the starred equation later.

For now, note that $$\overrightarrow{BT}=(r,s-1,t),$$so $$BT^2=-(a^2(s-1)t+b^2rt+c^2r(s-1)).$$Let's expand this: $$-(\mathbf{a^2st+b^2rt+c^2rs}-a^2t-c^2r).$$Since $(r,s,t)$ lies on the circumcircle, the bolded things sum to 0, so this is just $$BT^2=a^2t+c^2r.$$Let's now go back to the starred equation: $$b^2=2(c^2(1-r)+a^2(1-t)).$$Expanding, this becomes $$b^2=2(a^2+c^2-a^2t-c^2r)$$$$b^2=2(a^2+c^2-BT^2)$$$$BT^2=a^2+c^2-\frac{1}{2}b^2.$$It is well known that $$BM^2=\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2,$$so our answer is $\sqrt{2}$ and we are done.
This post has been edited 1 time. Last edited by john0512, Jan 26, 2023, 3:42 AM
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Ibrahim_K
62 posts
#54
Y by
Similar bary bash:
$A=(1,0,0),B=(0,1,0),C=(0,0,1),M=(\frac{1}{2},0,\frac{1}{2}),T=(m,n,k)$. Then equation of $PT$ and $QT$ is:
$$PT:(n+k)x-m(y+z)=0 \implies P=(m,1-m,0) \qquad QT:(m+n)z-k(x+y)=0 \implies Q=(0,1-k,k)$$Let $(BPMQ):-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz)=0.$ By plugging coordinates of $M,P$and $Q$ we get:
$$u+w=\frac{b^2}{2} \qquad c^2m=c^2-u \qquad a^2k=a^2-w$$Also since $T \in (ABC)$ we have $a^2nk+b^2mk+c^2mn=0$. Thus by distance formula:
$$BT^2=-a^2(n-1)k-b^2mk-c^2m(n-1)=-a^2nk-b^2mk-c^2mn+a^2k+c^2m=a^2k+c^2m=a^2+c^2-(u+w)=a^2+c^2-\frac{b^2}{2}$$On the other hand $BM^2=\frac{1}{2}(a^2+c^2-\frac{b^2}{2}).$ Hence $\frac{BT}{BM}=\sqrt 2 \qquad \blacksquare$
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awesomeming327.
1736 posts
#55 • 1 Y
Y by GeoKing
Let $X$ be $BM$ intersect $(ABC)$. Let $Z$ be center of $BPTQ$ and $Y$ be $BZ$ intersect $(BPQ)$. Let $(ABC)$ and $(BPQ)$ intersect again at $E$ then $E$ is center of spiral similarity taking $ATXCM$ to $PYMQZ$. Clearly, $\angle ZMB$ is the angle of the spin about $E$, since $ZM$ is taken to $MX$. Since $ZY$ is taken to $MT$, $\angle ZTM$ is equal to that. Thus, $BM$ is tangent to $ZMT$ and so the answer is just $\sqrt{2}$.
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vsamc
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#57
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Solution
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popop614
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#58 • 1 Y
Y by GeoKing
Great. The answer is $\sqrt{2}$ only.
Let $E$ and $F$ lie on $BA$ and $BC$ such that $M$ is the $B$-Dumpty point of $BEF$, and let $K$ be the reflection of $B$ over $M$, hence on $(BEF)$.

We now claim that $T$ lies on $EF$. Let $P'$ and $Q'$ denote the images of $P$ and $Q$ under a scale 2 homothety at $B$, so it suffices to prive $EF$ bisects $P'Q'$. But applying phantom points we observe that the intersection of $P'Q'$ and $EF$ is on $(KEP')$ and $(KFQ')$. The result is true after a very brief law of sines computation.

Take an inversion at $B$ with radius $BM\sqrt{2}$. $BEF$ becomes through a line antiparallel to $EF$ through $M$, whence $(E, A)$ amd $(F, C)$ swap. Hence line $EF$ maps to $(ABC)$. But this implies if $T$ lies on said circumcircle, $BT = BM\sqrt{2}$, as desired.
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cursed_tangent1434
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#59
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Actually trivial by bary. The answer is $\sqrt{2}$ only. We set $A=(1,0,0)$ and etc. Then, $M=(1:0:1)$. Now, let $T=(r,s,t)$ for real numbers $r,s,t$ such that $r+s+t=1$. Now, since $\overline{PT} \parallel \overline{BC}$, $P=(r,t_1,0)$ for some real number $t_1$ which then implies $t_1=s+t$ so $P=(r,s+t,0)$. Similarly, $Q=(0,r+s,t)$. Now, by Stewart's theorem, we can compute $BM^2 = \frac{2a^2-b^2+2c^2}{4}$. We consider the displacement vector $\overrightarrow{BT} = (r,s-1,t)=(r,-(r+t),t)$. Then, by the barycentric distance formula, we have that
\[BT^2 = a^2(r+t)t-b^2rt+c^2(r+t)t = a^2t+c^2r\]where the second equality follows from the fact that $a^2st+b^2rt+c^2rs=0$ since $T$ lies on the circumcircle. Now, since $BPMQ$ is cyclic, we consider the equation of this circle
\[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]where $v=0$ immediately since $B$ lies on the circle. Further, since $P$ lies on the circle we have
\begin{align*}
    -c^2r(s+t) + ur &=0\\
    u &= c^2(s+t)
\end{align*}similarly, we also have that since $Q$ lies on the circle, $w=a^2(r+s)$. Thus, the equation of the circle is simply,
\[-a^2yz-b^2xz-c^2xy+(c^2(s+t)x+a^2(r+s)z)(x+y+z)=0\]which since $M$ lies on this circle implies,
\begin{align*}
    -b^2 + 2(c^2(s+t)+a^2(r+s)) &=0\\
    -b^2 + 2(c^2 (1-r)+ a^2 (1-t)) &= 0\\
    c^2 + a^2 - (a^2t+c^2r) &= \frac{b^2}{2}\\
    a^2t + c^2 r &= \frac{2a^2 - b^2 + 2c^2}{2}
\end{align*}But this is precisely the length of $BT^2$! So, we have that
\[BT^2 = a^2t + c^2 r =  \frac{2a^2 - b^2 + 2c^2}{2} = 2\left(\frac{2a^2-b^2+2c^2}{4}\right) = 2BM^2\]Thus, $\frac{BT}{BM} = \sqrt{2}$ which finishes the problem.
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Aiden-1089
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#60
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I changed it to being $A$-centered for convenience, $P$ lies on $AB$ and $Q$ lies on $AC$.

Let $(A)$ be the circle with centre $A$ and radius $0$. Denote by $Pow_{\omega}(P)$ as the power of a point $P$ wrt circle $\omega$.
Define $f(X)=Pow_{(A)}(X)-Pow_{(ABC)}(X)$.
By linearity of pop, we have that $f(A)+f(T)=f(P)+f(Q)$. So $$AT^2=PA^2-(-PA \cdot PB)+QA^2-(-QA \cdot QB)=PA \cdot AB + QA \cdot AC.$$Now let $R \neq M$ be the second intersection of $\omega$ with $BC$, then $$PB \cdot AB + QC \cdot AC = BM \cdot BR + CM \cdot CR = \frac{BC^2}{2}.$$It follows that $AT^2+\frac{BC^2}{2}=AB^2+AC^2 \implies AT^2=AB^2+AC^2-\frac{BC^2}{2}$. By Stewart's theorem, $\frac{BC}{2}(AB^2 +AC^2)=BC \cdot (AM^2+ \frac{BC^2}{4}) \implies 2AM^2=AB^2+AC^2-\frac{BC^2}{2}=AT^2$.
Hence $\frac{AT}{AM}=\sqrt{2}$, so we are done. $\square$
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cj13609517288
1924 posts
#61
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WHAT

We employ barycentric coordinates wrt $ABC$. The equation of the circle is
\[-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z)=0\]where $u+w=\frac{b^2}{2}$.

Let $P=(p,1-p,0)$ and $Q=(0,1-q,q)$. Then
\[-c^2p(1-p)+up=0\Longrightarrow c^2(1-p)=u\]\[-a^2q(1-q)+wq=0\Longrightarrow a^2(1-p)=w\]Therefore,
\[c^2(1-p)+a^2(1-q)=\frac{b^2}{2}\Longrightarrow c^2p+a^2q=c^2+a^2-\frac{b^2}{2}.\]Also, note that $T=(p,1-p-q,q)$, so
\[a^2(1-p-q)q+b^2pq+c^2(1-p-q)p=0\]\[-a^2(-p-q)q-b^2pq-c^2(-p-q)p=a^2q+c^2p=c^2+a^2-\frac{b^2}{2}.\]That looks suspiciously like the distance formula! Indeed, we get that
\[BT=\sqrt{c^2+a^2-\frac{b^2}{2}}.\]Also, it is well known that
\[BM=\frac{1}{\sqrt2}\cdot\sqrt{c^2+a^2-\frac{b^2}{2}}.\]Therefore, $\frac{BT}{BM}=\boxed{\sqrt2}$.
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SomeonesPenguin
129 posts
#62 • 1 Y
Y by zzSpartan
bary
This post has been edited 1 time. Last edited by SomeonesPenguin, Nov 7, 2024, 3:41 PM
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eg4334
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#63
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Use bary wrt to $ABC$. The equation of $\omega$ is $-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z) = 0$ for $u+w=\frac{b^2}{2}$. Then its trivial to find that $Q = (0, \frac{w}{a^2}, 1 - \frac{w}{a^2}), P = (1 - \frac{u}{c^2}, \frac{u}{c^2}, 0)$. But $P+Q=B+T$, so $T = (1 - \frac{u}{c^2}, \frac{w}{a^2} + \frac{u}{c^2} - 1, 1 - \frac{w}{a^2})$. For simplicity, let this be $T = (x, y, z)$. Then $a^2yz+b^2xz+c^2xy=0$. But by bary distance formula \begin{align*}
\frac{BT}{BM} &= \sqrt{\frac{-a^2(y-1)z - b^2xz - c^2x(y-1)}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \sqrt{\frac{c^2x+a^2z}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \sqrt{\frac{c^2+a^2-(u+w)}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \frac{a^2+c^2-\frac{b^2}{2}}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}} \\
&= \boxed{\sqrt{2}}
\end{align*}
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ja.
23 posts
#64 • 1 Y
Y by navi_09220114
Let $N$ be the midpoint of $PQ$. Then, as $P$ moves, the shape of $MPNQ$ is constant, and the locus of $P$ is a line so the locus of $N$ Is a line. Then, the locus of $T$ must be a line too. Let $\omega_1$ be the circle through $M$ tangent to $BC$ at $B$, and $\omega_2$ is tangent to $BA$ at $B$. Take $P=B$, $Q=B$ gives that $T$ lies on line $DE$ where $D$ is the intersection of $\omega_1$ with $BA$ and $E$ is the intersection of $\omega_2$ with $BC$.

Let $B'$ be the reflection of $B$ in $M$, then $\angle{AB'M}=\angle{MBC}=\angle{ADM}$ hence $AMB'D$ cyclic, then $BA\cdot BD=2BM^2$. Similarly, we will have $BC\cdot BE=2BM^2$.

Finally, consider the inversion at $B$ with radius $BM\sqrt2$, then $(BAC)$ is swapped with $DE$, so $T$ is fixed. Therefore, $BT=BM\sqrt2$.
This post has been edited 1 time. Last edited by ja., Apr 25, 2025, 8:09 AM
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