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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.

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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
Linearity in a specific function
youochange   2
N a minute ago by youochange
$f(x-c)+c=f(x)$ $and f:\mathbb Z \to \mathbb Z$

Does this mean f(x) is linear?
2 replies
youochange
18 minutes ago
youochange
a minute ago
Serbian selection contest for the BMO 2025 - P1
OgnjenTesic   3
N 7 minutes ago by EeEeRUT
Given is triangle $ABC$ with centroid $T$, such that $\angle BAC + \angle BTC = 180^\circ$. Let $G$ and $H$ be the second points of intersection of lines $CT$ and $BT$ with the circumcircle of triangle $ABC$, respectively. Prove that the line $GH$ is tangent to the Euler circle of triangle $ABC$.

Proposed by Andrija Živadinović
3 replies
OgnjenTesic
Apr 7, 2025
EeEeRUT
7 minutes ago
Hard combi
EeEApO   8
N 38 minutes ago by navier3072
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
8 replies
EeEApO
May 8, 2025
navier3072
38 minutes ago
Reflections of AB, AC with respect to BC and angle bisector of A
falantrng   29
N 41 minutes ago by cursed_tangent1434
Source: BMO 2024 Problem 1
Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the
$A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points
$E$ and $F$ respectively. A line through $D$ meets $AC$ and $AB$ at $G$ and $H$ respectively such that $G$
lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of
$\triangle EDG$ and $\triangle FDH$ are tangent to each other.
29 replies
falantrng
Apr 29, 2024
cursed_tangent1434
41 minutes ago
JBMO Combinatorics vibes
Sadigly   1
N 44 minutes ago by Royal_mhyasd
Source: Azerbaijan Senior NMO 2018
Numbers $1,2,3...,100$ are written on a board. $A$ and $B$ plays the following game: They take turns choosing a number from the board and deleting them. $A$ starts first. They sum all the deleted numbers. If after a player's turn (after he deletes a number on the board) the sum of the deleted numbers can't be expressed as difference of two perfect squares,then he loses, if not, then the game continues as usual. Which player got a winning strategy?
1 reply
Sadigly
Yesterday at 9:53 PM
Royal_mhyasd
44 minutes ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   3
N an hour ago by cursed_tangent1434
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
3 replies
1 viewing
parmenides51
Dec 11, 2018
cursed_tangent1434
an hour ago
Six variables
Nguyenhuyen_AG   2
N an hour ago by arqady
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
2 replies
Nguyenhuyen_AG
Yesterday at 5:09 AM
arqady
an hour ago
Brilliant guessing game on triples
Assassino9931   2
N an hour ago by Mirjalol
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
2 replies
Assassino9931
Saturday at 9:46 AM
Mirjalol
an hour ago
ISI UGB 2025 P5
SomeonecoolLovesMaths   4
N an hour ago by Shiny_zubat
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
4 replies
SomeonecoolLovesMaths
Yesterday at 11:15 AM
Shiny_zubat
an hour ago
ISI UGB 2025 P2
SomeonecoolLovesMaths   6
N an hour ago by quasar_lord
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.
6 replies
SomeonecoolLovesMaths
Yesterday at 11:16 AM
quasar_lord
an hour ago
ISI UGB 2025 P6
SomeonecoolLovesMaths   3
N an hour ago by Shiny_zubat
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
3 replies
SomeonecoolLovesMaths
Yesterday at 11:18 AM
Shiny_zubat
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   5
N an hour ago by Royal_mhyasd
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)$ is a perfect square, then $pq + 1$ is also a perfect square.
5 replies
AlperenINAN
Yesterday at 7:51 PM
Royal_mhyasd
an hour ago
d+2 pts in R^d can partition
EthanWYX2009   0
3 hours ago
Source: Radon's Theorem
Show that: any set of $d + 2$ points in $\mathbb R^d$ can be partitioned into two sets whose convex hulls intersect.
0 replies
EthanWYX2009
3 hours ago
0 replies
hard inequality omg
tokitaohma   4
N 3 hours ago by arqady
1. Given $a, b, c > 0$ and $abc=1$
Prove that: $ \sqrt{a^2+1} + \sqrt{b^2+1} + \sqrt{c^2+1} \leq \sqrt{2}(a+b+c) $

2. Given $a, b, c > 0$ and $a+b+c=1 $
Prove that: $ \dfrac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}} + \dfrac{\sqrt{b^2+2bc}}{\sqrt{c^2+2a^2}} + \dfrac{\sqrt{c^2+2ca}}{\sqrt{a^2+2b^2}} \geq \dfrac{1}{a^2+b^2+c^2} $
4 replies
tokitaohma
Yesterday at 5:24 PM
arqady
3 hours ago
Find all possible values of BT/BM
va2010   54
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 $\frac{BT}{BM}$.
54 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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va2010
1276 posts
#1 • 13 Y
Y by mathmaths, tenplusten, Davi-8191, anantmudgal09, mathematicsy, Jalil_Huseynov, selenium_e, Adventure10, Mango247, lian_the_noob12, Rounak_iitr, Funcshun840, ehuseyinyigit
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 $\frac{BT}{BM}$.
This post has been edited 1 time. Last edited by v_Enhance, Jul 7, 2016, 9:23 PM
Reason: improve title
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v_Enhance
6877 posts
#2 • 19 Y
Y by PRO2000, Davi-8191, Ferid.---., Alexandros2233, B.J.W.T, BobaFett101, WizardMath, Imayormaynotknowcalculus, Cindy.tw, PIartist, v4913, hsiangshen, primesarespecial, suh, selenium_e, Adventure10, Mango247, Asynchrone, endless_abyss
whooo let's go bary :D

Denote by $X$ the second intersection of $(BPMQ)$ with $\overline{AC}$. Let $T = (u,v,w)$ thus $a^2vw+b^2wu+c^2uv=0$. Then $\overline{PT} \parallel \overline{BC} \implies P = (u:w+v:0)$. Analogously, $Q = (0:u+v,w)$. So, \[ AX = \frac{AB \cdot AP}{AM} = \frac{2c^2}{b} (v+w) \quad\text{and}\quad CX = \frac{CB \cdot CQ}{CM} = \frac{2a^2}{b} (v+u). \]Adding these implies that $\frac{1}{2} b^2 = (v+w)c^2 + (v+u)a^2$.

However, by barycentric distance formula, \[	BT^2 = -a^2(v-1)w-b^2wv-c^2u(v-1) = a^2w+c^2u + \underbrace{-a^2vw-b^2wu-c^2uv}_{=0}. \]Thus, adding gives $\frac{1}{2} b^2 + BT^2 = a^2+c^2$, so $BT^2 = a^2+c^2 - \frac{1}{2} b^2 = 2BM^2$, thus $BT/BM =\sqrt2$ is the only possible value.
This post has been edited 1 time. Last edited by v_Enhance, Jul 7, 2016, 9:01 PM
Reason: smiley
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TelvCohl
2312 posts
#3 • 13 Y
Y by baladin, Alexandros2233, jt314, naw.ngs, enhanced, r_ef, Cindy.tw, A64298347, Jalil_Huseynov, selenium_e, Adventure10, This_deserves_a_like, Funcshun840
Let $ N $ be the midpoint of $ PQ $ and let $ E $ $ \equiv $ $ BM $ $ \cap $ $ \odot (ABC), $ $ S $ $ \equiv $ $ BN $ $ \cap $ $ \odot (BPQ). $ Since $$ \left\{\begin{array}{cc} \measuredangle MPQ = \measuredangle MBC = \measuredangle EAC, \measuredangle SPQ = \measuredangle SBC = \measuredangle TAC \\\\  \measuredangle MQP = \measuredangle MBA = \measuredangle ECA, \measuredangle SQP = \measuredangle SBA = \measuredangle TCA \end{array}\right\| \Longrightarrow ACETM \stackrel{+}{\sim} PQMSN ,$$so $ \measuredangle BMN $ $ = $ $ \measuredangle (EM, MN) $ $ = $ $ \measuredangle (TM,SN) $ $ = $ $ \measuredangle MTB $ $ \Longrightarrow $ $ BM $ is tangent to $ \odot (MNT) $ at $ M, $ hence we conclude that $$ \frac{1}{2} \cdot {BT}^2 = BN \cdot BT = {BM}^2 \Longrightarrow \frac{BT}{BM} = \sqrt{2}. $$
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mjuk
196 posts
#4 • 3 Y
Y by PcelicaMaja, Adventure10, Rounak_iitr
Let $D=AC\cap PQ$. Let $N$ be midpoint of $PQ$ and let $S$ be a point such that $\triangle ACS\stackrel{+}{\sim}\triangle PQT$.
Let $R$ be Miquel point of $CAPQ\implies R\in \odot ABC$, $R\in \odot CDQ$. $R$ is center of spiral similarity sending $CQ\rightarrow AP$, so it's also center of spiral similarity sending $MN\rightarrow AP\implies R\in \odot DMN$.
Let $T'=MS\cap BT$. Obviously $\triangle NPT\sim \triangle MAS\implies  \angle DNT'=\angle DMT'\implies T'\in \odot DMN$.
$\angle BT'R=\angle NT'R= \angle NDR=\angle QDR=\angle QCR=\angle BCR\implies T'\in \odot ABC\implies T\equiv T'$
$\angle MDR=\angle CDR=\angle BQR=\angle BMR \implies BM$ touches $\odot TMN$
$\implies BM^2=BN\cdot BT=\frac{BT^2}{2}\implies \frac{BT}{BM}=\sqrt{2} \blacksquare$
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navi_09220114
479 posts
#5 • 2 Y
Y by Chokechoke, Adventure10
Answer. $\sqrt{2}$.

Solution. Let us prove a well known spiral similarity lemma.

Lemma. Let $AB$ and $CD$ be two segments, and let lines $AC$ and $BD$ meet at $X$. Let circumcircle of $ABX$ and $CDX$ meet again at $O$. Then $O$ is the center of the spiral similarity that carries $AB$ to $CD$.

Proof of Lemma. Since $ABOX$ and $CDXO$ are cyclic. we have $\angle OBD= \angle OAC$ and $\angle OCA=\angle ODB$. It follows that $\triangle AOC$ and $\triangle BOD$ are similar, thus the result.

Back to the main proof. Suppose that $T$ lies on $(ABC)$. Let $PQ\cap BT=N$, $(ABC)\cap (APQ)=X\neq A$, $PQ\cap AC=Y$. Clearly since $BPTQ$ is a parallelogram, we have $BT=2BN$. We will prove that points $X, Y$ lies on $(TMN)$.

By the lemma, $X$ is the center of spiral similarity that maps segment $PQ$ to $AC$. Note that this spiral similarity also maps the midpoint of $PQ$ to midpoint of $AC$, which is $N$ to $M$. So it also maps segment $PN$ to $AM$, which implies $\triangle XMA\sim\triangle XNP$. Similarly, $X$ is also the center of spiral similarity that maps $AP$ to $BQ$. Notice that this spiral similarity also maps $A$ to $M$ and $P$ to $N$ because $M$ and $N$ are midpoints of $AP$ and $BQ$ respectively, so we also conclude that $\triangle XPA\sim\triangle XNM$. (Also known as the Averaging Principle.)

So using directed angles, $\angle (XM, MN)=\angle (XA, AP)=\angle (XT, TB)=\angle (XT, TN)$, so $X$ lies on $(TMN)$. Likewise, $\angle (YN, NX)=\angle (PN, NX)=\angle (AM, MX)=\angle (YM,MX)$, so $Y$ lies on $(TMN)$. Now, note that $\angle (BM,MX)=\angle (BA,AX)=\angle (PA,AX)=\angle (NM,MX)$, so $BM$ is tangent to circle $(TMN)$.

This means $2BM^2=2BN\cdot BT=BT^2$, which gives $\displaystyle \frac{BT}{BM}=\sqrt{2}$, as desired. Q.E.D
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EulerMacaroni
851 posts
#6 • 2 Y
Y by Zaro23, Adventure10
Let $X$ be the Miquel point of complete quadrilateral $\{PC, CA, AQ, QP\}$ with $R$ as the midpoint of $PQ$. Since spiral similarities are linear functions in the plane, it follows that $X$ is also the spiral center sending $AQ$ to $MR$, so $G\equiv PQ\cap AC$ gives $X\in\odot(GRM)$. Then $$\angle BMX=\angle BPX=\pi-\angle XPC=\pi-\angle XRM$$so $BM$ is tangent to $\odot(XRM)$. Since $T\in \odot(XRM)$, $BM^2=BR\cdot BT=\frac{BT^2}{2}$, so the answer is $\sqrt{2}$.
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WizardMath
2487 posts
#7 • 1 Y
Y by Adventure10
Bary was the first thing that came to my mind when I saw this one in the TST. Since v_Enhance already posted the bash I won't do it again but just remark that this is an excellent bary tutorial problem.
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anantmudgal09
1980 posts
#8 • 3 Y
Y by TheOneYouWant, Adventure10, Mango247
Here is my approach.

Firstly observe that since $B,P,M,Q$ are concyclic, the midpoint $K$ of $PQ$ varies on a line as $P,Q$ vary on $BA,BC$ respectively. Indeed, this follows since triangle $MPQ$ has a fixed shape and spiral similarity moves each linear combination uniformly.

Let $(BAM)$ meet $BC$ again at $X$ and $(BCM)$ meet $BA$ again at $Y$. Let $U,V$ be the midpoints of $AX,CY$. Then, the locus of $K$ is the line $UV$. Let $L,N$ be the midpoints of $BA,BC$. Consider the circle $\gamma=(BLN)$ and $\omega=(B,\frac{BM}{\sqrt{2}})$. We claim that the line $UV$ is the radical axis of $\omega$ and $\gamma$. This shall establish the result; $\frac{BT}{BM}=\frac{1}{\sqrt{2}}$.

Indeed, we define the function $f$ from the Euclidean plane to the set of real numbers as follows: $f(Z)=p(Z,\gamma)-p(Z,\omega)$. It is clear that $f$ is a linear function in $Z$. We want to establish that $f(U)=f(V)=0$. Proving $f(U)=0$ suffices, since $f(V)=0$ shall follow analogously.

Notice that $f(U)=\frac{f(A)+f(X)}{2}$. We only need to ascertain ourselves that $f(A)=-f(X)$.

It is evident that $f(A)=-\frac{c^2}{2}+\frac{2a^2+2c^2-b^2}{8}$ and $f(X)=XB.XN-XB^2+\frac{2a^2+2c^2-b^2}{8}$. It is equivalent to showing that $XB.XN-XB.XB=-\frac{2a^2-b^2}{4}$. Notice that $XB-XN=-(BX+XN)=-BN=-\frac{a}{2}$ and $XB=a-\frac{b^2}{2a}$ and hence the claim holds.

Comment. This problem came in India TST 2016 and Romania TST 2016 as per my knowledge.
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kapilpavase
595 posts
#9 • 8 Y
Y by PRO2000, tenplusten, Ankoganit, Orkhan-Ashraf_2002, Jalil_Huseynov, Kaimiaku, Adventure10, CyclicISLscelesTrapezoid
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(9.cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
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pen dotstyle = black; /* point style */ 
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angle chasing yields $\Delta BTC \sim \Delta AQT$ so that $$ \dfrac{BT}{AQ}= \dfrac{TC}{QT}= \dfrac{BC}{AT}$$
Now take $X,Y,Z$ to be the midpts of $AB,AC,PQ$ resp. Similar angle chasing gives $\Delta PMQ \sim \Delta AYM$ so that $$\dfrac{AC}{PM}=\dfrac{AB}{MQ}=\dfrac{2AM}{PQ}$$
multiplying these two gives $$\dfrac{2AM.BC}{AT.PQ}=\dfrac{BT.AC}{AQ.PM}=\dfrac{TC.AB}{QT.MQ}=\dfrac{BT.AC+TC.AB}{AQ.PM+QT.MQ}=\dfrac{AT.BC}{AM.PQ}$$
so that $2AM^2=AT^2$ as desired :)
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Number1
355 posts
#12 • 2 Y
Y by Adventure10, Mango247
Rename $A$ and $B$.
https://www.geogebra.org/files/00/03/80/64/material-3806449.png?v=1468097902

Let tangent at $A$ on $\omega $ cuts cicumcircle $\mathcal{K}$ of $ABC$ at $A'$ and let $AM$ cut $\mathcal{K}$ at $M'$.
Let $R$ halves $PQ$ and let $\mathcal{K}$ meets $\omega $ second time at $S$. Then $S$ is center of spiral similarity $\mathcal{P}$, which maps $ \omega \longmapsto \mathcal{K} $ and
\begin{eqnarray*}
%  \omega &\longmapsto &\mathcal{K} \\
  P &\longmapsto & B\\
  Q &\longmapsto & C \\
  A &\longmapsto & A' \\
  R &\longmapsto & M \\
  M &\longmapsto & M'
\end{eqnarray*}Say $AR$ meets $A'M$ at $T'$. Since $\mathcal{P}$ send $AR$ to $A'M$ and $SA$ to $SA'$ we have $\angle AT'A' = \angle (AR,AM')= \angle ASA'$, thus $T'\in \mathcal{K}$, and so $T'=T$.

Finally since $\mathcal{P}$ preserves angles we have
$$ \angle RMA \stackrel{\mathcal{P}}{=} \angle MM'A' = \angle AM'A' = \angle ATA' =\angle RTM$$and this means $AT$ is tangent on circumcircle $\triangle MTR$, so $AM^2 = AT \cdot AR = {AT^2 \over 2} \Longrightarrow {AT \over AM} =\sqrt{2}$.
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Ankoganit
3070 posts
#13 • 2 Y
Y by Adventure10, Mango247
This is also India TST 2016 Day 3 Problem 2.
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rterte
209 posts
#15 • 2 Y
Y by Adventure10, Mango247
Can we use Cartesian coordinate on this problem?
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tenplusten
1000 posts
#17 • 4 Y
Y by kapilpavase, adorefunctionalequation, Adventure10, Mango247
kapilpavase wrote:
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(9.cm); 
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 = 2., xmax = 11., ymin = 0., ymax = 8.;  /* image dimensions */
pen ffxfqq = rgb(1.,0.4980392156862745,0.); 
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draw(circle((6.331211478628279,3.9090896183427657), 3.6128252550638846), blue); 
draw(circle((5.406292594169632,4.696263381502455), 2.503407971114143), ffxfqq); 
draw((4.532022448266124,7.042047510918392)--(6.747139816608675,0.320286204767108)); 
draw((6.747139816608675,0.320286204767108)--(3.4209710969677025,3.1712879644593737)); 
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draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); 
draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); 
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label("$P$", (3.533779574206499,3.3381932857108896), NE * labelscalefactor); 
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label("$Q$", (7.982289670898113,4.376178974938934), NE * labelscalefactor); 
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label("$X$", (3.948973849897716,4.791373250630153), NE * labelscalefactor); 
dot((7.117251789159322,4.826136647295648),linewidth(3.pt) + dotstyle); 
label("$Y$", (7.24087132144951,4.998970388475762), NE * labelscalefactor); 
dot((5.639581132437399,3.68116685784275),linewidth(3.pt) + dotstyle); 
label("$Z$", (5.758034622552305,3.8720144973138844), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]



angle chasing yields $\Delta BTC \sim \Delta AQT$ so that $$ \dfrac{BT}{AQ}= \dfrac{TC}{QT}= \dfrac{BC}{AT}$$
Now take $X,Y,Z$ to be the midpts of $AB,AC,PQ$ resp. Similar angle chasing gives $\Delta PMQ \sim \Delta AYM$ so that $$\dfrac{AC}{PM}=\dfrac{AB}{MQ}=\dfrac{2AM}{PQ}$$
multiplying these two gives $$\dfrac{2AM.BC}{AT.PQ}=\dfrac{BT.AC}{AQ.PM}=\dfrac{TC.AB}{QT.MQ}=\dfrac{BT.AC+TC.AB}{AQ.PM+QT.MQ}=\dfrac{AT.BC}{AM.PQ}$$
so that $2AM^2=AT^2$ as desired :)

Really GREAT Solution. İn the last step you used Ptolemy's theorem.
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Ferid.---.
1008 posts
#18 • 2 Y
Y by Adventure10, Mango247
v_Enhance wrote:
Let $T = (u,v,w)$ thus $a^2vw+b^2wu+c^2uv=0$. Then $\overline{PT} \parallel \overline{BC} \implies P = (u:w+v:0)$. .
Teacher Evan,How can we get $P=(u:w+v;0)$ and why $T(u,v,w)?$
Can you tell me,please.
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Ferid.---.
1008 posts
#20 • 2 Y
Y by Adventure10, Mango247
Please,help me.
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