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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
Inequality in triangle
Nguyenhuyen_AG   0
23 minutes ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
23 minutes ago
0 replies
D,E,F are collinear.
TUAN2k8   2
N 27 minutes ago by TUAN2k8
Source: Own
Help me with this:
2 replies
TUAN2k8
May 28, 2025
TUAN2k8
27 minutes ago
Combinatorial identity
MehdiGolafshan   4
N 34 minutes ago by watery
Let $n$ is a positive integer. Prove that
$$\sum_{k=0}^{n-1}\frac{1}{k+1}\binom{n-1}{k} = \frac{2^n-1}{n}.$$
4 replies
MehdiGolafshan
Jan 16, 2023
watery
34 minutes ago
JBMO Shortlist 2023 G7
Orestis_Lignos   7
N an hour ago by tilya_TASh
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
7 replies
Orestis_Lignos
Jun 28, 2024
tilya_TASh
an hour ago
Reflected point lies on radical axis
Mahdi_Mashayekhi   5
N an hour ago by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
an hour ago
Find the value
sqing   18
N an hour ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
an hour ago
Number Theory
fasttrust_12-mn   14
N 2 hours ago by Namisgood
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
14 replies
fasttrust_12-mn
Aug 15, 2024
Namisgood
2 hours ago
find question
mathematical-forest   5
N 2 hours ago by Jupiterballs
Are there any contest questions that seem simple but are actually difficult? :-D
5 replies
mathematical-forest
Thursday at 10:19 AM
Jupiterballs
2 hours ago
Own made functional equation
Primeniyazidayi   10
N 2 hours ago by Phat_23000245
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
10 replies
Primeniyazidayi
May 26, 2025
Phat_23000245
2 hours ago
Tough inequality
TUAN2k8   4
N 3 hours ago by Phat_23000245
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
Phat_23000245
3 hours ago
Guess period of function
a1267ab   9
N 3 hours ago by HamstPan38825
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
9 replies
a1267ab
Dec 14, 2024
HamstPan38825
3 hours ago
Inequality with abc=1
tenplusten   11
N 3 hours ago by sqing
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
11 replies
tenplusten
May 15, 2016
sqing
3 hours ago
Central sequences
EeEeRUT   13
N 4 hours ago by v_Enhance
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
13 replies
EeEeRUT
Apr 16, 2025
v_Enhance
4 hours ago
Interesting inequality
sqing   0
4 hours ago
Source: Own
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+  c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
0 replies
sqing
4 hours ago
0 replies
Find all possible values of BT/BM
va2010   54
N Thursday at 5:39 PM by lpieleanu
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
lpieleanu
Thursday at 5:39 PM
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
6882 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
487 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 */
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); 
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draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); 
draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); 
draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); 
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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.); 
 /* draw figures */
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)); 
draw((4.532022448266124,7.042047510918392)--(3.1442802912498613,2.207332899377859)); 
draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); 
draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); 
draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); 
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label("$A$", (4.660735465368374,7.223225436821573), NE * labelscalefactor); 
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label("$T$", (6.855333779736237,0.4911468238282515), NE * labelscalefactor); 
dot((3.4209710969677025,3.1712879644593737),linewidth(3.pt) + dotstyle); 
label("$P$", (3.533779574206499,3.3381932857108896), NE * labelscalefactor); 
dot((7.858191167907094,4.191045751226126),linewidth(3.pt) + dotstyle); 
label("$Q$", (7.982289670898113,4.376178974938934), NE * labelscalefactor); 
dot((3.838151369757993,4.624690205148125),linewidth(3.pt) + dotstyle); 
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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