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An interesting geometry
k.vasilev   19
N 8 minutes ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
8 minutes ago
Bosnia and Herzegovina JBMO TST 2016 Problem 3
gobathegreat   3
N an hour ago by Sh309had
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2016
Let $O$ be a center of circle which passes through vertices of quadrilateral $ABCD$, which has perpendicular diagonals. Prove that sum of distances of point $O$ to sides of quadrilateral $ABCD$ is equal to half of perimeter of $ABCD$.
3 replies
gobathegreat
Sep 16, 2018
Sh309had
an hour ago
Power Of Factorials
Kassuno   180
N an hour ago by maromex
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
180 replies
1 viewing
Kassuno
Jul 17, 2019
maromex
an hour ago
100 card with 43 having odd integers on them
falantrng   7
N an hour ago by Just1
Source: Azerbaijan JBMO TST 2018, D2 P4
In the beginning, there are $100$ cards on the table, and each card has a positive integer written on it. An odd number is written on exactly $43$ cards. Every minute, the following operation is performed: for all possible sets of $3$ cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by $2^{2018}.$ Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by $2^{2018}.$
7 replies
falantrng
Aug 1, 2023
Just1
an hour ago
GCD and LCM operations
BR1F1SZ   1
N an hour ago by WallyWalrus
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
1 reply
BR1F1SZ
Saturday at 11:24 PM
WallyWalrus
an hour ago
x^2+4 = y^5
Valentin Vornicu   14
N an hour ago by Rayvhs
Source: Balkan MO 1998, Problem 4
Prove that the following equation has no solution in integer numbers: \[ x^2 + 4 = y^5.  \] Bulgaria
14 replies
Valentin Vornicu
Apr 24, 2006
Rayvhs
an hour ago
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
2 hours ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
2 hours ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N 3 hours ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
3 hours ago
the same prime factors
andria   5
N 3 hours ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
3 hours ago
A sharp estimation of the product
mihaig   0
3 hours ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
3 hours ago
0 replies
xf(x) + f ^2(y) +2 xf(y) perfect square for all positive integers x,y
parmenides51   9
N 3 hours ago by Gaunter_O_Dim_of_math
Source: Balkan BMO Shortlist 2017 N2
Find all functions $f :Z_{>0} \to Z_{>0}$ such that the number $xf(x) + f ^2(y) + 2xf(y)$ is a perfect square for all positive integers $x,y$.
9 replies
parmenides51
Aug 1, 2019
Gaunter_O_Dim_of_math
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
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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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This post has been edited 4 times. Last edited by mjuk, Jul 7, 2016, 9:36 PM
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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 */
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real labelscalefactor = 0.5; /* changes label-to-point distance */
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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 */ 
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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); 
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dot((4.532022448266124,7.042047510918392),linewidth(3.pt) + dotstyle); 
label("$A$", (4.660735465368374,7.223225436821573), NE * labelscalefactor); 
dot((3.1442802912498613,2.207332899377859),linewidth(3.pt) + dotstyle); 
label("$B$", (3.266868968405002,2.389177798416677), NE * labelscalefactor); 
dot((9.70248113005252,2.6102257836729033),linewidth(3.pt) + dotstyle); 
label("$C$", (9.821007177530646,2.774715340129951), NE * labelscalefactor); 
dot((6.423380710651191,2.4087793415253813),linewidth(3.pt) + dotstyle); 
label("$M$", (6.529109705978852,2.596774936262286), NE * labelscalefactor); 
dot((6.747139816608675,0.320286204767108),linewidth(3.pt) + dotstyle); 
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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#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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#20 • 2 Y
Y by Adventure10, Mango247
Please,help me.
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