Find all possible values of BT/BM

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va2010

1276 posts

va2010

#1 h Jul 7, 2016, 8:59 PM • 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

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v_Enhance

#2 h Jul 7, 2016, 9:00 PM • 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

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

TelvCohl

#3 h Jul 7, 2016, 9:12 PM • 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

mjuk

#4 h Jul 7, 2016, 9:16 PM • 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

navi_09220114

#5 h Jul 8, 2016, 3:37 AM • 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

EulerMacaroni

#6 h Jul 8, 2016, 6:19 AM • 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

WizardMath

#7 h Jul 8, 2016, 9:16 AM • 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

anantmudgal09

#8 h Jul 8, 2016, 10:49 AM • 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

kapilpavase

#9 h Jul 8, 2016, 1:35 PM • 8 Y

Y by PRO2000, tenplusten, Ankoganit, Orkhan-Ashraf_2002, Jalil_Huseynov, Kaimiaku, Adventure10, CyclicISLscelesTrapezoid

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2., xmax = 11., ymin = 0., ymax = 8.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((6.331211478628279,3.9090896183427657), 3.6128252550638846), blue); draw(circle((5.406292594169632,4.696263381502455), 2.503407971114143), ffxfqq); draw((4.532022448266124,7.042047510918392)--(6.747139816608675,0.320286204767108)); draw((6.747139816608675,0.320286204767108)--(3.4209710969677025,3.1712879644593737)); draw((4.532022448266124,7.042047510918392)--(3.1442802912498613,2.207332899377859)); draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); draw(circle((5.431616963447202,5.47556856463058), 1.8064126275319419), linetype("4 4")); draw((7.858191167907094,4.191045751226126)--(6.747139816608675,0.320286204767108)); /* dots and labels */ dot((4.532022448266124,7.042047510918392),linewidth(3.pt) + dotstyle); label("$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

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Number1

355 posts

Number1

#12 h Jul 9, 2016, 9:10 PM • 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

Ankoganit

#13 h Jul 22, 2016, 9:37 AM • 2 Y

Y by Adventure10, Mango247

This is also India TST 2016 Day 3 Problem 2.

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rterte

209 posts

rterte

#15 h Oct 9, 2016, 2:03 AM • 2 Y

Y by Adventure10, Mango247

Can we use Cartesian coordinate on this problem?

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tenplusten

1000 posts

tenplusten

#17 h Dec 4, 2016, 12:27 PM • 4 Y

Y by kapilpavase, adorefunctionalequation, Adventure10, Mango247

kapilpavase wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2., xmax = 11., ymin = 0., ymax = 8.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((6.331211478628279,3.9090896183427657), 3.6128252550638846), blue); draw(circle((5.406292594169632,4.696263381502455), 2.503407971114143), ffxfqq); draw((4.532022448266124,7.042047510918392)--(6.747139816608675,0.320286204767108)); draw((6.747139816608675,0.320286204767108)--(3.4209710969677025,3.1712879644593737)); draw((4.532022448266124,7.042047510918392)--(3.1442802912498613,2.207332899377859)); draw((3.1442802912498613,2.207332899377859)--(9.70248113005252,2.6102257836729033)); draw((4.532022448266124,7.042047510918392)--(9.70248113005252,2.6102257836729033)); draw((3.4209710969677025,3.1712879644593737)--(7.858191167907094,4.191045751226126)); draw(circle((5.431616963447202,5.47556856463058), 1.8064126275319419), linetype("4 4")); draw((7.858191167907094,4.191045751226126)--(6.747139816608675,0.320286204767108)); /* dots and labels */ dot((4.532022448266124,7.042047510918392),linewidth(3.pt) + dotstyle); label("$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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Ferid.---.

1008 posts

Ferid.---.

#18 h Feb 18, 2017, 2:18 PM • 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

Ferid.---.

#20 h Feb 19, 2017, 3:49 PM • 2 Y

Y by Adventure10, Mango247

Please,help me.

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bjump

1028 posts

bjump

#48 h Aug 24, 2022, 8:27 PM

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

3302 posts

fuzimiao2013

#50 h Sep 1, 2022, 3:31 AM • 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

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Knty2006

#51 h Dec 24, 2022, 9:13 AM

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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

Taco12

#52 h Jan 16, 2023, 12:43 AM • 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

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john0512

#53 h Jan 26, 2023, 3:36 AM

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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

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Ibrahim_K

#54 h May 21, 2023, 3:36 PM

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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.

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awesomeming327.

#55 h May 24, 2023, 11:17 PM • 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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vsamc

#57 h Jul 22, 2023, 2:25 PM

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Solution

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popop614

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popop614

#58 h Dec 29, 2023, 5:26 PM • 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

634 posts

cursed_tangent1434

#59 h May 2, 2024, 8:34 AM

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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

293 posts

Aiden-1089

#60 h May 11, 2024, 5:58 PM

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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

1919 posts

cj13609517288

#61 h Sep 5, 2024, 1:38 PM

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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

SomeonesPenguin

#62 h Nov 5, 2024, 6:32 PM • 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

637 posts

eg4334

#63 h Mar 6, 2025, 1:36 AM

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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.

22 posts

ja.

#64 h Apr 25, 2025, 8:08 AM • 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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