Let ABC be an acute triangle and let M be the midpoint of AC. A circle ω 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 BTBM.
Problem
Source: 2015 ISL G4
Tags: geometry
08.07.2016 00:00
whooo let's go bary Denote by X the second intersection of (BPMQ) with ¯AC. Let T=(u,v,w) thus a2vw+b2wu+c2uv=0. Then ¯PT∥¯BC⟹P=(u:w+v:0). Analogously, Q=(0:u+v,w). So, AX=AB⋅APAM=2c2b(v+w)andCX=CB⋅CQCM=2a2b(v+u).Adding these implies that 12b2=(v+w)c2+(v+u)a2. However, by barycentric distance formula, BT2=−a2(v−1)w−b2wv−c2u(v−1)=a2w+c2u+−a2vw−b2wu−c2uv⏟=0.Thus, adding gives 12b2+BT2=a2+c2, so BT2=a2+c2−12b2=2BM2, thus BT/BM=√2 is the only possible value.
08.07.2016 00:12
Let N be the midpoint of PQ and let E ≡ BM ∩ ⊙(ABC), S ≡ BN ∩ ⊙(BPQ). Since {∡MPQ=∡MBC=∡EAC,∡SPQ=∡SBC=∡TAC∡MQP=∡MBA=∡ECA,∡SQP=∡SBA=∡TCA‖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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08.07.2016 00:16
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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08.07.2016 06:37
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
08.07.2016 09:19
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 XRMso 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}.
08.07.2016 12:16
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.
08.07.2016 13:49
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.
08.07.2016 16:35
[asy][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][/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
10.07.2016 00:10
Rename A and B. 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 RTMand 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}.
22.07.2016 12:37
This is also India TST 2016 Day 3 Problem 2.
09.10.2016 05:03
Can we use Cartesian coordinate on this problem?
04.12.2016 15:27
kapilpavase wrote: [asy][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][/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.
18.02.2017 17:18
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.
19.02.2017 18:49
Please,help me.
27.05.2017 14:18
Pure Bary: No synthetic observation whatsoever! So B = (0, 1, 0) and M = (\frac {1} {2}, 0, \frac {1} {2}). Take the arbitrary circle through B and M, which has equation -a^2yz-b^2zx-c^2xy + (ux+vy+wz)(x+y+z) = 0. We will now find relations of these constants. Plugging in B yields v = 0, and then plugging in M implies \frac {-b^2} {4} + \frac {u + w} {2} = 0, so u + w = \frac {b^2} {2}. To find P, note this equation intersects AB, or the line z = 0, at B and P. z = 0 implies in our circle equation that -c^2xy + (ux)(x + y) = 0. Let P be normalized coordinates (x, y, 0), so that x + y = 1, and thus we get c^2xy = ux. x is not 0 lest P be congruent to B, so dividing by x we have y = \frac {u} {c^2}. Thus P = (1 - \frac {u} {c^2}, \frac {u} {c^2}, 0), and by similar arguments, Q = (0, \frac {w} {a^2}, 1 - \frac {w} {a^2}). BPTQ is a parallelogram, so B + T = P + Q. Thus solving for T, T = (1 - \frac {u} {c^2}, \frac {u} {c^2} + \frac {w} {a^2} - 1, 1 - \frac {w} {a^2}). T lies on the circumcircle of \triangle ABC, so it satisfies a^2yz + b^2zx + c^2xy = 0. Thus, a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1) \cdot (1 - \frac {w} {a^2}) + b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) + c^2 \cdot (1 - \frac {u} {c^2}) \cdot ({\frac {u} {c^2} + \frac {w} {a^2} - 1}) = 0. Keep this equation in mind , and call it f(T). Now look at {TB}^2. The displacement vector TB is (1 - \frac {u} {c^2}, \frac {u} {c^2} + \frac {w} {a^2} - 2, 1 - \frac {w} {a^2}). The distance formula is -a^2yz-b^2zx-c^2zy,, and plugging this in, {BT}^2 = -a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 2) \cdot (1 - \frac {w} {a^2}) -b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) -c^2 \cdot (1 - \frac {u} {c^2}) \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 2). This equals: -a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1 -1) \cdot (1 - \frac {w} {a^2}) -b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) -c^2 \cdot (1 - \frac {u} {c^2}) \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1 - 1) = -[a^2 \cdot (\frac {u} {c^2} + \frac {w} {a^2} - 1) \cdot (1 - \frac {w} {a^2}) + b^2 \cdot (1 - \frac {w} {a^2}) \cdot (1 - \frac {u} {c^2}) + c^2 \cdot (1 - \frac {w} {a^2}) \cdot ({\frac {u} {c^2} + \frac {u} {c^2} - 1})] + (-a^2)(1 - \frac {w} {a^2}) + (-c^2)(1 - \frac {u} {c^2}). But the part in brackets is just f(T) = 0! Thus {BT}^2 = (-a^2)(1 - \frac {w} {a^2}) + (-c^2)(1 - \frac {u} {c^2}) = -a^2-c^2+u+w, but remember u+w = \frac {b^2} {2}, so {BT}^2 = -a^2-c^2+\frac {b^2} {2}. We now find {BM}^2 using the distance formula and displacement vector MB = (\frac {1} {2}, -1, \frac {1} {2}). Thus plugging in, {BM}^2 = -a^2 \cdot \frac{1} {2} + b^2 \cdot \frac{1} {4} - c^2 \cdot \frac {1} {2}. Magically, {BT}^2 = 2 \cdot {BM}^2, and we conclude \frac {BT} {BM} = \sqrt 2. Done!
27.05.2017 23:28
Different solution from post #8. IMO ShortList 2015 G4 wrote: 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}. Henceforth \Gamma denotes the circumcircle of triangle ABC. Let K be the midpoint of \overline{PQ}, L=\overline{TM} \cap \Gamma, N=\overline{BT} \cap \omega and R=\overline{BM} \cap \Gamma with B \not \in \{L, R, N\}. Let X \ne B be the intersection of \omega and \Gamma; then it is the spiral center for the similarity \mathcal{S}: \omega \mapsto \Gamma. Notice that P \mapsto A, Q \mapsto C so K \mapsto M. Also, M \mapsto R since \triangle PMQ \sim \triangle ARC and N \mapsto T since \triangle PNQ \sim \triangle ATC. As B, K, N are collinear; B \mapsto L. Therefore, \measuredangle BMK=\measuredangle LNM=\measuredangle LNB=\measuredangle MTB,so \overline{BM} is tangent to the circumcircle of triangle MKT. Finally, we have BM^2=BK \cdot BT=\frac{1}{2}BT^2 \Longrightarrow \frac{BT}{BM}=\sqrt{2}.
17.04.2019 00:50
A-index everything. Let M' be the midpoint of \overline{PQ}, U be the second intersection of line AM with \odot (ABC), and A' be the second intersection of the tangent to \omega at A with \odot (ABC). Lemma: Let \mathcal{S} be the spiral similarity sending PQ\to BC. Then \mathcal{S} sends R\in \omega to AR\cap \odot (ABC). Proof: Just angle chasing. Thus A\to A',M'\to M,M\to U under \mathcal{S}. Then the preimage of T under \mathcal{S} is AT\cap \omega, so AM'\mathcal{S}^{-1}(T) maps to A'MT and so A'MT is a line. Then PoP at M gives \triangle A'MU\sim \triangle AMT. But \triangle AM'M\sim \triangle A'MU because of \mathcal{S}, so we conclude \triangle AM'M\sim \triangle AMT\implies AM^2=AM'\cdot AT\implies \tfrac{AT}{AM}=\sqrt{2}.
21.11.2019 09:45
As P varies linearly on BA, we have that Q varies linearly on BC due to spiral similarity. This means that T=P+Q-B varies along a line. In particular, this line passes through X and Y, where X is the intersection of the circle through B and M tangent to BC, and Y is defined similarly (this is by taking the limiting cases of P=B and Q=B). Thus, the T given in the problem is found by intersecting line XY with (BAC). We now apply a \sqrt{BA\cdot BC} inversion followed by a reflection about the angle bisector of \angle ABC to get the following new problem. Restated Problem wrote: Let BAC be a triangle, and let K be the harmonic conjugate of B in AC. Let X and Y be on BA and BC such that BYKX is a parallelogram, and let T be an intersection of (BAC) and AC. Find all possible values of BK/BT. [asy][asy] unitsize(2.5inches); pair B=dir(100); pair A=dir(195); pair C=dir(-15); pair H=foot(B,A,C); pair K=2*foot(0,2*A*C/(A+C),B)-B; pair X=extension(K,B+K-A,B,C); pair Y=extension(K,B+K-C,B,A); pair U=foot(H,B,C); pair V=foot(H,A,B); pair HH=2*circumcenter(B,X,Y)-B; pair T1=intersectionpoints(circumcircle(B,X,Y),A--C)[0]; pair T2=intersectionpoints(circumcircle(B,X,Y),A--C)[1]; pair M=(A+C)/2; draw(circumcircle(A,B,C)); draw(circumcircle(B,X,Y)); draw(B--H); draw(A--B--C--cycle); draw(K--X); draw(K--Y); draw(U--H--V); draw(H--HH); dot("A",A,dir(A)); dot("B",B,dir(B)); dot("C",C,dir(C)); dot("H",H,dir(H)); dot("K",K,dir(K)); dot("U",U,dir(U)); dot("V",V,dir(V)); dot("X",X,dir(X)); dot("Y",Y,dir(Y)); dot("H'",HH,dir(HH)); dot("T_1",T1,dir(240)); dot("T_2",T2,dir(-60)); dot("M",M,dir(M)); [/asy][/asy] We'll solve this restated problem. Let H be the foot from B to AC, and let H' denote the second intersection of BH with (BXY). Let U and V be the feet from H to BA and BC, and let \{T_1,T_2\}=(BXY)\cap AC. Note that the diameter line of \triangle BUV is BH, which is the altitude line of \triangle BAC. Thus, UV and AC are anti-parallel. Furthermore, by the law of sines on \triangle BYK, we have YK=\frac{BK}{\sin B}\sin\angle ABK=\frac{BK\cdot AK}{AC},and similarly XK=\frac{BK\cdot CK}{AC}. Thus, \frac{BY\cdot BA}{BX\cdot BC}=\frac{XK\cdot BA}{YK\cdot BC}=\frac{BA\cdot CK}{AK\cdot BC}=1,as (AC;KB)=-1, so YXCA is cyclic, so YX is also anti-parallel to AC. This implies that BH' is a diameter of (BXY) and that UV\parallel XY. Let \alpha=BX/BU=BY/BV=BH'/BH be the scale factor in the similarity between \triangle BUV and \triangle BXY. Note that BU=BC-BC\cos^2 C=BC\sin^2 C. Thus, \alpha=\frac{YK}{BC\sin^2 C}=\frac{BK\cdot AK}{AC\cdot BC\cdot AB^2}(2R)^2,where R is the circumradius of (BAC). We also have the similarity \triangle BT_1H\sim\triangle BH'T_1 with scale factor \sqrt{\alpha}, so we have BT_1^2=BH^2\cdot\alpha. Thus, BT_1^2=\frac{BK\cdot AK\cdot BH^2}{AC\cdot BC\cdot AB^2}(2R)^2,By computing the area of \triangle BAC in two ways, we have AB\cdot BC\cdot CA=(2R)\cdot AC\cdot BH,so it's not hard to see that BT_1^2=\frac{BK\cdot AK\cdot BC}{AC}.Thus, (BK/BT_1)^2=\frac{BK\cdot AC}{AK\cdot BC}.Note that AK and AM are isogonal, so \triangle BAK\sim\triangle BCM, so \frac{BK\cdot CM}{AK\cdot BC}=1, so (BK/BT_1)^2=2, so the answer is \boxed{\sqrt{2}}.
15.03.2020 04:47
We use bary wrt ABC. A=(1,0,0),B=(0,1,0),C=(0,0,1), and M=(1/2,0,1/2). Let T=(k,j,m), so that a^2jm + b^2km + c^2kj=0. We will now get P and Q in terms of T, and then enforce that BPMQ is cyclic to get an equation. Let Q=(0,r,1-r). Since QT\parallel AB, then [BQA]=[BTA]. So \begin{vmatrix} 0&1&0\\0&r&1-r\\1&0&0\end{vmatrix} = \begin{vmatrix}0&1&0\\k&j&m\\1&0&0\end{vmatrix} \implies r=1-m \implies Q=(0,1-m,m). Similarly, P=(k,1-k,0). Now, M\in (BPQ). It is not hard to get that the equation of (BPQ) is -a^2yz-b^2xz-c^2xy + (c^2(1-k)x+a^2(1-m)z)(x+y+z) = 0.Since M=(1:0:1) is on this circle, -b^2 + 2(c^2(1-k) + a^2(1-m))=0 \implies c^2k + a^2m = \frac{2a^2+2c^2-b^2}{2}. The displacement vector for BT is \overleftrightarrow{BT} = (k,j-1,m), so \begin{align*} BT^2 &= -a^2(j-1)m - b^2km - c^2(j-1)k \\ &= (-a^2jm-b^2km-c^2jk) + (a^2m+c^2k) \\ &= \frac{2a^2+2c^2-b^2}{2}. \end{align*}It is well-known that BM^2 = \frac{2a^2+2c^2-b^2}{4} \implies \frac{BT}{BM} = \boxed{\sqrt2}.
24.08.2022 02:33
24.08.2022 02:37
wtf why why? Why? WHY? WHY?!!!!!!!!!!
24.08.2022 23:27
DottedCaculator wrote:
How do you learn to make diagrams like this also this got liked by tapir orz orz orz
01.09.2022 06:31
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
24.12.2022 12:13
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}
16.01.2023 03:43
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.
26.01.2023 06:36
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.
21.05.2023 18:36
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,Pand Q we get: u+w=\frac{b^2}{2} \qquad c^2m=c^2-u \qquad a^2k=a^2-wAlso 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
25.05.2023 02:17
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}.
22.07.2023 17:25
29.12.2023 20:26
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.
02.05.2024 11:34
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^2rwhere 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)=0where 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)=0which 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^2Thus, \frac{BT}{BM} = \sqrt{2} which finishes the problem.
11.05.2024 20:58
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
05.09.2024 16:38
WHAT We employ barycentric coordinates wrt ABC. The equation of the circle is -a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z)=0where 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)=wTherefore, 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}.
05.11.2024 21:32