Circles W1,W2 meet at Dand P. A and B are on W1,W2 respectively, such that AB is tangent to W1 and W2. Suppose D is closer than P to the line AB. AD meet circle W2 for second time at C. Let M be the midpoint of BC. Prove that ∠DPM=∠BDC.
Problem
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Tags: geometry, geometric transformation
30.04.2010 22:23
it is as cute as the second day problem let us suppose DP intersect AB at K KA*KA=KB*KB=KD*KP so K is the middle so KM is the middle line in the triangle ABC let us show that MPC=KPB then the problem will be solved DCB=f(angle)=KMB=KPB=DBK note MPC=x so BPM=k-x (BDC=k) now KPMB cyclic (KMB=KPB) and DPCB cyclic so DBC+DPC=180 but also KBM+KPM=180 so KPB=MPC so we are done
09.05.2010 17:13
I think too easy for a second round.
23.05.2010 05:34
Actually if BD∩W1=C′, then △APB, △BPC, △C′PA are spirally similar, with spiral similarity center P (apparently).
26.08.2010 07:50
15.09.2018 03:12
[asy][asy] import graph; size(16cm); 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 = -14.12, xmax = 7.68, ymin = -8.39, ymax = 7.19; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-8.18,0.53), 3.7643060449437424), linewidth(.7) + wrwrwr); draw(circle((-1.3,0.67), 5.2619388061816155), linewidth(.7) + wrwrwr); draw((xmin, 0.24443676978465095*xmin + 6.404624783463565)--(xmax, 0.24443676978465095*xmax + 6.404624783463565), linewidth(.7) + wrwrwr); /* line */ draw((xmin, -0.22971799065596446*xmin + 2.102230014008526)--(xmax, -0.22971799065596446*xmax + 2.102230014008526), linewidth(.7) + wrwrwr); /* line */ draw((-2.549426614613997,5.781451176984303)--(3.935360294483731,1.1982069546524587), linewidth(.7) + wrwrwr); draw((-5.78,3.43)--(-5.664012332643493,-2.269965367234024), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(0.692966839934867,3.4898290658183813), linewidth(.7) + wrwrwr); draw((-2.549426614613997,5.781451176984303)--(-5.78,3.43), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(-2.549426614613997,5.781451176984303), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(3.935360294483731,1.1982069546524587), linewidth(.7) + wrwrwr); draw((-9.073819622641716,4.186649625296442)--(-5.664012332643493,-2.269965367234024), linewidth(.7) + wrwrwr); draw((-5.811623118627857,4.984050401140373)--(-5.78,3.43), linewidth(.7) + wrwrwr); /* dots and labels */ dot((-8.18,0.53),linewidth(2pt) +dotstyle); label("W1", (-8.1,0.73), NE * labelscalefactor); dot((-5.78,3.43),linewidth(2pt) +dotstyle); label("D", (-5.7,3.63), NE * labelscalefactor); dot((-1.3,0.67),linewidth(2pt) +dotstyle); label("W2", (-1.22,0.87), NE * labelscalefactor); dot((-5.664012332643493,-2.269965367234024),linewidth(2pt) + dotstyle); label("P", (-5.58,-2.11), NE * labelscalefactor); dot((-2.549426614613997,5.781451176984303),linewidth(2pt) + dotstyle); label("B", (-2.46,5.95), NE * labelscalefactor); dot((-9.073819622641716,4.186649625296442),linewidth(2pt) + dotstyle); label("A", (-9,4.35), NE * labelscalefactor); dot((3.935360294483731,1.1982069546524587),linewidth(2pt) + dotstyle); label("C", (4.02,1.35), NE * labelscalefactor); dot((0.692966839934867,3.4898290658183813),linewidth(2pt) + dotstyle); label("M", (0.78,3.65), NE * labelscalefactor); dot((-5.811623118627857,4.984050401140373),linewidth(2pt) + dotstyle); label("E", (-5.74,5.15), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let E:=PD∩AB. E lies on radical axis of W1 and W2, so EA=EB We have ∠ABP=∠BCP, because tangency. Also, since triangles EAD, EPA similar, we have ∠PBC=∠PDC=∠EDA=∠EAP=∠BAP. Thus, PCB and PBA, and since E is midpoint of AB, same spiral similarity that brings CB to BA also takes M to E, implying the conclusion as ∠BDC=∠BPC=∠MPD.
08.01.2020 00:51
iran 2010 2nd Round P3 wrote: Circles W1,W2 meet at Dand P. A and B are on W1,W2 respectively, such that AB is tangent to W1 and W2. Suppose D is closer than P to the line AB. AD meet circle W2 for second time at C. Let M be the midpoint of BC. Prove that ∠DPM=∠BDC. Notice that D is the P−Humpty Point of △BPA, so if PD∩AB=R then AR=RB. Hence RM‖. So, by Homothety at B we get that \odot(BXM),\odot(BDC) are internally tangent at B where X=BD\cap RM and RB is the common tangent to both the circles. Hence, RD\cdot RP=RB^2=RX\cdot RM\implies \angle DPM=\angle RXD=\angle BDC. \blacksquare
29.12.2020 00:03
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.05015421010951cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -3.9231536695241593, xmax = 12.827103347325025, ymin = -9.04805769080714, ymax = 2.446955861661247; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A = (-0.49061490038429356,-0.59975490629623), B = (5.522980152527406,1.1431233059414534), O_1 = (0.44165357896052615,-3.816436987699729), O_2 = (6.7524473745649995,-3.0990078723867294), D = (2.699762430277838,-1.3431572732305677), P = (3.197304676053955,-5.719738109826762), C = (11.163846035069993,-3.3154066762272723), M = (8.343413093798699,-1.0861416851429095); /* draw figures */ draw(A--B, linewidth(0.8)); draw(circle(O_1, 3.3490548114956624), linewidth(0.8)); draw(circle(O_2, 4.416703135168685), linewidth(0.8)); draw(B--C, linewidth(0.8)); draw(A--C, linewidth(0.8)); draw((2.516182626071557,0.27168419982261194)--P, linewidth(0.8)); draw(P--M, linewidth(0.8)); draw(P--C, linewidth(0.8)); draw(D--B, linewidth(0.8)); draw(P--B, linewidth(0.8)); draw(circle((4.945276180299027,-2.486596594688917), 3.6754058873219146), linewidth(2.) + dotted + red); draw(M--(2.516182626071557,0.27168419982261194), linewidth(0.8)); /* dots and labels */ dot(A); label("A", (-0.4373972450487933,-0.46671076795747596), NE * labelscalefactor); dot(B); label("B", (5.576197807862907,1.2761674442802073), NE * labelscalefactor); dot(O_1); label("O_1", (0.320954343482107,-3.5533347774165795), NE * labelscalefactor); dot(O_2); label("O_2", (6.800203880579448,-2.9679405687260596), NE * labelscalefactor); dot(D,uuuuuu); label("D", (2.7556620750813132,-1.2117579426545009), NE * labelscalefactor,uuuuuu); dot(P,linewidth(4.pt) + uuuuuu); label("P", (3.2479253869347047,-5.615518921667274), S* labelscalefactor,uuuuuu); dot(C,uuuuuu); label("C", (11.217269273426096,-3.180811190068067), NE * labelscalefactor,uuuuuu); dot(M,uuuuuu); label("M", (8.396733540644501,-0.9589740798108675), NE * labelscalefactor,uuuuuu); dot((2.516182626071557,0.27168419982261194),uuuuuu); label("E", (2.569400281407057,0.41138054507830346), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] we can see that BMPE is cyclic since EM||AC so: \color{green} \angle DPM=\angle BME+\angle BEM =\angle BAC+\angle BCA=\angle BAC+\angle ABD=\angle BDC so we are done \color{magenta} \huge{\boxed{\mathbf{Q.E.D}}}
29.03.2022 17:16
Let DP meet AB at S. \angle ABP = \angle BCP and BAP = \angle DAP + \angle DPA = \angle PDC = \angle PBC so triangles ABP and BPC are similar. we know S is midpoint of AB so BSP and CMP are similar. \angle DPM = \angle CPB = \angle BDC. we're Done.