Circles $Q_1$ and $Q_2$ are tangent to each other externally at $T$. Points $A$ and $E$ are on $Q_1$, lines $AB$ and $DE$ are tangent to $Q_2$ at $B$ and $D$, respectively, lines $AE$ and $BD$ meet at point $P$. Prove that (1) $\frac{AB}{AT}=\frac{ED}{ET}$; (2) $\angle ATP + \angle ETP = 180^{\circ}$. [asy][asy]import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-6,xmax=5.94,ymin=-3.19,ymax=3.43; pair Q_1=(-2.5,-0.5), T=(-1.5,-0.5), Q_2=(0.5,-0.5), A=(-2.09,0.41), B=(-0.42,1.28), D=(-0.2,-2.37), P=(-0.52,2.96); D(CR(Q_1,1)); D(CR(Q_2,2)); D(A--B); D((-3.13,-1.27)--D); D(P--(-3.13,-1.27)); D(P--D); D(T--(-3.13,-1.27)); D(T--A); D(T--P); D(Q_1); MP("Q_1",(-2.46,-0.44),NE*lsf); D(T); MP("T",(-1.46,-0.44),NE*lsf); D(Q_2); MP("Q_2",(0.54,-0.44),NE*lsf); D(A); MP("A",(-2.22,0.58),NE*lsf); D(B); MP("B",(-0.35,1.45),NE*lsf); D((-3.13,-1.27)); MP("E",(-3.52,-1.62),NE*lsf); D(D); MP("D",(-0.17,-2.31),NE*lsf); D(P); MP("P",(-0.47,3.02),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
Problem
Source: 2012 China Girl's Mathematical Olympiad
Tags: trigonometry, geometry, similar triangles
13.08.2012 22:04
14.08.2012 06:10
2. If we let $BA$ intersect $DE$ at $C$, then by Menalaus $\frac{AP}{PE}=\frac{AB}{BC}\frac{CD}{DE}$. But since $CB=CD$, we have $\frac{AP}{PE}=\frac{AB}{DE}=\frac{AT}{TE}$, so $TP$ bisects $\angle ATE$ externally.
14.08.2012 06:18
prob.1 I suppose we can find a temporary fraction to solve the problem. The graph's very familiar to me but I just couldnt remind it ...
17.08.2012 16:16
Alternative Solution to (1): Let $r_1$ be the radius of $Q_1$ and let $r_2$ be the radius of $Q_2$. $AB = \sqrt{AQ_2^2 - r_2^2} = \sqrt{AT^2 + r_2^2 + 2(AT)(r_2)(\cos \angle ATQ_1)- r_2^2} = \sqrt{AT^2 + 2(AT)(r_2)(\cos \angle ATQ_1)}$ Similarly, $ED = \sqrt{ET^2 + 2(ET)(r_2)(\cos \angle ETQ_1)}$. So $\frac{AB}{AT} = \frac{ED}{ET} \Leftrightarrow \frac{AT}{ET} = \frac{AB}{ED} \Leftrightarrow \frac{\sqrt{2(AT)(r_2)(\cos \angle ATQ_1)}}{\sqrt{2(ET)(r_2)(\cos \angle ETQ_1)}} = \frac{AT}{ET}$ $\Leftrightarrow \frac{AT}{ET} = \frac{\cos \angle ATQ_1}{\cos \angle ETQ_1}$. Let $X$ be the midpoint of $AT$ and let $Y$ be the midpoint of $ET$. $\frac{\cos \angle ATQ_1}{\cos \angle ETQ_1} = \frac{XT}{YT} = \frac{AT}{ET}$ so we are done.
26.03.2020 11:32
Since I'm stubborn (as is evident given the hour), here's an overkill solution to (1) that took way too long to find. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(250); 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 = -8.993860671376272, xmax = 8.97550972000206, ymin = -5.227108234930132, ymax = 5.334888361780027; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ pair T = (-0.3,0.81), B = (2.57467651723,4.6334356276), A = (-2.5051095590,3.16491166528), P = (-2.0045820554,-1.45716283446), Q = (4.8902186715,-0.09991127735); draw(circle((-2.66,0.81), 2.36),rgb(0.1,0.3,0.7)); draw(circle((3.68,0.81), 3.98),rgb(0.1,0.3,0.7)); draw(circle((0.16350893354008234,3.4538981262937014), 2.6842201537300947),rgb(0.9,0.2,0.2)); draw(A--B--T--A,rgb(0.1,0.8,0.3)); draw(A--P--T,rgb(0.7,0.1,0.7)); draw(B--Q--T,rgb(0.9,0.5,0.1)); draw(A--Q,rgb(0.8,0.8,0.05)); /* dots and labels */ dot("$A$",A,1.2*NW); dot("$B$",B,1.2*dir(60)); dot("$T$",T,1.2*dir(310)); dot("$P$",P,1.2*S); dot("$Q$",Q,SE); dot("$R$",intersectionpoint(A--Q,B--T),1.2*S); /* end of picture */ [/asy][/asy] Extend $BT$ past $T$ to intersect circle $Q_1$ again at $P$. Then via homothety, \[ \angle ABT = \tfrac12\widehat{BT} = \tfrac12\widehat{TP} = \angle PAT, \]so the circumcircle of $\triangle BAT$ is tangent to $AP$. Now let the tangents to $\odot(ABT)$ at $B$ and $T$ intersect at $Q$. Observe that $BT$ is the polar of $Q$ with respect to $\odot(ABT)$. Since $P$ lies on this polar, duality yields that $Q$ lies on the polar of $P$ with respect to $\odot(ABT)$. However, $AP$ is tangent to $\odot(ABT)$, and so this polar is $AQ$ -- the $A$-symmedian of $\triangle ABT$. Thus, if $R$ is the intersection point of $AQ$ with $BT$, the points $B$, $R$, $T$, and $P$ form a harmonic division, whence \[ \frac{TP}{PB} = \frac{TR}{RB} = \left(\frac{TA}{AB}\right)^2. \]So $\tfrac{BA}{AT} = \sqrt{\tfrac{PB}{BT}}$, which is constant by homothety.
27.03.2020 17:16
another solution for this: 1)extend $ET,AT$ to intersect circle $Q_2$ AT $E'.A'$ we know that $A'E'||AE$ then compute the power of $E,T$ then we get (1). 2)use sine theorem on triangle $\triangle ETP $,$\triangle ATP $,$\triangle ABP $,$\triangle EDP $
29.01.2022 10:39
1) Let AT meet $Q_2$ at X and ET meet $Q_2$ at Y. we want to prove AB.ET = AT.ED or (AB.ET)^2 = (AT.ED)^2 or AX.ET = AT.EY . Note that AET and XYT are similar so AT/AX = ET/EY. 2) Let AB and CD meet at Z. Note that ZB = ZD so by Menalaus theorem we have AP/PE = AB/DE which AB/DE = AT/ET by last part so AP/PE = AT/ET so PT is angle bisector of ∠ATY so ∠PTE + ∠PTA = ∠PTE + ∠PTY = 180. we're Done.