Right triangle $ ABC,$ with $ \angle A=90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.
Problem
Source: China Girls Mathematical Olympiad 2009, Problem 2
Tags: geometry, circumcircle, incenter, angle bisector
18.08.2009 19:19
18.08.2009 19:48
let angle CAE=y, ACB=x then ACF=90+y-x, EDF=90+y, AFE=90+x-2y, AEF=90-x hence OEC=y then EC/ CF=AE/AF=sin(90+x-2y)/cos x we get EM/MC=(cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx) = tan ECO / tan y so tan ECO = tany( (cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx)) = tan(90+y-x) so ECO = 90+y-x hence ACF,ECO are vertical angles which means A,C,O collinear.
19.08.2009 08:34
this is fun, better than number 6.
20.08.2009 07:35
21.08.2009 01:09
Let $ AF$ meet the circuncircle of $ EDF$ again at $ E'$. Angle chasing shows that $ AEE'$ is isosceles, that is, $ E$ and $ E'$ are symmetrical wrt to $ AC$ and the result follows.
21.08.2009 07:24
We solve it just by angles chasing : denote $ {\angle ABC = N}$;$ {\angle BCA=90-N}$ ;$ {\angle ACE= M}$; then we get :$ {\angle EAC=180-M- N}$ $ {\longrightarrow}$ $ {\angle EAF =360-2M-2N}$;and $ {\angle BCE=\angle BCE=M-(90-N)=M+N-90}$; $ {\longrightarrow}$ $ {\angle EDF =270-M-N}$ ;$ {\angle EOF=[180-(270-M-N)]*2=2M+2N-180}$; thus $ {\angle EAF+\angle EOF=180}$; $ {\longrightarrow}$ $ {A,E,F,O}$ are con-cyclic; since $ {EO=FO}$ ;$ {O}$ is on the angle bisector of $ {\angle EAF}$; that is $ {A,C,O}$ are collinear.....$ {Q.E.D.}$[ ]
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12.12.2009 10:12
Quote: Right triangle $ ABC,$ with $ \angle A = 90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA > EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear. Here is my solution Since $ \angle OED$ $ =90^{\circ}-$ $ \frac {1}{2}$ $ \angle EOD$ $ =90^{\circ}-$ $ \angle DFE$ $ =\angle DCF$ $ =\angle DBE$, implying $ OE$ is a tangent from $ O$ to $ (\Gamma)$. The same argument holds for $ OD$, which yields that $ (O)$ and $ (\Gamma)$ are orthogonal. But in the other hand, due to the fact that $ (AB,AC,AE,AS)=-1$, then if $ W$ is the intersection of $ AB$ with $ EF$ then $ (EFCW)=-1$, which implies $ (BE,BF,BC,BW)=-1$ or we can say $ (BE,BD,BC,BA)=-1$ $ \Longrightarrow$ $ ADCE$ is a harmonic quadrilateral. As the result, $ AC$ with then tangents at $ D$ and $ E$ will concur, hence, $ AC,$ $ OD,$ $ OE$ will concur at $ O$, which implies $ A,$ $ C,$ $ O$ are collinear. $ \square$
30.04.2022 09:03
Notice $$-1=(E,F;C,\overline{AB}\cap\overline{EF})\stackrel{B}=(E,D;C,A)$$by the Right Angles and Bisectors Lemma. Since $$\angle CEO=90-\tfrac{1}{2}\angle FOE=90-\angle EDB=90-\angle EAB=\angle CAE,$$we know $\overline{OE}$ is tangent to $\Gamma$ at $E.$ Similarly, $O=\overline{DD}\cap\overline{EE}$ so if $A'=\overline{OC}\cap\Gamma,$ we see $(A',C;DE)=-1.$ Hence, $A=A'.$ $\square$