2007 IMO Problems/Problem 2

Problem

Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of $\angle DAB$.

Solution

[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch);  path circle = Circle(origin, 1); draw(circle);  pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5); //G = bisectorpoint(C, B, D); pair Ee = rotate(38,C)*D; pair E = IP(C--Ee, circle,1); pair Gg = rotate(76,C)*D; path circle2 = Circle(E, length(C-E)); pair G = IP(C--Gg, circle2, 1); pair F = IP(C--D, circle2, 1); pair Bb = rotate(-104,C)*D; pair B = IP(C--Bb, circle, 1);  pair A = extension((-1,B.y),(1,B.y),G,F); draw(circle2, dashed); draw(A--G); draw(C--D--A--B); draw(G--B);  draw(E--F); draw(E--C); draw(E--G);  dot("$C$", C, dir(30)); dot("$D$", D, SW); dot("$G$", G, SE);  dot("$E$", E, SW); dot("$F$", F, SW); dot("$A$", A, SW); dot("$B$", B, SE);  draw(D--E,dashed); draw(B--E,dashed);  [/asy]

Since $\angle{DAF}=\angle{CGF}$, $\angle{BAF}=\angle{CFG}$, it suffices to prove $CF=CG$.

Let $\angle{FCE}=\alpha$, $\angle{GCE}=\beta$, $\angle{CDE}=\gamma$. We have: \[CF=2CE\cos{\alpha}, CG=2CE\cos{\beta}\] so, \[\dfrac{CF}{CG}=\dfrac{\cos{\alpha}}{\cos{\beta}}\] Meantime, using Law of Sines on $\triangle{DEC}$, we have, \[\dfrac{CE}{\sin{\gamma}}=\dfrac{DC}{\sin{(180-\alpha-\gamma)}}=\dfrac{DC}{\sin{(\alpha+\gamma)}}\] Using Law of Sines on $\triangle{BEG}$, and notice that $\angle{CBE}=\angle{CDE}=\gamma$, we have, \[\dfrac{CE}{\sin{\gamma}}=\dfrac{BG}{\sin{(180-\beta-\gamma)}}=\dfrac{BG}{\sin{(\beta+\gamma)}}\] so, \[\dfrac{DC}{BG}=\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}\] Since $\triangle{GFC} \sim \triangle{GAB}$, and $DC=AB$, we have, $\dfrac{DC}{BG}=\dfrac{AB}{BG}=\dfrac{CF}{CG}$. Hence, \[\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}=\dfrac{\cos{\alpha}}{\cos{\beta}}\] or, \[\sin{(\alpha+\gamma)}\cos{\beta}=\sin{(\beta+\gamma)}\cos{\alpha}\] \[\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\alpha+\gamma-\beta)})=\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\beta+\gamma-\alpha)})\] \[\sin{(\alpha+\gamma-\beta)}=\sin{(\beta+\gamma-\alpha)}\] There are two possibilities: (1) $\alpha+\gamma-\beta = \beta+\gamma-\alpha$, or (2) $\alpha+\gamma-\beta = 180 - (\beta+\gamma-\alpha)$. However, (2) would mean $\gamma=90$, then $EC$ would be a diameter, and $EF < EC$ because $F$ is inside the circle, so (2) is not valid. From condition (1), we have $\alpha=\beta$, therefore $CF=CG$. $\square$

Solution by Mathdummy

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2007 IMO (Problems) • Resources)
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