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
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions

Page

Toolbox

Search

2007 IMO Problems/Problem 2

Problem

Solution