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Difference between revisions of "2007 IMO Problems/Problem 2"

m (Created page with "== Problem == Consider five points <math>A,B,C,D</math>, and <math>E</math> such that <math>ABCD</math> is a parallelogram and <math>BCED</math> is a cyclic quadrilateral. Let <...")
 
 
(One intermediate revision by one other user not shown)
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line <math>BC</math> at <math>G</math>. Suppose also that <math>EF=EG=EC</math>. Prove that <math>\ell</math> is the bisector of <math>\angle DAB</math>.
 
line <math>BC</math> at <math>G</math>. Suppose also that <math>EF=EG=EC</math>. Prove that <math>\ell</math> is the bisector of <math>\angle DAB</math>.
  
−
== Solution ==
+
==Solution==
 +
<center><asy>
 +
import cse5;
 +
import graph;
 +
import olympiad;
 +
dotfactor = 3;
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unitsize(1.5inch);
 +
 
 +
path circle = Circle(origin, 1);
 +
draw(circle);
 +
 
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pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5);
 +
//G = bisectorpoint(C, B, D);
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pair Ee = rotate(38,C)*D;
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pair E = IP(C--Ee, circle,1);
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pair Gg = rotate(76,C)*D;
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path circle2 = Circle(E, length(C-E));
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pair G = IP(C--Gg, circle2, 1);
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pair F = IP(C--D, circle2, 1);
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pair Bb = rotate(-104,C)*D;
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pair B = IP(C--Bb, circle, 1);
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pair A = extension((-1,B.y),(1,B.y),G,F);
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draw(circle2, dashed);
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draw(A--G); draw(C--D--A--B); draw(G--B);  draw(E--F); draw(E--C); draw(E--G);
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dot("$C$", C, dir(30)); dot("$D$", D, SW); dot("$G$", G, SE);
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dot("$E$", E, SW); dot("$F$", F, SW); dot("$A$", A, SW); dot("$B$", B, SE);
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draw(D--E,dashed); draw(B--E,dashed);
 +
 
 +
</asy></center>
 +
Since <math>\angle{DAF}=\angle{CGF}</math>, <math>\angle{BAF}=\angle{CFG}</math>, it suffices to prove <math>CF=CG</math>.
 +
 
 +
Let <math>\angle{FCE}=\alpha</math>, <math>\angle{GCE}=\beta</math>, <math>\angle{CDE}=\gamma</math>. We have:
 +
<cmath>CF=2CE\cos{\alpha}, CG=2CE\cos{\beta}</cmath>
 +
so,
 +
<cmath>\dfrac{CF}{CG}=\dfrac{\cos{\alpha}}{\cos{\beta}}</cmath>
 +
Meantime, using Law of Sines on <math>\triangle{DEC}</math>, we have,
 +
<cmath>\dfrac{CE}{\sin{\gamma}}=\dfrac{DC}{\sin{(180-\alpha-\gamma)}}=\dfrac{DC}{\sin{(\alpha+\gamma)}}</cmath>
 +
Using Law of Sines on <math>\triangle{BEG}</math>, and notice that <math>\angle{CBE}=\angle{CDE}=\gamma</math>, we have,
 +
<cmath>\dfrac{CE}{\sin{\gamma}}=\dfrac{BG}{\sin{(180-\beta-\gamma)}}=\dfrac{BG}{\sin{(\beta+\gamma)}}</cmath>
 +
so,
 +
<cmath>\dfrac{DC}{BG}=\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}</cmath>
 +
Since <math>\triangle{GFC} \sim \triangle{GAB}</math>, and <math>DC=AB</math>, we have, <math>\dfrac{DC}{BG}=\dfrac{AB}{BG}=\dfrac{CF}{CG}</math>. Hence,
 +
<cmath>\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}=\dfrac{\cos{\alpha}}{\cos{\beta}}</cmath>
 +
or,
 +
<cmath>\sin{(\alpha+\gamma)}\cos{\beta}=\sin{(\beta+\gamma)}\cos{\alpha}</cmath>
 +
<cmath>\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\alpha+\gamma-\beta)})=\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\beta+\gamma-\alpha)})</cmath>
 +
<cmath>\sin{(\alpha+\gamma-\beta)}=\sin{(\beta+\gamma-\alpha)}</cmath>
 +
There are two possibilities: (1) <math>\alpha+\gamma-\beta = \beta+\gamma-\alpha</math>, or (2) <math>\alpha+\gamma-\beta = 180 - (\beta+\gamma-\alpha)</math>. However, (2) would mean <math>\gamma=90</math>, then <math>EC</math> would be a diameter, and <math>EF < EC</math> because <math>F</math> is inside the circle, so (2) is not valid. From condition (1), we have <math>\alpha=\beta</math>, therefore <math>CF=CG</math>. <math>\square</math>
 +
 
 +
Solution by Mathdummy
 +
 
 +
{{alternate solutions}}
 +
 
 +
{{IMO box|year=2007|num-b=1|num-a=3}}
 +
 
 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 00:45, 24 August 2018

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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