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Difference between revisions of "2005 IMO Problems/Problem 1"

(Created page with "Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A_2</math> on <math>BC</math>, <math>B_1</math>, <math>B_2</math> on <math>CA</math>...")
 
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==Problem==
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Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A_2</math> on <math>BC</math>, <math>B_1</math>, <math>B_2</math> on <math>CA</math> and <math>C_1</math>, <math>C_2</math> on <math>AB</math>, such that they are the vertices of a convex hexagon <math>A_1A_2B_1B_2C_1C_2</math> with equal side lengths. Prove that the lines <math>A_1B_2, B_1C_2</math> and <math>C_1A_2</math> are concurrent.
 
Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A_2</math> on <math>BC</math>, <math>B_1</math>, <math>B_2</math> on <math>CA</math> and <math>C_1</math>, <math>C_2</math> on <math>AB</math>, such that they are the vertices of a convex hexagon <math>A_1A_2B_1B_2C_1C_2</math> with equal side lengths. Prove that the lines <math>A_1B_2, B_1C_2</math> and <math>C_1A_2</math> are concurrent.
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==Solution==
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Let <math>D = C_2A_1 \cap A_2B_1</math>, and similarly define <math>E</math> and <math>F</math>. We claim that <math>DEF</math> is equilateral.
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Proof: Note that the vectors <math>\overrightarrow{A_1A_2}</math>, <math>\overrightarrow{A_2B_1}</math>, <math>\overrightarrow{B_1B_2}</math>, <math>\overrightarrow{B_2C_1}</math>, <math>\overrightarrow{C_1C_2}</math>, <math>\overrightarrow{C_2A_1}</math> add to 0. But we also have:
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<cmath>\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0</cmath>
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since these are at the sides of an equilateral triangle.
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This means that
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<cmath>\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0.</cmath>
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These three unit vectors thus form an equilateral triangle.
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Now <math>\angle A_1DB_1 = 60^{\circ} = \angle A_1CB_1</math>, so quadrilateral <math>A_1DCB_1</math> is cyclic.
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So <math>\angle CA_1D = \angle CB_1D</math> and <math>\angle C_2A_1A_2 = \angle A_2B_1B_2</math>.
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This means that triangles <math>C_2A_1A_2</math> and <math>A_2B_1B_2</math> are congruent, so <math>C_2A_2 = A_2B_2</math>.
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But also, <math>C_2C_1 = C_1B_2</math>, so <math>C_1A_2</math> is the perpendicular bisector of <math>A_1B_1C_1</math>.
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Similarly, <math>A_1B_2</math> and <math>B_1C_2</math> are also perpendicular bisectors. Therefore, they concur.
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==See Also==
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{{IMO box|year=2005|before=First Problem|num-a=2}}

Latest revision as of 05:36, 16 April 2025

Problem

Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2, B_1C_2$ and $C_1A_2$ are concurrent.

Solution

Let $D = C_2A_1 \cap A_2B_1$, and similarly define $E$ and $F$. We claim that $DEF$ is equilateral.

Proof: Note that the vectors $\overrightarrow{A_1A_2}$, $\overrightarrow{A_2B_1}$, $\overrightarrow{B_1B_2}$, $\overrightarrow{B_2C_1}$, $\overrightarrow{C_1C_2}$, $\overrightarrow{C_2A_1}$ add to 0. But we also have: \[\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0\] since these are at the sides of an equilateral triangle. This means that \[\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0.\] These three unit vectors thus form an equilateral triangle.

Now $\angle A_1DB_1 = 60^{\circ} = \angle A_1CB_1$, so quadrilateral $A_1DCB_1$ is cyclic. So $\angle CA_1D = \angle CB_1D$ and $\angle C_2A_1A_2 = \angle A_2B_1B_2$. This means that triangles $C_2A_1A_2$ and $A_2B_1B_2$ are congruent, so $C_2A_2 = A_2B_2$. But also, $C_2C_1 = C_1B_2$, so $C_1A_2$ is the perpendicular bisector of $A_1B_1C_1$. Similarly, $A_1B_2$ and $B_1C_2$ are also perpendicular bisectors. Therefore, they concur.

See Also

2005 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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