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Difference between revisions of "2001 JBMO Problems"

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Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
 
Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
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[[2001 JBMO Problems/Problem 1|Solution]]
  
 
==Problem 2==
 
==Problem 2==
  
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Let <math>ABC</math> be a triangle with <math>\angle C = 90^\circ</math> and <math>CA \ne CB</math>.  Let <math>CH</math> be an altitude and <math>CL</math> be an interior angle bisector.  Show that for <math>X \ne C</math> on the line <math>CL,</math> we have <math>\angle XAC \ne \angle XBC</math>.  Also show that for <math>Y \ne C</math> on the line <math>CH</math> we have <math>\angle XAC \ne \angle XBC</math>.
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Let <math>ABC</math> be a triangle with <math>\angle C = 90^\circ</math> and <math>CA \ne CB</math>.  Let <math>CH</math> be an altitude and <math>CL</math> be an interior angle bisector.  Show that for <math>X \ne C</math> on the line <math>CL,</math> we have <math>\angle XAC \ne \angle XBC</math>.  Also show that for <math>Y \ne C</math> on the line <math>CH</math> we have <math>\angle YAC \ne \angle YBC</math>.
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[[2001 JBMO Problems/Problem 2|Solution]]
  
 
==Problem 3==
 
==Problem 3==
  
 
Let <math>ABC</math> be an equilateral triangle and <math>D,E</math> on the sides <math>[AB]</math> and <math>[AC]</math> respectively.  If <math>DF,EF</math> (with <math>F \in AE, G \in AD</math>) are the interior angle bisectors of the angles of the triangle <math>ADE</math>, prove that the sum of the areas of the triangles <math>DEF</math> and <math>DEG</math> is at most equal with the area of the triangle <math>ABC</math>.  When does the equality hold?
 
Let <math>ABC</math> be an equilateral triangle and <math>D,E</math> on the sides <math>[AB]</math> and <math>[AC]</math> respectively.  If <math>DF,EF</math> (with <math>F \in AE, G \in AD</math>) are the interior angle bisectors of the angles of the triangle <math>ADE</math>, prove that the sum of the areas of the triangles <math>DEF</math> and <math>DEG</math> is at most equal with the area of the triangle <math>ABC</math>.  When does the equality hold?
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Bonus Question:
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In the above problem, prove that <math>DF/EG = AD/AE</math>.
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- Proposed by Kris17
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[[2001 JBMO Problems/Problem 3|Solution]]
  
 
==Problem 4==
 
==Problem 4==
  
 
Let <math>N</math> be a convex polygon with 1415 vertices and perimeter 2001.  Prove that we can find 3 vertices of <math>N</math> which form a triangle of area smaller than 1.
 
Let <math>N</math> be a convex polygon with 1415 vertices and perimeter 2001.  Prove that we can find 3 vertices of <math>N</math> which form a triangle of area smaller than 1.
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[[2001 JBMO Problems/Problem 4|Solution]]
  
 
==See Also==
 
==See Also==
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{{JBMO box|year=2001|before=[[2000 JBMO]]|num-a=[[2002 JBMO]]|five=}}
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{{JBMO box|year=2001|before=[[2000 JBMO]]|after=[[2002 JBMO]]|five=}}

Latest revision as of 17:09, 8 December 2018

Problem 1

Solve the equation $a^3 + b^3 + c^3 = 2001$ in positive integers.

Solution

Problem 2

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \ne CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \ne C$ on the line $CL,$ we have $\angle XAC \ne \angle XBC$. Also show that for $Y \ne C$ on the line $CH$ we have $\angle YAC \ne \angle YBC$.

Solution

Problem 3

Let $ABC$ be an equilateral triangle and $D,E$ on the sides $[AB]$ and $[AC]$ respectively. If $DF,EF$ (with $F \in AE, G \in AD$) are the interior angle bisectors of the angles of the triangle $ADE$, prove that the sum of the areas of the triangles $DEF$ and $DEG$ is at most equal with the area of the triangle $ABC$. When does the equality hold?


Bonus Question:

In the above problem, prove that $DF/EG = AD/AE$.

- Proposed by Kris17

Solution

Problem 4

Let $N$ be a convex polygon with 1415 vertices and perimeter 2001. Prove that we can find 3 vertices of $N$ which form a triangle of area smaller than 1.

Solution

See Also

2001 JBMO (Problems • Resources)
Preceded by
2000 JBMO 		Followed by
2002 JBMO
1 • 2 • 3 • 4
All JBMO Problems and Solutions


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