2001 JBMO Problems: Difference between revisions

Latest revision as of 17:09, 8 December 2018

Problem 1

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

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

@@ Line 19: / Line 19: @@
 Bonus Question:
-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,EG</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 <math>DF/EG = AD/AE</math>.
+In the above problem, prove that <math>DF/EG = AD/AE</math>.
-Proposed by Kris17
+- Proposed by Kris17
 [[2001 JBMO Problems/Problem 3|Solution]]

2001 JBMO (Problems • Resources)
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2001 JBMO Problems: Difference between revisions

Latest revision as of 17:09, 8 December 2018

Contents

Problem 1

Problem 2

Problem 3

Problem 4

See Also