KöMaL - Mathematical and Physical Journal for Secondary Schools
Hungarian version Information Contest Journal Articles News
Conditions
Entry form to the contest
Problems and solutions
Results of the competition
Problems of the previous years

 

 

Order KöMaL!

Competitions Portal

B. 3802. Given are seven real numbers in such a way that the sum of any three of them is less than the sum of the remaining four. Show that each number is positive.

(3 points)

Deadline expired on 15 April 2005.


Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás. Legyen a 7 szám a_1\le a_2\le \ldots \le a_7. Ekkor

a2+a3+a4\lea5+a6+a7.

Ha tehát a1\le0 lenne, akkor

a1+a2+a3+a4\lea5+a6+a7

is teljesülne, ami viszont ellentmond a feltételeknek. Ezért hát a1>0, következésképpen ai\gea1>0 is igaz minden 1\lei\le7 esetén.


Statistics on problem B. 3802.
171 students sent a solution.
3 points:100 students.
2 points:60 students.
1 point:7 students.
0 point:4 students.


  • Problems in Mathematics of KöMaL, March 2005

  • Our web pages are supported by: Ericsson   Google   Emberi ErĹ‘forrás TámogatáskezelĹ‘   Emberi ErĹ‘források MinisztĂ©riuma   OktatáskutatĂł Ă©s FejlesztĹ‘ IntĂ©zet   Nemzeti TehetsĂ©g Program     Nemzeti
Kulturális Alap   ELTE