Difference between revisions of "1976 IMO Problems"

 
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===Problem 1===
 
===Problem 1===
 
In a convex quadrilateral (in the plane) with the area of <math>32 \text{ cm}^{2}</math> the sum of two opposite sides and a diagonal is <math>16 \text{ cm}</math>. Determine all the possible values that the other diagonal can have.
 
In a convex quadrilateral (in the plane) with the area of <math>32 \text{ cm}^{2}</math> the sum of two opposite sides and a diagonal is <math>16 \text{ cm}</math>. Determine all the possible values that the other diagonal can have.
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[[1976 IMO Problems/Problem 1|Solution]]
 
===Problem 2===
 
===Problem 2===
 
Let <math>P_{1}(x) = x^{2} - 2</math> and <math>P_{j}(x) = P_{1}(P_{j - 1}(x))</math> for <math>j= 2,\ldots</math> Prove that for any positive integer n the roots of the equation <math>P_{n}(x) = x</math> are all real and distinct.
 
Let <math>P_{1}(x) = x^{2} - 2</math> and <math>P_{j}(x) = P_{1}(P_{j - 1}(x))</math> for <math>j= 2,\ldots</math> Prove that for any positive integer n the roots of the equation <math>P_{n}(x) = x</math> are all real and distinct.
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[[1976 IMO Problems/Problem 2|Solution]]
 
===Problem 3===
 
===Problem 3===
 
A box whose shape is a parallelepiped can be completely filled with cubes of side <math>1.</math> If we put in it the maximum possible number of cubes, each of volume <math>2</math>, with the sides parallel to those of the box, then exactly <math>40</math> percent from the volume of the box is occupied. Determine the possible dimensions of the box.
 
A box whose shape is a parallelepiped can be completely filled with cubes of side <math>1.</math> If we put in it the maximum possible number of cubes, each of volume <math>2</math>, with the sides parallel to those of the box, then exactly <math>40</math> percent from the volume of the box is occupied. Determine the possible dimensions of the box.
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[[1976 IMO Problems/Problem 3|Solution]]
 
==Day 2==
 
==Day 2==
 
===Problem 4===
 
===Problem 4===
 
Find the largest number obtainable as the product of positive integers whose sum is <math>1976</math>.
 
Find the largest number obtainable as the product of positive integers whose sum is <math>1976</math>.
  
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[[1976 IMO Problems/Problem 4|Solution]]
 
===Problem 5===
 
===Problem 5===
 
Let a set of <math>p</math> equations be given,  
 
Let a set of <math>p</math> equations be given,  
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</cmath>
 
</cmath>
 
with coefficients <math>a_{ij}</math> satisfying <math>a_{ij}=-1</math>, <math>0</math>, or <math>+1</math> for all <math>i=1,\dots, p</math>, and <math>j=1,\dots, q</math>. Prove that if <math>q=2p</math>, there exists a solution <math>x_1, \dots, x_q</math> of this system such that all <math>x_j</math> (<math>j=1,\dots, q</math>) are integers satisfying <math>|x_j|\le q</math> and <math>x_j\ne 0</math> for at least one value of <math>j</math>.
 
with coefficients <math>a_{ij}</math> satisfying <math>a_{ij}=-1</math>, <math>0</math>, or <math>+1</math> for all <math>i=1,\dots, p</math>, and <math>j=1,\dots, q</math>. Prove that if <math>q=2p</math>, there exists a solution <math>x_1, \dots, x_q</math> of this system such that all <math>x_j</math> (<math>j=1,\dots, q</math>) are integers satisfying <math>|x_j|\le q</math> and <math>x_j\ne 0</math> for at least one value of <math>j</math>.
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[[1976 IMO Problems/Problem 5|Solution]]
  
 
===Problem 6===
 
===Problem 6===
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<cmath>3\log_2[u_n]=2^n-(-1)^n,</cmath>
 
<cmath>3\log_2[u_n]=2^n-(-1)^n,</cmath>
 
where <math>[x]</math> is the integral part of <math>x</math>.
 
where <math>[x]</math> is the integral part of <math>x</math>.
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[[1976 IMO Problems/Problem 6|Solution]]
  
 
* [[1976 IMO]]  
 
* [[1976 IMO]]  
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1976 IMO 1976 Problems on the Resources page] * [[IMO Problems and Solutions, with authors]]  
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* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1976 IMO 1976 Problems on the Resources page]  
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* [[IMO Problems and Solutions, with authors]]  
 
* [[Mathematics competition resources]] {{IMO box|year=1976|before=[[1975 IMO]]|after=[[1977 IMO]]}}
 
* [[Mathematics competition resources]] {{IMO box|year=1976|before=[[1975 IMO]]|after=[[1977 IMO]]}}

Latest revision as of 16:25, 29 January 2021

Problems of the 18th IMO 1976 in Austria.

Day 1

Problem 1

In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.

Solution

Problem 2

Let $P_{1}(x) = x^{2} - 2$ and $P_{j}(x) = P_{1}(P_{j - 1}(x))$ for $j= 2,\ldots$ Prove that for any positive integer n the roots of the equation $P_{n}(x) = x$ are all real and distinct.

Solution

Problem 3

A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent from the volume of the box is occupied. Determine the possible dimensions of the box.

Solution

Day 2

Problem 4

Find the largest number obtainable as the product of positive integers whose sum is $1976$.

Solution

Problem 5

Let a set of $p$ equations be given, \[\begin{array}{ccccccc} a_{11}x_1&+&\cdots&+&a_{1q}x_q&=&0,\\ a_{21}x_1&+&\cdots&+&a_{2q}x_q&=&0,\\ &&&\vdots&&&\\ a_{p1}x_1&+&\cdots&+&a_{pq}x_q&=&0,\\ \end{array}\] with coefficients $a_{ij}$ satisfying $a_{ij}=-1$, $0$, or $+1$ for all $i=1,\dots, p$, and $j=1,\dots, q$. Prove that if $q=2p$, there exists a solution $x_1, \dots, x_q$ of this system such that all $x_j$ ($j=1,\dots, q$) are integers satisfying $|x_j|\le q$ and $x_j\ne 0$ for at least one value of $j$.

Solution

Problem 6

For all positive integral $n$, $u_{n+1}=u_n(u_{n-1}^2-2)-u_1$, $u_0=2$, and $u_1=2\frac12$. Prove that \[3\log_2[u_n]=2^n-(-1)^n,\] where $[x]$ is the integral part of $x$.

Solution

1976 IMO (Problems) • Resources
Preceded by
1975 IMO
1 2 3 4 5 6 Followed by
1977 IMO
All IMO Problems and Solutions