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Difference between revisions of "1978 IMO Problems"

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Problems of the 20th [[IMO]] 1978 in Romania.
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==Day I==
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===Problem 1===
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Let <math> m</math> and <math> n</math> be positive integers such that <math> 1 \le m < n</math>. In their decimal representations, the last three digits of <math> 1978^m</math> are equal, respectively, to the last three digits of <math> 1978^n</math>. Find <math> m</math> and <math> n</math> such that <math> m + n</math> has its least value.
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[[1978 IMO Problems/Problem 1|Solution]]
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===Problem 2===
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We consider a fixed point <math>P</math> in the interior of a fixed sphere<math>.</math> We construct three segments <math>PA, PB,PC</math>, perpendicular two by two<math>,</math> with the vertexes <math>A, B, C</math> on the sphere<math>.</math> We consider the vertex <math>Q</math> which is opposite to <math>P</math> in the parallelepiped (with right angles) with <math>PA, PB, PC</math> as edges<math>.</math> Find the locus of the point <math>Q</math> when <math>A, B, C</math> take all the positions compatible with our problem.
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[[1978 IMO Problems/Problem 2|Solution]]
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===Problem 3===
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Let <math>0<f(1)<f(2)<f(3)<\ldots</math> a sequence with all its terms positive<math>.</math> The <math>n^{\text{th}}</math> positive integer which doesn't belong to the sequence is <math>f(f(n))+1.</math> Find <math>f(240).</math>
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[[1978 IMO Problems/Problem 3|Solution]]
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==Day II==
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===Problem 4===
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In a triangle <math>ABC</math> we have <math>AB = AC.</math> A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides <math>AB, AC</math> in the points <math>P,</math> respectively <math>Q.</math> Prove that the midpoint of <math>PQ</math> is the center of the inscribed circle of the triangle <math>ABC.</math>
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[[1978 IMO Problems/Problem 4|Solution]]
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===Problem 5===
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Let <math>f</math> be an injective function from <math>{1,2,3,\ldots}</math> in itself. Prove that for any <math>n</math> we have: <math>\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.</math>
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[[1978 IMO Problems/Problem 5|Solution]]
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===Problem 6===
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An international society has its members from six different countries. The list of members contain <math>1978</math> names, numbered <math>1, 2, \dots, 1978</math>. Prove that there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country.
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[[1978 IMO Problems/Problem 6|Solution]]
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* [[1978 IMO]]
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* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1978 IMO 1978 Problems on the Resources page]
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* [[IMO Problems and Solutions, with authors]]
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* [[Mathematics competition resources]] {{IMO box|year=1978|before=[[1977 IMO]]|after=[[1979 IMO]]}}

Latest revision as of 15:56, 29 January 2021

Problems of the 20th IMO 1978 in Romania.

Day I

Problem 1

Let $m$ and $n$ be positive integers such that $1 \le m < n$. In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$. Find $m$ and $n$ such that $m + n$ has its least value.

Solution

Problem 2

We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

Solution

Problem 3

Let $0<f(1)<f(2)<f(3)<\ldots$ a sequence with all its terms positive$.$ The $n^{\text{th}}$ positive integer which doesn't belong to the sequence is $f(f(n))+1.$ Find $f(240).$

Solution

Day II

Problem 4

In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

Solution

Problem 5

Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$

Solution

Problem 6

An international society has its members from six different countries. The list of members contain $1978$ names, numbered $1, 2, \dots, 1978$. Prove that there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country.

Solution

1978 IMO (Problems) • Resources
Preceded by
1977 IMO 		1 • 2 • 3 • 4 • 5 • 6 Followed by
1979 IMO
All IMO Problems and Solutions
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