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Difference between revisions of "1973 IMO Problems/Problem 1"

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==Problem==
 
==Problem==
 +
 
Point <math>O</math> lies on line <math>g;</math> <math>\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}</math> are unit vectors such that points <math>P_1, P_2, \cdots, P_n</math> all lie in a plane containing <math>g</math> and on one side of <math>g.</math> Prove that if <math>n</math> is odd, <cmath>\left|\overrightarrow{OP_1}+\overrightarrow{OP_2}+\cdots+ \overrightarrow{OP_n}\right|\ge1.</cmath> Here <math>\left|\overrightarrow{OM}\right|</math> denotes the length of vector <math>\overrightarrow{OM}.</math>
 
Point <math>O</math> lies on line <math>g;</math> <math>\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}</math> are unit vectors such that points <math>P_1, P_2, \cdots, P_n</math> all lie in a plane containing <math>g</math> and on one side of <math>g.</math> Prove that if <math>n</math> is odd, <cmath>\left|\overrightarrow{OP_1}+\overrightarrow{OP_2}+\cdots+ \overrightarrow{OP_n}\right|\ge1.</cmath> Here <math>\left|\overrightarrow{OM}\right|</math> denotes the length of vector <math>\overrightarrow{OM}.</math>
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==Solution==
 
==Solution==
 +
 
We prove it by induction on the number <math>2n+1</math> of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the <math>2n-1</math> vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm <math>\ge 1</math> betwen two with norm <math>1</math>. The sum of the two vectors of norm <math>1</math> makes an angle of <math>\le\frac\pi 2</math> with the vector of norm <math>\ge 1</math>, so their sum has norm <math>\ge 1</math>, and we're done.
 
We prove it by induction on the number <math>2n+1</math> of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the <math>2n-1</math> vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm <math>\ge 1</math> betwen two with norm <math>1</math>. The sum of the two vectors of norm <math>1</math> makes an angle of <math>\le\frac\pi 2</math> with the vector of norm <math>\ge 1</math>, so their sum has norm <math>\ge 1</math>, and we're done.
  
 
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p357939]
 
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p357939]
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 +
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==Remarks (added by pf02, May 2025)==
 +
 +
1. The "solution" given above is so incomplete that it can not
 +
be called a solution.  It simply shoves the difficulty of the
 +
problem into the phrase "... so their sum has norm <math>\ge 1</math>, and
 +
we're done."  Indeed, we don't know that their sum has norm
 +
<math>\ge 1</math>.  This is exactly what we have to prove.  To make this
 +
difficulty clear, imagine <math>n = 3</math> (I am referring to <math>n</math> from
 +
the statement of the problem, not the one from the "solution",
 +
which should have been "k", so that <math>n = 2k - 1</math>; so the parameter
 +
from the proof is <math>k = 3</math>).  Take
 +
<math>P_1 = (\cos \epsilon_1, \sin \epsilon_1)</math>,
 +
<math>P_2 = (\cos (\pi/3), \sin (\pi/3))</math>,
 +
<math>P_3 = (\cos (\pi - \epsilon_2), \sin (\pi - \epsilon_2))</math>,
 +
where <math>\epsilon_1, \epsilon_2</math> are very small positive numbers
 +
(and, assume <math>O = (0, 0)</math>, and <math>g</math> given by <math>y = 0</math>.)
 +
Then <math>\overrightarrow{OP_1} + \overrightarrow{OP_3}</math> has a
 +
very small norm, and its direction depends on <math>\epsilon_1, \epsilon_2</math>.
 +
It is not clear at all that adding
 +
<math>\overrightarrow{OP_1} + \overrightarrow{OP_3}</math> to <math>\overrightarrow{OP_2}</math>
 +
has norm <math>\ge 1</math>.  (It is true, but it needs a proof!)
 +
 +
2. Below, I will give a proof, which follows the idea from the proof
 +
given above in the sense that it uses induction, but it fills in the
 +
details of the essential step.
 +
 +
3. Then, I will give a second proof, which uses the same argument,
 +
but it proves the problem directly, without using induction.
 +
 +
 +
==Solution 2==
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 +
 +
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 +
 +
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[TO BE CONTINUED]
 +
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 13:02, 15 May 2025

Problem

Point $O$ lies on line $g;$ $\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}$ are unit vectors such that points $P_1, P_2, \cdots, P_n$ all lie in a plane containing $g$ and on one side of $g.$ Prove that if $n$ is odd, \[\left|\overrightarrow{OP_1}+\overrightarrow{OP_2}+\cdots+ \overrightarrow{OP_n}\right|\ge1.\] Here $\left|\overrightarrow{OM}\right|$ denotes the length of vector $\overrightarrow{OM}.$


Solution

We prove it by induction on the number $2n+1$ of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the $2n-1$ vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm $\ge 1$ betwen two with norm $1$. The sum of the two vectors of norm $1$ makes an angle of $\le\frac\pi 2$ with the vector of norm $\ge 1$, so their sum has norm $\ge 1$, and we're done.

The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


Remarks (added by pf02, May 2025)

1. The "solution" given above is so incomplete that it can not be called a solution. It simply shoves the difficulty of the problem into the phrase "... so their sum has norm $\ge 1$, and we're done." Indeed, we don't know that their sum has norm $\ge 1$. This is exactly what we have to prove. To make this difficulty clear, imagine $n = 3$ (I am referring to $n$ from the statement of the problem, not the one from the "solution", which should have been "k", so that $n = 2k - 1$; so the parameter from the proof is $k = 3$). Take $P_1 = (\cos \epsilon_1, \sin \epsilon_1)$, $P_2 = (\cos (\pi/3), \sin (\pi/3))$, $P_3 = (\cos (\pi - \epsilon_2), \sin (\pi - \epsilon_2))$, where $\epsilon_1, \epsilon_2$ are very small positive numbers (and, assume $O = (0, 0)$, and $g$ given by $y = 0$.) Then $\overrightarrow{OP_1} + \overrightarrow{OP_3}$ has a very small norm, and its direction depends on $\epsilon_1, \epsilon_2$. It is not clear at all that adding $\overrightarrow{OP_1} + \overrightarrow{OP_3}$ to $\overrightarrow{OP_2}$ has norm $\ge 1$. (It is true, but it needs a proof!)

2. Below, I will give a proof, which follows the idea from the proof given above in the sense that it uses induction, but it fills in the details of the essential step.

3. Then, I will give a second proof, which uses the same argument, but it proves the problem directly, without using induction.


Solution 2

[TO BE CONTINUED]


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1973 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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