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Difference between revisions of "1973 IMO Problems/Problem 1"
Line 33: Line 33:
 
has norm <math>\ge 1</math>.  (It is true, but it needs a proof!)
 
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
+
2. Below, I will give a proof, which follows the idea from the "solution"
 
given above in the sense that it uses induction, but it fills in the
 
given above in the sense that it uses induction, but it fills in the
 
details of the essential step.
 
details of the essential step.

Revision as of 13:06, 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 "solution" 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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