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Difference between revisions of "1973 IMO Problems/Problem 1"
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==Remarks (added by pf02, May 2025)==
 
==Remarks (added by pf02, May 2025)==
  
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1. The "solution" given above is so incomplete that it can not
+
The "solution" given above is somewhat incomplete.  It simply
−
be called a solution.  It simply shoves the difficulty of the
+
shoves the difficulty of the problem into the phrase "... so their
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problem into the phrase "... so their sum has norm <math>\ge 1</math>, and
+
sum has norm <math>\ge 1</math>, and we're done."  Indeed, while true, this
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we're done."  Indeed, we don't know that their sum has norm
+
is not obvious, and it needs a proof.
−
<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
 
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the statement of the problem, not the one from the "solution",
 
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which should have been "k", so that <math>n = 2k - 1</math>; so the parameter
 
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from the proof is <math>k = 3</math>).  Take
 
−
<math>P_1 = (\cos \epsilon_1, \sin \epsilon_1)</math>,
 
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<math>P_2 = (\cos (\pi/3), \sin (\pi/3))</math>,
 
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<math>P_3 = (\cos (\pi - \epsilon_2), \sin (\pi - \epsilon_2))</math>,
 
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where <math>\epsilon_1, \epsilon_2</math> are very small positive numbers
 
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(and, assume <math>O = (0, 0)</math>, and <math>g</math> given by <math>y = 0</math>.)
 
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Then <math>\overrightarrow{OP_1} + \overrightarrow{OP_3}</math> has a
 
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very small norm, and its direction depends on <math>\epsilon_1, \epsilon_2</math>.
 
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It is not clear at all that adding
 
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<math>\overrightarrow{OP_1} + \overrightarrow{OP_3}</math> to <math>\overrightarrow{OP_2}</math>
 
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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 "solution"
+
Below, I will fill in the gaps in the first "solution".  Then,
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given above in the sense that it uses induction, but it fills in the
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I will give two more solutions.
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details of the essential step.
 
  
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3. Then, I will give a second proof, which uses the same argument,
+
 
−
but it proves the problem directly, without using induction.
+
==Solution with gaps filled in==
 +
 
 +
Let <math>n = 2k - 1</math>. We give a proof by induction on <math>k</math>.
 +
 
 +
For <math>k = 0</math>, we have one vector <math>\overrightarrow{OP_1}</math> whose
 +
norm (i.e. length) is <math>1</math>, so we are done.
 +
 
 +
Assume the statement is true for <math>0, 1, \dots, k - 1</math> and prove
 +
that it is true for <math>k</math>.  For ease of visualization, assume the
 +
plane to have <math>x, y</math> axes, <math>g</math> to be the line <math>y = 0</math>, and
 +
<math>O = (0, 0)</math>.  Let <math>n = 2k + 1</math>.  Assume the vectors are determined
 +
by <math>P_i = (\cos \alpha_i, \sin \alpha_1)</math>, such that
 +
<math>0 < \alpha_1 \le \alpha_2 \le \dots \le \alpha_n < \pi</math>.
 +
 
 +
Let <math>Q_1 = P_1, Q_3 = P_n</math> and <math>Q_2</math> be determined by
 +
<math>\overrightarrow{OQ_2} = \overrightarrow{OP_2} + \dots + \overrightarrow{OP_{n-1}}</math>
 +
(like in the solution above).  Since <math>\overrightarrow{OQ_2}</math> is the
 +
sum of <math>n - 2 = 2k - 1</math> unit vectors on one side of <math>y = 0</math>, its norm
 +
is <math>\ge 1</math> (because of the induction hypothesis).  Let us say that
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<math>Q_2 = (a \cos \beta, a \sin \beta)</math>..  We want to show that the norm
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<math>\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| \ge 1</math>.
 +
 
 +
Since <math>\overrightarrow{OQ_1}, \overrightarrow{OQ_3}</math> are unit vectors
 +
whose angles with the positive direction of the <math>x</math>-axis are
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<math>0 < \alpha_1 \le \alpha_n < \pi</math> the direction of
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<math>\overrightarrow{OQ_1} + \overrightarrow{OQ_3}</math> is <math>\frac{\alpha_1 + \alpha_n}{2}</math>.
 +
Since <math>\alpha_1 \le \beta \le \alpha_n</math>, the angle between
 +
<math>\overrightarrow{OQ_1} + \overrightarrow{OQ_3}</math> and <math>\overrightarrow{OQ_2}</math>
 +
is <math>\le \frac{\pi}{2}</math>.
 +
 
 +
Now use that the norm of <math>\overrightarrow{OQ_2}</math> is <math>a \ge 1</math>, and the
 +
general fact that if the angle between two vectors
 +
<math>\overrightarrow{u}, \overrightarrow{v}</math> is <math>\le \frac{\pi}{2}</math>, then
 +
<math>\left| \overrightarrow{u} + \overrightarrow{v} \right| \ge
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\max \{ \left| \overrightarrow{u} \right|, \left| \overrightarrow{v} \right| \}</math>.
 +
 
 +
It follows that
 +
<math>\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| =
 +
\left| (\overrightarrow{OQ_1} + \overrightarrow{OQ_3}) + \overrightarrow{OQ_2} \right| \ge
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\max \left\{ \left| \overrightarrow{OQ_1} + \overrightarrow{OQ_3} \right|, a \right\} \ge
 +
a \ge 1</math>.
 +
 
 +
To finish the proof, let us show this general fact about
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<math>\overrightarrow{u}, \overrightarrow{v}</math> forming an angle <math>\le \frac{\pi}{2}</math>.
 +
Let <math>\overrightarrow{u} = \overrightarrow{OA}, \overrightarrow{v} = \overrightarrow{OB}</math>
 +
and <math>\overrightarrow{u} + \overrightarrow{v} = \overrightarrow{OC}</math>.
 +
 
 +
Let <math>a, b, c</math> be the lengths of the segments <math>OA, OB, OC</math>.  Consider
 +
the triangle <math>\triangle OAC</math>.  The cosine formula in <math>\triangle OAC</math>
 +
yields <math>c^2 = a^2 + b^2 - 2ab\cos C</math>.  Since <math>\angle C</math> is
 +
obtuse, we have <math>\cos C \le 0</math>, so <math>c^2 \ge a^2 + b^2 \ge \max \{a^2, b^2\}</math>.
 +
This implies <math>c \ge \max \{ a, b \}</math>.
  
  
 
==Solution 2==
 
==Solution 2==
 +
 +
 +
  
  

Revision as of 18:29, 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)

The "solution" given above is somewhat incomplete. It simply shoves the difficulty of the problem into the phrase "... so their sum has norm $\ge 1$, and we're done." Indeed, while true, this is not obvious, and it needs a proof.

Below, I will fill in the gaps in the first "solution". Then, I will give two more solutions.


Solution with gaps filled in

Let $n = 2k - 1$. We give a proof by induction on $k$.

For $k = 0$, we have one vector $\overrightarrow{OP_1}$ whose norm (i.e. length) is $1$, so we are done.

Assume the statement is true for $0, 1, \dots, k - 1$ and prove that it is true for $k$. For ease of visualization, assume the plane to have $x, y$ axes, $g$ to be the line $y = 0$, and $O = (0, 0)$. Let $n = 2k + 1$. Assume the vectors are determined by $P_i = (\cos \alpha_i, \sin \alpha_1)$, such that $0 < \alpha_1 \le \alpha_2 \le \dots \le \alpha_n < \pi$.

Let $Q_1 = P_1, Q_3 = P_n$ and $Q_2$ be determined by $\overrightarrow{OQ_2} = \overrightarrow{OP_2} + \dots + \overrightarrow{OP_{n-1}}$ (like in the solution above). Since $\overrightarrow{OQ_2}$ is the sum of $n - 2 = 2k - 1$ unit vectors on one side of $y = 0$, its norm is $\ge 1$ (because of the induction hypothesis). Let us say that $Q_2 = (a \cos \beta, a \sin \beta)$.. We want to show that the norm $\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| \ge 1$.

Since $\overrightarrow{OQ_1}, \overrightarrow{OQ_3}$ are unit vectors whose angles with the positive direction of the $x$-axis are $0 < \alpha_1 \le \alpha_n < \pi$ the direction of $\overrightarrow{OQ_1} + \overrightarrow{OQ_3}$ is $\frac{\alpha_1 + \alpha_n}{2}$. Since $\alpha_1 \le \beta \le \alpha_n$, the angle between $\overrightarrow{OQ_1} + \overrightarrow{OQ_3}$ and $\overrightarrow{OQ_2}$ is $\le \frac{\pi}{2}$.

Now use that the norm of $\overrightarrow{OQ_2}$ is $a \ge 1$, and the general fact that if the angle between two vectors $\overrightarrow{u}, \overrightarrow{v}$ is $\le \frac{\pi}{2}$, then $\left| \overrightarrow{u} + \overrightarrow{v} \right| \ge \max \{ \left| \overrightarrow{u} \right|, \left| \overrightarrow{v} \right| \}$.

It follows that $\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| = \left| (\overrightarrow{OQ_1} + \overrightarrow{OQ_3}) + \overrightarrow{OQ_2} \right| \ge \max \left\{ \left| \overrightarrow{OQ_1} + \overrightarrow{OQ_3} \right|, a \right\} \ge a \ge 1$.

To finish the proof, let us show this general fact about $\overrightarrow{u}, \overrightarrow{v}$ forming an angle $\le \frac{\pi}{2}$. Let $\overrightarrow{u} = \overrightarrow{OA}, \overrightarrow{v} = \overrightarrow{OB}$ and $\overrightarrow{u} + \overrightarrow{v} = \overrightarrow{OC}$.

Let $a, b, c$ be the lengths of the segments $OA, OB, OC$. Consider the triangle $\triangle OAC$. The cosine formula in $\triangle OAC$ yields $c^2 = a^2 + b^2 - 2ab\cos C$. Since $\angle C$ is obtuse, we have $\cos C \le 0$, so $c^2 \ge a^2 + b^2 \ge \max \{a^2, b^2\}$. This implies $c \ge \max \{ a, b \}$.


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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