1973 IMO Problems/Problem 1

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 $P_i = (\cos \alpha_i, \sin \alpha_i)$ , 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

This solution starts exactly like the previous one, till the line

$Q_2 = (a \cos \beta, a \sin \beta)$ . We want to show that the norm $\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| \ge 1$ .

Then, we continue as follows:

The end point of $\overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} = (\cos \alpha_1 + a \cos \beta + \cos \alpha_n, \sin \alpha_1 + a \sin \beta + \sin \alpha_n)$ , and the square of the norm of this vector is $S(\alpha_1, \alpha_n, \beta, a) = (\cos \alpha_1 + a \cos \beta + \cos \alpha_n)^2 + (\sin \alpha_1 + a \sin \beta + \sin \alpha_n)^2$ .

After some computations, we have $S(\alpha_1, \alpha_n, \beta, a) = 2 + a^2 + 2a\cos (\alpha_n - \beta) + 2a\cos (\beta - \alpha_1) + 2\cos (\alpha_n - \alpha_1)$ .

Since $\cos$ is a decreasing function from $1$ to $-1$ on the interval $[0, \pi]$ , it follows that $S(\alpha_1, \alpha_n, \beta, a) \ge S(0, \pi, \beta, a)$ .

Thus $S(\alpha_1, \alpha_n, \beta, a) \ge 2 + a^2 + 2a\cos (\pi - \beta) + 2a\cos (\beta) + 2\cos (\pi) = a^2$ .

It follows that the norm $\left| \overrightarrow{OQ_1} + \overrightarrow{OQ_2} + \overrightarrow{OQ_3} \right| \ge a \ge 1$ .

[Solution by pf02, May 2025]

Solution 3

For ease of visualization, assume the plane to have $x, y$ axes, $g$ to be the line $y = 0$ , and $O = (0, 0)$ . Assume the vectors are determined by $P_i = (\cos \alpha_i, \sin \alpha_i)$ , such that $0 < \alpha_1 \le \alpha_2 \le \dots \le \alpha_n < \pi$ .

We are asked to show that $\left| \sum_{i=1}^n \overrightarrow{OP_i} \right| \ge 1$ . Before embarking on the computations which prove this inequality, let us give the idea of the proof. Let $n = 2k + 1$ . We will prove formally that by "moving" $P_1, P_n$ to their limiting positions on $y = 0$ (i.e. replacing $P_1$ by $(1, 0)$ and $P_n$ by $(-1, 0)$ ) the norm of the sum of vectors will be decreased. Repeat this argument for $P_2, P_{n-1}$ , etc. (for a total of $k$ times). We are left with one vector (the middle vector), which has norm $1$ . Now let us turn this idea into a formal proof.

The point determining the sum of vectors is $P = (\sum_{i=1}^n \cos \alpha_i, \sum_{i=1}^n \sin \alpha_i)$ .

The square of the norm of the sum of vectors is $S(\alpha_1, \dots, \alpha_n) = \left| \overrightarrow{OP} \right|^2 = \left( \sum_{i=1}^n \cos \alpha_i \right)^2 + \left( \sum_{i=1}^n \sin \alpha_i \right)^2$ .

After some computations, we get $S(\alpha_1, \dots, \alpha_n) = n + 2 \sum_{1 \le i < j \le n} \cos (\alpha_j - \alpha_i)$ .

Denote $T(\alpha_1, \dots, \alpha_n) = 2 \sum_{1 \le i < j \le n} \cos (\alpha_j - \alpha_i)$ . When $n = 1$ , we define $T(\alpha_1) = 0$ . (This makes sense both formally, and as an accommodation to $S(\alpha_1) = 1$ .)

We have $T(\alpha_1, \dots, \alpha_n) = 2 \cos (\alpha_n - \alpha_1) + 2 \sum_{1 < i < n} \cos (\alpha_n - \alpha_i) + 2 \sum_{1 < j < n} \cos (\alpha_j - \alpha_1) + 2 \sum_{2 \le i < j \le n-1} \cos (\alpha_j - \alpha_i)$ .

Since $\cos$ is a decreasing function from $1$ to $-1$ on the interval $[0, \pi]$ , it follows that $T(\alpha_1, \alpha_2, \dots, \alpha_{n-1}, \alpha_n) \ge T(0, \alpha_2, \dots, \alpha_{n-1}, \pi)$ .

But $T(0, \alpha_2, \dots, \alpha_{n-1}, \pi) = 2 \cos (\pi - 0) + 2 \sum_{1 < i < n} \cos (\pi - \alpha_i) + 2 \sum_{1 < j < n} \cos (\alpha_j - 0) + 2 \sum_{2 \le i < j \le n-1} \cos (\alpha_j - \alpha_i) =$

$= T(\alpha_2, \dots, \alpha_{n-1}) - 2$

In short, we showed that $T(\alpha_1, \dots, \alpha_n) \ge T(\alpha_2, \dots, \alpha_{n-1}) - 2$ .

Now, if $n = 2k + 1$ , and we apply this argument $k$ times, we get $T(\alpha_1, \dots, \alpha_n) \ge -2k$ .

Thus, $S(\alpha_1, \dots, \alpha_n) = n + T(\alpha_1, \dots, \alpha_n) \ge 2k + 1 - 2k = 1$ .

[Solution by pf02, May 2025]

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

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