IMO 1973 Problem 1
The reviewer identified a false claim in the previous proof:
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 22m51s
Problem
Point $O$ lies on line $g$; $\overrightarrow{OP_1}, \overrightarrow{OP_2}, \ldots, \overrightarrow{OP_n}$ are unit vectors such that points $P_1, P_2, \ldots, P_n$ all lie in a plane containing $g$ and on one side of $g$. Prove that if $n$ is odd, $$ \left\vert \overrightarrow{OP_1} + \overrightarrow{OP_2} + \ldots + \overrightarrow{OP_n} \right\vert \geq 1 $$ Here $\left\vert \overrightarrow{OM} \right\vert$ denotes the length of vector $\overrightarrow{OM}$.
Exploration
The reviewer identified a false claim in the previous proof:
$\theta_i-\theta_{m+1}\in[-\pi/2,\pi/2].$
Before repairing the argument, test this statement directly for small cases.
For $n=3$, take
$\theta_1=0,\qquad \theta_2=\frac{2\pi}{3},\qquad \theta_3=\pi.$
After rotation by $-\theta_2$, the shifted angles become
$-\frac{2\pi}{3},\quad 0,\quad \frac{\pi}{3}.$
Since
$-\frac{2\pi}{3}<-\frac{\pi}{2},$
the previous claim is false.
Now test whether the desired inequality itself still holds in this example. The sum is
$$\left(1-\frac12-1,\frac{\sqrt3}{2}\right) = \left(-\frac12,\frac{\sqrt3}{2}\right),$$
whose magnitude equals
$$\sqrt{\frac14+\frac34}=1.$$
Thus the theorem remains true even though the incorrect intermediate claim fails.
The previous strategy rotated around the median angle and attempted to show that the rotated sum has real part at least $1$. That idea may still work if the pair contributions are handled correctly.
Test the corrected pair inequality on small cases.
For $n=1$, there are no pairs, and the rotated sum is simply $(1,0)$.
For $n=3$, with angles
$0,\frac{2\pi}{3},\pi,$
the paired contribution to the first coordinate is
$$\cos\left(-\frac{2\pi}{3}\right) + \cos\left(\frac{\pi}{3}\right) = -\frac12+\frac12=0.$$
The middle vector contributes $1$, so the total first coordinate is exactly $1$.
For $n=5$, take
$0,0,\frac{\pi}{2},\pi,\pi.$
After rotation by $-\frac{\pi}{2}$, the shifted angles are
$$-\frac{\pi}{2}, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{\pi}{2}.$$
Each pair contributes
$$\cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 0.$$
Again the middle vector contributes $1$.
Now test an asymmetric configuration where the paired contribution is strictly positive. Take
$$\theta_1=0,\qquad \theta_2=\frac{\pi}{4},\qquad \theta_3=\pi.$$
After rotation by $-\frac{\pi}{4}$, the pair contributes
$$\cos\left(-\frac{\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt2}{2}-\frac{\sqrt2}{2} = 0.$$
Take instead
$$\theta_1=0,\qquad \theta_2=\frac{\pi}{3},\qquad \theta_3=\frac{5\pi}{6}.$$
After rotation by $-\frac{\pi}{3}$, the pair contributes
$$\cos\left(-\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{2}\right) = \frac12.$$
These tests suggest the corrected claim is:
For every pair,
$$\cos(\theta_i-\theta_{m+1}) + \cos(\theta_{n+1-i}-\theta_{m+1}) \ge0.$$
The proof must establish this directly, without falsely claiming each cosine is individually nonnegative.
Problem Understanding
The vectors are unit vectors whose directions all lie inside some interval of length $\pi$. After choosing coordinates appropriately, every vector can be written as
$$(\cos\theta_i,\sin\theta_i), \qquad 0\le\theta_i\le\pi.$$
The statement fails for even $n$ because opposite vectors can cancel in pairs. For example,
$$(1,0)+(-1,0)=0.$$
The odd case asserts that complete cancellation is impossible. The geometric reason is that after centering the configuration around the median direction, one vector remains unpaired, and every symmetric pair contributes a nonnegative amount in that central direction.
The previous proof failed because it tried to prove something stronger than necessary, namely that every paired cosine term is individually nonnegative. Counterexamples show that this is false. The corrected proof only needs the sum of the two cosine terms to be nonnegative.
Key Observations
Lemma 1. If two unit vectors make angles $\alpha$ and $\beta$, then
$$|u+v| = 2\cos\frac{\alpha-\beta}{2},$$
provided $|\alpha-\beta|\le\pi$.
Proof.
Write
$$u=(\cos\alpha,\sin\alpha), \qquad v=(\cos\beta,\sin\beta).$$
Then
\begin{align*}
|u+v|^2
&=
(\cos\alpha+\cos\beta)^2
(\sin\alpha+\sin\beta)^2 \
&=
2+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta).
\end{align*}
Using
$$\cos\alpha\cos\beta+\sin\alpha\sin\beta = \cos(\alpha-\beta),$$
we obtain
$$|u+v|^2 = 2+2\cos(\alpha-\beta).$$
The identity
$$1+\cos t = 2\cos^2\frac t2$$
gives
$$|u+v|^2 = 4\cos^2\frac{\alpha-\beta}{2}.$$
Since $|\alpha-\beta|\le\pi$,
$$\cos\frac{\alpha-\beta}{2}\ge0,$$
hence
$$|u+v| = 2\cos\frac{\alpha-\beta}{2}.$$
∎
Lemma 2. If
$$a\le0\le b \qquad\text{and}\qquad b-a\le\pi,$$
then
$$\cos a+\cos b\ge0.$$
Proof.
Use the identity
$$\cos a+\cos b = 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}.$$
Because
$$a\le0\le b,$$
the number
$$\frac{a+b}{2}$$
lies in the interval $[-\pi/2,\pi/2]$. Therefore
$$\cos\frac{a+b}{2}\ge0.$$
Also,
$$0\le b-a\le\pi,$$
so
$$-\frac{\pi}{2} \le \frac{a-b}{2} \le 0,$$
which implies
$$\cos\frac{a-b}{2}\ge0.$$
The product of these two cosine factors is nonnegative, proving
$$\cos a+\cos b\ge0.$$
∎
Solution
Choose coordinates so that the line $g$ is the $x$-axis and all vectors lie in the closed upper half-plane. Write
$$\overrightarrow{OP_i} = (\cos\theta_i,\sin\theta_i), \qquad 0\le\theta_i\le\pi.$$
Reorder the angles so that
$$0\le\theta_1\le\theta_2\le\cdots\le\theta_n\le\pi.$$
Since $n$ is odd, write
$$n=2m+1.$$
Let
$$S=\sum_{i=1}^n \overrightarrow{OP_i}.$$
Rotate the coordinate system by angle $-\theta_{m+1}$. Rotation preserves vector lengths, so it suffices to prove that the rotated vector sum has magnitude at least $1$.
After rotation, the vectors have angles
$$\theta_i-\theta_{m+1}, \qquad 1\le i\le n.$$
The middle vector has angle $0$, hence contributes $(1,0)$.
For each $i=1,\dots,m$, pair the vectors with shifted angles
$$a_i=\theta_i-\theta_{m+1}, \qquad b_i=\theta_{n+1-i}-\theta_{m+1}.$$
Because
$$\theta_i\le\theta_{m+1}\le\theta_{n+1-i},$$
we have
$$a_i\le0\le b_i.$$
Also,
$$b_i-a_i = \theta_{n+1-i}-\theta_i \le\pi,$$
since all original angles lie in $[0,\pi]$.
Lemma 2 now gives
$$\cos a_i+\cos b_i\ge0.$$
The first coordinate of the rotated sum therefore equals
$$1+\sum_{i=1}^m(\cos a_i+\cos b_i),$$
which is at least $1$.
Hence the rotated sum vector has first coordinate at least $1$. Its Euclidean norm is therefore at least $1$. Rotation preserves norms, so
$$\left| \overrightarrow{OP_1} + \cdots + \overrightarrow{OP_n} \right| \ge1.$$
This completes the proof.
∎
Verification of Key Steps
The reviewer identified the false statement
$$\theta_i-\theta_{m+1}\in[-\pi/2,\pi/2].$$
The corrected proof does not use this claim.
Check the repaired argument in the counterexample that invalidated the original proof:
$$\theta_1=0, \qquad \theta_2=\frac{2\pi}{3}, \qquad \theta_3=\pi.$$
After rotation by $-\frac{2\pi}{3}$,
$$a_1=-\frac{2\pi}{3}, \qquad b_1=\frac{\pi}{3}.$$
The individual cosine values are
$$\cos a_1=-\frac12, \qquad \cos b_1=\frac12.$$
The first cosine is negative, so the old argument fails. The corrected argument uses the sum:
\begin{align*}
\cos a_1+\cos b_1
&=
2\cos\frac{a_1+b_1}{2}
\cos\frac{a_1-b_1}{2} \
&=
2\cos\left(-\frac{\pi}{6}\right)
\cos\left(-\frac{\pi}{2}\right) \
&=0.
\end{align*}
Thus the pair contributes a nonnegative amount exactly as required.
Check the extremal example
$$0,0,\frac{\pi}{2},\pi,\pi.$$
After rotation by $-\frac{\pi}{2}$, the shifted angles are
$$-\frac{\pi}{2}, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{\pi}{2}.$$
Each pair contributes
$$\cos\left(-\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 0.$$
The middle vector contributes $1$, so the total first coordinate equals $1$. The total second coordinate equals $0$, giving equality in the theorem.
The argument also handles the degenerate case $n=1$. There are no pairs, and the rotated sum equals $(1,0)$.
Alternative Approaches
A complex-number formulation gives a shorter proof. Write
$$z_i=e^{i\theta_i}, \qquad 0\le\theta_i\le\pi.$$
After ordering the angles and rotating by $e^{-i\theta_{m+1}}$, the sum becomes
$$\sum_{i=1}^n e^{i(\theta_i-\theta_{m+1})}.$$
Its real part equals
$$1+\sum_{i=1}^m \left( \cos a_i+\cos b_i \right),$$
with the same notation as in the main proof. Lemma 2 shows every paired contribution is nonnegative, so the real part is at least $1$. Since the modulus of a complex number is at least the absolute value of its real part, the modulus of the sum is at least $1$.
Another approach starts from
$$|S|^2 = n+2\sum_{i<j}\cos(\theta_i-\theta_j).$$
The angles may be paired symmetrically around the median angle, and the cosine terms regrouped into nonnegative blocks. This eventually reduces to the same trigonometric inequality used in Lemma 2, but the vector formulation is more transparent because it isolates the geometric role of the median direction.