1973 IMO Problems/Problem 1: Difference between revisions

Revision as of 17:29, 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.

@@ Line 13: / Line 13: @@
 ==Remarks (added by pf02, May 2025)==
-. 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
-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
-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
-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!)
-. 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,
-given above in the sense that it uses induction, but it fills in the
+I will give two more solutions.
-details of the essential step.
-. 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
+<math>Q_2 = (a \cos \beta, a \sin \beta)</math>..  We want to show that the norm
+<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
+<math>0 < \alpha_1 \le \alpha_n < \pi</math> the direction of
+<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
+\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
+\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
+<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==

1973 IMO (Problems) • Resources
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