riddle of the day

The equator contains infinitely many places, each with its own temperature. An exact match seems impossible to guarantee without measuring them all, yet the geometry of the circle forces one to occur.

Across the Equator

Strategist
Pure logic

Riddle statement

Consider the temperature along Earth's equator at one fixed instant.

Assume that temperature varies continuously: moving a very small distance along the equator can change the temperature only by a very small amount.

Two points on the equator are opposite if they are half a revolution apart; equivalently, they are the endpoints of the same diameter of Earth.

Must there necessarily be a pair of opposite points with exactly the same temperature?

Justify your answer.

Show solution

Solution

Answer: yes. At every instant, there must be at least one pair of opposite points on the equator with the same temperature, provided that temperature varies continuously.

1. Represent the points on the equator

Parametrize the equator by an angle:

\[ 0\leq \theta<2\pi. \]

Let:

\[ T(\theta) \]

be the temperature at angular position \(\theta\).

A complete revolution returns to the same point, so:

\[ T(\theta+2\pi)=T(\theta). \]

The point opposite the position \(\theta\) has angular position:

\[ \theta+\pi. \]

2. Compare each point with its opposite

Define:

\[ f(\theta)=T(\theta)-T(\theta+\pi). \]

This function records which of the two opposite points is warmer:

  • if \(f(\theta)>0\), the point \(\theta\) is warmer;
  • if \(f(\theta)<0\), it is colder;
  • if \(f(\theta)=0\), the temperatures are equal.

Because \(T\) is continuous, \(f\) is continuous as well.

3. Move halfway around the equator

Evaluate the function at the opposite point:

\[ \begin{aligned} f(\theta+\pi) &=T(\theta+\pi)-T(\theta+2\pi)\\ &=T(\theta+\pi)-T(\theta)\\ &=-\bigl(T(\theta)-T(\theta+\pi)\bigr)\\ &=-f(\theta). \end{aligned} \]

Therefore:

\[ \boxed{f(\theta+\pi)=-f(\theta)}. \]

After half a revolution, the same two temperatures are being compared in the reverse order, so their difference changes sign.

4. Use continuity

Choose any starting position \(\theta_0\).

If \(f(\theta_0)=0\)

Then:

\[ T(\theta_0)=T(\theta_0+\pi), \]

and the required pair has already been found.

If \(f(\theta_0)>0\)

The identity above gives:

\[ f(\theta_0+\pi)=-f(\theta_0)<0. \]

As the continuous function \(f\) travels from a positive value to a negative one along the semicircle from \(\theta_0\) to \(\theta_0+\pi\), the intermediate value theorem guarantees a point \(c\) satisfying:

\[ f(c)=0. \]

If \(f(\theta_0)<0\)

Then:

\[ f(\theta_0+\pi)>0, \]

and the same argument again produces a point \(c\) for which:

\[ f(c)=0. \]

5. Interpret the zero

The equality:

\[ f(c)=0 \]

means:

\[ T(c)-T(c+\pi)=0. \]

Hence:

\[ T(c)=T(c+\pi). \]

The positions \(c\) and \(c+\pi\) are opposite points on the equator with equal temperatures.

Therefore:

\[ \boxed{ \text{there is always a pair of opposite points with equal temperatures} }. \]

6. Why continuity matters

The proof depends on the difference being unable to jump directly from a positive value to a negative value without passing through zero.

If arbitrary discontinuous changes were allowed, that passage through zero would no longer be guaranteed. Continuity is therefore an essential assumption.

7. The broader mathematical idea

This result is the one-dimensional case of a family of theorems about antipodal points. It is also a particularly elegant application of the intermediate value theorem.

Key idea: exchanging a point with its opposite reverses the temperature difference exactly. Continuity forces that difference to equal zero somewhere in between.