Lesson · Counting and extremes
Counting and extremes
A few geometry problems contain no congruent triangles at all. Points are scattered on a line, a square is cut into pieces, four points sit in general position - and what settles the question is a parity, an angle budget, or the extreme member of a finite family.
Parity and invariants
When a quantity is built from many segments, it is often easier to control its parity or its residue than its value. Two habits:
- compare the two sides of a claimed equality modulo something - if one side is always even and the other always odd, the equality is impossible and no computation was needed;
- on a line, distances combine additively: if \(P\) lies outside the segment \(AB\), then \(|PA - PB| = AB\), while for \(P\) inside \(PA + PB = AB\). A hypothesis about positions becomes an arithmetic constraint.
Example
A grasshopper sits at the origin of a line and makes \(n\) jumps, each of length \(1\), each either left or right. Its final position is a sum \(\varepsilon_1 + \cdots + \varepsilon_n\) with every \(\varepsilon_i = \pm 1\). Changing one sign changes the sum by \(2\), so the sum always has the same parity as \(n\): after an odd number of jumps the grasshopper cannot be back at the origin, however it chooses its directions.
Principle
A sum of \(n\) terms, each \(+t\) or \(-t\), is \(t\) times a number of the same parity as \(n\). Whenever a geometric quantity decomposes into equal pieces with signs, that single remark decides whether it can vanish.
The angle budget
Angles in a figure are a finite resource, and counting them proves impossibility:
- the interior angles of a convex \(n\)-gon add up to \((n-2)\cdot 180^\circ\), and its exterior angles always add up to \(360^\circ\);
- around an interior point of a dissection the angles add up to \(360^\circ\), and at a point in the interior of an edge to \(180^\circ\);
- a triangle has at most one angle that is at least \(90^\circ\).
Example
No convex polygon has four acute interior angles. An acute interior angle means an exterior angle greater than \(90^\circ\); four of those would already exceed the \(360^\circ\) that all exterior angles share.
Principle
To prove something is impossible, find the budget it overspends. To prove something is possible, no counting argument will do - you must construct an example and check it.
Extremes and the four-point dichotomy
Principle
In a finite configuration, take the extreme member: the largest angle, the longest segment, the pair of points that are closest, the vertices of the convex hull. The extreme choice comes with an inequality attached to it for free - everything else in the family is smaller.
For four points in the plane, no three collinear, there are exactly two shapes:
Each case hands you an angle sum: the four interior angles of the quadrilateral add to \(360^\circ\), and the three angles at an interior point also add to \(360^\circ\). In both cases some angle of the figure is at least \(90^\circ\) - and an averaging argument like that is usually the whole solution of an extremal problem.
Warm-up
Example
Five points lie inside a square of side \(2\). Prove that two of them are at distance at most \(\sqrt{2}\).
Five points, and a bound you must produce
Nothing is known about the points, so the argument cannot look at them individually. Split the square instead.
Cut the square
Divide it into four unit squares. Five points in four regions: some region contains two of them.
How far apart can two points in a unit square be?
At most the diagonal, \(\sqrt{2}\). That is the required bound. This is the pigeonhole principle in geometric clothing: the regions are the boxes, and the diameter of a box is the estimate.