Math
Choosing 5 of 13 Equally Spaced Points Guarantees an Isosceles Triangle
Step-by-step mathematics solution: Choosing 5 of 13 Equally Spaced Points Guarantees an Isosceles Triangle
As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of Calculus, 7th Edition.
Problem restated
On a circle we mark the vertices of a regular 13‑gon (so the 13 points are equally spaced).
Take any 5 of those vertices. Show that among the 5 chosen points there are three that are the
vertices of an isosceles triangle (i.e. two of the three sides have the same length).
Full solution
1. Put the problem in a convenient language
Number the 13 equally spaced points
[ 0,1,2,\dots ,12 \pmod{13} ]
clockwise around the circle.
If we walk clockwise from a point (i) to a point (j) we pass a certain number of steps;
the chord length between (i) and (j) depends only on the shorter of the two arc lengths,
so it is the integer
[ \ell(i,j)=\min{\,|j-i|,\;13-|j-i|\,}\in{1,2,\dots ,6}. ]
Thus two chords are equal iff the corresponding step numbers are equal.
2. Order the chosen points and look at the gaps
Let the five chosen vertices be written in clockwise order
[ P_{1},P_{2},P_{3},P_{4},P_{5}. ]
Define the gaps (numbers of steps between successive chosen points)
[ g_{1}=|P_{2}-P_{1}|,\; g_{2}=|P_{3}-P_{2}|,\; g_{3}=|P_{4}-P_{3}|,\; g_{4}=|P_{5}-P_{4}|,\; g_{5}=13-|P_{5}-P_{1}|. ]
All (g_{k}) are positive integers and, because we have gone once all the way round the circle,
[ \boxed{g_{1}+g_{2}+g_{3}+g_{4}+g_{5}=13}\tag{1} ]
3. When do we obtain an isosceles triangle?
Look at a vertex, say (P_{2}).
The two sides that meet at (P_{2}) are the chords joining (P_{2}) with its neighbours
(P_{
Original question: Choosing 5 of 13 Equally Spaced Points Guarantees an Isosceles Triangle on Mathematics Stack Exchange, licensed CC BY-SA.