Reasoning · Grade 4-2 One-Stroke Drawing

Problem

Draw a figure in one stroke

A one-stroke drawing never lifts the pen and never repeats a line. The four figures are a pentagon with all its diagonals, a square inside a square with only two corner pairs joined, three circles overlapping in pairs, and a house with a door. Hidden crossings count as points too. Pick every figure that can be drawn in one stroke.
A B C D
Your answer
How to solve
Strategy Make a Systematic List — Trying to trace each figure by trial and error would take forever and would never prove that a figure is impossible. Instead I work out the rule on tiny figures first, where I can see everything: at every point the pen passes through in the middle of the journey it uses up two lines, one to arrive and one to leave, so a point with an odd number of lines has to be where the pen starts or where it stops. That gives the test. Then I make a careful list, figure by figure and point by point, of how many lines meet at each point - marking every crossing point on the diagram first, because those are the ones people forget - and count the odd ones. A quick real trace with a pencil confirms the two that pass.
1STEP 1

Work out the rule on the smallest cases

Small cases give the rule: 0 or 2 odd points.

odd points = 0 or 2 → one-stroke drawing is possible
2STEP 2

Mark every point, including the hidden crossings

Mark every point, hidden crossings included.

A: 5 + 5 = 10 points, C: 3 × 2 = 6 points
3STEP 3

Count figure A: pentagon with all its diagonals

A has 0 odd points, so it works.

corners: 2 + 2 = 4 (even), crossings: 4 (even) → 0 odd points
4STEP 4

Count figure B: square inside a square with two links

B has 4 odd points, so it fails.

2+1 = 3 at four corners → 4 odd points ＞ 2
5STEP 5

Count figure C: three overlapping circles

C also has 0 odd points, so it works.

each crossing: 4 arc-ends (even) → 0 odd points
6STEP 6

Count figure D: the house

D has 6 odd points, so it fails.

6 points with 3 lines → 6 odd points ＞ 2
7STEP 7

Trace the two winners to be sure

Tracing confirms only A and C.

Answer
A, C
0, 4, 0, 6
The answer is a list of letters, which is what the question asked for. The odd-point counts came out 0, 4, 0 and 6, and every one of them is an even number - which it has to be, because every line has two ends, so the total of all the line-counts is twice the number of lines and can never leave an odd number of odd points behind. That is a real check on the counting, and it passes for all four figures. A second check: adding up the lines at every point of figure D gives 3 x 6 + 2 x 6 = 30, so figure D has 30 divided by 2 = 15 lines, which is exactly what you get by counting them in the picture. Finally, the two figures the test rejected are rejected for the same visible reason - loose connectors in B and T-junctions in D - while the two it accepted, A and C, really can be traced by hand.
Takeaway

At every point your pen passes through it uses two lines, one in and one out - so count the lines at each point, and a figure works in one stroke only if no more than two points are left with an odd number.

  • Work out the rule on the smallest cases
  • Mark every point, including the hidden crossings
  • Count figure A: pentagon with all its diagonals
  • Count figure B: square inside a square with two links
  • Count figure C: three overlapping circles
  • Count figure D: the house
  • Trace the two winners to be sure