Grade 5 Olympiad · The Pigeonhole Principle

Partition Shapes to Force Closeness or Small Area

Turn a square, triangle, road, hexagon, or circle into geometric drawers. An overloaded region then forces nearby points or a small polygon.

Geometric drawersDistance boundsArea bounds10 missions

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Geometry becomes a drawer problem

In this lesson, the points are the objects and the subregions are the drawers. Use distinct points, including boundary points unless a problem says otherwise. Distances are straight-line distances; diagrams are not for measuring. A diameter is the greatest distance between two points of a region. The partition is chosen so that two or three points in one region already give the desired distance or area bound.

Four-step plan
Partition the shape → count regions → force an overloaded region → translate its size into a distance or area conclusion.

Two engines

some region has at least ⌈N ÷ k⌉ points

polygon area ≤ region area
point distance ≤ region diameter

For a strict “less than” conclusion, boundary assignments must be handled carefully.

Identify the objects, drawers, and forced occupancy

The shape matters only after the counting structure is clear.

Not complete

Balanced worst case

Points N

Regions k

4forced in one

13to force 4

13objects

4drawers

Translate the story

For every geometry problem in this lesson:

The objects are

The drawers are

The partition is useful only when each drawer has a known area or diameter.

Checkpoint

13 points in 4 regions force at least how many in one region?

Need a hint? Start here

The points are objects; the regions receive them.

Show the reasoning

Thirteen points in four regions force at least ceil(13÷4)=4 in one region. For a general k regions, 3k+1 points force four in one.

Worked example 3: overload one quarter of a square

Place nine distinct points in or on a square of side1unit, with no three collinear. Divide it into four equal small squares.

Not complete

Drag the nine points

Boundary rule: the center lines belong to the region on their right or below. Every point belongs to exactly one drawer.

Occupancy ledger

9points

4small squares

3largest count

1/4area each

Because 9 = 4×2 + 1, at least one quarter-square contains three points. The worked example excludes three collinear points so those three form a genuine triangle.

Checkpoint

Number of drawers

Points forced into one drawer

Area of each drawer in square units (enter a decimal)

Need a hint? Start here

Divide the unit square into four equal quarter-squares.

Show the reasoning

Nine points in four drawers force three into one. Each drawer has area 1/4. Assign shared boundaries consistently; the worked example assumes no three points are collinear.

Prove that a triangle occupies at most half its rectangle

This converts the overloaded quarter-square into the worked example area bound.

Not complete

Move three vertices (drag or arrow keys)

0.000triangle ÷ square

0.500maximum

0.000of whole unit square

0.125original bound

Proof, one step at a time

The two shaded strips together have height h₁ + h₂ ≤ H. The drawing supports the proof; do not measure it.

1. Order the vertices from top to bottom: A, B, C. Draw a horizontal line through B to meet AC at D. This splits triangle ABC into triangles ABD and BDC.

2. Both triangles have the same horizontal base BD. Their heights are the vertical gaps h₁ from A to B and h₂ from B to C. Their combined area is ½ × BD × (h₁ + h₂).

3. BD is no wider than the rectangle width W, and h₁ + h₂ is no greater than its height H. So the area is at most ½WH. If all three vertices are at the same height, the area is zero and the bound still holds.

triangle ≤ 1/2 × quarter-square = 1/2 × 1/4 = 1/8.

Checkpoint

Greatest fraction of a rectangle occupied by a triangle (enter a decimal)

Guaranteed triangle area bound in the whole square of side1 (enter a decimal)

Need a hint? Start here

Find the horizontal segment BD in the labelled proof.

Show the reasoning

The two smaller triangles share base BD and have heights h₁ and h₂. Their area is ½BD(h₁+h₂)≤½WH. Inside a quarter-square this is at most ½×¼=⅛.

Guided Practice 3: partition an equilateral triangle

Place ten distinct points in or on an equilateral triangle of side1unit. Divide it into nine equal small equilateral triangles.

Not complete

Nine geometric drawers

Number of points10

Why the distance is controlled

9drawers

2forced together

1/3small side

1/3diameter

Assign shared boundaries to one neighboring triangle only. Two points in the same small equilateral triangle are at distance at most its side length, 1/3.

Checkpoint

Small triangles

Points forced into one

Distance bound is 1 /

Need a hint? Start here

The subdivision must control side length, not just area.

Show the reasoning

Nine equilateral triangles of side 1/3 cover the unit equilateral triangle. Ten points force two in one small triangle, at distance at most 1/3.

Partition a line segment to force nearby points

Exercise 3 becomes a one-dimensional drawer model.

Not complete

General interval laboratory

Segment length L

Desired distance at most d

Points placed

Computed guarantee

100regions

1max width

2forced in one

101minimum points

original exercise. Planting 101 trees along a 100-metre roadside forces two trees at distance no more than 1 metre.

We use half-open intervals so a tree on a boundary is assigned to exactly one drawer.

Checkpoint

One-metre drawers along 100 m

Trees needed to force a pair

Maximum forced distance

Need a hint? Start here

Put the trees into one-metre intervals.

Show the reasoning

Use [0,1), [1,2), …, [99,100]. These are 100 disjoint drawers covering the roadside. With 101 trees, two share one, so their distance is at most 1 metre.

Test 23 Question 2: force four points into one quarter-square

Thirteen points give a small four-point hull.

Not complete

Counting step

Partition the unit square into four quarter-squares. Then:

⌈13 ÷ 4⌉ = 4

4points

3points

Convention used on this page. Because the printed test does not forbid collinear points, the “quadrilateral area” is interpreted as the area of the four-point convex hull. The convex hull is the region enclosed by a tight rubber band around the points. One point can lie inside the triangle formed by the others, so the hull need not have four corners. Collinear points give area0.

Drag four points inside one quarter-square

0.000of small square

0.000of whole square

0.250upper bound

4hull vertices

Checkpoint

Points forced into one quarter

Area bound in the whole square of side1 (enter a decimal)

Area object used by this page

Need a hint? Start here

Use the region containing all four points.

Show the reasoning

Thirteen points in four quarter-squares force four in one. Their convex hull lies inside that area-1/4 square. A hull may be a triangle if one point lies inside the other three; this is why the convention matters.

Choose regions by area or by diameter

A practice problem and a strict-distance assessment transfer.

Not complete

Area-1 equilateral triangle

Nine points enter four equal-area triangles, so three lie in one. Their triangle has area at most:

1/4

Exercise 8

Regular hexagon of side 1

Partition the hexagon into 24 side-1/2 triangles. Thus:

24 + 1 = 25 points

Test 23 Question 7

Why the distance is strictly less than ½. In a closed equilateral triangle of side ½, distance equals ½ only between two different vertices. To prevent that case, give each grid vertex its own different triangle drawer, as listed below. Give every nonvertex point on a shared edge to the triangle with the smaller drawer number; outer-edge points go to their only adjacent triangle. Interior points go to their triangle. Every point now has exactly one drawer, and no drawer owns two vertices. Thus any two points in one drawer have distance < ½. With 25 points and 24 drawers, such a pair is forced.

This proves that 25 is sufficient; this argument alone does not prove it is the least possible count.

Checkpoint

Nine points in four area drawers force

Their area bound is 1 /

Least point count that exceeds the 24 drawers in our hexagon proof, for distance < 1/2

Need a hint? Start here

For a strict distance, check what happens at vertices.

Show the reasoning

Nine points force three into one of four area-1/4 triangles. For the hexagon, the vertex ownership table gives 24 drawers each excluding the equality case for distance 1/2. Thus 25 points suffice.

Use six sectors inside a disk of radius1

The same partition supports a distance guarantee and an area guarantee.

Not complete

Six 60° sectors

The seven-point witness is the center plus the six vertices of a regular hexagon. No pair is closer than 1.

Two conclusions

8points force distance < 1

6sector drawers

3of 13 in one

π/6sector area

Distance. If the center is absent, 8 noncenter points enter 6 sectors. If the center is present, the other 7 enter 6 sectors. Either way, two noncenter points lie in one half-open 60° sector and are less than 1 apart.

Area. Assign the center, if present, to just one sector. Thirteen points then force three into one sector. Their straight-sided triangle has area strictly less than the curved sector’s area, π/6.

Test 23 Question 8 Exercise 11

Checkpoint

Minimum points for a pair at distance < 1

Among 13 points, forced into one sector

Sector area is π /

Need a hint? Start here

Separate the center when proving a strict distance.

Show the reasoning

Give noncenter points to six sectors that include one bounding ray and exclude the other. With eight points, at least seven are noncenter, so two share a sector and are less than 1 apart. Seven can fail: six regular-hexagon vertices and the center. Thirteen points also force three into a sector of area π/6.

Try a fresh problem

A square of side 2 contains nine points, with no three collinear. Our four-square partition guarantees a triangle of area at most how many square units?

Your answer

Why do you halve a small square’s area, rather than its side length?

This fresh problem has its own checkpoint. Your written explanation is saved for comparison and is not automatically graded.

Worked explanation for the fresh problem

Divide into four squares of side 1 and area 1. One contains three points. Their triangle has area at most half of 1, or 0.5 square units.

Geometric pigeonhole workshop

Correct all eight to complete the workshop.

Not complete

0 / 8 correctAnswer all eight questions.

Need a hint? Start here

Revisit the mission connected to each question.

Show the reasoning

Answers:points;3;1/2;2;101;4;25;8. Keep area and distance separate. The hexagon answer is the sufficient count from24drawers;the disk answer is minimal because seven points can still avoid a distance below1.

Exit ticket and certificate

Correct all five answers and complete all ten missions to earn the certificate.

Not complete

Nine points in a side-1square, no three collinear: denominator of the triangle-area bound from our proof,1 / ?

Ten points in a side-1equilateral triangle: denominator of the distance bound from our nine-triangle proof,1 / ?

Minimum trees along a straight100-metre segment forcing two at distance at most1metre

Least point count exceeding the 24 drawers in our proof for a hexagon of side1 and distance <1/2

Minimum number of distinct points in a disk of radius1 forcing distance <1

Chapter 23 Achievement

Geometric Drawer & Closeness Architect

This certificate recognizes

A determined learner

for partitioning shapes into useful drawers and proving distance and area guarantees with the Pigeonhole Principle.

Lesson 23.4 completed

Need a hint? Start here

Solve each item without using the answer shown in an earlier example.

Show the reasoning

Answers:8;3;101;25;8. Half of a quarter-square is1/8;the small equilateral triangles have side1/3;100intervals need101trees;24hexagon drawers need25points;the disk needs8and seven can fail.