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Problem-Solving Strategies / LEVEL 3 · DIFFICULTY 3/5

Change the Representation

Use residues, bijections and shared-altitude geometry.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. MINI QUEST PS 3.1Rule Out a PossibilityRead the lesson
  2. MINI QUEST PS 3.2Build a BijectionRead the lesson
  3. MINI QUEST PS 3.3Share an AltitudeRead the lesson
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STAGE PS 3.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Rule Out a Possibility

Useful preparation: Patterns and Induction

Goal: Understand and apply rule out a possibility.

Before you begin: Patterns and Induction

Understand the idea

A remainder classification can disprove a claim without finding the alleged object. Every integer is congruent to zero, one or two modulo three, so its square has only two possible residues.

integer squares are 0 or 1 modulo 3

Choose and carry out a method

Square each residue class. Compare the proposed total with the resulting list; a residue of two cannot be a square.

Check the reasoning

Passing this test does not prove a number is a square. It is a necessary condition used to rule out impossible cases.

WORKED EXAMPLE 1

An integer N equals 5. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.

  1. Classify all integer inputs by their remainder modulo 3.
  2. 0²≡0, 1²≡1 and 2²≡1 modulo 3; 5≡2.
  3. The remainder is 2. A square cannot have this residue.

2

WORKED EXAMPLE 2

An integer N equals 8. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.

  1. Classify all integer inputs by their remainder modulo 3.
  2. 0²≡0, 1²≡1 and 2²≡1 modulo 3; 8≡2.
  3. The remainder is 2. A square cannot have this residue.

2

Common pitfalls

Possible mix-up: Remainder one guarantees a square.

It passes one necessary test but does not establish squareness.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Find a nonsquare that still has remainder one modulo three.

Preview the eight practice prompts
  1. An integer N equals 14. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  2. An integer N equals 17. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  3. An integer N equals 20. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  4. An integer N equals 23. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  5. An integer N equals 26. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  6. An integer N equals 29. What is its remainder modulo 3? Explain to yourself why this rules out N being a perfect square; enter the remainder only.
  7. A square array is claimed to contain exactly 32 dots. What is the total’s remainder modulo 3, which shows that an integer-sided square array is impossible? New context
  8. A square array is claimed to contain exactly 35 dots. What is the total’s remainder modulo 3, which shows that an integer-sided square array is impossible? New context
Open stage PS 3.1 in the student workspace →

STAGE PS 3.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Build a Bijection

Useful preparation: Rule Out a Possibility

Goal: Understand and apply build a bijection.

Before you begin: Rule Out a Possibility

Understand the idea

A bijection matches objects in two collections in both directions without duplication. Integer pairs with a<b match two-element subsets because sorting gives a unique ordered representation.

{(a,b):1≤a<b≤n} ↔ two-element subsets of {1,…,n}

Choose and carry out a method

Describe how to turn one counted object into the other and how to reverse the process. Then use the simpler collection’s count.

Check the reasoning

Both directions matter: every target must arise and no two source objects may map to the same target.

WORKED EXAMPLE 1

How many integer pairs (a,b) satisfy 1≤a<b≤6?

  1. Match each allowed pair to a two-element subset.
  2. Choose two distinct labels without order: C(6,2)=6·5/2.
  3. There are 15 pairs. Strict inequality excludes equal labels and prevents reversed duplicates.

15

WORKED EXAMPLE 2

How many integer pairs (a,b) satisfy 1≤a<b≤7?

  1. Match each allowed pair to a two-element subset.
  2. Choose two distinct labels without order: C(7,2)=7·6/2.
  3. There are 21 pairs. Strict inequality excludes equal labels and prevents reversed duplicates.

21

Common pitfalls

Possible mix-up: Count both (a,b) and (b,a).

The strict inequality selects exactly one order per pair.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Give the forward and reverse correspondence for a round-robin match.

Preview the eight practice prompts
  1. How many integer pairs (a,b) satisfy 1≤a<b≤9?
  2. How many integer pairs (a,b) satisfy 1≤a<b≤10?
  3. How many integer pairs (a,b) satisfy 1≤a<b≤11?
  4. How many integer pairs (a,b) satisfy 1≤a<b≤12?
  5. How many integer pairs (a,b) satisfy 1≤a<b≤13?
  6. How many integer pairs (a,b) satisfy 1≤a<b≤14?
  7. 15 players each meet every other player exactly once. How many matches take place? New context
  8. 16 players each meet every other player exactly once. How many matches take place? New context
Open stage PS 3.2 in the student workspace →

STAGE PS 3.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Share an Altitude

Useful preparation: Build a Bijection

Goal: Understand and apply share an altitude.

Before you begin: Build a Bijection

ABDCh
ABD and ADC share the same perpendicular height h. Their area ratio equals BD:DC.

Understand the idea

Two triangles with bases on the same line and a common opposite vertex have equal perpendicular heights. Their areas therefore inherit the ratio of their bases.

shared height ⇒ area₁/area₂=base₁/base₂

Choose and carry out a method

Express the desired base as a fraction of the whole base, then multiply the whole triangle’s area by that fraction.

Check the reasoning

The pieces must reconstruct the original triangle. The area ratio uses matching base segments, not unrelated slanted side lengths.

WORKED EXAMPLE 1

Point D lies on BC with BD:DC=3:6. Triangle ABC has area 27. Find the area of triangle ABD.

  1. Triangles sharing an altitude have areas proportional to their bases.
  2. [ABD]/[ABC]=3/(3+6); multiply by 27.
  3. The area is 9. The other piece has area 18, and the two sum correctly.

9

WORKED EXAMPLE 2

Point D lies on BC with BD:DC=4:7. Triangle ABC has area 33. Find the area of triangle ABD.

  1. Triangles sharing an altitude have areas proportional to their bases.
  2. [ABD]/[ABC]=4/(4+7); multiply by 33.
  3. The area is 12. The other piece has area 21, and the two sum correctly.

12

Common pitfalls

Possible mix-up: The segment ratio a:b means the first piece is a/b of the whole.

It is a/(a+b) of the whole.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why is no side-length or height calculation needed?

Preview the eight practice prompts
  1. Point D lies on BC with BD:DC=6:9. Triangle ABC has area 45. Find the area of triangle ABD.
  2. Point D lies on BC with BD:DC=7:10. Triangle ABC has area 51. Find the area of triangle ABD.
  3. Point D lies on BC with BD:DC=8:11. Triangle ABC has area 57. Find the area of triangle ABD.
  4. Point D lies on BC with BD:DC=9:12. Triangle ABC has area 63. Find the area of triangle ABD.
  5. Point D lies on BC with BD:DC=10:13. Triangle ABC has area 69. Find the area of triangle ABD.
  6. Point D lies on BC with BD:DC=11:14. Triangle ABC has area 75. Find the area of triangle ABD.
  7. A triangular field of area 81 is divided from its top vertex to a point splitting the opposite edge in ratio 12:15. Find the area beside the first edge segment. New context
  8. A triangular field of area 87 is divided from its top vertex to a point splitting the opposite edge in ratio 13:16. Find the area beside the first edge segment. New context
Open stage PS 3.3 in the student workspace →
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