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Patterns - Magic Squares and Triangles

Grade 5CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Magic Square is a grid where the sum of numbers in every row, column, and diagonal is the same. This constant total is called the Magic Sum.

A 3x3 Magic Square showing numbers 1 to 9 where rows, columns, and diagonals add up to 15.
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A Magic Triangle is formed by placing numbers in circles on the sides of a triangle. The sum of the numbers on each side of the triangle is equal.

Structure of a magic triangle with circles on the vertices and mid-sides.
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In a 3×33 \times 3 magic square, the middle number is always exactly one-third of the Magic Sum. For example, if the sum is 3030, the middle number must be 1010.

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The numbers at the corners of a Magic Triangle are counted twice when summing up all the sides. To find a specific sum, strategic placement of the largest or smallest numbers at the corners is key.

📐Formulae

Magic Sum of 3×3 Square=3×Middle Number\text{Magic Sum of } 3 \times 3 \text{ Square} = 3 \times \text{Middle Number}

Total Sum of all sides (Triangle)=(Sum of all numbers used)+(Sum of corner numbers)\text{Total Sum of all sides (Triangle)} = (\text{Sum of all numbers used}) + (\text{Sum of corner numbers})

Magic Constant (n × n square)=n(n2+1)2\text{Magic Constant (n } \times \text{ n square)} = \frac{n(n^2 + 1)}{2}

Missing Number=Magic Sum−(Sum of known numbers in that line)\text{Missing Number} = \text{Magic Sum} - (\text{Sum of known numbers in that line})

💡Examples

Problem 1:

Complete a 3×33 \times 3 magic square using numbers from 46 to 54. The Magic Sum is 150.

Solution:

  1. Find the middle number of the sequence 46, 47, 48, 49, 50, 51, 52, 53, 54. The middle number is 5050.
  2. Place 5050 in the center cell.
  3. Check the rule: Magic Sum=3×Middle Number⇒150=3×50\text{Magic Sum} = 3 \times \text{Middle Number} \Rightarrow 150 = 3 \times 50. This is correct.
  4. Place the remaining numbers such that each row, column, and diagonal adds to 150.
  5. Top row: 53+46+51=15053 + 46 + 51 = 150
  6. Middle row: 48+50+52=15048 + 50 + 52 = 150
  7. Bottom row: 49+54+47=15049 + 54 + 47 = 150
  8. Check diagonal: 53+50+47=15053 + 50 + 47 = 150.

Explanation:

By placing the median of the sequence in the center and balancing the largest numbers with the smallest numbers on opposite sides, we satisfy the magic sum requirement for all directions.

Problem 2:

Arrange the numbers 1, 2, 3, 4, 5, and 6 in a Magic Triangle so that the sum of each side is 9.

Solution:

  1. Sum of all numbers provided: 1+2+3+4+5+6=211 + 2 + 3 + 4 + 5 + 6 = 21.
  2. Target sum for 3 sides: 9×3=279 \times 3 = 27.
  3. Find the sum of corner numbers: Total target−Sum of numbers=27−21=6\text{Total target} - \text{Sum of numbers} = 27 - 21 = 6.
  4. Identify three numbers from the set {1, 2, 3, 4, 5, 6} that add up to 6. These are 1,2, and 31, 2, \text{ and } 3.
  5. Place 1, 2, and 3 at the corners of the triangle.
  6. Find the middle numbers for each side:
    • Side between corners 1 and 2: 9−(1+2)=69 - (1 + 2) = 6.
    • Side between corners 2 and 3: 9−(2+3)=49 - (2 + 3) = 4.
    • Side between corners 3 and 1: 9−(3+1)=59 - (3 + 1) = 5.
  7. The sides are (1,6,2)(1, 6, 2), (2,4,3)(2, 4, 3), and (3,5,1)(3, 5, 1). All sum to 9.

Explanation:

The 'extra' sum needed to reach the side totals comes from the corner numbers being counted twice. By calculating that the corners must sum to 6, we correctly identify 1, 2, and 3 as the vertex numbers.

Problem 3:

Fill in the missing numbers in the 3×33 \times 3 Magic Square below so that every row, column, and diagonal adds up to 7575. The numbers already present are 2020 and 2525 in the middle row, and 3030 in the bottom middle cell.

Magic square with sum 75 filled with numbers 21 to 29.

Solution:

  1. Identify the middle number: Since the Magic Sum is 7575, the middle number is 75÷3=2575 \div 3 = 25.
  2. Calculate top-middle: Bottom-middle is 3030 and center is 2525. To make the middle column sum to 7575: 75−(30+25)=2075 - (30 + 25) = 20.
  3. Calculate middle-left: Center is 2525 and middle-right is 2020. To make the middle row sum to 7575: 75−(25+20)=3075 - (25 + 20) = 30.
  4. Continue using the sum of 7575 for remaining cells.

Explanation:

We used the property that the center number is 13\frac{1}{3} of the Magic Sum (75÷3=2575 \div 3 = 25) and then solved for individual lines where two numbers were known.

Problem 4:

Arrange the numbers 2,3,4,5,6,72, 3, 4, 5, 6, 7 in a Magic Triangle such that the sum of each side is 1212.

Magic triangle with numbers 2, 7, 3 on one side, 3, 5, 4 on the second, and 4, 6, 2 on the third, all summing to 12.

Solution:

  1. Place the numbers 2,3,42, 3, 4 at the corners.
  2. Side 1: Between corner 22 and 33, place 77. Total: 2+7+3=122+7+3 = 12.
  3. Side 2: Between corner 33 and 44, place 55. Total: 3+5+4=123+5+4 = 12.
  4. Side 3: Between corner 44 and 22, place 66. Total: 4+6+2=124+6+2 = 12.

Explanation:

To achieve a smaller sum like 1212, the smallest numbers (2,3,42, 3, 4) are placed at the corners as they are shared by two sides.