Problem
Magic star built from paired sums
A five-pointed star has ten circles. The five outer circles hold 8, 6, 9, 12 and 10. The five inner circles are empty. Fill them so each side of the star totals 26.
Reasoning · Grade 5-2 Varieties of Magic Squares
Your answer
Watch out Solving one side at a time stalls — every side has two blanks, so take all five together.
How to solve
Strategy Change Focus / Count the Complement — Every side of the star has two empty circles on it, so no side can be finished in one subtraction — chasing the circles one at a time gets stuck immediately. The way through is to stop looking at single sides and add all five side totals at once. Because each circle lies on exactly two sides, that grand total counts every number in the star exactly twice, which hands over the sum of the five inner circles in one move. After that the star unlocks: one inner circle is pinned down and the rest fall in a ring, one subtraction each.
1STEP 1
Name the five empty circles and write down what each side says
Note what each side's two blanks must add to.
A+B = 26-(8+12) = 6, B+C = 26-(10+9) = 7, C+D = 26-(6+12) = 8, D+E = 26-(8+9) = 9, A+E = 26-(10+6) = 10
Tracing each of the five straight segments with a finger is the whole geometry of this problem — a side is a straight line, so the circles on it are exactly the four the line goes through. Once each side is written as a pair total, the star becomes pure arithmetic.
4.G.A.1Draw A Diagram2STEP 2
Notice that no single side can be finished
No side finishes on its own.
A+B = 6, B+C = 7, C+D = 8, D+E = 9, A+E = 10 (every one still has two blanks)
Recognising that you are stuck is a real step. When every equation has two blanks, the useful move is to change what you are looking at, not to try harder on one line.
3.NBT.A.2Make A Systematic List3STEP 3
Add all five sides at once — every circle gets counted twice
Adding all five sides makes the ten circles total 65.
5 × 26 = 130 = 2 × (sum of all ten circles) → sum of all ten = 130 ÷ 2 = 65
This is the whole idea of the problem. Nothing was measured or guessed — just counting the same numbers a second time on purpose, then dividing by two to undo the doubling. Multiplying 5 x 26 and halving 130 is Grade 4 arithmetic.
4.OA.A.3Change Focus Count The Complement4STEP 4
Subtract the numbers you can already see
Removing the visible numbers leaves 20.
8+6+9+12+10 = 45, A+B+C+D+E = 65-45 = 20
Once you know the total of everything, the part you cannot see is just the whole minus the part you can. That is the same 'total minus known' move used for a missing bar in a bar graph.
3.NBT.A.2Change Focus Count The Complement5STEP 5
Use two of the pair totals to pin down one circle
Two pair totals pin one circle at 6.
(A+B) + (C+D) + E = 20 → 6 + 8 + E = 20 → E = 6
Because there is an odd number of inner circles, any two of the pair totals always leave exactly one circle uncovered — so one number is always reachable this way. That is why a five-pointed star works out and a four-sided ring would not.
4.OA.A.3Change Focus Count The ComplementTwo of the pair totals set against each other pin down a single circle.
Why?
Both pair totals are true statements, so subtracting one from the other keeps a true statement while cancelling what they share.
🧱Do the same to both sides and equality holdsA balanced scale stays level when you take the same amount off both pans or add the same amount to both.
Why?
Each pair total is its two circles added together, so what remains after the cancelling is one circle on its own.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
6STEP 6
Walk round the ring, one subtraction at a time
Walking round gives 3, 5, 2 and 4.
D = 9-6 = 3, C = 8-3 = 5, B = 7-5 = 2, A = 6-2 = 4; A+E = 4+6 = 10 ✓
The five pair totals form a loop, so once one circle is known the rest follow like dominoes. The loop closing on the correct value at the end is a free check that no subtraction slipped.
3.NBT.A.2Look For A Pattern7STEP 7
Check all five sides
All five sides total 26.
8+4+2+12 = 26, 10+4+6+6 = 26, 8+6+3+9 = 26, 10+2+5+9 = 26, 6+3+5+12 = 26
Five short additions confirm the answer directly on the picture, so you never have to trust the algebra on its own.
3.NBT.A.2Make A Systematic ListAnswer
4, 6, 3, 5, 2
65 − 45 = 20
The five answers 4, 6, 3, 5, 2 add to 20, exactly the inner total the double-counting argument predicted, and 20 + 45 = 65 is half of 5 x 26 = 130, so the bookkeeping is consistent. The sizes are sensible too: each side already carries two outer numbers averaging about 9, so the two inner circles only have to supply about 26 - 18 = 8 between them, and indeed the inner numbers are all small, between 2 and 6. All five side totals were re-added on the picture and every one is 26. The five pair totals form a single odd-length loop, which means the answer is forced — there is no second way to fill the star.
Takeaway
When every line has two blanks, add all the lines together — each circle gets counted twice, and that doubling hands you the missing total in one step.
- Name the five empty circles and write down what each side says
- Notice that no single side can be finished
- Add all five sides at once — every circle gets counted twice
- Subtract the numbers you can already see
- Use two of the pair totals to pin down one circle
- Walk round the ring, one subtraction at a time
- Check all five sides