Magic Triangles math puzzle worksheet for students to solve by placing numbers 1-6 in circles so each side sums to 10.

We are given a Magic Triangle puzzle where we must place the numbers 1 through 6, each used exactly once, into the six circles of a triangle such that:

- Each side of the triangle adds up to 10.
- There are three sides, and each side has three circles (including the corners).

Let’s analyze the structure.

---

Step 1: Understand the Structure

The triangle has 6 circles:
- 3 at the corners (shared by two sides),
- 3 along the edges (not corners, one per side).

Label the positions as follows:

```
A
/ \
B---C
/ \ / \
D---E---F
```

But actually, in this diagram, it's a triangle with three vertices and three edge centers, so more accurately:

Let’s label the positions like this:

```
O
/ \
O---O
/ \ / \
O---O---O
```

So there are 6 circles:
- Top vertex: T
- Left bottom vertex: L
- Right bottom vertex: R
- Middle of left side: M_L
- Middle of right side: M_R
- Middle of base: M_B

Now, the three sides are:

1. Left side: T → M_L → L
2. Right side: T → M_R → R
3. Base: L → M_B → R

Each side must sum to 10.

We need to assign numbers 1–6, each used once, to these six positions.

---

Step 2: Total Sum of Numbers

Sum of numbers 1 to 6:

$$
1 + 2 + 3 + 4 + 5 + 6 = 21
$$

Now, each corner is shared by two sides, while each middle (edge center) is on only one side.

Let’s denote:
- Let T, L, R be the corner values (each appears in two side sums).
- Let M_L, M_R, M_B be the middle values (each appears in one side sum).

Now, if we add up all three side sums:

Each side must sum to 10 → total sum of all sides = $ 10 + 10 + 10 = 30 $

But when we add the three side sums, we count:
- Each corner twice (since it's on two sides),
- Each middle once.

So:
$$
\text{Total} = 2(T + L + R) + (M_L + M_R + M_B) = 30
$$

But:
$$
T + L + R + M_L + M_R + M_B = 21 \quad \text{(sum of all numbers)}
$$

Let:
- $ C = T + L + R $ (sum of corners)
- $ M = M_L + M_R + M_B $ (sum of middles)

Then:
- $ 2C + M = 30 $
- $ C + M = 21 $

Subtract the second equation from the first:
$$
(2C + M) - (C + M) = 30 - 21 \Rightarrow C = 9
$$

So:
- Sum of corners = 9
- Sum of middles = $ 21 - 9 = 12 $

Now, we need to find three distinct numbers from 1–6 that sum to 9 for the corners, and the remaining three will sum to 12 for the middles.

---

Step 3: Find Possible Corner Triples That Sum to 9

List all combinations of three distinct numbers from 1–6 that sum to 9:

- 1 + 2 + 6 = 9 ✔
- 1 + 3 + 5 = 9 ✔
- 2 + 3 + 4 = 9 ✔

These are the only ones.

So possible corner sets: {1,2,6}, {1,3,5}, {2,3,4}

Now, let’s test each possibility.

---

Try Option 1: Corners = {1,2,6}, Middles = {3,4,5}

Sum of middles = 3+4+5=12 ✔

Let’s try assigning.

Suppose:
- T = 6
- L = 1
- R = 2

Then:
- Left side: T + M_L + L = 6 + M_L + 1 = 7 + M_L = 10 ⇒ M_L = 3
- Right side: T + M_R + R = 6 + M_R + 2 = 8 + M_R = 10 ⇒ M_R = 2 → but 2 is already used (R), and 2 not in middles — wait, middles are 3,4,5 → M_R must be 2? No! Invalid.

Try different assignment.

Let’s suppose:
- T = 1
- L = 6
- R = 2

Then:
- Left side: T + M_L + L = 1 + M_L + 6 = 7 + M_L = 10 ⇒ M_L = 3 ✔
- Right side: T + M_R + R = 1 + M_R + 2 = 3 + M_R = 10 ⇒ M_R = 7 ✘ Not possible.

Try:
- T = 1
- L = 2
- R = 6

Then:
- Left: 1 + M_L + 2 = 3 + M_L = 10 ⇒ M_L = 7 ✘

Try:
- T = 6
- L = 2
- R = 1

Left: 6 + M_L + 2 = 8 + M_L = 10 ⇒ M_L = 2 → but 2 is used (L), and 2 not in middles → invalid.

Try:
- T = 2
- L = 6
- R = 1

Left: 2 + M_L + 6 = 8 + M_L = 10 ⇒ M_L = 2 → again, 2 is used → invalid.

Try:
- T = 2
- L = 1
- R = 6

Left: 2 + M_L + 1 = 3 + M_L = 10 ⇒ M_L = 7 ✘

Try:
- T = 6
- L = 1
- R = 2 → already tried → M_R = 2 → conflict

Seems no valid assignment for {1,2,6}?

Wait — what if we try:
- T = 6
- L = 3 → but 3 not in corners → can't do.

Wait — corners are only {1,2,6}. So only those three.

Try:
- T = 1
- L = 6
- R = 2 → tried

Wait — maybe try:
- T = 1
- L = 2
- R = 6 → left: 1 + M_L + 2 = 3 + M_L = 10 → M_L = 7 → invalid

No way seems to work.

Wait — perhaps try:
- T = 6
- L = 1
- R = 2 → left: 6 + M_L + 1 = 7 + M_L = 10 → M_L = 3 ✔
- Right: 6 + M_R + 2 = 8 + M_R = 10 → M_R = 2 → but 2 is used at R → conflict

But 2 is already used at R → cannot use again.

Similarly, if we try:
- T = 6
- L = 2
- R = 1 → left: 6 + M_L + 2 = 8 + M_L = 10 → M_L = 2 → conflict (2 already used)

So no assignment works for {1,2,6}?

Wait — maybe try:
- T = 2
- L = 6
- R = 1 → left: 2 + M_L + 6 = 8 + M_L = 10 → M_L = 2 → conflict (T=2)

Same issue.

So no solution with corners {1,2,6}? Maybe try another set.

---

Try Option 2: Corners = {1,3,5}, Middles = {2,4,6}

Sum: 1+3+5=9 ✔, 2+4+6=12 ✔

Try assignments.

Try:
- T = 1
- L = 3
- R = 5

Then:
- Left side: T + M_L + L = 1 + M_L + 3 = 4 + M_L = 10 → M_L = 6 ✔
- Right side: T + M_R + R = 1 + M_R + 5 = 6 + M_R = 10 → M_R = 4 ✔
- Base: L + M_B + R = 3 + M_B + 5 = 8 + M_B = 10 → M_B = 2 ✔

Now check:
- Used corners: 1,3,5
- Middles: 6,4,2 → all in {2,4,6} ✔
- All numbers used once ✔

So this works!

Let’s write it down:

```
T = 1
/ \
M_L=6 M_R=4
/ \ / \
L=3 M_B=2 R=5
```

But wait — the base is L → M_B → R = 3 → 2 → 5 → sum = 3+2+5 = 10 ✔

Left side: T=1 → M_L=6 → L=3 → 1+6+3 = 10 ✔

Right side: T=1 → M_R=4 → R=5 → 1+4+5 = 10 ✔

Perfect!

So the configuration is:

```
1
/ \
6 4
/ \ / \
3---2---5
```

Now, check if all numbers 1–6 are used: 1,2,3,4,5,6 ✔

Each side sums to 10 ✔

So this is a valid solution.

But are there others?

Let’s see.

This was with corners = {1,3,5}, assigned as:
- T=1, L=3, R=5

But we could permute.

Try:
- T=1, L=5, R=3

Then:
- Left: 1 + M_L + 5 = 6 + M_L = 10 → M_L = 4
- Right: 1 + M_R + 3 = 4 + M_R = 10 → M_R = 6
- Base: 5 + M_B + 3 = 8 + M_B = 10 → M_B = 2

Middles: 4,6,2 → same set ✔

So another solution:

```
1
/ \
4 6
/ \ / \
5---2---3
```

Also valid.

Other permutations possible.

But the problem just asks to put numbers in the circles — so one solution suffices.

Now, check if other corner sets work.

---

Try Option 3: Corners = {2,3,4}, Middles = {1,5,6}

Sum: 2+3+4=9 ✔, 1+5+6=12 ✔

Try:
- T=2, L=3, R=4

Then:
- Left: 2 + M_L + 3 = 5 + M_L = 10 → M_L = 5 ✔
- Right: 2 + M_R + 4 = 6 + M_R = 10 → M_R = 4 → but 4 is already used (R) → conflict

Try:
- T=2, L=4, R=3

Left: 2 + M_L + 4 = 6 + M_L = 10 → M_L = 4 → conflict (L=4)

Try:
- T=3, L=2, R=4

Left: 3 + M_L + 2 = 5 + M_L = 10 → M_L = 5 ✔
Right: 3 + M_R + 4 = 7 + M_R = 10 → M_R = 3 → conflict (T=3)

Try:
- T=3, L=4, R=2

Left: 3 + M_L + 4 = 7 + M_L = 10 → M_L = 3 → conflict

Try:
- T=4, L=2, R=3

Left: 4 + M_L + 2 = 6 + M_L = 10 → M_L = 4 → conflict

Try:
- T=4, L=3, R=2

Left: 4 + M_L + 3 = 7 + M_L = 10 → M_L = 3 → conflict

All seem to have conflicts.

Try:
- T=2, L=4, R=3 → tried

Wait — what if:
- T=4, L=3, R=2 → left: 4 + M_L + 3 = 7 + M_L = 10 → M_L = 3 → conflict

No good.

So only the corner set {1,3,5} works.

Wait — try:
- T=2, L=3, R=4 → left: 2 + M_L + 3 = 5 + M_L = 10 → M_L = 5 ✔
- Right: 2 + M_R + 4 = 6 + M_R = 10 → M_R = 4 → conflict

No.

Try:
- T=3, L=4, R=2 → left: 3 + M_L + 4 = 7 + M_L = 10 → M_L = 3 → conflict

No.

So only {1,3,5} as corners works.

---

Final Answer

One valid solution:

```
1
/ \
6 4
/ \ / \
3---2---5
```

Check:
- Left side: 1 + 6 + 3 = 10 ✔
- Right side: 1 + 4 + 5 = 10 ✔
- Base: 3 + 2 + 5 = 10 ✔
- Numbers 1–6 used once ✔

---

How to Fill the Diagram

From the image, the triangle has:

- Top circle: 1
- Left middle: 6
- Right middle: 4
- Bottom-left: 3
- Bottom-middle: 2
- Bottom-right: 5

So fill as:

```
[1]
/ \
[6] [4]
/ / \
[3]---[2]---[5]
```

✔ This solves the puzzle.

---

Summary

Solution:

- Top: 1
- Left middle: 6
- Right middle: 4
- Bottom-left: 3
- Bottom-middle: 2
- Bottom-right: 5

Each side sums to 10.

✔ Valid and complete.