# A Three Pegs Question

Three pegs on a plane form an isosceles right triangle with a leg of length 3. The pegs may move to an arbitrary distance but on a line parallel to the line formed by the other two. Is it possible to eventually get the three pegs at the vertices of a right triangle with legs 2 and 4?

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Solution

Three pegs on a plane form an isosceles right triangle with a leg of length 3. The pegs may move to an arbitrary distance but on a line parallel to the line formed by the other two. Is it possible to eventually get the three pegs at the vertices of a right triangle with legs 2 and 4?

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### Solution

No, this is not possible.

To see why, consider the triangle formed by the three pegs. The displacement of a vertex along the line parallel to the opposite side does not change the area of the triangle. In other words, the area of the triangle formed by the three pegs is invariant under any possible puzzle move. Initially, the area was 3×3/2 = 4.5. This is different from 2×4/2 = 4. Therefore, the pegs can never settle at the vertices of the other triangle.

### References

1. A Decade of the Berkeley Mathematical Circle, The American Experience, Volume I, Z. Stankova, Tom Rike (eds), AMS/MSRI, 2008, p. 96

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