# Solution

(related to Problem: The Spot on the Table)

The ordinary schoolboy would correctly treat this as a quadratic equation. Here is the actual arithmetic. Double the product of the two distances from the walls. This gives us $144,$ which is the square of $12.$ The sum of the two distances is $17.$ If we add these two numbers, $12$ and $17,$ together, and also subtract one from the other, we get the two answers that $29$ or $5$ was the radius, or half-diameter, of the table. Consequently, the full diameter was $58$ in. or $10$ in. But a table of the latter dimensions would be absurd, and not at all in accordance with the illustration. Therefore the table must have been $58$ in. in diameter. In this case, the spot was on the edge nearest to the corner of the room — to which the boy was pointing. If the other answer were admissible, the spot would be on the edge farthest from the corner of the room.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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