The ordinary schoolboy would correctly treat
this as a quadratic equation.
Here is the actual arithmetic.
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
If the other answer were admissible, the spot would be on the
edge farthest from the corner of the room.