A circular table stands in a corner, touching both walls.
A certain point on the table’s edge is 90 cm from one wall and 80 cm
from the other.
Find the diameter of the table.
Adapted from Nonroutine Problems in Algebra, Geometry, and Trigonometry, 1965, by Steven Jerome Bryant et al.
Per a diagram of the situation, if r is the radius of the table:
The points of tangency of the table and the two walls are r cm from the corner, and of course the "certain point" is r cm from the center of the table, and the table's center is also r cm from each wall. Then:
(r-80)^2 + (r-90)^2 = r^2
r^2 - 160r + 6400 + r^2 - 180r + 8100 = r^2
r^2 - 340r + 14500 = 0
r = (340 +/- sqrt(115600 - 58000)) / 2 = 290 or 50
The answer 50 refers to a differing interpretation of the diagram (or so I thought when first writing this sentence).
The correct answer for the diameter of the table is 2*290 = 580 cm.
Sounds like a big table: Over 19 feet in diameter.
In fact, the radius of 50 cm is sounding better now. We were not told that the "certain point" was the nearest such point to the corner, as I had been assuming. It could indeed be a point that's not in the inaccessible pseudotriangle (quasitriangle?) in the corner.
In that case, the actual answer is that the diameter is 100 cm. That's only 39.37 inches.
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Posted by Charlie
on 2016-05-31 16:07:57 |
solution after a false assumption
