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.
let the table have radius r and the sides of the corner be the x and y axis.
Then the table is centered at (r,r) and the equation of the table is
(x-r)^2+(y-r)^2=r^2
x^2-2xr+r^2+y^2-2yr+r^2=r^2
x^2+y^2+r^2-2(x+y)r=0
solving for r we get
r=[2(x+y)+-sqrt(4(x+y)^2-4(x^2+y^2))]/2
r=[2(x+y)+-sqrt(8xy)]/2
r=x+y+-sqrt(2xy)
no we are given a single point on the table is (80,90)
which gives
r=80+90+-sqrt(2*80*90)
r=170+-sqrt(16*9*100)
r=170+-4*3*10
r=170+-120
r=290 or r=50
the difference between these two solutions can be summarized by looking at the line connecting the points at which the table touches the corner. with r=290, the given point is on the side towards the corner and with r=50 the point is on the side away from the corner.
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Posted by Daniel
on 2016-05-31 12:34:17 |
solution
