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.
My grade 8 students came up with four variant solutions, many like Daniel's with coordinates and the origin either at the corner or at the center of the circle.
Here's one that doesn't use coordinates:
Draw lines through the center of the circle and perpendicular to the two walls. Draw lines through the given point and perpendicualr to the wall. A rectangle is formed whose opposite corners are the center of the circle and the given point.
The dimensions of the rectangle are (r80) by (r90) and its diagonal is r.
By Pythagorean theorem (r80)^2 + (r90)^2 = r^2
r^2  340r + 14500 = 0
Solve to get
r=290 or r=50

A more complicate alternate solution I devised was to use the triangle whose corners are
A = the center of the circle
B = the given point
C = Corner of the room
AC = r*sqrt(2)
BC = sqrt(14500)
AB = r
angle C can be found as 45 arctan(80/90)
From which you can get Cos(C)
The law of Cosines eventually leads to the same quadratic equation as above.
This was a very well timed problem. Thanks Ady.

Posted by Jer
on 20160531 13:22:04 