There are two rectangles 1 × 2 and 3 × 4 with parallel sides centered at a common point O. Points F and G are selected on the boundary of the inner and outer rectangle, respectively. What is the maximum possible area of triangle FOG?
There are two configurations that fit the description.
1. The lengths 1 and 3 are parallel.
Choose a corner of the smaller rectangle and either of two opposing corners of the larger. Both triangles have area 5. (Choosing either of the other corners gives area 1.)
2. The lengths 1 and 4 are parallel. Here as before choosing either of one pair opposing corners gives area 6 [Edit: 5.5 it's correct on my paper but I entered it here wrong.] (and choosing the others gives 2.5)
So the maximum is 6. [Edit: 5.5 but see below]
Incidentally, choosing corners gives two sides of the triangle sqrt(5) and 5. The angle between them is fixed at 45 degrees in the first and arctan(5.5) in the second. If the rectangles need no have parallel sides to each other, we could tilt the larger rectangle by arctan(2/11) = 10.3 degrees and get a right triangle with maximum area 5sqrt(5)/2 = 5.59
Edit:
These number are all off by a factor of 4. I was using graph paper and made 1 square = 1/2 unit. I forgot to correct for this when I typed it up.
So the answer to this problem is 5.5/4 = 1.375 in agreement with Larry.
The absolute max is 5sqrt(5)/8 = 1.40
Edited on December 13, 2024, 11:42 am
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Posted by Jer
on 2024-12-11 15:14:56 |
Solution
