△ ABC has area 1 unit, with |AB|≤|AC|<|BC|
Determine the minimum length of |AC|.
Using Brian's valuable insight that ABC must be isosceles, say the short sides are x, and the long side is (x+y).
Then from Heron,
1=1/4 sqrt(((x+y)+(x)+(x))(-(x+y)+(x)+(x))((x+y)-(x)+(x))((x+y)+(x)-(x))), or 4 = sqrt((x - y) (x + y)^2 (3 x + y)).
But if so, then since x and y are positive real quantities, amazingly WolframAlpha gives x=sqrt(2) and y=(2-sqrt(2)), giving (x+y) as 2, the same result as Brian.
Edited on October 6, 2022, 4:49 am
|
|
Posted by broll
on 2022-10-06 04:23:38 |
Possible Solution
