The triangle 5,12,13 has an area A=30 and a perimeter P=30, so A/P is 1.
The triangle 9,75,78 has an area A=324 and a perimeter P=162, so A/P is 2.
Find the smallest and largest integer-sided triangles where A/P is 10.
Sides can be parameterized as 2ab, a^2-b^2, a^2+b^2.
A=ab*(a^2-b^2)
P=2ab+2a^2
For the problem, A=10P. Substituting and simplifying gives b(a-b)=20.
Then we have parameter values (a,b)=(21,1),(12,2),(9,4),(9,5),(12,10),(21,20)
with corresponding sides (x,y,z)=(42,440,442),(48,140,148),(72,65,97),(90,56,106),(240,44,244),(840,41,841).
The smallest resulting triangle is (72,65,97) with A=2340 and P=234.
The largest triangle is (840,41,841) with A=17220 and P=1722.
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Posted by xdog
on 2024-01-04 09:20:44 |
the case for right triangles
