Two non-congruent inter-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is 8:7. Find the minimum possible value of their common perimeter.
When I did, I got a very ugly radical, but AIME answers are always whole numbers. Solution please?
Triangles are: 233-233-210 and 218-218-240
Perimeter = 676, area = 21840
240:210 = 8:7
Best way is stick together 2 pythagorean triangles to get an isosceles triangle:
like 233-233-210 from 2 "105-208-233", thus 208 = height.
NOTE that I'm assuming that (somehow!) your problem means:
2 isoceles triangles with integer sides and different bases have the same integer area and integer perimeter.
The ratio of the bases is 8:7. Find the smallest case.
Like Serena, I've never heard of "inter-sided" triangles!
EDIT: unable to add a new post...(wonder why)...so I'll go this way:
on Soroban's solution, a bit easier if 7k is used as height instead of h, and 8k instead of k
Hello, Mrdavid445!
Two non-congruent integer-sided isosceles triangles have the same perimeter and the same area.
The ratio of the lengths of the bases of the two triangles is 8:7.
Find the minimum possible value of their common perimeter.
The two triangles looks like this:
Triangle has equal sides , base , and altitudeCode:* * *|* a * | * a b * | * b * |h * * |k * * | * * | * *-------+-------* *----+----* : - - 8c - - - : : - 7c - :
Triangle has equal sides , base and altitude
Their perimeters are equal: . .[1]
Their areas are equal: .
Square both sides: .
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. . . . . . . . . . . . . . .
Substitute [1]: .
n . . . . . .
n . . . .
n . . . . . . . . .
. . . . . . . . . . . .
n . . . . . . . . . .
Hence: .
Substitute into [1]: .
For the least integral dimensions, let
Hence: .
Perimeter of triangle is: .
Perimeter of triangle is: .