Isosceles triangles
For some value an isosceles triangle has at least one angle of degrees and at least one angle of degrees.
For each of such triangles, let denote the smallest angle in degrees. Then what is the sum of all possible 's?
The answer is 126.
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1 solution
There are three possible triangles that can be formed.
Triangle 1:
The base angles of the isosceles triangle are 2 d − 4 0 degrees and the vertex angle is d + 2 0 degrees.
Therefore: 2 × ( 2 d − 4 0 ) + ( d + 2 0 ) = 1 8 0 ⇔ 5 d − 6 0 = 1 8 0 ⇔ 5 d = 2 4 0 ⇔ d = 4 8
In this case the smallest angle is α = 2 d − 4 0 = 2 × 4 8 − 4 0 = 5 6 degrees.
Triangle 2:
The base angles of the isosceles triangle are d + 2 0 degrees and the vertex angle is 2 d − 4 0 degrees.
Therefore: 2 × ( d + 2 0 ) + ( 2 d − 4 0 ) = 1 8 0 ⇔ 4 d = 1 8 0 ⇔ d = 4 5
In this case the smallest angle is α = 2 d − 4 0 = 2 × 4 5 − 4 0 = 5 0 degrees.
Triangle 3:
The base angles of the isosceles triangle are d + 2 0 and 2 d − 4 0 degrees.
Therefore: d + 2 0 = 2 d − 4 0 ⇔ d = 6 0
In this case the base angles are d + 2 0 = 2 d − 4 0 = 8 0 degrees and the smallest angle is α = 1 8 0 − 2 × 8 0 = 2 0 degrees.
In conclusion: the sum of the possible angles α is 5 6 + 5 0 + 2 0 = 1 2 6 .