Tanya, you like triangles.
a = sqrt(c^2-b^2);
test_1 a=sqrt(3) is longer than b=1
the good solution would be
a= min(sqrt(c^2 -b^2),b)
Kudos for finding this solution - it did indeed meet all the test cases, but wasn't quite what I had in mind! I will add a new test.
Back to basics 22 - Rotate a matrix
Scoring for oriented dominoes
Spot the outlier
Find nth maximum
Make a full wave rectifier
Area of an Isoceles Triangle
Length of the hypotenuse
Is this triangle right-angled?
Find a Pythagorean triple
Side of an equilateral triangle
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