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P.o.t.W. #8 Solution

Consider the right triangle OAT where O is the origin and T is the point where the circle is tangent to the graph of \(y = k\left| x \right| such that \(OA = a and \(AT = b . Because the slope of the right side of the graph of \(y = k\left| x \right| is \(k , then \(\frac{a}{b} = \frac{k}{1} . Now consider right triangle CTO. Triangles OAT and CTO are similar (corresponding sides proportional); so,...

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