Problem: Let I be the incenter of △ABC and M be the midpoint of the side AB. Find the least possible value of CIM if CI=MI.
Solution
Solution: We may assume that AC<BC. Since ACI and AMI are acute, then △ACI≅△AMI. Hence AC=AM and AIC = AIM, i.e. CIM = 360 - 2 AIC = 180 - ABC. Note that ABC is maximal if BC is tangent to the circle with center A and radius AM.
Then ACB = 90, ABC = 30, and hence the least possible value of CIM is 150∘.
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Source: MathNet,
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