Rmo 1993

In $\triangle CAD$, we have $CA < CD + DA$. In $\triangle CBD$, we have $CB < CD + DB$. Summing these gives $CA + CB < 2CD + (DA + DB) = 2CD + AB$. This doesn't immediately give the result. Better Strategy: Use the Angle Bisector Theorem property or area relations. Alternatively, assume the contrary: $CD \ge \frac12(CA + CB)$. Or use the property that the angle bisector lies strictly between the median and the altitude (or specific sides). A simpler approach: We know that the side opposite the largest angle is the longest. In $\triangle CDA$ and $\triangle CDB$, the angles at $D$ sum to $180$. Using the triangle inequality on $\triangle CDA$ and $\triangle CDB$ individually: $CA < CD + DA$ $CB < CD + DB$ $CA + CB < 2CD + (DA + DB)$. This path is circular.

So solution set: ( t \in [2,3] ).

By the early 1990s, India had begun to formalize its national Olympiad structure. The RMO was designed to transition students from standard school curricula toward the high-level problem-solving required at the national (INMO) and international levels. The 1993 session was held across various regions, with some specific regional variations such as the Madhya Pradesh RMO 1993 . rmo 1993

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