Rmo 1993 [new] Info

[ \sqrtx+3-4\sqrtx-1 ;+; \sqrtx+8-6\sqrtx-1 ;=; 1. ]

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

Let's use the "Angle Bisector Length formula" or basic geometry. Consider triangle $ABC$. Assume $CA \ge CB$. Then $\angle A \le \angle B$. Since $\angle ACD = \angle BCD = C/2$. If $C$ is large, $CD$ is small. Use the "hinge" method. Consider $\triangle CAD$ and $\triangle CBD$. We have $CD < CA$ and $CD < CB$ ONLY if $\angle A > \angle CDA$ and $\angle B > \angle CDB$. But $\angle CDA + \angle CDB = 180^\circ$. One of these angles must be obtuse (unless $CD \perp AB$). If $\angle CDA$ is obtuse, then $CA > CD$. If $\angle CDB$ is acute, then $CB > CD$ is not guaranteed by obtuseness. Actually, if $\angle CDB > 90$, $CB > CD$. If $\angle CDB < 90$, $CB$ is not necessarily greater than $CD$. However, since $CDA + CDB = 180$, at least one is $\ge 90$. Thus $CD < \max(CA, CB)$. If $CD < \max(CA, CB)$, then $CD < \frac12(CA + CB)$ holds ONLY if $CD$ is not huge. Wait, is the inequality $CD < \frac12(CA+CB)$ always true? Yes. Proof: $CD$ is a side of $\triangle ADC$ and $\triangle BDC$. $CD < AC$ and $CD < BC$ are not simultaneously guaranteed. However, $CD$ is the side opposite angle $A$ in $\triangle ADC$ and opposite $B$ in $\triangle BDC$. Since $\angle ADC + \angle BDC = 180$, one is $\le 90$. WLOG let $\angle ADC \ge 90$. Then $\angle ADC > \angle A$ (since $\angle A < 90$ in a triangle where one angle is $\ge 90$... wait, triangle can be obtuse). Correct logic: In $\triangle ADC$, $CD / \sin A = CA / \sin(\angle CDA)$. In $\triangle BDC$, $CD / \sin B = CB / \sin(\angle CDB)$. $CD = CA \frac\sin A\sin(\angle CDA)$. $CD = CB \frac\sin B\sin(\angle CDB)$. Summing: $2CD = CA \frac\sin A\sin(\angle CDA) + CB \frac\sin B\sin(\angle CDB)$. Note that $\sin(\angle CDA) = \sin(\angle CDB) = \sin(\angle ADC)$. Let $\delta = \sin(\angle CDA)$. $2CD = \frac1\delta(CA \sin A + CB \sin B)$. Since $AB = c$, $c = CD(\frac\sin(A+C/2)\sin A + \dots)$. We want to prove $2CD < CA + CB$. From above, $2CD = \fracCA \sin A + CB \sin B\delta$. Since $\delta \le 1$, this suggests $2CD \ge CA \sin A + CB \sin B$, which doesn't help. [ \sqrtx+3-4\sqrtx-1 ;+; \sqrtx+8-6\sqrtx-1 ;=; 1

( x = t^2 + 1 ), with ( t \in [2,3] ). So ( x \in [5, 10] ). The 1993 session was held across various regions,

The RMO 1993 paper remains a relevant practice set for modern aspirants. While the "bag of tricks" required is relatively small (basic identities and inequalities), the paper tests the ability to apply these tricks elegantly. The geometry section is particularly instructive regarding the power of auxiliary constructions.