Let's go back to reflection. Reflect $A$ across the line $CD$. Let $A'$ be the image. $A'$ lies on the line $CB$ (since $CD$ is bisector). Then $CA' = CA$. Connect $A'$ to $D$. $DA' = DA$. In $\triangle BDA'$, sides are $BD, DA', BA'$. $BA' = |CB - CA'| = |CB - CA|$. Triangle inequality in $\triangle BDA'$: $BD + DA' > BA'$. $BD + DA > |CB - CA|$. Note: $CD$ is the median of $\triangle ADA'$? No, perpendicular bisector. $CD \perp AA'$. In right triangle $CDA$, $CD < CA$. (Hypotenuse). So $CD < CA$. Similarly, in right triangle $CDA'$, $CD < CA' = CA$. So $CD$ is strictly less than $CA$. Does this imply $CD < \frac12(CA+CB)$? Since $CD < CA$, and $CD < CA$ (true for reflection), Wait, the reflection creates $\triangle ADA'$ with perpendicular bisector $CD$. So $CD$ is the altitude. In $\triangle CAD$, $\angle CDA$ is the angle. Since $CD \perp AA'$, $\triangle CDA$ is a right triangle? No, $A$ is reflected across $CD$, so $AA' \perp CD$. Thus $\triangle CDA$ is a right triangle with hypotenuse $CA$. Thus . If $\triangle CDA$ is a right triangle, then $\angle CDA = 90$. Then $\angle CDB = 90$. This implies $CD$ is an altitude. In that case, $CD < CB$ is also true. So $2CD < CA + CB$. Is $CD < CA$ always true? Only if $\triangle CDA$ is right angled? No. Reflection of $A$ across $CD$ puts $A'$ on the extension of $CB$. Since $CD \perp AA'$, $\triangle CDA$ is right angled? Yes, the line connecting a point and its reflection is perpendicular to the mirror line. So $CD \perp AA'$. Thus $\triangle CDA$ is a right triangle with hypotenuse $CA$. Therefore $CD < CA$ is always true. By symmetry, reflecting $B$ across $CD$, we get $CD < CB$. Therefore, $2CD < CA + CB$, implying $CD < \frac12(CA + CB)$. Q.E.D.
, given specific altitudes and lengths. Another complex geometry task involved calculating the sum of radii ( rmo 1993
Two circles $C_1$ and $C_2$ intersect at distinct points $A$ and $B$. Let $P$ be a point on $C_1$ outside $C_2$. Let the lines $PA$ and $PB$ meet $C_2$ again at $Q$ and $R$ respectively. Prove that the line $QR$ is the external angle bisector of $\angle QPR$ (or that $QR$ bisects the angle formed by $PR$ and the tangent at $P$). Let's go back to reflection
The Romanian Mathematical Olympiad final round (for high school) in 1993 had two or three papers (each with 3–4 problems), covering: $A'$ lies on the line $CB$ (since $CD$ is bisector)
Problem 8 of the set involved a logic puzzle regarding the ages of "foresworn" individuals, where the solution hinged on finding a sum that was a multiple of 9. Legacy and Resources for Preparation
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 .