Rmo 1993 -

(RMO 1993, Grade 9 or 10, likely): Solve in real numbers:

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. rmo 1993

) for circles touching various sides of a rectangle, challenging students to prove that (RMO 1993, Grade 9 or 10, likely): Solve

( x = t^2 + 1 ), with ( t \in [2,3] ). So ( x \in [5, 10] ). $A'$ lies on the line $CB$ (since $CD$ is bisector)