Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Find the answer Italy

Problem:

A blacksmith is building a horizontal iron railing made up of many vertical bars, parallel to one another, each of which is placed at a distance of 18 cm18~\mathrm{cm} from the 2 neighboring ones. The blacksmith connects the ends of each pair of adjacent bars with a bar curved into a circular arc, lying in the plane of the bars, whose highest point is at a distance of 33 cm3 \sqrt{3}~\mathrm{cm} from the line (dashed in the figure) that passes through the upper ends of all the bars, and which is perpendicular to the bars themselves. How long is each of the small bars used to build the arcs?

Figure 1

Pick one

Solution

Solution:

The answer is (D). Referring to the figure, let AA and BB be the ends of adjacent bars, VV the vertex of the arc A B\text{A B}, MM the midpoint of ABA B and OO the center of the circle to which the arc AB^\widehat{A B} belongs. It must hold that OB=OVO B=O V, so setting OM=xO M=x, by the Pythagorean Theorem, it must hold that:
OM2+MB2=OV2=(OM+MV)2x2+81=(x+33)2=x2+63x+27 \begin{gathered} O M^{2}+M B^{2}=O V^{2}=(O M+M V)^{2} \\ x^{2}+81=(x+3 \sqrt{3})^{2}=x^{2}+6 \sqrt{3} x+27 \end{gathered}
Figure 2
It follows that x=33x=3 \sqrt{3}, hence OM=MV,OB=2⋅OMO M=M V, O B=2 \cdot O M, and the angle AOB^\widehat{A O B} has amplitude equal to 2π3\frac{2 \pi}{3}. The arc AB\text{AB} therefore has length equal to:
2π3⋅OB=4π3 \frac{2 \pi}{3} \cdot O B=4 \pi \sqrt{3}

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.