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2021 WSMO Accuracy Round Problems/Problem 6

Revision as of 10:03, 11 July 2022 by Pinkpig (talk | contribs)

Problem

Roy is baking a circular three tier cake. All of the tiers are centered around the same point. Each tier's radius is $\frac{3}{4}$ of the radius of the tier below it, but the height of each tier stays constant. Roy wants to ice the cake, but only on the curved surfaces of the cake and the top of the smallest tier. The diameter of the lowest tier is $128$ centimeters and its height is $10$ centimeters. If the surface area that is iced can be expressed as $m\pi,$ find $m.$

Solution 1

We will consider the three tiers separately. The bottom tier has a diameter of 128 and and a height of 10. Therefore, the surface area of the iced part of the bottom tier is $128\pi\cdot10=1280\pi.$ The middle tier has a diameter of $128\cdot\frac{3}{4}=96$ and a height of 10. Therefore, the surface area of the iced part of the middle tier is $96\pi\cdot10=960\pi.$ The top tier has a diameter of $96\cdot\frac{3}{4}=72$ and a height of 10. Therefore, the surface area of the iced part of the side of the top tier is $72\pi\cdot10=720\pi.$ Also, the radius of the top tier is $\frac{72}{2}=36.$ Thus, the surface area of the iced part of the top of the top tier is $36^2\pi=1296\pi.$ We conclude that the final answer is \[1280\pi+960\pi+720\pi+1296\pi=4256\pi\Longrightarrow\boxed{4256}.\] ~pinkpig

Solution 2

So there is a 3 tier cake i.e. 3 cylinders of the same height but different radii stacked over each other. The LOWERMOST Tier: Given Diameter = 128 => Radius = 64 and Height = 10 So Surface Area Iced = 2πrh = 2*π*64*10 = 1280π

The MIDDLE Tier : Radius = 3/4*64 = 48 and Height = 10 So Surface Area Iced = 2πrh=2*π*48*10 = 960π

The TOPMOST Tier : Radius = 3/4*48 = 36 and Height = 10 So Surface Area Iced = 2πrh=2*π*36*10 = 720π

The Top Layer of TOPMOST Tier : Radius = 36 So Surface Area Iced = πr^2 = π*36*36 = 2592π

Thus Total Surface Area Iced = 1280π+960π+720π+2592π = 5552π

SO m=5552

~AshutoshVerma

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