CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

Round 1: Problem 5 Solution

leave a comment »

Cross Section

Cross Section

Pictured at the left is the cross-section of the three spheres. To solve the problem, let us note that the spheres of radii 3 and 5 are similar, and they both have the same center. Therefore, any area traced by the radius-1 ball on the outer sphere will project itself onto the inner ball. We must then discern at what ratio the projection will be in. To determine the ratio, consider what type of inking is occurring on the outer sphere. The black ball colors in an area of the outer sphere’s surface. Therefore the amount covered, in this case an area of 1, should be proportionate with the surface areas of the two spheres. For a sphere of radius r the surface area A = 4 π r2. For our purposes then we need only to compare r2 for each of the spheres, arriving at our answer of 9/25.

Written by Administrator

March 16, 2009 at 00:04

Posted in Round 1

Tagged with ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: