Difference between revisions of "2013 AMC 12A Problems/Problem 18"

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Revision as of 16:00, 3 July 2013

Problem

Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

$\textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$

Solution

You have a regular hexagon with side lengths 2 and six spheres on each vertice radius 1 that are internally tangent, therefore drawing radius's through them all would create this regular hexagon.

There is a larger sphere the 6 spheres are internally tangent to, with center in the center of the hexagon. To find the radius of the larger sphere we must first, either by prior knowledge or by deducing from the angle sum that the hexagon can be split into 6 equilateral triangles from it's vertices and then the radius is $2+1=3$

The 8th sphere is now, when thinking about it in 3D sitting on top of the 6 spheres which is the only possibility for it to tangent all the 6 small spheres externally and the larger sphere internally. The ring of the 6 small spheres is symmetrical and the 8th sphere will be resting with it's center alligned with the diameter of the large sphere.

We can therefore now create a triangle with the horizontal component 2, as it is from the vertice of the hexagon to the center of the hexagon. The vertical component is from the center of the large sphere to the center of the 8th sphere. This length equals 3, the radius of the large sphere, take away the radius of the 8th sphere, we can call it r, since the radius of the large sphere will include the diameter of the 8th sphere if we subtract radius we will reach the center. The last component is the hypotenuse of the right angled triangle. This consists of the radius of the small sphere - 1 - and the radius of the 8th sphere - r -.

We therefore now have a right angled triangle which when applied pythagoras states $2^2+(3-r)^2=(1+r)^2$ Expanding brackets gives us $4+9-6r+r^2=1+2r+r^2$ here we can cancel out $r^2$ Isolating the r's $12=8r$ and then finally we have the answer: $r=\frac{12}{8}=\frac{3}{2}$

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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