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Difference between revisions of "2013 AMC 10A Problems/Problem 22"

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==Solution 1==
 
==Solution 1==
  
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Set up an isosceles triangle between the center of the 8th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the 7th and 8th spheres, and the points of tangency between the 7th sphere and 2 of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of r (the radius of sphere 8) and h (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of h and r.
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Set up an isosceles triangle between the center of the <math>8</math>th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the <math>7</math>th and <math>8</math>th spheres, and the points of tangency between the <math>7</math>th sphere and <math>2</math> of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of <math>r</math> (the radius of sphere <math>8</math>) and <math>h</math> (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of <math>h</math> and <math>r</math>.
  
 
<math>(1+r)^2=2^2+h^2</math>
 
<math>(1+r)^2=2^2+h^2</math>
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==Solution 2==
 
==Solution 2==
  
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We have a regular hexagon with side lengths 2 and six spheres on each vertex with radius 1 that are internally tangent, therefore drawing radii going through all of them would create this regular hexagon.
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We have a regular hexagon with side length <math>2</math> and six spheres on each vertex with radius <math>1</math> that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon.
  
−
There is a larger sphere which 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, that the radius is <math>2+1=3</math>
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Imagine a 2D overhead view.  There is a larger sphere which the <math>6</math> spheres are internally tangent to, with the 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 <math>6</math> equilateral triangles from its vertices, that the radius is two more than the small radius, or <math>3</math>. 
  
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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 aligned with the diameter of the large sphere.
 
  
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We can therefore now create a triangle with the horizontal component 2, as it is from the vertex of the hexagon to the center of the hexagon.  
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<asy>
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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.
+
draw(circle((0.5,0.866025404),0.5));
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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 -.
+
draw(circle((-0.5,0.866025404),0.5));
 +
draw(circle((1,0),0.5));
 +
draw(circle((-1,0),0.5));
 +
draw(circle((0.5,-0.866025404),0.5));
 +
draw(circle((-0.5,-0.866025404),0.5));
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draw(circle((0,0),1.5));
  
−
We therefore now have a right angled triangle which when applied Pythagoras states <math>2^2+(3-r)^2=(1+r)^2</math>
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draw((-0.5,0.866025404)--(0.5,0.866025404));
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Expanding brackets gives us <math>4+9-6r+r^2=1+2r+r^2</math> here we can cancel out <math>r^2</math>
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draw((-1,0)--(1,0));
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Isolating the r's <math>12=8r</math>
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draw((-0.5,-0.866025404)--(0.5,-0.866025404));
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and then finally we have the answer: <math>r=\frac{12}{8}=\frac{3}{2}</math>
+
 
 +
draw((-1,0)--(-0.5,0.866025404));
 +
draw((-0.5,-0.866025404)--(0.5,0.866025404));
 +
draw((0.5,-0.866025404)--(1,0));
 +
 
 +
draw((0.5,0.866025404)--(1,0));
 +
draw((-0.5,0.866025404)--(0.5,-0.866025404));
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draw((-1,0)--(-0.5,-0.866025404));
 +
</asy>
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 +
diagram made by erics118
 +
 
 +
 
 +
Now imagine the figure in <math>3</math> dimensions.  The 8th sphere must be sitting atop of the <math>6</math> spheres, which is the only possibility for it to be tangent to all the <math>6</math> small spheres externally and the larger sphere internally. The ring of <math>6</math> small spheres is symmetrical and the 8th sphere will be resting atop it with its center aligned with the diameter of the large sphere.
 +
 
 +
We can now create a right triangle with the legs being the line from a vertex of the hexagon to the center of the hexagon and the line from the center of the 7th sphere to the center of the 8th sphere.  Let the radius of our 8th sphere be <math>r</math>.  As previously mentioned, the distance from the center of the hexagon to one of its vertices is <math>2</math>.  The distance between the centers will be <math>3-r</math>.  The hypotenuse will be <math>r+1</math>. 
 +
 
 +
We now have a right triangle.  Applying the Pythagorean Theorem, <math>2^2+(3-r)^2=(1+r)^2</math>.  Expanding and solving for <math>r</math>, we find <math>r=\frac{12}{8}=\boxed{\textbf{(B)}\frac{3}{2}}</math>.
 +
 
 +
==Video Solution by Richard Rusczyk==
 +
https://www.youtube.com/watch?v=83JgSTi_0VE
 +
 
 +
hint: turn a 3D geometry problem into a 2D problem by taking cross-sections! Much easier to visualize.
 +
 
 +
What type of cross-section...? Well, that's in the video :D
 +
 
 +
~ dolphin7
  
 
==See Also==
 
==See Also==

Latest revision as of 21:58, 21 August 2023

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?


$\textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2$

Solution 1

Set up an isosceles triangle between the center of the $8$th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the $7$th and $8$th spheres, and the points of tangency between the $7$th sphere and $2$ of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of $r$ (the radius of sphere $8$) and $h$ (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of $h$ and $r$.

$(1+r)^2=2^2+h^2$

$(3\sqrt{2})^2=3^2+(h+r)^2$

$r = \boxed{\textbf{(B) }\frac{3}{2}}$

Solution 2

We have a regular hexagon with side length $2$ and six spheres on each vertex with radius $1$ that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon.

Imagine a 2D overhead view. There is a larger sphere which the $6$ spheres are internally tangent to, with the 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 its vertices, that the radius is two more than the small radius, or $3$.


[asy] draw(circle((0.5,0.866025404),0.5)); draw(circle((-0.5,0.866025404),0.5)); draw(circle((1,0),0.5)); draw(circle((-1,0),0.5)); draw(circle((0.5,-0.866025404),0.5)); draw(circle((-0.5,-0.866025404),0.5)); draw(circle((0,0),1.5)); draw((-0.5,0.866025404)--(0.5,0.866025404)); draw((-1,0)--(1,0)); draw((-0.5,-0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,0.866025404)); draw((-0.5,-0.866025404)--(0.5,0.866025404)); draw((0.5,-0.866025404)--(1,0)); draw((0.5,0.866025404)--(1,0)); draw((-0.5,0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,-0.866025404)); [/asy]

diagram made by erics118


Now imagine the figure in $3$ dimensions. The 8th sphere must be sitting atop of the $6$ spheres, which is the only possibility for it to be tangent to all the $6$ small spheres externally and the larger sphere internally. The ring of $6$ small spheres is symmetrical and the 8th sphere will be resting atop it with its center aligned with the diameter of the large sphere.

We can now create a right triangle with the legs being the line from a vertex of the hexagon to the center of the hexagon and the line from the center of the 7th sphere to the center of the 8th sphere. Let the radius of our 8th sphere be $r$. As previously mentioned, the distance from the center of the hexagon to one of its vertices is $2$. The distance between the centers will be $3-r$. The hypotenuse will be $r+1$.

We now have a right triangle. Applying the Pythagorean Theorem, $2^2+(3-r)^2=(1+r)^2$. Expanding and solving for $r$, we find $r=\frac{12}{8}=\boxed{\textbf{(B)}\frac{3}{2}}$.

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=83JgSTi_0VE

hint: turn a 3D geometry problem into a 2D problem by taking cross-sections! Much easier to visualize.

What type of cross-section...? Well, that's in the video :D

~ dolphin7

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 10 Problems and Solutions
2013 AMC 12A (Problems • Answer Key • Resources)
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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