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Difference between revisions of "1996 AHSME Problems/Problem 27"

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(→‎Solution 2)
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Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a two-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).
 
Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a two-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).
  
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The spheres now become circles with centers at <math>(1,0)</math> and (21/2,0)<math>.  They have radii </math>9/2<math> and </math>4<math>, respectively.
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The spheres now become circles with centers at <math>(1,0)</math> and <math>(21/2,0)</math>.  They have radii <math>9/2</math> and <math>4</math>, respectively.
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Let circle </math>A<math> be the circle centered on </math>(1,0)<math> and circle </math>B<math> be the one centered on </math>(21/2,0)<math>.
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Let circle <math>A</math> be the circle centered on <math>(1,0)</math> and circle <math>B</math> be the one centered on <math>(21/2,0)</math>.
  
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The point on circle </math>A<math> closest to the center of circle </math>B<math> is </math>(11/2,0)<math>.  The point on circle B closest to the center of circle </math>A<math> is </math>(13/2,0)$.
+
The point on circle <math>A</math> closest to the center of circle <math>B</math> is <math>(11/2,0)</math>.  The point on circle B closest to the center of circle <math>A</math> is <math>(13/2,0)</math>.
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=26|num-a=28}}
 
{{AHSME box|year=1996|num-b=26|num-a=28}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:06, 30 December 2018

Problem

Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?

$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution 1

The two equations of the balls are

\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]

\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$, and the second ball goes from $1 \pm 4.5$. The only integer value that $z$ can be is $z=5$.

Plugging that in to both equations, we get:

\[x^2 + y^2 \le \frac{23}{4}\]

\[x^2 + y^2 \le \frac{17}{4}\]

The second inequality implies the first inequality, so the only condition that matters is the second inequality.

From here, we do casework, noting that $|x|, |y| \le 3$:

For $x=\pm 2$, we must have $y=0$. This gives $2$ points.

For $x = \pm 1$, we can have $y\in \{-1, 0, 1\}$. This gives $2\cdot 3 = 6$ points.

For $x = 0$, we can have $y \in \{-2, -1, 0, 1, 2\}$. This gives $5$ points.

Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{D}$.

Solution 2

Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a two-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).

The spheres now become circles with centers at $(1,0)$ and $(21/2,0)$. They have radii $9/2$ and $4$, respectively. Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(21/2,0)$.

The point on circle $A$ closest to the center of circle $B$ is $(11/2,0)$. The point on circle B closest to the center of circle $A$ is $(13/2,0)$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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