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

(Solution 2)
(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).
  
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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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>6</math>, respectively.
 
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>.
 
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>.
  
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>.
+
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>(9/2,0)</math>.
 +
 
 +
Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent dome shapes on either end.  All
  
 
==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 22:11, 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 $6$, 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 $(9/2,0)$.

Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent dome shapes on either end. All

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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All AHSME Problems and Solutions

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