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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 2-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 2-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>(\frac{21}{2},0)</math>.  They have radii <math>9/2</math> and <math>6</math>, respectively.
+
The spheres now become circles with centers at <math>(1,0)</math> and <math>(\frac{21}{2},0)</math>.  They have radii <math>\frac{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>(\frac{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>(9/2,0)</math>.
+
The point on circle <math>A</math> closest to the center of circle <math>B</math> is <math>(\frac{11}{2},0)</math>.  The point on circle B closest to the center of circle <math>A</math> is <math>(\frac{11}{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.  Because the tops of the domes are at <math>(0,0,9/2)</math> and <math>(0,0,11/2)</math>, respectively, the lattice points inside the area of intersection must have z-value <math>5</math> (because <math>5</math> is the only integer between <math>9/2</math> and <math>11/2</math>).  Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle.  The radius of the circle will be the distance from the z-axis.
+
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.  Because the tops of the domes are at <math>(0,0,\frac{9}{2})</math> and <math>(0,0,\frac{11}{2})</math>, respectively, the lattice points inside the area of intersection must have z-value <math>5</math> (because <math>5</math> is the only integer between <math>\frac{9}{2}</math> and <math>\frac{11}{2}</math>).  Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle.  The radius of the circle will be the distance from the z-axis.
  
 
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles <math>A</math> and <math>B</math> and third point at the first quadrant intersection of the circles.
 
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles <math>A</math> and <math>B</math> and third point at the first quadrant intersection of the circles.
  
−
We know all three side lengths of this triangle:  <math>9/2</math> (the radius of circle <math>A</math>), <math>6</math> (the radius of circle <math>B</math>), and <math>19/2</math> (the distance between the centers of circles <math>A</math> and <math>B</math>).  Using Heron's formula:
+
We know all three side lengths of this triangle:  <math>\frac{9}{2}</math> (the radius of circle <math>A</math>), <math>6</math> (the radius of circle <math>B</math>), and <math>\frac{19}{2}</math> (the distance between the centers of circles <math>A</math> and <math>B</math>).  We can now find the area of the triangle using Heron's formula:
 +
 
 +
<cmath>s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20</cmath>
 +
 
 +
<cmath>A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}</cmath>
  
 
==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 21:25, 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 2-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 $(\frac{21}{2},0)$. They have radii $\frac{9}{2}$ and $6$, respectively. Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$.

The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$. The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$.

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. Because the tops of the domes are at $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$, respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.

Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles.

We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$), $6$ (the radius of circle $B$), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$). We can now find the area of the triangle using Heron's formula:

\[s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20\]

\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}\]

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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