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

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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.  Let's call that altitude <math>h</math>.
 
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.  Let's call that altitude <math>h</math>.
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\begin{asy}
 
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include graph;
 
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size(1inch)
 
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A=(1,0);
 
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B=(10.5,0);
 
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C=(5,2.2);
 
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draw(A--B--C--A);
 
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label("<math>A</math>",A,W);
 
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\end{asy}
 
  
 
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:
 
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:
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Finally, looking at a circle with radius slightly larger than <math>2</math>, we see that there are <math>\boxed{13}</math> lattice points within it <math>\boxed{D}</math>.
 
Finally, looking at a circle with radius slightly larger than <math>2</math>, we see that there are <math>\boxed{13}</math> lattice points within it <math>\boxed{D}</math>.
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==Solution 3==
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Note that the spheres are on the same <math>x</math> and <math>y</math> axis. Therefore, we can draw the spheres so that only the <math>x</math>  and <math>z</math> axis are featured.
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[[File:1996AHSMEP271.jpg|300px|center]]
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<math>A = (1, 0, 5)</math>. <math>\sqrt{1^2+4^2} = \sqrt{17}<\frac92</math>, <math>A</math> is inside the smaller sphere. <math>\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6</math>, <math>A</math> is inside the larger sphere. Point <math>A</math> is inside the intersection.
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<math>B = (2, 0, 5)</math>, <math>\sqrt{2^2+4^2} = \sqrt{20}<\frac92</math>, <math>B</math> is inside the smaller sphere. <math>\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6</math>, <math>B</math> is inside the larger sphere. Point <math>B</math> is inside the intersection.
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<math>C = (3, 0, 5)</math>, <math>\sqrt{3^2+4^2} = \sqrt{25}>\frac92</math>, <math>C</math> is outside the smaller sphere. <math>\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6</math>, <math>C</math> is outside the larger sphere. Point <math>C</math> is outside the intersection.
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The intersection of <math>2</math> spheres is a circle, by drawing the circle flat we can see that there are <math>4</math> more points within the intersection.
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We can see that points with the same <math>x</math>-coordinates with <math>A</math> are still inside the circle. <math>B</math> is very close to the circle, the points with the same <math>x</math>-coordinates with <math>B</math> are not necessarily inside the circle. 
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For example, <math>D = (2,1,5)</math>, <math>\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92</math>, <math>D</math> is outside the smaller sphere. <math>\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6</math>, <math>D</math> is inside the larger sphere. Point <math>D</math> is outside the intersection.
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[[File:1996AHSMEP272.jpg|200px|center]]
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Therefore, the answer is <math>9+4 = \boxed{\textbf{(D) } 13}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==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}}

Latest revision as of 10:42, 30 September 2023

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 (minimal 3D geometry)

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 while 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. Let's call that altitude $h$.

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:

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

\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}\]

Using the area of the triangle, we can find that altitude $h$ from the x-axis:

\[\sqrt{110}=\frac{h*\frac{19}{2}}{2}\]

\[h=\frac{4*\sqrt{110}}{19}\]

Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.

Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.

The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$, a little greater than $2$. We know that $h>2$ because $4\sqrt{110}>4*10$, and $\frac{40}{19}>2$.

Finally, looking at a circle with radius slightly larger than $2$, we see that there are $\boxed{13}$ lattice points within it $\boxed{D}$.

Solution 3

Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.

1996AHSMEP271.jpg

$A = (1, 0, 5)$. $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$, $A$ is inside the smaller sphere. $\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6$, $A$ is inside the larger sphere. Point $A$ is inside the intersection.

$B = (2, 0, 5)$, $\sqrt{2^2+4^2} = \sqrt{20}<\frac92$, $B$ is inside the smaller sphere. $\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6$, $B$ is inside the larger sphere. Point $B$ is inside the intersection.

$C = (3, 0, 5)$, $\sqrt{3^2+4^2} = \sqrt{25}>\frac92$, $C$ is outside the smaller sphere. $\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6$, $C$ is outside the larger sphere. Point $C$ is outside the intersection.

The intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.

We can see that points with the same $x$-coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$-coordinates with $B$ are not necessarily inside the circle.

For example, $D = (2,1,5)$, $\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92$, $D$ is outside the smaller sphere. $\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6$, $D$ is inside the larger sphere. Point $D$ is outside the intersection.

1996AHSMEP272.jpg

Therefore, the answer is $9+4 = \boxed{\textbf{(D) } 13}$.

~isabelchen

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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