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

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.

$A = (1, 0, 5)$ . $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$ , $A$ 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.

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

~isabelchen

@@ Line 33: / Line 33: @@
 Thus, there are <math>\boxed{13}</math> possible points, giving answer <math>\boxed{D}</math>.
-==Solution 2==
+==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 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 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 (21/2,0).  They have radii 9/2 and 4, 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 A be the circle centered on (1,0) and circle B be the one centered on (21/2,0).
+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 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).
+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,\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.  Let's call that altitude <math>h</math>.
+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})}=\sqrt{110}</cmath>
+Using the area of the triangle, we can find that altitude <math>h</math> from the x-axis:
+<cmath>\sqrt{110}=\frac{h*\frac{19}{2}}{2}</cmath>
+<cmath>h=\frac{4*\sqrt{110}}{19}</cmath>
+Remember, the altitude <math>h</math> 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 <math>h</math> of the circle in a 2-dimensional plane is <math>\frac{4\sqrt{110}}{19}</math>, a little greater than <math>2</math>.  We know that <math>h>2</math> because <math>4\sqrt{110}>4*10</math>, and <math>\frac{40}{19}>2</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>.
+==Solution 3==
+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.
+[[File:1996AHSMEP271.jpg|300px|center]]
+<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.
+<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.
+<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.
+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.
+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.
+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.
+[[File:1996AHSMEP272.jpg|200px|center]]
+Therefore, the answer is <math>9+4 = \boxed{\textbf{(D) } 13}</math>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 ==See also==
 {{AHSME box|year=1996|num-b=26|num-a=28}}
 {{MAA Notice}}

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

Page

Toolbox

Search