Difference between revisions of "1996 AHSME Problems/Problem 27"
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==Solution 2== | ==Solution 2== | ||
− | Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a | + | 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>(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>(21/2,0)</math>. They have radii <math>9/2</math> and <math>6</math>, respectively. | ||
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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>(9/2,0)</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. | + | 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. |
+ | |||
+ | 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: | ||
==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 22:20, 30 December 2018
Contents
Problem
Consider two solid spherical balls, one centered at with radius
, and the other centered at
with radius
. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
Solution 1
The two equations of the balls are
Note that along the axis, the first ball goes from
, and the second ball goes from
. The only integer value that
can be is
.
Plugging that in to both equations, we get:
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that :
For , we must have
. This gives
points.
For , we can have
. This gives
points.
For , we can have
. This gives
points.
Thus, there are possible points, giving answer
.
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 and
. They have radii
and
, respectively.
Let circle
be the circle centered on
and circle
be the one centered on
.
The point on circle closest to the center of circle
is
. The point on circle B closest to the center of circle
is
.
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 and
, respectively, the lattice points inside the area of intersection must have z-value
(because
is the only integer between
and
). 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 and
and third point at the first quadrant intersection of the circles.
We know all three side lengths of this triangle: (the radius of circle
),
(the radius of circle
), and
(the distance between the centers of circles
and
). Using Heron's formula:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.