Difference between revisions of "2022 AIME II Problems/Problem 3"
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A right square pyramid with volume <math>54</math> has a base with side length <math>6.</math> The five vertices of the pyramid all lie on a sphere with radius <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | A right square pyramid with volume <math>54</math> has a base with side length <math>6.</math> The five vertices of the pyramid all lie on a sphere with radius <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Diagram== | ||
+ | |||
+ | <asy> | ||
+ | import three; | ||
+ | import settings; | ||
+ | leftbutton=new string[] {"rotate","zoom","shift","pan"}; | ||
+ | middlebutton=new string[] {""}; | ||
+ | rightbutton=new string[] {"zoom","rotateX","rotateY","rotateZ"}; | ||
+ | wheelup=new string[] {"zoomin"}; | ||
+ | wheeldown=new string[] {"zoomout"}; | ||
+ | import three; | ||
+ | pen p = rgb("00FF80")+opacity(0.25); | ||
+ | pen q = rgb("FF8000")+opacity(0.75); | ||
+ | triple A = (-3,-3,0); | ||
+ | triple B = (-3,3,0); | ||
+ | triple C = (3,3,0); | ||
+ | triple D = (3,-3,0); | ||
+ | triple F = (0,0,9/2); | ||
+ | surface sphere = shift((0,0,1/4))*scale3(17/4)*unitsphere; | ||
+ | draw(sphere,p); | ||
+ | draw(A--B--C--D--A--F--B--F--C--F--D--F--A,q); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(F); | ||
+ | </asy> | ||
==Solution 1== | ==Solution 1== | ||
Line 10: | Line 38: | ||
Since the volume is <math>54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h</math>, where <math>h=\frac{9}{2}</math> is the height of this pyramid, we have: <math>l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2</math> according to the Pythagorean theorem. | Since the volume is <math>54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h</math>, where <math>h=\frac{9}{2}</math> is the height of this pyramid, we have: <math>l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2</math> according to the Pythagorean theorem. | ||
− | Solve this equation will give us <math>l = \frac{17}{4},</math> therefore <math>m+n=\boxed{ | + | Solve this equation will give us <math>l = \frac{17}{4},</math> therefore <math>m+n=\boxed{021}.</math> |
− | |||
− | |||
==Solution 2== | ==Solution 2== | ||
− | To start, we find the height of the pyramid. By the volume of a pyramid formula, we have <cmath>\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.</cmath> Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is <math>\frac12 \cdot 6\sqrt2=3\sqrt2</math>, we have <cmath>x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.</cmath> Taking the cross section of the pyramid and transforming the problem into <math>2</math>-d, it suffices to find the radius of the circumcircle of a triangle of side lengths <math>\frac{3\sqrt{17}}2</math>, <math>\frac{3\sqrt{17}}2</math>, <math>6\sqrt2</math>. This turns out to be easy by the formula <math>R=\frac{abc}{4A}</math>, and through computing this value (the work has been left out) we find that <math>R=\frac{17}4</math>, so our answer is <math>\boxed{\textbf{ | + | To start, we find the height of the pyramid. By the volume of a pyramid formula, we have <cmath>\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.</cmath> Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is <math>\frac12 \cdot 6\sqrt2=3\sqrt2</math>, we have <cmath>x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.</cmath> Taking the cross section of the pyramid and transforming the problem into <math>2</math>-d, it suffices to find the radius of the circumcircle of a triangle of side lengths <math>\frac{3\sqrt{17}}2</math>, <math>\frac{3\sqrt{17}}2</math>, <math>6\sqrt2</math>. This turns out to be easy by the formula <math>R=\frac{abc}{4A}</math>, and through computing this value (the work has been left out) we find that <math>R=\frac{17}4</math>, so our answer is <math>\boxed{\textbf{021}}</math>. |
~A1001 | ~A1001 | ||
+ | |||
+ | ==Solution 3 (Coord bash)== | ||
+ | By the volume of a pyramid formula, we have that the height of the pyramid is <math>\frac{9}{2}</math>. Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the <math>xy</math> plane. Then, the vertices of the base would be <math>(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)</math> in some order. Also, let the vertex be <math>(0,0,\frac{9}{2})</math>. Recall that the formula for a sphere is <math>(x-a)^2+(y-b)^2+(z-c)^2=r^2</math> where the center is <math>(a,b,c)</math> and the radius is <math>r</math>. Symmetry gives that <math>a=b=0</math>. Plug in <math>(3,3,0)</math> and <math>(0,0,\frac{9}{2})</math> and you get the system of equation | ||
+ | |||
+ | <math>18+c^2=r^2</math> | ||
+ | |||
+ | <math>(\frac{9}{2}-c)^2=r^2</math> | ||
+ | |||
+ | Solving gives <math>c=1/4</math> and <math>r=17/4</math>, so our answer is <math>17+4=\boxed{021}</math>.~[[Ddk001]] | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=UJAYW6YNFVU | https://www.youtube.com/watch?v=UJAYW6YNFVU | ||
+ | |||
+ | ==Video Solution by Power of Logic== | ||
+ | https://youtu.be/YieIiYCDytM | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=2|num-a=4}} | {{AIME box|year=2022|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:33, 29 January 2024
Contents
Problem
A right square pyramid with volume has a base with side length
The five vertices of the pyramid all lie on a sphere with radius
, where
and
are relatively prime positive integers. Find
.
Diagram
Solution 1
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same:
. Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is , where
is the height of this pyramid, we have:
according to the Pythagorean theorem.
Solve this equation will give us therefore
Solution 2
To start, we find the height of the pyramid. By the volume of a pyramid formula, we have Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is
, we have
Taking the cross section of the pyramid and transforming the problem into
-d, it suffices to find the radius of the circumcircle of a triangle of side lengths
,
,
. This turns out to be easy by the formula
, and through computing this value (the work has been left out) we find that
, so our answer is
.
~A1001
Solution 3 (Coord bash)
By the volume of a pyramid formula, we have that the height of the pyramid is . Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the
plane. Then, the vertices of the base would be
in some order. Also, let the vertex be
. Recall that the formula for a sphere is
where the center is
and the radius is
. Symmetry gives that
. Plug in
and
and you get the system of equation
Solving gives and
, so our answer is
.~Ddk001
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=UJAYW6YNFVU
Video Solution by Power of Logic
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.