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Difference between revisions of "2022 AIME II Problems/Problem 3"
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==Solution==
 
==Solution==
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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 <math>\frac{6}{\sqrt{2}} = 3\sqrt{2}</math>). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: <math>l</math>. 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 base (square) at the (base's) center.
 +
 +
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.
 +
Then, according to pythagorean theorem, we have: <math>l^2=(\frac{9}{2}-l)^2+(3\sqrt{2})^2</math>.
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 +
Solve this equation will give us <math>l = \frac{17}{4}</math>. Therefore, <math>m+n=\boxed{021}</math>
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~DSAERF-CALMIT
  
 
==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}}

Revision as of 22:58, 17 February 2022

Problem

A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

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 $\frac{6}{\sqrt{2}} = 3\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. 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 base (square) at the (base's) center.

Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$, where $h=\frac{9}{2}$ is the height of this pyramid. Then, according to pythagorean theorem, we have: $l^2=(\frac{9}{2}-l)^2+(3\sqrt{2})^2$.

Solve this equation will give us $l = \frac{17}{4}$. Therefore, $m+n=\boxed{021}$

~DSAERF-CALMIT

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

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