Difference between revisions of "2011 AIME I Problems/Problem 13"
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| + | == Problem == | ||
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers, and <math>r+s+t<{1000}</math>. Find <math>r+s+t</math>. | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers, and <math>r+s+t<{1000}</math>. Find <math>r+s+t</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2011|n=I|num-b=11|num-a=13}} | ||
Revision as of 07:40, 29 March 2011
Problem
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled 
. The three vertices adjacent to vertex are at heights 10, 11, and 12 above the plane. The distance from vertex to the plane can be expressed as 
, where 
, 
, and 
are positive integers, and 
. Find 
.
See also
| 2011 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
