Difference between revisions of "2021 AIME I Problems/Problem 6"
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Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>? | Segments <math>\overline{AB}, \overline{AC},</math> and <math>\overline{AD}</math> are edges of a cube and <math>\overline{AG}</math> is a diagonal through the center of the cube. Point <math>P</math> satisfies <math>PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},</math> and <math>PG=36\sqrt{7}</math>. What is <math>PA</math>? | ||
− | ==Solution== | + | ==Solution 1== |
First scale down the whole cube by 12. Let point P have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations | First scale down the whole cube by 12. Let point P have coordinates <math>(x, y, z)</math>, A have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations | ||
<cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2</cmath> | <cmath>(s-x)^2+y^2+z^2=(5\sqrt{10})^2</cmath> | ||
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~JHawk0224 | ~JHawk0224 | ||
− | ==(Solution 1 with slight simplification)== | + | ==Solution 2 (Solution 1 with slight simplification)== |
Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, <cmath>2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.</cmath> Subtracting the fourth equation gives, <cmath>2(x^2 + y^2 + z^2) = 575 - 63</cmath> <cmath>x^2 + y^2 + z^2 = 256</cmath> <cmath>\sqrt{x^2 + y^2 + z^2} = 16.</cmath> Since point <math>A = (0,0,0), PA = 16</math>, and since we scaled the answer is <math>16 \cdot 12 = \boxed{192}</math> | Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, <cmath>2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.</cmath> Subtracting the fourth equation gives, <cmath>2(x^2 + y^2 + z^2) = 575 - 63</cmath> <cmath>x^2 + y^2 + z^2 = 256</cmath> <cmath>\sqrt{x^2 + y^2 + z^2} = 16.</cmath> Since point <math>A = (0,0,0), PA = 16</math>, and since we scaled the answer is <math>16 \cdot 12 = \boxed{192}</math> | ||
~Aaryabhatta1 | ~Aaryabhatta1 |
Revision as of 18:13, 12 March 2021
Contents
[hide]Problem
Segments and are edges of a cube and is a diagonal through the center of the cube. Point satisfies and . What is ?
Solution 1
First scale down the whole cube by 12. Let point P have coordinates , A have coordinates , and be the side length. Then we have the equations These simplify into Adding the first three equations together, we get . Subtracting these, we get , so . This means . However, we scaled down everything by 12 so our answer is . ~JHawk0224
Solution 2 (Solution 1 with slight simplification)
Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, Subtracting the fourth equation gives, Since point , and since we scaled the answer is ~Aaryabhatta1
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.