Difference between revisions of "2021 AIME I Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | 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> | + | 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>BP=60\sqrt{10}</math>, <math>CP=60\sqrt{5}</math>, <math>DP=120\sqrt{2}</math>, and <math>GP=36\sqrt{7}</math>. Find <math>AP.</math> |
==Solution 1== | ==Solution 1== |
Revision as of 12:55, 11 October 2022
Contents
Problem
Segments and
are edges of a cube and
is a diagonal through the center of the cube. Point
satisfies
,
,
, and
. Find
Solution 1
First scale down the whole cube by . Let point
have coordinates
, point
have coordinates
, and
be the side length. Then we have the equations
These simplify into
Adding the first three equations together, we get
.
Subtracting this from the fourth equation, we get
, so
. This means
. However, we scaled down everything by
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
Solution 3
Let be the vertex of the cube such that
is a square.
By the British Flag Theorem, we can easily we can show that
and
Hence, adding the two equations together, we get
. Substituting in the values we know, we get
.
Thus, we can solve for , which ends up being
.
Solution 4
For all points in space, define the function
by
. Then
is linear; let
be the center of
. Then since
is linear,
where
denotes the side length of the cube. Thus
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=vaRfI0l4s_8
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.