Difference between revisions of "2021 AIME I Problems/Problem 6"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Solution 1 with Slight Simplification)) |
MRENTHUSIASM (talk | contribs) m (→Solution 3) |
||
Line 35: | Line 35: | ||
==Solution 3== | ==Solution 3== | ||
− | Let E be the vertex of the cube such that ABED is a square. | + | Let <math>E</math> be the vertex of the cube such that <math>ABED</math> is a square. |
By the British Flag Theorem, we can easily we can show that | By the British Flag Theorem, we can easily we can show that | ||
<cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | <cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> |
Revision as of 19:40, 12 December 2021
Contents
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 . 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
.
(Lokman GÖKÇE)
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.