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>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== | + | ==Solution 1== |
+ | First scale down the whole cube by <math>12</math>. Let point <math>P</math> have coordinates <math>(x, y, z)</math>, point <math>A</math> have coordinates <math>(0, 0, 0)</math>, and <math>s</math> be the side length. Then we have the equations | ||
+ | <cmath>\begin{align*} | ||
+ | (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ | ||
+ | x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ | ||
+ | x^2+y^2+(s-z)^2&=\left(10\sqrt{2}\right)^2, \\ | ||
+ | (s-x)^2+(s-y)^2+(s-z)^2&=\left(3\sqrt{7}\right)^2. | ||
+ | \end{align*}</cmath> | ||
+ | These simplify into | ||
+ | <cmath>\begin{align*} | ||
+ | s^2+x^2+y^2+z^2-2sx&=250, \\ | ||
+ | s^2+x^2+y^2+z^2-2sy&=125, \\ | ||
+ | s^2+x^2+y^2+z^2-2sz&=200, \\ | ||
+ | 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. | ||
+ | \end{align*}</cmath> | ||
+ | Adding the first three equations together, we get <math>3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575</math>. | ||
+ | Subtracting this from the fourth equation, we get <math>2(x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>PA=16</math>. However, we scaled down everything by <math>12</math> so our answer is <math>16*12=\boxed{192}</math>. | ||
− | ==See | + | ~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, <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>\begin{align*} | ||
+ | 2(x^2 + y^2 + z^2) &= 575 - 63 \\ | ||
+ | x^2 + y^2 + z^2 &= 256 \\ | ||
+ | \sqrt{x^2 + y^2 + z^2} &= 16. | ||
+ | \end{align*}</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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let <math>E</math> be the vertex of the cube such that <math>ABED</math> is a square. | ||
+ | Using the [[British Flag Theorem]], we can easily show that | ||
+ | <cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath> | ||
+ | and | ||
+ | <cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath> | ||
+ | Hence, by adding the two equations together, we get <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. Substituting in the values we know, we get <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>. | ||
+ | |||
+ | Thus, we can solve for <math>PA</math>, which ends up being <math>\boxed{192}</math>. | ||
+ | |||
+ | (Lokman GÖKÇE) | ||
+ | [[File:2021 AIME I 6b.jpg|size:100px]] | ||
+ | |||
+ | ==Solution 4== | ||
+ | For all points <math>X</math> in space, define the function <math>f:\mathbb{R}^{3}\rightarrow\mathbb{R}</math> by <math>f(X)=PX^{2}-GX^{2}</math>. Then <math>f</math> is linear; let <math>O=\frac{2A+G}{3}</math> be the center of <math>\triangle BCD</math>. Then since <math>f</math> is linear, | ||
+ | <cmath>\begin{align*} | ||
+ | 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\ | ||
+ | \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\ | ||
+ | \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, | ||
+ | \end{align*}</cmath> | ||
+ | where <math>x</math> denotes the side length of the cube. Thus | ||
+ | <cmath>\begin{align*} | ||
+ | 36\text{,}000+18\text{,}000+28\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ | ||
+ | 82\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ | ||
+ | 73\text{,}728&=2PA^{2} \\ | ||
+ | 36\text{,}864&=PA^{2} \\ | ||
+ | PA&=\boxed{192}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=I|num-b=5|num-a=7}} | {{AIME box|year=2021|n=I|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:50, 12 January 2024
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.
Using the British Flag Theorem, we can easily show that
and
Hence, by 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
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.