Difference between revisions of "2024 AIME I Problems/Problem 14"
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Our answer is <math>20 + 21 + 63 = \boxed{104}.</math> | Our answer is <math>20 + 21 + 63 = \boxed{104}.</math> | ||
− | - spectraldragon8 | + | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] |
==Solution 3(Formula Abuse)== | ==Solution 3(Formula Abuse)== |
Revision as of 22:33, 4 February 2024
Contents
Problem
Let be a tetrahedron such that
,
, and
. There exists a point
inside the tetrahedron such that the distances from
to each of the faces of the tetrahedron are all equal. This distance can be written in the form
, where
,
, and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal \begin{equation*} \mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}. \end{equation*} Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is \begin{equation*} h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}. \end{equation*} Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for cones, \begin{align*} \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ &=\frac13Sr\cdot4. \end{align*} Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).
Solution by Quantum-Phantom
Solution 2
Inscribe tetrahedron in an rectangular prism as shown above.
By the Pythagorean theorem, we note
Solving yields and
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of
We know that the distance from all faces must be the same, so we only need to find the distance from the center to plane
.
Let and
We obtain that the plane of
can be marked as
or
and the center of the prism is
Using the Point-to-Plane distance formula, our distance is
Our answer is
Solution 3(Formula Abuse)
We use the formula for the volume of iscoceles tetrahedron.
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find .
From this, we find and can find the area of
as
Let be the distance we want to find. By taking the sum of (equal) volumes
We have
Plugging in and simplifying, we get
for an answer of
~AtharvNaphade
Solution 4
Let be perpendicular to
that meets this plane at point
.
Let
,
, and
be heights to lines
,
, and
with feet
,
, and
, respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as , is
.
Hence, by using this area, we can compute ,
and
.
We have
,
, and
.
Because , we have
. Recall that
.
Hence,
. Hence,
.
Analogously, and
.
We introduce a function for
that is equal to 1 (resp. -1) if point
and the opposite vertex of side
are on the same side (resp. opposite sides) of side
.
The area of is
\begin{align*}
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
\end{align*}
Denote .
The above equation can be organized as
\begin{align*}
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
This can be further reorganized as \begin{align*} B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2} & = \epsilon_{CD} \sqrt{B^2 - 41 AH^2} + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . \end{align*}
Taking squares on both sides and reorganizing terms, we get \begin{align*} & 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\ & = \epsilon_{CD} \epsilon_{BD} \sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} . \end{align*}
Taking squares on both sides and reorganizing terms, we get
Taking squares on both sides, we finally get \begin{align*} AH & = \frac{20B}{189} \\ & = \frac{40A}{189}. \end{align*}
Now, we plug this solution to Equation (1). We can see that ,
.
This indicates that
is out of
. To be specific,
and
are on opposite sides of
,
and
are on the same side of
, and
and
are on the same side of
.
Now, we compute the volume of the tetrahedron , denoted as
. We have
.
Denote by the inradius of the inscribed sphere in
.
Denote by
the incenter.
Thus, the volume of
can be alternatively calculated as
\begin{align*}
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
& = \frac{1}{3} r \cdot 4A .
\end{align*}
From our two methods to compute the volume of and equating them, we get
\begin{align*}
r & = \frac{10A}{189} \\
& = \frac{20 \sqrt{21}}{63} .
\end{align*}
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5(A quicker method to compute the height from
to plane
)
We put the solid to a 3-d coordinate system. Let ,
.
We put
on the
plane.
Now ,we compute the coordinates of
.
Applying the law of cosines on , we get
.
Thus,
.
Thus,
.
Denote with
.
Because , we have
\[
x^2 + y^2 + z^2 = 89 \hspace{1cm} (1)
\]
Because , we have
\[
\left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2)
\]
Because , we have
\[
\left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2
+ z^2 = 80 \hspace{1cm} (3)
\]
Now, we compute ,
and
.
Taking , we get
\[
2 \sqrt{80} x = 128 .
\]
Thus, .
Taking , we get
\[
2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y
= 50 .
\]
Thus, .
Plugging and
into Equation (1), we get
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn.org
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.