Difference between revisions of "2024 AIME I Problems/Problem 15"
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{{AIME box|year=2024|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2024|n=I|num-b=14|after=Last Problem}} | ||
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Revision as of 19:03, 4 February 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (constrained optimization with Lagrangian multiplier)
- 4 Solution 3 (Vieta's Formula and Rational Root Theroem)
- 5 Solution 3a (Derivative)
- 6 Video Solution 1 by OmegaLearn.org (super short)
- 7 Video Solution 2 (constrained optimization with Lagrangian multiplier)
- 8 See also
Problem
Let be the set of rectangular boxes with surface area
and volume
. Let
be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of
. The value of
can be written as
, where
and
are relatively prime positive integers. Find
.
Solution 1
Observe that the "worst" possible box is one of the maximum possible length.
By symmetry, the height and the width are the same in this antioptimal box. (If the height and width weren't the same, the extra difference between them could be used to make the length longer.) Thus, let the width and height be of length and the length be
.
We're given that the volume is ; thus,
. We're also given that the surface area is
; thus,
.
From the first equation, we can get . We do a bunch of algebra:
\begin{align*} L&=\dfrac{23}{a^2} \\ 27&=a^2+2aL \\ &=a^2+2a\left(\dfrac{23}{a^2}\right) \\ &=a^2+\dfrac{46}a \\ 27a&=a^3+46 \\ a^3-27a+46&=0. \\ \end{align*}
We can use the Rational Root Theorem and test a few values. It turns out that works. We use synthetic division to divide by
:
As we expect, the remainder is , and we are left with the polynomial
. We can now simply use the quadratic formula and find that the remaining roots are
. We want the smallest
to maximize
, and it turns out that
is in fact the smallest root. Thus, we let
. Substituting this into
, we find that
. However, this is not our answer! This is simply the length of the box; we want the radius of the sphere enclosing it. We know that the diameter of the sphere is the diagonal of the box, and the 3D Pythagorean Theorem can give us the space diagonal. Applying it, we find that the diagonal has length
. This is the diameter; we halve it to find the radius,
. We then square this and end up with
, giving us an answer of
.
~Technodoggo
Solution 2 (constrained optimization with Lagrangian multiplier)
Denote by ,
,
the length, width, and height of a rectangular box.
We have
\begin{align*}
xy + yz + zx & = \frac{54}{2} \hspace{1cm} (1) \\
xyz & = 23 \hspace{1cm} (2)
\end{align*}
We have \begin{align*} 4 r^2 & = x^2 + y^2 + z^2 \\ & = \left( x + y + z \right)^2 - 2 \cdot \left( xy + yz + zx \right) \\ & = \left( x + y + z \right)^2 - 54 . \end{align*}
Therefore, we solve the following constrained optimization problem: \begin{align*} \max_{x,y,z} \ & x + y + z \\ \mbox{subject to } & (1), (2) \end{align*}
First, we prove that an optimal solution must have at least two out of ,
,
that are the same.
Denote by and
lagrangian multipliers of constraints (1) and (2), respectively.
Consider the following Lagrangian:
\begin{align*}
\max_{x,y,z, \lambda, \eta} & x + y + z + \lambda \left( xy + yz + zx - 27 \right)
+ \eta \left( xyz - 23 \right) .
\end{align*}
Taking first-order-condition with respect to ,
,
, respectively, we get
\begin{align*}
1 + \lambda \left( y + z \right) + \eta yz & = 0 \hspace{1cm} (3) \\
1 + \lambda \left( z + x \right) + \eta zx & = 0 \hspace{1cm} (4) \\
1 + \lambda \left( x + y \right) + \eta xy & = 0 \hspace{1cm} (5)
\end{align*}
Suppose there is an optimal solution with ,
,
that are all distinct.
Taking , we get
Because , we have
Analogously, we have \begin{align*} \lambda + \eta x & = 0 \hspace{1cm} (7) \end{align*}
Taking , we get
.
Because
, we have
. Plugging this into (6), we get
.
However, the solution that is a contradiction with (3).
Therefore, in an optimal solution, we cannot have
,
, and
to be all distinct.
W.L.O.G, in our remaining analysis, we assume an optimal solution satisfies .
Therefore, we need to solve the following two-variable optimization problem: \begin{align*} \max_{x,y} \ & x + 2y \\ \mbox{subject to } & 2 xy + y^2 = 27 \\ & xy^2 = 23 \end{align*}
Replacing with
by using the constraint
, we solve the following single-variable optimization problem:
\begin{align*}
\max_y \ & \frac{23}{y^2} + 2y \hspace{1cm} (8) \\
\mbox{subject to } & \frac{46}{y} + y^2 = 27 \hspace{1cm} (9)
\end{align*}
By solving (9), we get and
.
Plugging into (8), we get
.
Plugging into (8), we get
.
We have .
Therefore, the maximum value of
is
.
Therefore, \begin{align*} r^2 & = \frac{1}{4} \left( \left( x + y + z \right)^2 - 54 \right) \\ & = \frac{1}{4} \left( \left( \frac{39}{4} \right)^2 - 54 \right) \\ & = \frac{657}{64} . \end{align*}
Therefore, the answer is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Vieta's Formula and Rational Root Theroem)
First, let's list the conditions:
Denote by ,
,
the length, width, and height of a rectangular box.
\begin{align*}
2(lw+wh+hl)&=54\\
lw+wh+hl&=27.
\end{align*}
Applying the Pythagorean theorem, we can establish that
\begin{align*} (2r)^2&=(l^2+w^2+h^2)\\ 4r^2&=(l^2+w^2+h^2)\\ 4r^2&=(l+w+h)^2-2(lw+wh+hl)\\ 4r^2&=(l+w+h)^2-54. \end{align*}
We can spot Vieta's formula hidden inside this equation and call this . Now we have three equations:
Let there be a cubic equation. . Its roots are
,
and
. We can use our formulas from before to derive
and
.
We can now rewrite the equation from before:
To find the maximum we need the maximum
. This only occurs when this equation has double roots illustrated with graph below.
WLOG we can set .
Thus:
We can substitute and form a depressed cubic equation with
.
\begin{align*}
lw^2&=23\\
l&=\frac{23}{w^2}\\
2\left(\frac{23}{w^2}\right)w+w^2&=27\\
\frac{46}{w}+w^2&=27\\
w^2+\frac{46}{w}-27&=0\\
w^3 -27w+46&=0.
\end{align*}
Based on Rational Root Theorem the possible rational roots are
A quick test reveals that is a root of the equation. Comparing coefficients we can factorize the equation into:
Besides , we derive another positive root using the quadratic formula,
But to maximize the
we need to pick the smaller
, which is
.
Substituting this into , we find that
.
Applying it to our equation above:
\begin{align*}
4r^2&=(l+w+h)^2-54\\
4r^2&=(l+2w)^2-54\\
4r^2&=\left(\dfrac{23}4+2\cdot2\right)^2-54\\
4r^2&=\left(\dfrac{39}4\right)^2-54\\
4r^2&=\left(\dfrac{1521}{16}\right)-54\\
4r^2&=\left(\dfrac{657}{16}\right)\\
r^2&=\left(\dfrac{657}{64}\right).
\end{align*}
.
Solution 3a (Derivative)
to find the maximum m for
rewrite m as function of x and calculate derivatives to get maximum value,
when x = 2,
the rest is similar to solution 3
Video Solution 1 by OmegaLearn.org (super short)
Video Solution 2 (constrained optimization with Lagrangian multiplier)
https://www.youtube.com/watch?v=KjEy2Ju2z8A
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.