Difference between revisions of "2022 AIME I Problems/Problem 15"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>a=1-x</math>, <math>b=1-y</math>, and <math>c=1-z</math>. Then, | ||
+ | <cmath>\begin{align*} | ||
+ | \sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} &= 1 \hspace{15mm}(1) \\ | ||
+ | \sqrt{(1-b)(1+c)} + \sqrt{(1+b)(1-c)} &= \sqrt2 \hspace{11.5mm}(2) \\ | ||
+ | \sqrt{(1-a)(1+c)} + \sqrt{(1+a)(1-c)} &= \sqrt3. \hspace{10.5mm}(3). | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Notice that <math>\frac{1-a}{2}+\frac{1+a}{2}=1</math>, <math>\frac{1-b}{2}+\frac{1+b}{2}=1</math>, and <math>\frac{1-c}{2}+\frac{1+c}{2}=1</math>. Let <math>\sin^2(\alpha)=(1-a)/2</math>, <math>\sin^2(\beta)=(1-b)/2</math>, and <math>\sin^2(\gamma)=(1-c)/2</math> where <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are real. Substituting into <math>(1)</math>, <math>(2)</math>, and <math>(3)</math> yields | ||
+ | <cmath>\begin{align*} | ||
+ | \sin(\alpha + \beta) &= 1/2 \\ | ||
+ | \sin(\alpha + \gamma) &= \sqrt2/2 \\ | ||
+ | \sin(\beta + \gamma) &= \sqrt3/2. | ||
+ | \end{align*}</cmath> | ||
+ | Thus, | ||
+ | <cmath>\begin{align*} | ||
+ | \alpha + \beta &= 30^{\circ} \\ | ||
+ | \alpha + \gamma &= 45^{\circ} \\ | ||
+ | \beta + \gamma &= 60^{\circ}, | ||
+ | \end{align*}</cmath> | ||
+ | so <math>(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})</math>. Hence, | ||
+ | |||
+ | <cmath> abc = (1-2\sin^2(\alpha))(1-2\sin^2(\beta))(1-2\sin^2(\gamma))=\cos(15^{\circ})\cos(45^{\circ})\cos(75^{\circ})=\frac{\sqrt{2}}{8}, </cmath> | ||
+ | so <math>(abc)^2=(\sqrt{2}/8)^2=\frac{1}{32}</math>, for a final answer of <math>\boxed{033}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:37, 18 February 2022
Contents
Problem
Let
and
be positive real numbers satisfying the system of equations:
Then
can be written as
where
and
are relatively prime positive integers. Find
Solution 1 (geometric interpretation)
First, we note that we can let a triangle exist with side lengths ,
, and opposite altitude
. This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be
for symmetry purposes. So, we note that if the angle opposite the side with length
has a value of
, then the altitude has length
and thus
so
and the triangle side with length
is equal to
.
We can symmetrically apply this to the two other triangles, and since by law of sines, we have is the circumradius of that triangle. Hence. we calculate that with
, and
, the angles from the third side with respect to the circumcenter are
, and
. This means that by half angle arcs, we see that we have in some order,
,
, and
(not necessarily this order, but here it does not matter due to symmetry), satisfying that
,
, and
. Solving, we get
,
, and
.
We notice that
- kevinmathz
Solution 2 (pure algebraic trig, easy to follow)
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor ,
, or
from each term (on the left sides), since each of
,
, and
are positive real numbers. After factoring out accordingly from each terms one of
,
, or
, the system should look like this:
This should give off tons of trigonometry vibes. To make the connection clear,
,
, and
is a helpful substitution:
From each equation
can be factored out, and when every equation is divided by 2, we get:
which simplifies to (using the Pythagorean identity
):
which further simplifies to (using sine addition formula
):
Without loss of generality, taking the inverse sine of each equation yields a simple system:
giving solutions
,
,
. Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
,
, and
. When plugging into the expression
, noting that
helps to simplify this expression into:
Now, all the cosines in here are fairly standard: ,
,
and
. With some final calculations:
This is our answer in simplest form
, so
-Oxymoronic15
solution 3
Let , rewrite those equations
;
square both sides, get three equations:
Getting that
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since , the final answer is
the final answer is
~bluesoul
Solution 4
Denote ,
,
.
Hence, the system of equations given in the problem can be written as
Each equation above takes the following form:
Now, we simplify this equation by removing radicals.
Denote and
.
Hence, the equation above implies
Hence, .
Hence,
.
Because and
, we get
.
Plugging this into the equation
and simplifying it, we get
Therefore, the system of equations above can be simplified as
Denote .
The system of equations above can be equivalently written as
Taking , we get
Thus, we have either or
.
:
.
Equation (2') implies .
Plugging and
into Equation (2), we get contradiction. Therefore, this case is infeasible.
:
.
Plugging this condition into (1') to substitute , we get
Taking , we get
Taking (4) + (5), we get
Hence, .
Therefore,
Therefore, the answer is .
\end{solution}
~Steven Chen (www.professorchenedu.com)
Solution 5
Let ,
, and
. Then,
Notice that ,
, and
. Let
,
, and
where
,
, and
are real. Substituting into
,
, and
yields
Thus,
so
. Hence,
so
, for a final answer of
.
~ Leo.Euler
Video Solution
https://www.youtube.com/watch?v=ihKUZ5itcdA
~Steven Chen (www.professorchenedu.com)
See Also
2022 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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.