Difference between revisions of "2022 AIME I Problems/Problem 15"
(→Solution 6 (Geometric)) |
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We define some points: | We define some points: | ||
<cmath>\bar {O} = (0,0), \bar {A} = (1.0), | <cmath>\bar {O} = (0,0), \bar {A} = (1.0), | ||
− | \bar{M} = (\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}),</cmath> | + | \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),</cmath> |
− | <cmath>\bar {X} = (\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}), | + | <cmath>\bar {X} = \left(\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}\right), |
− | \bar {Y'} = (\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}),</cmath> | + | \bar {Y'} = \left(\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}\right),</cmath> |
− | <cmath>\bar {Y} = (\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}), | + | <cmath>\bar {Y} = \left(\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}\right), |
− | \bar {Z} = (\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}).</cmath> | + | \bar {Z} = \left(\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}\right).</cmath> |
Notice, that <cmath>\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1</cmath> and each points lies in the first quadrant. | Notice, that <cmath>\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1</cmath> and each points lies in the first quadrant. | ||
− | We use | + | We use given equations and get some scalar products: |
− | <cmath>\vec {XO} \cdot \vec {YO} = \frac {1}{2} = \cos \angle XOY \implies \angle XOY = 60 ^\circ,</cmath> | + | <cmath>(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies \angle XOY = 60 ^\circ,</cmath> |
− | <cmath>\vec {XO} \cdot \vec {ZO} = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies \angle XOZ = 30^\circ,</cmath> | + | <cmath>(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies \angle XOZ = 30^\circ,</cmath> |
− | <cmath>\vec {Y'O} \cdot \vec {ZO} = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies \angle Y'OZ = 45^\circ.</cmath> | + | <cmath>(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies \angle Y'OZ = 45^\circ.</cmath> |
So <math> \angle YOZ = \angle XOY – \angle XOZ = 60 ^\circ – 30 ^\circ = 30 ^\circ, | So <math> \angle YOZ = \angle XOY – \angle XOZ = 60 ^\circ – 30 ^\circ = 30 ^\circ, | ||
\angle Y'OY = \angle Y'OZ + \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.</math> | \angle Y'OY = \angle Y'OZ + \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.</math> | ||
Line 273: | Line 273: | ||
<cmath>1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2 = \cos^2 \angle XOA – \sin^2 \angle XOA = \cos 2 \angle XOA = \cos 135^\circ,</cmath> | <cmath>1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2 = \cos^2 \angle XOA – \sin^2 \angle XOA = \cos 2 \angle XOA = \cos 135^\circ,</cmath> | ||
<cmath>1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ - \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =</cmath> | <cmath>1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ - \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =</cmath> | ||
− | <cmath>\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2 | + | <cmath>\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2 = \frac {1}{32} \implies \boxed{\textbf{033}}.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 11:09, 12 September 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
.
Remark
The motivation for the trig substitution is that if , then
, and when making the substitution in each equation of the initial set of equations, we obtain a new equation in the form of the sine addition formula.
~ Leo.Euler
Solution 6 (Geometric)
We define some points:
Notice, that
and each points lies in the first quadrant.
We use given equations and get some scalar products:
So
Points and
are simmetric with respect to
so
or
vladimir.shelomovskii@gmail.com, vvsss
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.