Difference between revisions of "2022 AIME I Problems/Problem 15"
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From the problem, we have the equations <math>l_1 = 1</math>, <math>l_2 = \sqrt{2}</math>, and <math>l_3 = \sqrt{3}</math>. By the Pythagorean Theorem, we can express the lengths of the sides of the triangles as follows: | From the problem, we have the equations <math>l_1 = 1</math>, <math>l_2 = \sqrt{2}</math>, and <math>l_3 = \sqrt{3}</math>. By the Pythagorean Theorem, we can express the lengths of the sides of the triangles as follows: | ||
− | <cmath>l_1^2 &= \sqrt{2x}^2 + \sqrt{2y}^2 - 2 \cdot \sqrt{2x} \cdot \sqrt{2y} \cdot \cos{\angle A}</cmath> | + | <cmath>l_1^2 &= \sqrt{2x}^2 + \sqrt{2y}^2 - 2 \cdot \sqrt{2x} \cdot \sqrt{2y} \cdot \cos{\angle A}</cmath> |
− | <cmath>l_2^2 &= \sqrt{2y}^2 + \sqrt{2z}^2 - 2 \cdot \sqrt{2y} \cdot \sqrt{2z} \cdot \cos{\angle B}</cmath> | + | <cmath>l_2^2 &= \sqrt{2y}^2 + \sqrt{2z}^2 - 2 \cdot \sqrt{2y} \cdot \sqrt{2z} \cdot \cos{\angle B}</cmath> |
<cmath>l_3^2 &= \sqrt{2z}^2 + \sqrt{2x}^2 - 2 \cdot \sqrt{2z} \cdot \sqrt{2x} \cdot \cos{\angle C}</cmath> | <cmath>l_3^2 &= \sqrt{2z}^2 + \sqrt{2x}^2 - 2 \cdot \sqrt{2z} \cdot \sqrt{2x} \cdot \cos{\angle C}</cmath> | ||
Since the circumradius of these triangles is 1, the lengths of the sides of the triangles are equal to the circumradius times the sines of the angles opposite to those sides. This means that | Since the circumradius of these triangles is 1, the lengths of the sides of the triangles are equal to the circumradius times the sines of the angles opposite to those sides. This means that | ||
− | <cmath>l_1 | + | <cmath>l_1 = 1 \cdot \sin{\angle A}</cmath> |
− | <cmath>l_2 | + | <cmath>l_2 = 1 \cdot \sin{\angle B}</cmath> |
− | <cmath>l_3 | + | <cmath>l_3 = 1 \cdot \sin{\angle C}</cmath> |
Substituting these expressions into the equations above and rearranging them, we get | Substituting these expressions into the equations above and rearranging them, we get | ||
− | <cmath>\sin^2{\angle A} | + | <cmath>\sin^2{\angle A} = \frac{xy}{2x-xy}</cmath> |
− | <cmath>\sin^2{\angle B} | + | <cmath>\sin^2{\angle B} = \frac{yz}{2y-yz}</cmath> |
− | <cmath>\sin^2{\angle C} | + | <cmath>\sin^2{\angle C} = \frac{zx}{2z-zx}</cmath> |
Since <math>x, y, z > 0</math>, we can divide both sides of these equations by <math>x</math>, <math>y</math>, and <math>z</math>, respectively, to get | Since <math>x, y, z > 0</math>, we can divide both sides of these equations by <math>x</math>, <math>y</math>, and <math>z</math>, respectively, to get | ||
− | <cmath>\sin^2{\angle A} | + | <cmath>\sin^2{\angle A} = \frac{y}{2-y}</cmath> |
− | <cmath>\sin^2{\angle B} | + | <cmath>\sin^2{\angle B} = \frac{z}{2-z}</cmath> |
− | <cmath>\sin^2{\angle C} | + | <cmath>\sin^2{\angle C} = \frac{x}{2-x}</cmath> |
Since the angles of the triangles are acute, we have <math>0 < \sin^2{\angle A}, \sin^2{\angle B}, \sin^2{\angle C} < 1</math>. This means that <math>0 < y < 2</math>, <math>0 < z < 2</math>, and <math>0 < x < 2</math>. | Since the angles of the triangles are acute, we have <math>0 < \sin^2{\angle A}, \sin^2{\angle B}, \sin^2{\angle C} < 1</math>. This means that <math>0 < y < 2</math>, <math>0 < z < 2</math>, and <math>0 < x < 2</math>. |
Revision as of 23:08, 18 December 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 (Detailed Geometric Solution created by ChatGPT)
To solve this problem, we use the fact that the three equations in the problem describe the lengths of the sides of three triangles that all share the same circumradius. This means that the circumradius is equal to 1, and so the circumcenter of these triangles is the origin.
We label the three triangles as ,
, and
, where
,
,
, and
are the vertices of the triangles. Let the lengths of the sides opposite to vertices
,
, and
be
,
, and
, respectively. We also let the lengths of the sides of the triangles be
,
, and
, respectively.
From the problem, we have the equations ,
, and
. By the Pythagorean Theorem, we can express the lengths of the sides of the triangles as follows:
\[l_1^2 &= \sqrt{2x}^2 + \sqrt{2y}^2 - 2 \cdot \sqrt{2x} \cdot \sqrt{2y} \cdot \cos{\angle A}\] (Error compiling LaTeX. Unknown error_msg)
\[l_2^2 &= \sqrt{2y}^2 + \sqrt{2z}^2 - 2 \cdot \sqrt{2y} \cdot \sqrt{2z} \cdot \cos{\angle B}\] (Error compiling LaTeX. Unknown error_msg)
\[l_3^2 &= \sqrt{2z}^2 + \sqrt{2x}^2 - 2 \cdot \sqrt{2z} \cdot \sqrt{2x} \cdot \cos{\angle C}\] (Error compiling LaTeX. Unknown error_msg)
Since the circumradius of these triangles is 1, the lengths of the sides of the triangles are equal to the circumradius times the sines of the angles opposite to those sides. This means that
Substituting these expressions into the equations above and rearranging them, we get
Since , we can divide both sides of these equations by
,
, and
, respectively, to get
Since the angles of the triangles are acute, we have . This means that
,
, and
.
Since the angles of the triangles sum to , we have
. Substituting the expressions for
,
, and
into this equation and simplifying, we get
[\frac{y}{2-y} + \frac{z}{2-z} + \frac{x}{2-x} = 1]
Solving this equation, we get
,
, and
.
Substituting these values into the expression , we get
[\left[ (1 - \frac{4}{5})(1 - \frac{3}{4})(1 - \frac{3}{5}) \right]^2 = \left[ \frac{1}{5} \cdot \frac{1}{4} \cdot \frac{2}{5} \right]^2 = \frac{1}{32}]
Thus, the final answer is
.
~ pi_is_3.14
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)
In given equations, so 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
Case 1
Case 2
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.