Difference between revisions of "2022 AIME I Problems/Problem 15"
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By law of sines, we have <math>\frac{2\sin(\theta)}{\sin(\theta)} = 2R</math>, with <math>R=1</math> as the circumradius, same for all 3 triangles. | By law of sines, we have <math>\frac{2\sin(\theta)}{\sin(\theta)} = 2R</math>, with <math>R=1</math> as the circumradius, same for all 3 triangles. | ||
− | The circumcircle's central angle to a side is <math>2 \arcsin(l/2), so the 3 triangles' < | + | The circumcircle's central angle to a side is <math>2 \arcsin(l/2)</math>, so the 3 triangles' <math>l=1, \sqrt{2}, \sqrt{3}</math>, have angles <math>120^{\circ}, 90^{\circ}, 60^{\circ}</math>, respectively. |
− | This means that by half angle arcs, we see that we have in some order, <math>x=2\sin^2(\alpha)</math>, <math> | + | This means that by half angle arcs, we see that we have in some order, <math>x=2\sin^2(\alpha)</math>, <math>y=2\sin^2(\beta)</math>, and <math>z=2\sin^2(\gamma)</math> (not necessarily this order, but here it does not matter due to symmetry), satisfying that <math>\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}</math>, <math>\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}</math>, and <math>\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}</math>. Solving, we get <math>\alpha=\frac{135^{\circ}}{2}</math>, <math>\beta=\frac{105^{\circ}}{2}</math>, and <math>\gamma=\frac{165^{\circ}}{2}</math>. |
We notice that <cmath>[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2</cmath> <cmath>=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare</cmath> | We notice that <cmath>[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2</cmath> <cmath>=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare</cmath> | ||
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\sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. | \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
This should give off tons of trigonometry vibes. To make the connection clear, <math>x = 2\cos^2 \alpha</math>, <math>y = 2\cos^2 \beta</math>, and <math>z = 2\cos^2 \theta</math> is a helpful substitution: | This should give off tons of trigonometry vibes. To make the connection clear, <math>x = 2\cos^2 \alpha</math>, <math>y = 2\cos^2 \beta</math>, and <math>z = 2\cos^2 \theta</math> is a helpful substitution: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ | \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ | ||
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\sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \alpha} + \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \theta} &= \sqrt3. | \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \alpha} + \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \theta} &= \sqrt3. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
From each equation <math>\sqrt{2}^2</math> can be factored out, and when every equation is divided by 2, we get: | From each equation <math>\sqrt{2}^2</math> can be factored out, and when every equation is divided by 2, we get: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ | \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ | ||
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\sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{2}. | \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
which simplifies to (using the Pythagorean identity <math>\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C} </math>): | which simplifies to (using the Pythagorean identity <math>\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C} </math>): | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ | \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ | ||
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\cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. | \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
which further simplifies to (using sine addition formula <math>\sin(a + b) = \sin a \cos b + \cos a \sin b</math>): | which further simplifies to (using sine addition formula <math>\sin(a + b) = \sin a \cos b + \cos a \sin b</math>): | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sin(\alpha + \beta) &= \frac{1}{2} \\ | \sin(\alpha + \beta) &= \frac{1}{2} \\ | ||
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\sin(\alpha + \theta) &= \frac{\sqrt3}{2}. | \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | ||
+ | Taking the inverse sine (<math>0\leq\theta\frac{\pi}{2}</math>) of each equation yields a simple system: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\alpha + \beta &= \frac{\pi}{6} \\ | \alpha + \beta &= \frac{\pi}{6} \\ | ||
\beta + \theta &= \frac{\pi}{4} \\ | \beta + \theta &= \frac{\pi}{4} \\ | ||
− | \alpha + \theta &= \frac{\pi}{3} | + | \alpha + \theta &= \frac{\pi}{3} |
+ | \end{align*}</cmath> | ||
+ | |||
+ | giving solutions: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \alpha &= \frac{\pi}{8} \\ | ||
+ | \beta &= \frac{\pi}{24} \\ | ||
+ | \theta &= \frac{5\pi}{24} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | ||
− | <cmath>\left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2 </cmath> | + | Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: |
− | Now, all the cosines in here are fairly standard: < | + | |
− | <cmath>(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \left(\frac{1}{2}\right)\left(\frac{ | + | |
− | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033} | + | <cmath>\begin{align*} |
+ | x &= 2\cos^2\left(\frac{\pi}{8}\right) \\ | ||
+ | y &= 2\cos^2\left(\frac{\pi}{24}\right) \\ | ||
+ | z &= 2\cos^2\left(\frac{5\pi}{24}\right) | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | When plugging into the expression <math>\left[ (1-x)(1-y)(1-z) \right]^2</math>, noting that <math>-\cos 2\phi = 1 - 2\cos^2 \phi\; \forall \; \phi \in \mathbb{C}</math> helps to simplify this expression into: | ||
+ | |||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 \\ | ||
+ | = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2 | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Now, all the cosines in here are fairly standard: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \cos \frac{\pi}{4} &= \frac{\sqrt{2}}{2} \\ | ||
+ | \cos \frac{\pi}{12} &=\frac{\sqrt{6} + \sqrt{2}}{4} & (= \cos{\frac{\frac{\pi}{6}}{2}} ) \\ | ||
+ | \cos \frac{5\pi}{12} &= \frac{\sqrt{6} - \sqrt{2}}{4} & (= \cos\left({\frac{\pi}{6} + \frac{\pi}{4}} \right) ) | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | With some final calculations: | ||
+ | |||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | &(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 \\ | ||
+ | =& | ||
+ | \left(\frac{1}{2}\right) | ||
+ | \left(\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)\right)^2 \\ | ||
+ | =&\frac{1}{2} \frac{4^2}{16^2} = \frac{1}{32} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}</math>. | ||
~Oxymoronic15 | ~Oxymoronic15 | ||
− | == | + | ==Solution 3 (substitution)== |
Let <math>1-x=a;1-y=b;1-z=c</math>, rewrite those equations | Let <math>1-x=a;1-y=b;1-z=c</math>, rewrite those equations | ||
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Therefore, the answer is <math>1 + 32 = \boxed{\textbf{(033) }}</math>. | Therefore, the answer is <math>1 + 32 = \boxed{\textbf{(033) }}</math>. | ||
− | |||
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
− | + | bu-bye | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sin(\alpha + \beta) &= 1/2 \\ | \sin(\alpha + \beta) &= 1/2 \\ | ||
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\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> | ||
− | Points <math>Y</math> and <math>Y'</math> are | + | Points <math>Y</math> and <math>Y'</math> are symmetric with respect to <math>OM.</math> |
<i><b>Case 1</b></i> | <i><b>Case 1</b></i> | ||
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~Math Gold Medalist | ~Math Gold Medalist | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/aa_VY4e4OOM?si=1lHSwY3v7RICoEpk | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
Latest revision as of 01:25, 1 February 2024
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, let define a triangle with side lengths ,
, and
, with altitude from
's equal to
.
, the left side of one equation in the problem.
Let be angle opposite the side with length
. Then the altitude has length
and thus
, so
and the side length
is equal to
.
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have , with
as the circumradius, same for all 3 triangles.
The circumcircle's central angle to a side is
, so the 3 triangles'
, have angles
, respectively.
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 ):
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:
When plugging into the expression , noting that
helps to simplify this expression into:
Now, all the cosines in here are fairly standard:
With some final calculations:
This is our answer in simplest form , so
.
~Oxymoronic15
Solution 3 (substitution)
Let , rewrite those equations
;
and solve for
Square both sides and simplify, to get three equations:
Square both sides again, and simplify to get three equations:
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since ,
and so 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 .
~Steven Chen (www.professorchenedu.com)
bu-bye
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 symmetric with respect to
Case 1
Case 2
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~Math Gold Medalist
Video Solution
https://youtu.be/aa_VY4e4OOM?si=1lHSwY3v7RICoEpk
~MathProblemSolvingSkills.com
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 |
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