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Difference between revisions of "2022 AIME I Problems/Problem 15"

(→‎See also)
Line 94: Line 94:
  
 
~bluesoul
 
~bluesoul
 +
 +
==Solution 4==
 +
 +
Denote <math>u = 1 - x</math>, <math>v = 1 - y</math>, <math>w = 1 - z</math>.
 +
Hence, the system of equations given in the problem can be written as
 +
<cmath>
 +
\begin{align*}
 +
\sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\
 +
\sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\
 +
\sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3)
 +
\end{align*}
 +
</cmath>
 +
 +
Each equation above takes the following form:
 +
<cmath>
 +
\[
 +
\sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} = k .
 +
\]
 +
</cmath>
 +
 +
Now, we simplify this equation by removing radicals.
 +
 +
Denote <math>p = \sqrt{(1-a)(1+b)}</math> and <math>q = \sqrt{(1+a)(1-b)}</math>.
 +
 +
Hence, the equation above implies
 +
<cmath>
 +
\[
 +
\left\{
 +
\begin{array}{l}
 +
p + q = k \\
 +
p^2 = (1-a)(1+b) \\
 +
q^2 = (1+a)(1-b)
 +
\end{array}
 +
\right..
 +
\]
 +
</cmath>
 +
 +
Hence, <math>q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)</math>.
 +
Hence, <math>q - p = \frac{q^2 - p^2}{p+q} = \frac{2}{k} (a-b)</math>.
 +
 +
Because <math>p + q = k</math> and <math>q - p = \frac{2}{k} (a-b)</math>, we get <math>q = \frac{a-b}{k} + \frac{k}{2}</math>.
 +
Plugging this into the equation <math>q^2 = (1+a)(1-b)</math> and simplifying it, we get
 +
<cmath>
 +
\[
 +
a^2 + \left( k^2 - 2 \right) ab + b^2 = k^2 - \frac{k^4}{4} .
 +
\]
 +
</cmath>
 +
 +
Therefore, the system of equations above can be simplified as
 +
<cmath>
 +
\begin{align*}
 +
u^2 - uv + v^2 & = \frac{3}{4} \\
 +
v^2 + w^2 & = 1 \\
 +
w^2 + wu + u^2 & = \frac{3}{4}  .
 +
\end{align*}
 +
</cmath>
 +
 +
Denote <math>w' = - w</math>.
 +
The system of equations above can be equivalently written as
 +
<cmath>
 +
\begin{align*}
 +
u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\
 +
v^2 + w'^2 & = 1 \hspace{1cm} (2') \\
 +
w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') .
 +
\end{align*}
 +
</cmath>
 +
 +
Taking <math>(1') - (3')</math>, we get
 +
<cmath>
 +
\[
 +
(v - w') (v + w' - u) = 0 .
 +
\]
 +
</cmath>
 +
 +
Thus, we have either <math>v - w' = 0</math> or <math>v + w' - u = 0</math>.
 +
 +
<math>\textbf{Case 1}</math>: <math>v - w' = 0</math>.
 +
 +
Equation (2') implies <math>v = w' = \pm \frac{1}{\sqrt{2}}</math>.
 +
 +
Plugging <math>v</math> and <math>w'</math> into Equation (2), we get contradiction. Therefore, this case is infeasible.
 +
 +
<math>\textbf{Case 2}</math>: <math>v + w' - u = 0</math>.
 +
 +
Plugging this condition into (1') to substitute <math>u</math>, we get
 +
<cmath>
 +
\[
 +
v^2 + v w' + w'^2 = \frac{3}{4} \hspace{1cm} (4) .
 +
\]
 +
</cmath>
 +
 +
Taking <math>(4) - (2')</math>, we get
 +
<cmath>
 +
\[
 +
v w' = - \frac{1}{4} . \hspace{1cm} (5) .
 +
\]
 +
</cmath>
 +
 +
Taking (4) + (5), we get
 +
<cmath>
 +
\[
 +
\left( v + w' \right)^2 = \frac{1}{2} .
 +
\]
 +
</cmath>
 +
 +
Hence, <math>u^2 = \left( v + w' \right)^2 = \frac{1}{2}</math>.
 +
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
\left[ (1-x)(1-y)(1-z) \right]^2
 +
& = u^2 (vw)^2 \\
 +
& = u^2 (vw')^2 \\
 +
& = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\
 +
& = \frac{1}{32} .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>1 + 32 = \boxed{\textbf{(033) }}</math>.
 +
\end{solution}
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
==Video Solution==
 +
 +
https://www.youtube.com/watch?v=ihKUZ5itcdA
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See also==
 
==See also==

Revision as of 23:14, 17 February 2022

Problem

Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1 (geometric interpretation)

First, we note that we can let a triangle exist with side lengths $\sqrt{2x}$, $\sqrt{2z}$, and opposite altitude $\sqrt{xz}$. 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 $l$ for symmetry purposes. So, we note that if the angle opposite the side with length $\sqrt{2x}$ has a value of $\sin(\theta)$, then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$ so $x=2\sin^2(\theta)$ and the triangle side with length $\sqrt{2x}$ is equal to $2\sin(\theta)$.

We can symmetrically apply this to the two other triangles, and since by law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R \to R=1$ is the circumradius of that triangle. Hence. we calculate that with $l=1, \sqrt{2}$, and $\sqrt{3}$, the angles from the third side with respect to the circumcenter are $120^{\circ}, 90^{\circ}$, and $60^{\circ}$. This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$, $x=2\sin^2(\beta)$, and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$, $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$, and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$. Solving, we get $\alpha=\frac{135^{\circ}}{2}$, $\beta=\frac{105^{\circ}}{2}$, and $\gamma=\frac{165^{\circ}}{2}$.

We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare\]

- 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 $\sqrt{x}$, $\sqrt{y}$, or $\sqrt{z}$ from each term (on the left sides), since each of $x$, $y$, and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\sqrt{x}$, $\sqrt{y}$, or $\sqrt{z}$, the system should look like this: \begin{align*} \sqrt{x}\cdot\sqrt{2-y} + \sqrt{y}\cdot\sqrt{2-x} &= 1 \\ \sqrt{y}\cdot\sqrt{2-z} + \sqrt{z}\cdot\sqrt{2-y} &= \sqrt2 \\ \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. \end{align*} This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\cos^2 \alpha$, $y = 2\cos^2 \beta$, and $z = 2\cos^2 \theta$ is a helpful substitution: \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 \beta}\cdot\sqrt{2-2\cos^2 \theta} + \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \beta} &= \sqrt2 \\ \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*} From each equation $\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get: \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 \beta}\cdot\sqrt{1-\cos^2 \theta} + \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \beta} &= \frac{\sqrt2}{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*} which simplifies to (using the Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C}$): \begin{align*} \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ \cos \beta\cdot\sin \theta + \cos \theta\cdot\sin \beta &= \frac{\sqrt2}{2} \\ \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. \end{align*} which further simplifies to (using sine addition formula $\sin(a + b) = \sin a \cos b + \cos a \sin b$): \begin{align*} \sin(\alpha + \beta) &= \frac{1}{2} \\ \sin(\beta + \theta) &= \frac{\sqrt2}{2} \\ \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. \end{align*} Without loss of generality, taking the inverse sine of each equation yields a simple system: \begin{align*} \alpha + \beta &= \frac{\pi}{6} \\ \beta + \theta &= \frac{\pi}{4} \\ \alpha + \theta &= \frac{\pi}{3}. \end{align*} giving solutions $\alpha = \frac{\pi}{8}$, $\beta = \frac{\pi}{24}$, $\theta = \frac{5\pi}{24}$. Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: $x = 2\cos^2\left(\frac{\pi}{8}\right)$, $y = 2\cos^2\left(\frac{\pi}{24}\right)$, and $z = 2\cos^2\left(\frac{5\pi}{24}\right)$. When plugging into the expression $\left[ (1-x)(1-y)(1-z) \right]^2$, noting that $-\cos 2\phi = 1 - 2\cos^2 \phi\; \forall \; \phi \in \mathbb{C}$ helps to simplify this expression into: 

\[\left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2\]

Now, all the cosines in here are fairly standard: $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\;$ $\cos \frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}$,$\;$ and $\cos \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$. With some final calculations: \[(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \left(\frac{1}{2}\right)\left(\frac{2 + \sqrt{3}}{4}\right)\left(\frac{2 - \sqrt{3}}{4}\right) = \frac{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)}{2\cdot4\cdot4} = \frac{1}{32}.\] This is our answer in simplest form $\frac{m}{n}$, so $m + n = 1 + 32 = \boxed{033}.$ 

-Oxymoronic15

solution 3

Let $1-x=a;1-y=b;1-z=c$, rewrite those equations

$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$;

$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$

$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$

square both sides, get three equations:

$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$

$2bc=2\sqrt{(1-b^2)(1-c^2)}$

$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$

Getting that $a^2+b^2-ab=\frac{3}{4}$

$b^2+c^2=1$

$a^2+c^2+ac=\frac{3}{4}$

Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$

Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$, $bc=\frac{1}{4}$

Since $a^2=b^2+c^2-2bc=\frac{1}{2}$, the final answer is $\frac{1}{4}*\frac{1}{4}*\frac{1}{2}=\frac{1}{32}$ the final answer is $\boxed{033}$

~bluesoul

Solution 4

Denote $u = 1 - x$, $v = 1 - y$, $w = 1 - z$. Hence, the system of equations given in the problem can be written as \begin{align*} \sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\ \sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\ \sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3) \end{align*}

Each equation above takes the following form: \[ \sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} = k . \]

Now, we simplify this equation by removing radicals.

Denote $p = \sqrt{(1-a)(1+b)}$ and $q = \sqrt{(1+a)(1-b)}$.

Hence, the equation above implies \[ \left\{ \begin{array}{l} p + q = k \\ p^2 = (1-a)(1+b) \\ q^2 = (1+a)(1-b) \end{array} \right.. \]

Hence, $q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)$. Hence, $q - p = \frac{q^2 - p^2}{p+q} = \frac{2}{k} (a-b)$.

Because $p + q = k$ and $q - p = \frac{2}{k} (a-b)$, we get $q = \frac{a-b}{k} + \frac{k}{2}$. Plugging this into the equation $q^2 = (1+a)(1-b)$ and simplifying it, we get \[ a^2 + \left( k^2 - 2 \right) ab + b^2 = k^2 - \frac{k^4}{4} . \]

Therefore, the system of equations above can be simplified as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \\ v^2 + w^2 & = 1 \\ w^2 + wu + u^2 & = \frac{3}{4} . \end{align*}

Denote $w' = - w$. The system of equations above can be equivalently written as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\ v^2 + w'^2 & = 1 \hspace{1cm} (2') \\ w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') . \end{align*}

Taking $(1') - (3')$, we get \[ (v - w') (v + w' - u) = 0 . \]

Thus, we have either $v - w' = 0$ or $v + w' - u = 0$.

$\textbf{Case 1}$: $v - w' = 0$.

Equation (2') implies $v = w' = \pm \frac{1}{\sqrt{2}}$.

Plugging $v$ and $w'$ into Equation (2), we get contradiction. Therefore, this case is infeasible.

$\textbf{Case 2}$: $v + w' - u = 0$.

Plugging this condition into (1') to substitute $u$, we get \[ v^2 + v w' + w'^2 = \frac{3}{4} \hspace{1cm} (4) . \]

Taking $(4) - (2')$, we get \[ v w' = - \frac{1}{4} . \hspace{1cm} (5) . \]

Taking (4) + (5), we get \[ \left( v + w' \right)^2 = \frac{1}{2} . \]

Hence, $u^2 = \left( v + w' \right)^2 = \frac{1}{2}$.

Therefore, \begin{align*} \left[ (1-x)(1-y)(1-z) \right]^2 & = u^2 (vw)^2 \\ & = u^2 (vw')^2 \\ & = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\ & = \frac{1}{32} . \end{align*}

Therefore, the answer is $1 + 32 = \boxed{\textbf{(033) }}$. \end{solution}

~Steven Chen (www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=ihKUZ5itcdA

~Steven Chen (www.professorchenedu.com)

See also

2021 AIME I (Problems • Answer Key • Resources)
Preceded by
2021 AIME I, II 		Followed by
2022 AIME II
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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