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Difference between revisions of "2022 AIME I Problems/Problem 15"
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\end{align*}</cmath>  
 
\end{align*}</cmath>  
 
Then <math>\left[ (1-x)(1-y)(1-z) \right]^2</math> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 
Then <math>\left[ (1-x)(1-y)(1-z) \right]^2</math> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 +
 +
==Solution 1 (easy to follow)==
 +
 +
Note that in each equation in this system, it is possible to factor <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math> from each term (on the left sides), since each of <math>x</math>, <math>y</math>, and <math>z</math> are positive real numbers. After factoring out accordingly from each terms one of <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math>, the system should look like this:
 +
<cmath>\begin{align*}
 +
\sqrt{x}\cdot\sqrt{2-y} + \sqrt{y}\cdot\sqrt{2-x} &= 1 \\
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\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*}</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:
 +
<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 \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*}</cmath>
 +
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*}
 +
\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*}</cmath>
 +
which simplifies to (using the Pythagorean identity <math>\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C} </math>):
 +
<cmath>\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*}</cmath>
 +
which further simplifies to (using sine addition formula <math>\sin(a + b) = \sin a \cos b + \cos a \sin b</math>):
 +
<cmath>\begin{align*}
 +
\sin(\alpha + \beta) &= \frac{1}{2} \\
 +
\sin(\beta + \theta) &= \frac{\sqrt2}{2} \\
 +
\sin(\alpha + \theta) &= \frac{\sqrt3}{2}.
 +
\end{align*}</cmath>
 +
Without loss of generality, we can take the inverse sine of each equation. The resulting system is simple:
 +
<cmath>\begin{align*}
 +
\alpha + \beta &= \frac{\pi}{6} \\
 +
\beta + \theta &= \frac{\pi}{4} \\
 +
\alpha + \theta &= \frac{\pi}{3}.
 +
\end{align*}</cmath>
 +
giving solutions <math>\alpha = \frac{\pi}{8}</math>, <math>\beta = \frac{\pi}{24}</math>, <math>\theta = \frac{5\pi}{24}</math>.

Revision as of 20:12, 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 (easy to follow)

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, we can take the inverse sine of each equation. The resulting system is simple: \begin{align*} \alpha + \beta &= \frac{\pi}{6} \\ \beta + \theta &= \frac{\pi}{4} \\ \alpha + \theta &= \frac{\pi}{3}. \end{align*} giving solutions $\alpha = \frac{\pi}{8}$, $\beta = \frac{\pi}{24}$, $\theta = \frac{5\pi}{24}$.

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