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

(→‎Solution 1 (easy to follow))
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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>
  
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==Solution 1 (easy to follow)==
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==Solution 1 (geometric interpretation)==
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Favorite problem on the test. Extremely clean. (Solution close to that in post #2)
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First, we note that we can let a triangle exist with side lengths <math>\sqrt{2x}</math>, <math>\sqrt{2z}</math>, and opposite altitude <math>\sqrt{xz}</math>. 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 <math>l</math> for symmetry purposes. So, we note that if the angle opposite the side with length <math>\sqrt{2x}</math> has a value of <math>\sin(\theta)</math>, then the altitude has length <math>\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}</math> and thus <math>\sin(\theta) = \sqrt{\frac{x}{2}}</math> so <math>x=2\sin^2(\theta)</math> and the triangle side with length <math>\sqrt{2x}</math> is equal to <math>2\sin(\theta)</math>.
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We can symmetrically apply this to the two other triangles, and since by law of sines, we have <math>\frac{2\sin(\theta)}{\sin(\theta)} = 2R \to R=1</math> is the circumradius of that triangle. Hence. we calculate that with <math>l=1, \sqrt{2}</math>, and <math>\sqrt{3}</math>, the angles from the third side with respect to the circumcenter are <math>120^{\circ}, 90^{\circ}</math>, and <math>60^{\circ}</math>. This means that by half angle arcs, we see that we have in some order, <math>x=2\sin^2(\alpha)</math>, <math>x=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>.
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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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==Solution 2 (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:
 
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:

Revision as of 20:37, 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)

Favorite problem on the test. Extremely clean. (Solution close to that in post #2)

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

Solution 2 (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 $\alpha = \frac{\pi}{8}$, $\beta = \frac{\pi}{24}$, $\theta = \frac{5\pi}{24}$. Since these unknowns are directly related to our original unknowns, we also have 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

(Still editing) - Oxymoronic15

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