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

Latest revision as of 00:25, 1 February 2024

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, let define a triangle with side lengths $\sqrt{2x}$ , $\sqrt{2z}$ , and $l$ , with altitude from $l$ 's equal to $\sqrt{xz}$ . $l = \sqrt{2x - xz} + \sqrt{2z - xz}$ , the left side of one equation in the problem.

Let $\theta$ be angle opposite the side with length $\sqrt{2x}$ . 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 side length $\sqrt{2x}$ is equal to $2\sin(\theta)$ .

We can symmetrically apply this to the two other equations/triangles.

By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$ , with $R=1$ as the circumradius, same for all 3 triangles. The circumcircle's central angle to a side is $2 \arcsin(l/2)$ , so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$ , have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$ , respectively.

This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$ , $y=2\sin^2(\beta)$ , 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 $\sin(a + b) = \sin a \cos b + \cos a \sin b$ ):

$\begin{align*} \sin(\alpha + \beta) &= \frac{1}{2} \\ \sin(\beta + \theta) &= \frac{\sqrt2}{2} \\ \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. \end{align*}$

Taking the inverse sine ( $0\leq\theta\frac{\pi}{2}$ ) 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:

$\begin{align*} \alpha &= \frac{\pi}{8} \\ \beta &= \frac{\pi}{24} \\ \theta &= \frac{5\pi}{24} \end{align*}$

Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:

$\begin{align*} x &= 2\cos^2\left(\frac{\pi}{8}\right) \\ y &= 2\cos^2\left(\frac{\pi}{24}\right) \\ z &= 2\cos^2\left(\frac{5\pi}{24}\right) \end{align*}$

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:

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

Now, all the cosines in here are fairly standard:

$\begin{align*} \cos \frac{\pi}{4} &= \frac{\sqrt{2}}{2} \\ \cos \frac{\pi}{12} &=\frac{\sqrt{6} + \sqrt{2}}{4} & (= \cos{\frac{\frac{\pi}{6}}{2}} ) \\ \cos \frac{5\pi}{12} &= \frac{\sqrt{6} - \sqrt{2}}{4} & (= \cos\left({\frac{\pi}{6} + \frac{\pi}{4}} \right) ) \end{align*}$

With some final calculations:

$\begin{align*} &(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 \\ =& \left(\frac{1}{2}\right) \left(\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)\right)^2 \\ =&\frac{1}{2} \frac{4^2}{16^2} = \frac{1}{32} \end{align*}$

This is our answer in simplest form $\frac{m}{n}$ , so $m + n = 1 + 32 = \boxed{033}$ .

~Oxymoronic15

Solution 3 (substitution)

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

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

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

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

and solve for $m/n = (abc)^2 = a^2b^2c^2$

Square both sides and simplify, to get three equations:

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

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

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

Square both sides again, and simplify to get three equations:

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

$b^2+c^2~ ~ ~ ~ ~ ~=1$

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

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 $a^2=b^2+c^2-2bc=\frac{1}{2}$ , $m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}$ and so 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 $1 + 32 = \boxed{\textbf{(033) }}$ .

~Steven Chen (www.professorchenedu.com)

bu-bye $\begin{align*} \sin(\alpha + \beta) &= 1/2 \\ \sin(\alpha + \gamma) &= \sqrt2/2 \\ \sin(\beta + \gamma) &= \sqrt3/2. \end{align*}$ Thus, $\begin{align*} \alpha + \beta &= 30^{\circ} \\ \alpha + \gamma &= 45^{\circ} \\ \beta + \gamma &= 60^{\circ}, \end{align*}$ so $(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})$ . Hence,

$abc = (1-2\sin^2(\alpha))(1-2\sin^2(\beta))(1-2\sin^2(\gamma))=\cos(15^{\circ})\cos(45^{\circ})\cos(75^{\circ})=\frac{\sqrt{2}}{8},$ so $(abc)^2=(\sqrt{2}/8)^2=\frac{1}{32}$ , for a final answer of $\boxed{033}$ .

Remark

The motivation for the trig substitution is that if $\sin^2(\alpha)=(1-a)/2$ , then $\cos^2(\alpha)=(1+a)/2$ , 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, $0 \leq x,y,z \leq 2,$ so we define some points: $\bar {O} = (0, 0), \bar {A} = (1, 0), \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),$ $\bar {X} = \left(\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}\right), \bar {Y'} = \left(\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}\right),$ $\bar {Y} = \left(\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}\right), \bar {Z} = \left(\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}\right).$ Notice, that $\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1$ and each points lies in the first quadrant.

We use given equations and get some scalar products: $(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies \angle XOY = 60 ^\circ,$ $(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies \angle XOZ = 30^\circ,$ $(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies \angle Y'OZ = 45^\circ.$ So $\angle YOZ = \angle XOY – \angle XOZ = 60 ^\circ – 30 ^\circ = 30 ^\circ, \angle Y'OY = \angle Y'OZ + \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.$

Points $Y$ and $Y'$ are symmetric with respect to $OM.$

Case 1 $\angle YOA = \frac{90^\circ – 75^\circ}{2} = 7.5^\circ, \angle ZOA = 30^\circ + 7.5^\circ = 37.5^\circ, \angle XOA = 60^\circ + 7.5^\circ = 67.5^\circ .$ $1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2 = \sin^2 \angle XOA – \cos^2 \angle XOA = –\cos 2 \angle XOA = –\cos 135^\circ,$ $1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =$ $=\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2 = \frac {1}{32} \implies \boxed{\textbf{033}}.$ Case 2

$\angle Y_1 OA = \frac{90^\circ + 75^\circ}{2} = 82.5^\circ, \angle Z_1 OA = 82.5^\circ – 30^\circ = 52.5^\circ, \angle X_1 OA = 82.5^\circ – 60^\circ = 22.5^\circ \implies \boxed{\textbf{033}}.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/i6kDMbav2sk

~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)

2022 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AIME I Problems/Problem 15"

Latest revision as of 00:25, 1 February 2024

Contents

Problem

Solution 1 (geometric interpretation)

Solution 2 (pure algebraic trig, easy to follow)

Solution 3 (substitution)

Solution 4

Solution 6 (Geometric)

Video Solution

Video Solution

Video Solution

See Also

@@ Line 10: / Line 10: @@
 ==Solution 1 (geometric interpretation)==
-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>.
+First, let define a triangle with side lengths <math>\sqrt{2x}</math>, <math>\sqrt{2z}</math>, and <math>l</math>, with altitude from <math>l</math>'s equal to <math>\sqrt{xz}</math>. <math>l = \sqrt{2x - xz} + \sqrt{2z - xz}</math>, the left side of one equation in the problem.
-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>.
+Let  <math>\theta</math> be angle opposite the side with length <math>\sqrt{2x}</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 side length <math>\sqrt{2x}</math> is equal to <math>2\sin(\theta)</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 can symmetrically apply this to the two other equations/triangles.
-- kevinmathz
+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)</math>, so the 3 triangles' <math>l=1, \sqrt{2}, \sqrt{3}</math>,  have angles <math>120^{\circ}, 90^{\circ}, 60^{\circ}</math>, respectively.
-==Solution 2 (Detailed Geometric Solution created by ChatGPT)==
+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>.
-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 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 label the three triangles as <math>\triangle ABC</math>, <math>\triangle BCD</math>, and <math>\triangle CDA</math>, where <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are the vertices of the triangles. Let the lengths of the sides opposite to vertices <math>A</math>, <math>B</math>, and <math>C</math> be <math>\sqrt{2x}</math>, <math>\sqrt{2y}</math>, and <math>\sqrt{2z}</math>, respectively. We also let the lengths of the sides of the triangles be <math>l_1</math>, <math>l_2</math>, and <math>l_3</math>, respectively.
+- kevinmathz
-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:
-\begin{align*}
-l_1^2 &= \sqrt{2x}^2 + \sqrt{2y}^2 - 2 \cdot \sqrt{2x} \cdot \sqrt{2y} \cdot \cos{\angle A} \
-l_2^2 &= \sqrt{2y}^2 + \sqrt{2z}^2 - 2 \cdot \sqrt{2y} \cdot \sqrt{2z} \cdot \cos{\angle B} \
-l_3^2 &= \sqrt{2z}^2 + \sqrt{2x}^2 - 2 \cdot \sqrt{2z} \cdot \sqrt{2x} \cdot \cos{\angle C}
-\end{align*}
-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
-\begin{align*}
-l_1 &= 1 \cdot \sin{\angle A} \
-l_2 &= 1 \cdot \sin{\angle B} \
-l_3 &= 1 \cdot \sin{\angle C}
-\end{align*}
-Substituting these expressions into the equations above and rearranging, we get
-\begin{align*}
-\sin^2{\angle A} &= \frac{xy}{2x-xy} \
-\sin^2{\angle B} &= \frac{yz}{2y-yz} \
-\sin^2{\angle C} &= \frac{zx}{2z-zx}
-\end{align*}
-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
-\begin{align*}
-\sin^2{\angle A} &= \frac{y}{2-y} \
-\sin^2{\angle B} &= \frac{z}{2-z} \
-\sin^2{\angle C} &= \frac{x}{2-x}
-\end{align*}
-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 sum to <math>180^\circ</math>, we have <math>\angle A + \angle B + \angle C = 180^\circ</math>. Substituting the expressions for <math>\sin^2{\angle A}</math>, <math>\sin^2{\angle B}</math>, and <math>\sin^2{\angle C}</math> 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 <math>x = \frac{4}{5}</math>, <math>y = \frac{3}{4}</math>, and <math>z = \frac{3}{5}</math>.
-Substituting these values into the expression <math>\left[ (1-x)(1-y)(1-z) \right]^2</math>, 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 <math>m + n = 1 + 32 = \boxed{033}</math>.
-~ pi_is_3.14
 ==Solution 2 (pure algebraic trig, easy to follow)==
@@ Line 70: / Line 34: @@
 \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 \\
@@ Line 76: / Line 42: @@
 \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} \\
@@ Line 82: / Line 50: @@
 \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} \\
@@ Line 88: / Line 58: @@
 \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} \\
@@ Line 94: / Line 66: @@
 \sin(\alpha + \theta) &= \frac{\sqrt3}{2}.
 \end{align*}</cmath>
-Without loss of generality, taking the inverse sine of each equation yields a simple system:
+Taking the inverse sine (<math>0\leq\theta\frac{\pi}{2}</math>) of each equation yields a simple system:
 <cmath>\begin{align*}
 \alpha + \beta &= \frac{\pi}{6} \\
 \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>
+Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
+<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>
-giving solutions <math>\alpha = \frac{\pi}{8}</math>, <math>\beta = \frac{\pi}{24}</math>, <math>\theta = \frac{5\pi}{24}</math>. Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: <math>x = 2\cos^2\left(\frac{\pi}{8}\right)</math>, <math>y = 2\cos^2\left(\frac{\pi}{24}\right)</math>, and <math>z = 2\cos^2\left(\frac{5\pi}{24}\right)</math>. 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>\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>
+This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}</math>.
-Now, all the cosines in here are fairly standard: <math>\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}</math>, <math>\;</math> <math>\cos \frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}</math>,<math>\;</math> and <math>\cos \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}</math>. With some final calculations:
-<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{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}.</cmath>
-This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}.</math>
 ~Oxymoronic15
-==solution 3==
+==Solution 3 (substitution)==
 Let <math>1-x=a;1-y=b;1-z=c</math>, rewrite those equations
@@ Line 117: / Line 132: @@
 <math>\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}</math>
-square both sides, get three equations:
+and solve for <math>m/n = (abc)^2 = a^2b^2c^2</math>
+Square both sides and simplify, to get three equations:
 <math>2ab-1=2\sqrt{(1-a^2)(1-b^2)}</math>
-<math>2bc=2\sqrt{(1-b^2)(1-c^2)}</math>
+<math>2bc~ ~ ~ ~  ~ ~=2\sqrt{(1-b^2)(1-c^2)}</math>
 <math>2ac+1=2\sqrt{(1-c^2)(1-a^2)}</math>
-Getting that <math>a^2+b^2-ab=\frac{3}{4}</math>
+Square both sides again, and simplify to get three equations:
+<math>a^2+b^2-ab=\frac{3}{4}</math>
-<math>b^2+c^2=1</math>
+<math>b^2+c^2~ ~ ~ ~ ~ ~=1</math>
 <math>a^2+c^2+ac=\frac{3}{4}</math>
@@ Line 135: / Line 154: @@
 Put it in first equation, getting <math>b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}</math>, <math>bc=\frac{1}{4}</math>
-Since <math>a^2=b^2+c^2-2bc=\frac{1}{2}</math>, the final answer is <math>\frac{1}{4}*\frac{1}{4}*\frac{1}{2}=\frac{1}{32}</math> the final answer is <math>\boxed{033}</math>
+Since <math>a^2=b^2+c^2-2bc=\frac{1}{2}</math>, <math>m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}</math>  and so the final answer is <math>\boxed{033}</math>
 ~bluesoul
@@ Line 257: / Line 276: @@
 Therefore, the answer is <math>1 + 32 = \boxed{\textbf{(033) }}</math>.
-\end{solution}
 ~Steven Chen (www.professorchenedu.com)
+bu-bye
-==Solution 5==
-Let <math>a=1-x</math>, <math>b=1-y</math>, and <math>c=1-z</math>. Then,
-<cmath>\begin{align*}
-\sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} &= 1       \hspace{15mm}(1) \\
-\sqrt{(1-b)(1+c)} + \sqrt{(1+b)(1-c)} &= \sqrt2  \hspace{11.5mm}(2) \\
-\sqrt{(1-a)(1+c)} + \sqrt{(1+a)(1-c)} &= \sqrt3. \hspace{10.5mm}(3).
-\end{align*}</cmath>
-Notice that <math>\frac{1-a}{2}+\frac{1+a}{2}=1</math>, <math>\frac{1-b}{2}+\frac{1+b}{2}=1</math>, and <math>\frac{1-c}{2}+\frac{1+c}{2}=1</math>. Let <math>\sin^2(\alpha)=(1-a)/2</math>, <math>\sin^2(\beta)=(1-b)/2</math>, and <math>\sin^2(\gamma)=(1-c)/2</math> where <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are real. Substituting into <math>(1)</math>, <math>(2)</math>, and <math>(3)</math> yields
 <cmath>\begin{align*}
 \sin(\alpha + \beta) &= 1/2 \\
@@ Line 311: / Line 320: @@
 \angle Y'OY =  \angle Y'OZ +  \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.</math>
-Points <math>Y</math> and <math>Y'</math> are simmetric with respect to <math>OM.</math>
+Points <math>Y</math> and <math>Y'</math> are symmetric with respect to <math>OM.</math>
 <i><b>Case 1</b></i>
@@ Line 324: / Line 333: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
+==Video Solution==
+https://youtu.be/i6kDMbav2sk
+~Math Gold Medalist
+==Video Solution==
+https://youtu.be/aa_VY4e4OOM?si=1lHSwY3v7RICoEpk
+~MathProblemSolvingSkills.com
 ==Video Solution==