AIME 2022 I · 第 15 题

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 (Condensed Algebraic Trig)

Let $\sin{A} = \sqrt{\frac{x}{2}}$, $\sin{B} = \sqrt{\frac{y}{2}}$, and $\sin{C} = \sqrt{\frac{z}{2}}$ for some $A,B,C \in (0^\circ,90^\circ)$; therefore, $\cos{A} = \sqrt{\frac{2-x}{2}}$, etc. Dividing all three given equations by $2$, we see that they are equivalent to \begin{align*} \sin(A+B) &= \sin(30^\circ) \\ \sin(B+C) &= \sin(45^\circ) \\ \sin(C+A) &= \sin(60^\circ), \end{align*} which produces $A=\frac{45}{2}^\circ$, $B=\frac{15}{2}^\circ$, and $C=\frac{75}{2}^\circ$. Thus our answer is \begin{align*} [(1-x)(1-y)(1-z)]^2 &= [(1-2\sin^2 A)(1-2\sin^2 B)(1-2\sin^2 C)]^2 \\ &= (\cos(2A)\cos(2B)\cos(2C))^2 \\ &= \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{2}\right)^2 \\ &= \frac{1}{32}, \end{align*} producing the answer of $\boxed{033}$.

Remark. Note that liberties have been taken in mathematical rigor in the above solution, especially when dealing with angles $A,B,C$ and the inverse sine function. However, these details do not affect the overall solution and may be carefully checked during the contest if desired; thus we feel justified in leaving them out to emphasize the elegance of the trigonometric substitution.

~Oxymoronic15 (edited for length by Viliciri)

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

$$\begin{aligned} \sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\ \sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\ \sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3) \end{aligned}$$ 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

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

$$\begin{aligned} u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\ v^2 + w'^2 & = 1 \hspace{1cm} (2') \\ w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') . \end{aligned}$$ 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,

$$\begin{aligned} \left[ (1-x)(1-y)(1-z) \right]^2 & = u^2 (vw)^2 \\ & = u^2 (vw')^2 \\ & = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\ & = \frac{1}{32} . \end{aligned}$$ Therefore, the answer is $1 + 32 = \boxed{\textbf{(033) }}$.

~Steven Chen (www.professorchenedu.com)

bu-bye

$$\begin{aligned} \sin(\alpha + \beta) &= 1/2 \\ \sin(\alpha + \gamma) &= \sqrt2/2 \\ \sin(\beta + \gamma) &= \sqrt3/2. \end{aligned}$$ Thus,

$$\begin{aligned} \alpha + \beta &= 30^{\circ} \\ \alpha + \gamma &= 45^{\circ} \\ \beta + \gamma &= 60^{\circ}, \end{aligned}$$ 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

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~Math Gold Medalist

Video Solution

https://youtu.be/aa_VY4e4OOM?si=1lHSwY3v7RICoEpk

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Video Solution

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

~Steven Chen (www.professorchenedu.com)