Difference between revisions of "2022 AIME I Problems/Problem 15"
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==Solution 1 (geometric interpretation)== | ==Solution 1 (geometric interpretation)== | ||
− | + | 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. | |
− | + | 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 can symmetrically apply this to the two other triangles | + | We can symmetrically apply this to the two other equations/triangles. |
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
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 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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- kevinmathz | - kevinmathz | ||
− | ==Solution 2 (easy to follow)== | + | ==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 <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: | ||
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\sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. | \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. | ||
\end{align*}</cmath> | \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: | 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*} | <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 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ | ||
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\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. | \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> | \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: | 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*} | <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 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ | ||
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\sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{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> | \end{align*}</cmath> | ||
+ | |||
which simplifies to (using the Pythagorean identity <math>\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C} </math>): | which simplifies to (using the Pythagorean identity <math>\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C} </math>): | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ | \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ | ||
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\cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. | \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
which further simplifies to (using sine addition formula <math>\sin(a + b) = \sin a \cos b + \cos a \sin b</math>): | which further simplifies to (using sine addition formula <math>\sin(a + b) = \sin a \cos b + \cos a \sin b</math>): | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sin(\alpha + \beta) &= \frac{1}{2} \\ | \sin(\alpha + \beta) &= \frac{1}{2} \\ | ||
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\sin(\alpha + \theta) &= \frac{\sqrt3}{2}. | \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | ||
+ | Taking the inverse sine (<math>0\leq\theta\frac{\pi}{2}</math>) of each equation yields a simple system: | ||
+ | |||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\alpha + \beta &= \frac{\pi}{6} \\ | \alpha + \beta &= \frac{\pi}{6} \\ | ||
\beta + \theta &= \frac{\pi}{4} \\ | \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> | \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> | ||
+ | |||
+ | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}</math>. | ||
+ | |||
+ | ~Oxymoronic15 | ||
+ | |||
+ | ==Solution 3 (substitution)== | ||
+ | Let <math>1-x=a;1-y=b;1-z=c</math>, rewrite those equations | ||
+ | |||
+ | <math>\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1</math>; | ||
+ | |||
+ | <math>\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}</math> | ||
+ | |||
+ | <math>\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}</math> | ||
+ | |||
+ | 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>2ac+1=2\sqrt{(1-c^2)(1-a^2)}</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>a^2+c^2+ac=\frac{3}{4}</math> | ||
+ | |||
+ | Subtract first and third equation, getting <math>(b+c)(b-c)=a(b+c)</math>, <math>a=b-c</math> | ||
+ | |||
+ | 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>, <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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | bu-bye | ||
+ | <cmath>\begin{align*} | ||
+ | \sin(\alpha + \beta) &= 1/2 \\ | ||
+ | \sin(\alpha + \gamma) &= \sqrt2/2 \\ | ||
+ | \sin(\beta + \gamma) &= \sqrt3/2. | ||
+ | \end{align*}</cmath> | ||
+ | Thus, | ||
+ | <cmath>\begin{align*} | ||
+ | \alpha + \beta &= 30^{\circ} \\ | ||
+ | \alpha + \gamma &= 45^{\circ} \\ | ||
+ | \beta + \gamma &= 60^{\circ}, | ||
+ | \end{align*}</cmath> | ||
+ | so <math>(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})</math>. Hence, | ||
+ | |||
+ | <cmath> 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}, </cmath> | ||
+ | so <math>(abc)^2=(\sqrt{2}/8)^2=\frac{1}{32}</math>, for a final answer of <math>\boxed{033}</math>. | ||
+ | |||
+ | <u><b>Remark</b></u> | ||
+ | |||
+ | The motivation for the trig substitution is that if <math>\sin^2(\alpha)=(1-a)/2</math>, then <math>\cos^2(\alpha)=(1+a)/2</math>, 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)== | ||
+ | [[File:2022 AIME I 15.png|400px|right]] | ||
+ | In given equations, <math>0 \leq x,y,z \leq 2,</math> so we define some points: | ||
+ | <cmath>\bar {O} = (0, 0), \bar {A} = (1, 0), | ||
+ | \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),</cmath> | ||
+ | <cmath>\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),</cmath> | ||
+ | <cmath>\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).</cmath> | ||
+ | Notice, that <cmath>\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</cmath> and each points lies in the first quadrant. | ||
+ | |||
+ | We use given equations and get some scalar products: | ||
+ | <cmath>(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies \angle XOY = 60 ^\circ,</cmath> | ||
+ | <cmath>(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies \angle XOZ = 30^\circ,</cmath> | ||
+ | <cmath>(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies \angle Y'OZ = 45^\circ.</cmath> | ||
+ | So <math> \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.</math> | ||
+ | |||
+ | Points <math>Y</math> and <math>Y'</math> are symmetric with respect to <math>OM.</math> | ||
+ | |||
+ | <i><b>Case 1</b></i> | ||
+ | <cmath>\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 .</cmath> | ||
+ | <cmath>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,</cmath> | ||
+ | <cmath>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 =</cmath> | ||
+ | <cmath>=\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2 = \frac {1}{32} \implies \boxed{\textbf{033}}.</cmath> | ||
+ | <i><b>Case 2</b></i> | ||
+ | |||
+ | <cmath>\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}}.</cmath> | ||
+ | |||
+ | '''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) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2022|n=I|num-b=14|after=Last Problem}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 01:25, 1 February 2024
Contents
Problem
Let
and
be positive real numbers satisfying the system of equations:
Then
can be written as
where
and
are relatively prime positive integers. Find
Solution 1 (geometric interpretation)
First, let define a triangle with side lengths ,
, and
, with altitude from
's equal to
.
, the left side of one equation in the problem.
Let be angle opposite the side with length
. Then the altitude has length
and thus
, so
and the side length
is equal to
.
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have , with
as the circumradius, same for all 3 triangles.
The circumcircle's central angle to a side is
, so the 3 triangles'
, have angles
, respectively.
This means that by half angle arcs, we see that we have in some order, ,
, and
(not necessarily this order, but here it does not matter due to symmetry), satisfying that
,
, and
. Solving, we get
,
, and
.
We notice that
- 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 ,
, or
from each term (on the left sides), since each of
,
, and
are positive real numbers. After factoring out accordingly from each terms one of
,
, or
, the system should look like this:
This should give off tons of trigonometry vibes. To make the connection clear, ,
, and
is a helpful substitution:
From each equation can be factored out, and when every equation is divided by 2, we get:
which simplifies to (using the Pythagorean identity ):
which further simplifies to (using sine addition formula ):
Taking the inverse sine () of each equation yields a simple system:
giving solutions:
Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
When plugging into the expression , noting that
helps to simplify this expression into:
Now, all the cosines in here are fairly standard:
With some final calculations:
This is our answer in simplest form , so
.
~Oxymoronic15
Solution 3 (substitution)
Let , rewrite those equations
;
and solve for
Square both sides and simplify, to get three equations:
Square both sides again, and simplify to get three equations:
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since ,
and so the final answer is
~bluesoul
Solution 4
Denote ,
,
.
Hence, the system of equations given in the problem can be written as
Each equation above takes the following form:
Now, we simplify this equation by removing radicals.
Denote and
.
Hence, the equation above implies
Hence, .
Hence,
.
Because and
, we get
.
Plugging this into the equation
and simplifying it, we get
Therefore, the system of equations above can be simplified as
Denote .
The system of equations above can be equivalently written as
Taking , we get
Thus, we have either or
.
:
.
Equation (2') implies .
Plugging and
into Equation (2), we get contradiction. Therefore, this case is infeasible.
:
.
Plugging this condition into (1') to substitute , we get
Taking , we get
Taking (4) + (5), we get
Hence, .
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
bu-bye
Thus,
so
. Hence,
so
, for a final answer of
.
Remark
The motivation for the trig substitution is that if , then
, 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, so we define some points:
Notice, that
and each points lies in the first quadrant.
We use given equations and get some scalar products:
So
Points and
are symmetric with respect to
Case 1
Case 2
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~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)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.