https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jzhao000&feedformat=atom AoPS Wiki - User contributions [en] 2020-11-26T23:56:40Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_USAMO_Problems/Problem_6&diff=118272 2019 USAMO Problems/Problem 6 2020-02-20T21:10:51Z <p>Jzhao000: </p> <hr /> <div>==Problem==<br /> Find all polynomials &lt;math&gt;P&lt;/math&gt; with real coefficients such that &lt;cmath&gt;\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)&lt;/cmath&gt;holds for all nonzero real numbers &lt;math&gt;x,y,z&lt;/math&gt; satisfying &lt;math&gt;2xyz=x+y+z&lt;/math&gt;.<br /> <br /> ==Solution==<br /> If &lt;math&gt;P(x)=c&lt;/math&gt; for a constant &lt;math&gt;c,&lt;/math&gt; then &lt;math&gt;\dfrac{c(x+y+z)}{xyz}=3c&lt;/math&gt;. We have &lt;math&gt;2c=3c.&lt;/math&gt; Therefore &lt;math&gt;c=0.&lt;/math&gt;<br /> <br /> Now consider the case of non-constant polynomials. <br /> First we have &lt;cmath&gt;xP(x)+yP(y)+zP(z)=xyz(P(x-y)+P(y-z)+P(z-x))&lt;/cmath&gt; for all nonzero real numbers &lt;math&gt;x,y,z&lt;/math&gt; satisfying &lt;math&gt;2xyz=x+y+z&lt;/math&gt;. Both sides of the equality are polynomials (of &lt;math&gt;x,y,z&lt;/math&gt;). They have the same values on the 2-dimensional surface &lt;math&gt;2xyz=x+y+z&lt;/math&gt;, except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with &lt;math&gt;z=0.&lt;/math&gt; Let &lt;math&gt;z=0, &lt;/math&gt; we have &lt;math&gt;y=-x&lt;/math&gt; and &lt;math&gt;x(P(x)-P(-x))=0.&lt;/math&gt; Therefore &lt;math&gt;P&lt;/math&gt; is an even function. <br /> <br /> (Here is a sketch of an elementary proof. Let &lt;math&gt;z=\dfrac{x+y}{2xy-1}.&lt;/math&gt; We have<br /> &lt;cmath&gt;xP(x)+yP(y)+\dfrac{x+y}{2xy-1}P(\dfrac{x+y}{2xy-1})=xy\dfrac{x+y}{2xy-1}(P(x-y)+P(y-\dfrac{x+y}{2xy-1})+P(\dfrac{x+y}{2xy-1}-x)).&lt;/cmath&gt;<br /> This is an equality of rational expressions. By multiplying &lt;math&gt;(2xy-1)^N&lt;/math&gt; on both sides for a sufficiently large &lt;math&gt;N&lt;/math&gt;, they become polynomials, say &lt;math&gt;A(x,y)=B(x,y)&lt;/math&gt; for all real &lt;math&gt;x, y&lt;/math&gt; with &lt;math&gt;x\ne 0, y\ne 0, x+y\ne 0&lt;/math&gt; and &lt;math&gt;2xy-1\ne 0.&lt;/math&gt; For a fixed &lt;math&gt;x,&lt;/math&gt; we have two polynomials (of &lt;math&gt;y&lt;/math&gt;) having same values for infinitely many &lt;math&gt;y&lt;/math&gt;. They must be identical. Let &lt;math&gt;y=0,&lt;/math&gt; we have &lt;math&gt;x^{N+1}(P(x)-P(-x))=0.&lt;/math&gt; )<br /> <br /> Notice that if &lt;math&gt;P(x)&lt;/math&gt; is a solution, then is &lt;math&gt;cP(x)&lt;/math&gt; for any constant &lt;math&gt;c.&lt;/math&gt; For simplicity, we assume the leading coefficient of &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt;: &lt;cmath&gt;P(x)=x^n+a_{n-2}x^{n-2}+\cdots +a_2x^2+a_0,&lt;/cmath&gt; where &lt;math&gt;n&lt;/math&gt; is a positive even number.<br /> <br /> Let &lt;math&gt;y=\dfrac{1}{x}&lt;/math&gt;, &lt;math&gt;z=x+\dfrac{1}{x}.&lt;/math&gt; we have &lt;cmath&gt;xP(x)+\dfrac{1}{x}P\left (\dfrac{1}{x}\right )+\left ( x+\dfrac{1}{x}\right ) P\left ( x+\dfrac{1}{x}\right ) =\left (x+\dfrac{1}{x}\right )\left ( P\left (x-\dfrac{1}{x}\right )+P(-x)+P\left (\dfrac{1}{x}\right )\right ).&lt;/cmath&gt;<br /> <br /> Simplify using &lt;math&gt;P(x)=P(-x),&lt;/math&gt;<br /> &lt;cmath&gt;\left (x+\dfrac{1}{x}\right ) \left (P\left (x+\dfrac{1}{x}\right )-P\left (x-\dfrac{1}{x}\right )\right )=\dfrac{1}{x}P(x)+xP\left (\dfrac{1}{x}\right ).&lt;/cmath&gt;<br /> <br /> Expand and combine like terms, both sides are of the form &lt;cmath&gt;c_{n-1}x^{n-1}+c_{n-3}x^{n-3}+\cdots+c_1x+c_{-1}x^{-1}+\cdots+c_{-n+1}x^{-n+1}.&lt;/cmath&gt;<br /> <br /> They have the same values for infinitely many &lt;math&gt;x.&lt;/math&gt; They must be identical. We just compare their leading terms. On the left hand side it is &lt;math&gt;2nx^{n-1}&lt;/math&gt;. There are two cases for the right hand sides: If &lt;math&gt;n&gt;2&lt;/math&gt;, it is &lt;math&gt;x^{n-1}&lt;/math&gt;; If &lt;math&gt;n=2&lt;/math&gt;, it is &lt;math&gt;(1+a_0)x.&lt;/math&gt; It does not work for &lt;math&gt;n&gt;2.&lt;/math&gt; When &lt;math&gt;n=2,&lt;/math&gt; we have &lt;math&gt;4=1+a_0.&lt;/math&gt; therefore &lt;math&gt;a_0=3.&lt;/math&gt;<br /> <br /> The solution: &lt;math&gt;P(x)=c(x^2+3)&lt;/math&gt; for any constant &lt;math&gt;c.&lt;/math&gt; <br /> <br /> -JZ<br /> ==See also==<br /> {{USAMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_12&diff=117869 2013 AIME II Problems/Problem 12 2020-02-17T18:38:34Z <p>Jzhao000: </p> <hr /> <div>==Problem 12==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all polynomials of the form &lt;math&gt;z^3 + az^2 + bz + c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers. Find the number of polynomials in &lt;math&gt;S&lt;/math&gt; such that each of its roots &lt;math&gt;z&lt;/math&gt; satisfies either &lt;math&gt;|z| = 20&lt;/math&gt; or &lt;math&gt;|z| = 13&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from [[Vieta's formulas]]. <br /> <br /> *Case 1: &lt;math&gt;f(z)=(z-r)(z-\omega)(z-\omega^*)&lt;/math&gt;, where &lt;math&gt;r\in \mathbb{R}&lt;/math&gt;, &lt;math&gt;\omega&lt;/math&gt; is nonreal, and &lt;math&gt;\omega^*&lt;/math&gt; is the complex conjugate of omega (note that we may assume that &lt;math&gt;\Im(\omega)&gt;0&lt;/math&gt;).<br /> <br /> The real root &lt;math&gt;r&lt;/math&gt; must be one of &lt;math&gt;-20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;-13&lt;/math&gt;, or &lt;math&gt;13&lt;/math&gt;. By Viète's formulas, &lt;math&gt;a=-(r+\omega+\omega^*)&lt;/math&gt;, &lt;math&gt;b=|\omega|^2+r(\omega+\omega^*)&lt;/math&gt;, and &lt;math&gt;c=-r|\omega|^2&lt;/math&gt;. But &lt;math&gt;\omega+\omega^*=2\Re{(\omega)}&lt;/math&gt; (i.e., adding the conjugates cancels the imaginary part). Therefore, to make &lt;math&gt;a&lt;/math&gt; is an integer, &lt;math&gt;2\Re{(\omega)}&lt;/math&gt; must be an integer. Conversely, if &lt;math&gt;\omega+\omega^*=2\Re{(\omega)}&lt;/math&gt; is an integer, then &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are clearly integers. Therefore &lt;math&gt;2\Re{(\omega)}\in \mathbb{Z}&lt;/math&gt; is equivalent to the desired property. Let &lt;math&gt;\omega=\alpha+i\beta&lt;/math&gt;.<br /> <br /> *Subcase 1.1: &lt;math&gt;|\omega|=20&lt;/math&gt;.<br /> In this case, &lt;math&gt;\omega&lt;/math&gt; lies on a circle of radius &lt;math&gt;20&lt;/math&gt; in the complex plane. As &lt;math&gt;\omega&lt;/math&gt; is nonreal, we see that &lt;math&gt;\beta\ne 0&lt;/math&gt;. Hence &lt;math&gt;-20&lt;\Re{(\omega)}&lt; 20&lt;/math&gt;, or rather &lt;math&gt;-40&lt;2\Re{(\omega)}&lt; 40&lt;/math&gt;. We count &lt;math&gt;79&lt;/math&gt; integers in this interval, each of which corresponds to a unique complex number on the circle of radius &lt;math&gt;20&lt;/math&gt; with positive imaginary part.<br /> <br /> *Subcase 1.2: &lt;math&gt;|\omega|=13&lt;/math&gt;.<br /> In this case, &lt;math&gt;\omega&lt;/math&gt; lies on a circle of radius &lt;math&gt;13&lt;/math&gt; in the complex plane. As &lt;math&gt;\omega&lt;/math&gt; is nonreal, we see that &lt;math&gt;\beta\ne 0&lt;/math&gt;. Hence &lt;math&gt;-13&lt;\Re{(\omega)}&lt; 13&lt;/math&gt;, or rather &lt;math&gt;-26&lt;2\Re{(\omega)}&lt; 26&lt;/math&gt;. We count &lt;math&gt;51&lt;/math&gt; integers in this interval, each of which corresponds to a unique complex number on the circle of radius &lt;math&gt;13&lt;/math&gt; with positive imaginary part.<br /> <br /> Therefore, there are &lt;math&gt;79+51=130&lt;/math&gt; choices for &lt;math&gt;\omega&lt;/math&gt;. We also have &lt;math&gt;4&lt;/math&gt; choices for &lt;math&gt;r&lt;/math&gt;, hence there are &lt;math&gt;4\cdot 130=520&lt;/math&gt; total polynomials in this case.<br /> <br /> *Case 2: &lt;math&gt;f(z)=(z-r_1)(z-r_2)(z-r_3)&lt;/math&gt;, where &lt;math&gt;r_1,r_2,r_3&lt;/math&gt; are all real.<br /> In this case, there are four possible real roots, namely &lt;math&gt;\pm 13, \pm20&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be the number of times that &lt;math&gt;13&lt;/math&gt; appears among &lt;math&gt;r_1,r_2,r_3&lt;/math&gt;, and define &lt;math&gt;q,r,s&lt;/math&gt; similarly for &lt;math&gt;-13,20&lt;/math&gt;, and &lt;math&gt;-20&lt;/math&gt;, respectively. Then &lt;math&gt;p+q+r+s=3&lt;/math&gt; because there are three roots. We wish to find the number of ways to choose nonnegative integers &lt;math&gt;p,q,r,s&lt;/math&gt; that satisfy that equation. By balls and urns, these can be chosen in &lt;math&gt;\binom{6}{3}=20&lt;/math&gt; ways.<br /> <br /> Therefore, there are a total of &lt;math&gt;520+20=\boxed{540}&lt;/math&gt; polynomials with the desired property.<br /> <br /> ==Solution Systematics==<br /> This combinatorics problem involves counting, and casework is most appropriate.<br /> There are two cases: either all three roots are real, or one is real and there are two imaginary roots.<br /> <br /> Case 1: Three roots are of the set &lt;math&gt;{13, -13, 20, -20}&lt;/math&gt;. By stars and bars, there is &lt;math&gt;\binom{6}{3}=20&lt;/math&gt; ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).<br /> <br /> Case 2: One real root: one of &lt;math&gt;13, -13, 20, -20&lt;/math&gt;. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of &lt;math&gt;20&lt;/math&gt; or &lt;math&gt;13&lt;/math&gt;. Call the root &lt;math&gt;a+bi&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of &lt;math&gt;(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)&lt;/math&gt; tells us that we just need &lt;math&gt;2a&lt;/math&gt; to be integral, because &lt;math&gt;a^2+b^2&lt;/math&gt; IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) <br /> Therefore, when the norm is &lt;math&gt;20&lt;/math&gt;, the &lt;math&gt;a&lt;/math&gt; term can range from &lt;math&gt;-19.5, -19, ...., 0, 0.5, ..., 19.5&lt;/math&gt; or &lt;math&gt;79&lt;/math&gt; solutions. When the norm is &lt;math&gt;13&lt;/math&gt;, the &lt;math&gt;a&lt;/math&gt; term has &lt;math&gt;51&lt;/math&gt; possibilities from &lt;math&gt;-12.5, -12, ..., 12.5&lt;/math&gt;. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, &lt;math&gt;4&lt;/math&gt;, and you get &lt;math&gt;520&lt;/math&gt; for this case.<br /> <br /> And &lt;math&gt;520+20=540&lt;/math&gt; and we are done.<br /> <br /> ==Comments==<br /> If the polynomial has one real root and two complex roots, then it can be factored as &lt;math&gt;(z-r)(z^2+pz+q), &lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is real with &lt;math&gt;|r|=13,20&lt;/math&gt; and &lt;math&gt;p,q&lt;/math&gt; are integers with &lt;math&gt;p^2 &lt;4q.&lt;/math&gt; The roots &lt;math&gt;z_1&lt;/math&gt; and &lt;math&gt;z_2&lt;/math&gt; are conjugates. We have &lt;math&gt;|z_1|^2=|z_2|^2=z_1z_2=q.&lt;/math&gt; So &lt;math&gt;q&lt;/math&gt; is either &lt;math&gt;20^2&lt;/math&gt; or &lt;math&gt;13^2&lt;/math&gt;. The only requirement for &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;p&lt;\sqrt{4q^2}=2\sqrt{q}.&lt;/math&gt; All such quadratic equations are listed as follows:<br /> <br /> &lt;math&gt;z^2+pz+20^2,&lt;/math&gt; where &lt;math&gt;p=0,\pm1,\pm2,\cdots,\pm 39,&lt;/math&gt;<br /> <br /> &lt;math&gt;z^2+pz+13^2,&lt;/math&gt; where &lt;math&gt;p=0,\pm1,\pm2,\cdots,\pm 25&lt;/math&gt;.<br /> <br /> Total of 130 equations, multiplied by 4 (the number of cases for real &lt;math&gt;r&lt;/math&gt;, we have 520 equations, as indicated in the solution.<br /> <br /> -JZ<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_5&diff=117798 2013 AIME I Problems/Problem 5 2020-02-15T14:52:39Z <p>Jzhao000: </p> <hr /> <div>== Problem ==<br /> The real root of the equation &lt;math&gt;8x^3 - 3x^2 - 3x - 1 = 0&lt;/math&gt; can be written in the form &lt;math&gt;\frac{\sqrta + \sqrtb + 1}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> We note that &lt;math&gt;8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3&lt;/math&gt;. Therefore, we have that &lt;math&gt;9x^3 = (x+1)^3&lt;/math&gt;, so it follows that &lt;math&gt;x\sqrt{9} = x+1&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;\frac{1}{\sqrt{9}-1} = \frac{\sqrt{81}+\sqrt{9}+1}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; be the real root of the given [[polynomial]]. Now define the cubic polynomial &lt;math&gt;Q(x)=-x^3-3x^2-3x+8&lt;/math&gt;. Note that &lt;math&gt;1/r&lt;/math&gt; must be a root of &lt;math&gt;Q&lt;/math&gt;. However we can simplify &lt;math&gt;Q&lt;/math&gt; as &lt;math&gt;Q(x)=9-(x+1)^3&lt;/math&gt;, so we must have that &lt;math&gt;(\frac{1}{r}+1)^3=9&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{r}=\sqrt{9}-1&lt;/math&gt;, and &lt;math&gt;r=\frac{1}{\sqrt{9}-1}&lt;/math&gt;. We can then multiply the numerator and denominator of &lt;math&gt;r&lt;/math&gt; by &lt;math&gt;\sqrt{81}+\sqrt{9}+1&lt;/math&gt; to rationalize the denominator, and we therefore have &lt;math&gt;r=\frac{\sqrt{81}+\sqrt{9}+1}{8}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> It is clear that for the algebraic degree of &lt;math&gt;x&lt;/math&gt; to be &lt;math&gt;3&lt;/math&gt; that there exists some cubefree integer &lt;math&gt;p&lt;/math&gt; and positive integers &lt;math&gt;m,n&lt;/math&gt; such that &lt;math&gt;a = m^3p&lt;/math&gt; and &lt;math&gt;b = n^3p^2&lt;/math&gt; (it is possible that &lt;math&gt;b = n^3p&lt;/math&gt;, but then the problem wouldn't ask for both an &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;). Let &lt;math&gt;f_1&lt;/math&gt; be the [[automorphism]] over &lt;math&gt;\mathbb{Q}[\sqrt{a}][\omega]&lt;/math&gt; which sends &lt;math&gt;\sqrt{a} \to \omega \sqrt{a}&lt;/math&gt; and &lt;math&gt;f_2&lt;/math&gt; which sends &lt;math&gt;\sqrt{a} \to \omega^2 \sqrt{a}&lt;/math&gt; (note : &lt;math&gt;\omega&lt;/math&gt; is a cubic [[Roots of unity|root of unity]]).<br /> <br /> Letting &lt;math&gt;r&lt;/math&gt; be the root, we clearly we have &lt;math&gt;r + f_1(r) + f_2(r) = \frac{3}{8}&lt;/math&gt; by Vieta's formulas. Thus it follows &lt;math&gt;c=8&lt;/math&gt;.<br /> Now, note that &lt;math&gt;\sqrt{a} + \sqrt{b} + 1&lt;/math&gt; is a root of &lt;math&gt;x^3 - 3x^2 - 24x - 64 = 0&lt;/math&gt;. Thus &lt;math&gt;(x-1)^3 = 27x + 63&lt;/math&gt; so &lt;math&gt;(\sqrt{a} + \sqrt{b})^3 = 27(\sqrt{a} + \sqrt{b}) + 90&lt;/math&gt;. Checking the non-cubicroot dimension part, we get &lt;math&gt;a + b = 90&lt;/math&gt; so it follows that &lt;math&gt;a + b + c = \boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> We have &lt;math&gt;cx-1=\sqrt{a}+\sqrt{b}.&lt;/math&gt; Therefore &lt;math&gt;(cx-1)^3=(\sqrt{a}+\sqrt{b})^3=a+b+3\sqrt{ab}(\sqrt{a}+\sqrt{b})=a+b+3\sqrt{ab}(cx-1).&lt;/math&gt; We have <br /> &lt;cmath&gt;c^3x^3-3c^2x^2-(3c\sqrt{ab}-3c)x-(a+b+1-3\sqrt{ab})=0.&lt;/cmath&gt;<br /> We will find &lt;math&gt;a,b,c&lt;/math&gt; so that the equation is equivalent to the original one. Let &lt;math&gt;\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt{ab}}{c^3}=\dfrac{1}{8}.&lt;/math&gt; Easily, &lt;math&gt;c=8, \sqrt{ab}=9,&lt;/math&gt; and &lt;math&gt;a+b=90.&lt;/math&gt; So &lt;math&gt;a + b + c = 90+8=\boxed{098}&lt;/math&gt;. <br /> <br /> -JZ<br /> <br /> == See Also ==<br /> {{AIME box|year=2013|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_13&diff=117792 2012 AIME I Problems/Problem 13 2020-02-15T03:42:10Z <p>Jzhao000: </p> <hr /> <div>==Problem 13==<br /> Three concentric circles have radii &lt;math&gt;3,&lt;/math&gt; &lt;math&gt;4,&lt;/math&gt; and &lt;math&gt;5.&lt;/math&gt; An equilateral triangle with one vertex on each circle has side length &lt;math&gt;s.&lt;/math&gt; The largest possible area of the triangle can be written as &lt;math&gt;a + \tfrac{b}{c} \sqrt{d},&lt;/math&gt; where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime, and &lt;math&gt;d&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d.&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Reinterpret the problem in the following manner. Equilateral triangle &lt;math&gt;ABC&lt;/math&gt; has a point &lt;math&gt;X&lt;/math&gt; on the interior such that &lt;math&gt;AX = 5,&lt;/math&gt; &lt;math&gt;BX = 4,&lt;/math&gt; and &lt;math&gt;CX = 3.&lt;/math&gt; A &lt;math&gt;60^\circ&lt;/math&gt; counter-clockwise rotation about vertex &lt;math&gt;A&lt;/math&gt; maps &lt;math&gt;X&lt;/math&gt; to &lt;math&gt;X'&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;C'.&lt;/math&gt; Note that angle &lt;math&gt;XAX'&lt;/math&gt; is &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;XA = X'A = 5&lt;/math&gt; which tells us that triangle &lt;math&gt;XAX'&lt;/math&gt; is equilateral and that &lt;math&gt;XX' = 5.&lt;/math&gt; We now notice that &lt;math&gt;XC = 3&lt;/math&gt; and &lt;math&gt;X'C = 4&lt;/math&gt; which tells us that angle &lt;math&gt;XCX'&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt; because there is a &lt;math&gt;3&lt;/math&gt;-&lt;math&gt;4&lt;/math&gt;-&lt;math&gt;5&lt;/math&gt; Pythagorean triple. Now note that &lt;math&gt;\angle ABC + \angle ACB = 120^\circ&lt;/math&gt; and &lt;math&gt;\angle XCA + \angle XBA = 90^\circ,&lt;/math&gt; so &lt;math&gt;\angle XCB+\angle XBC = 30^\circ&lt;/math&gt; and &lt;math&gt;\angle BXC = 150^\circ.&lt;/math&gt; Applying the law of cosines on triangle &lt;math&gt;BXC&lt;/math&gt; yields<br /> <br /> &lt;cmath&gt;BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}&lt;/cmath&gt;<br /> <br /> and thus the area of &lt;math&gt;ABC&lt;/math&gt; equals &lt;cmath&gt;BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.&lt;/cmath&gt;<br /> <br /> so our final answer is &lt;math&gt;3+4+25+9 = \boxed{041}.&lt;/math&gt;<br /> <br /> Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply <br /> &lt;cmath&gt;\frac{\frac{9\sqrt3+16\sqrt3+25\sqrt3}{4}+3(\frac{3*4}{2}}{2})=25\frac{\sqrt{3}}{4}+9&lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties:<br /> We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center &lt;math&gt;O&lt;/math&gt; of the circles lies in the interior of triangle &lt;math&gt;ABC&lt;/math&gt; or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let &lt;math&gt;\theta&lt;/math&gt; be the measure of angle &lt;math&gt;XAC&lt;/math&gt; so that angle &lt;math&gt;BAX&lt;/math&gt; has measure &lt;math&gt;60-\theta&lt;/math&gt;. Let &lt;math&gt;AB=BC=AC=x&lt;/math&gt;. The law of cosines on triangles &lt;math&gt;BAX&lt;/math&gt; and &lt;math&gt;XAC&lt;/math&gt; yields &lt;math&gt;\cos(60-\theta)=\frac{x^2+9}{10x}&lt;/math&gt; and &lt;math&gt;\cos\theta=\frac{x^2+16}{10x}&lt;/math&gt;. Solving this system will yield the value of &lt;math&gt;x&lt;/math&gt;. Since &lt;math&gt;\cos\theta=\frac{x^2+16}{10x}&lt;/math&gt; we have that &lt;math&gt;\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}&lt;/math&gt;. Substituting these into the equation &lt;math&gt;\frac{x^2+9}{10x}=\cos(60-\theta)=\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta&lt;/math&gt; we obtain &lt;math&gt;\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}&lt;/math&gt;. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain &lt;math&gt;x^4-50x^2+193=0&lt;/math&gt; so that by the quadratic formula &lt;math&gt;x^2=25\pm12\sqrt{3}&lt;/math&gt;. Under the hypothesis that &lt;math&gt;O&lt;/math&gt; lies in the interior of triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;x^2&lt;/math&gt; must be &lt;math&gt;25+12\sqrt{3}&lt;/math&gt;. To see this, note that the other value for &lt;math&gt;x^2&lt;/math&gt; is roughly &lt;math&gt;4.2&lt;/math&gt; so that &lt;math&gt;x\approx 2.05&lt;/math&gt;, but since &lt;math&gt;AX=5&lt;/math&gt; and &lt;math&gt;AX\leq x&lt;/math&gt; we have a contradiction. We then obtain the area as in Solution 1.<br /> <br /> Now, suppose &lt;math&gt;O&lt;/math&gt; does not lie in the interior of triangle &lt;math&gt;ABC&lt;/math&gt;. We then obtain convex quadrilateral &lt;math&gt;OBAC&lt;/math&gt; with diagonals &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;OA&lt;/math&gt; intersecting at &lt;math&gt;X&lt;/math&gt;. Here &lt;math&gt;AX=AB=AC=x&lt;/math&gt;. We may let &lt;math&gt;\alpha&lt;/math&gt; denote the measure of angle &lt;math&gt;CAX&lt;/math&gt; so that angle &lt;math&gt;XAB&lt;/math&gt; measures &lt;math&gt;60-\alpha&lt;/math&gt;. Note that the law of cosines on triangles &lt;math&gt;CXA&lt;/math&gt; and &lt;math&gt;BXA&lt;/math&gt; yield the same equations as in the first case with &lt;math&gt;\theta&lt;/math&gt; replaced with &lt;math&gt;\alpha&lt;/math&gt;. Thus we obtain again &lt;math&gt;x^2=25\pm12\sqrt{3}&lt;/math&gt;. If &lt;math&gt;x^2=25+12\sqrt{3}&lt;/math&gt; then &lt;math&gt;x\approx 6.8&lt;/math&gt;, but this is impossible since &lt;math&gt;AX\leq 5&lt;/math&gt; but the shortest possible distance from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;X&lt;/math&gt; is the height of equilateral triangle &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\approx6.8\sqrt{3}\approx5.8&lt;/math&gt;; a contradiction. Hence in this case &lt;math&gt;x^2=25-12\sqrt{3}&lt;/math&gt;. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).<br /> <br /> ===Solution 3===<br /> The problem basically asks for the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; such that it is equilateral and there is a point &lt;math&gt;O&lt;/math&gt; inside of the triangle which satisfies &lt;math&gt;AO = 3&lt;/math&gt;, &lt;math&gt;BO = 4&lt;/math&gt;, and &lt;math&gt;CO = 5&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB = BC = AC = s&lt;/math&gt;. We want the area of the triangle, which is just &lt;math&gt;[ABC] = \frac{s^{2}\sqrt{3}}{4}&lt;/math&gt;. Thus, we want to know &lt;math&gt;s^{2}&lt;/math&gt;, and then finding the area will be a matter of simple calculation.<br /> <br /> By law of cosines, &lt;math&gt;s^{2} = 3^{2} + 4^{2} - 2 \cdot 3 \cdot 4 \cdot \cos{\theta} = 25 - 24\cos{\theta}&lt;/math&gt;, where &lt;math&gt;\theta = \angle{AOB}&lt;/math&gt;.<br /> <br /> Now, if we plug this value in for &lt;math&gt;s^{2}&lt;/math&gt; in the area formula, we get &lt;math&gt;[ABC] = \frac{25}{4}\sqrt{3} - 6\sqrt{3}\cos{\theta}&lt;/math&gt;.<br /> <br /> Notice that, in order for this expression to be in the required answer form &lt;math&gt;a + \tfrac{b}{c} \sqrt{d}&lt;/math&gt;, &lt;math&gt;\cos{\theta}&lt;/math&gt; must involve &lt;math&gt;\sqrt{3}&lt;/math&gt;. Now, even minimal experience with simple trigonometric functions will instantly make you think of &lt;math&gt;\frac{\pi}{6}&lt;/math&gt; or &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;. The former doesn't work, since &lt;math&gt;\angle{AOB}&lt;/math&gt; is obtuse, so &lt;math&gt;\theta = \frac{5\pi}{6}&lt;/math&gt;.<br /> <br /> Thus, our area must be &lt;math&gt;[ABC] = \frac{25}{4}\sqrt{3} + 9&lt;/math&gt;, and so our answer is &lt;math&gt;\boxed{041}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{NOTE:}&lt;/math&gt;<br /> <br /> When we were using law of cosines, we instantly went to the sub-triangle &lt;math&gt;\bigtriangleup ABO&lt;/math&gt;. Why?<br /> <br /> Notice that &lt;math&gt;\angle{ABO}&lt;/math&gt; is bigger than both &lt;math&gt;\angle{BOC}&lt;/math&gt; and &lt;math&gt;\angle{AOC}&lt;/math&gt; because the sides which meet to form the angle are shorter for &lt;math&gt;\angle{ABO}&lt;/math&gt; than the other two triangles.<br /> <br /> Also notice that the three angles add up to &lt;math&gt;2\pi&lt;/math&gt; radians. One of our angles is &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt; by our reasoning/guessing above. This angle then must be the greatest angle out of the three, since the three angles all must be obtuse, and if one angle was &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;, and another was greater than &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;, then the third angle would be less than &lt;math&gt;\frac{\pi}{3}&lt;/math&gt;, leading to a contradiction, since we know that all of the three angles are obtuse.<br /> <br /> Thus, we know that our greatest angle &lt;math&gt;\angle{ABO} = \frac{5\pi}{6}&lt;/math&gt;. So, we instantly went to the sub-triangle &lt;math&gt;\bigtriangleup ABO&lt;/math&gt; when using law of cosines since the other sub-triangles will get us nowhere.<br /> <br /> ===Solution 4===<br /> &lt;center&gt;&lt;asy&gt;<br /> import olympiad;<br /> import cse5;<br /> import graph;<br /> <br /> dotfactor = 2;<br /> unitsize(0.3inch);<br /> <br /> pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4);<br /> pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B;<br /> pair X = extension(A,F,D,C);<br /> pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M;<br /> <br /> dot(&quot;$C$&quot;, C, dir(0)); dot(&quot;$A$&quot;, A, dir(90));dot(&quot;$B$&quot;, B, dir(180));<br /> dot(&quot;$D$&quot;, D, NE); dot(&quot;$E$&quot;, E, dir(90));dot(&quot;$F$&quot;, F, dir(270));<br /> dot(&quot;$M$&quot;, M, NE); dot(&quot;$N$&quot;, N, dir(270));dot(&quot;$L$&quot;, L, NW);<br /> dot(&quot;$X$&quot;, X, dir(250));<br /> draw(L--X); draw(M--X); draw(N--X); <br /> <br /> draw(A--B--C--cycle);<br /> draw(A--D--B); draw(B--F--C); draw(A--E--C);<br /> draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed);<br /> draw(L--M--N--cycle);<br /> <br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let's call the circle center &lt;math&gt;X&lt;/math&gt;. It has a distance of 3, 4, 5 to an equilateral triangle &lt;math&gt;LMN&lt;/math&gt;. Consider &lt;math&gt;X&lt;/math&gt;’s pedal triangle &lt;math&gt;ABC&lt;/math&gt;. Since &lt;math&gt;X&lt;/math&gt;’s antipedal triangle is equilateral, &lt;math&gt;X&lt;/math&gt; must be the one of the isogonic centers of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. We’ll take the one inside &lt;math&gt;ABC&lt;/math&gt;, i.e., the Fermat point, because it leads to larger &lt;math&gt;\triangle LMN&lt;/math&gt;. Now we construct the three equilateral triangles &lt;math&gt;ABD&lt;/math&gt;, &lt;math&gt;ACE&lt;/math&gt;, and &lt;math&gt;BCF&lt;/math&gt;, the same way the Fermat point is constructed. Then we have &lt;math&gt;\angle DXE = \angle EXF = \angle FXE = 120&lt;/math&gt;. Since &lt;math&gt;AEMCX&lt;/math&gt; is concyclic with &lt;math&gt;XM&lt;/math&gt;=4 as diameter, we have &lt;math&gt;AC=4\sin(60)&lt;/math&gt;. Similarly, &lt;math&gt;AB=3\sin(60)&lt;/math&gt;, and &lt;math&gt;BC=5\sin(60)&lt;/math&gt;. So &lt;math&gt;\triangle ABC&lt;/math&gt; is a 3-4-5 right triangle with &lt;math&gt;\angle BAC=90&lt;/math&gt;. With some more angle chasing we get <br /> &lt;cmath&gt;\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90&lt;/cmath&gt;<br /> &lt;cmath&gt;\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150&lt;/cmath&gt;<br /> By Law of Cosines, we have <br /> &lt;cmath&gt;LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3&lt;/cmath&gt;<br /> And the area follows.<br /> &lt;cmath&gt;[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.&lt;/cmath&gt; <br /> By Mathdummy<br /> <br /> ===Solution 5===<br /> Let &lt;math&gt;ABC&lt;/math&gt; be the equilateral triangle with &lt;math&gt;AB=BC=CA=x.&lt;/math&gt; Assume the coordinates of the vertices are &lt;math&gt;A(-\dfrac{x}{2},0), B(\dfrac{x}{2},0)&lt;/math&gt; and &lt;math&gt;C(0,\dfrac{\sqrt{3}}{2}x).&lt;/math&gt; Let &lt;math&gt;P(a,b)&lt;/math&gt; be such that &lt;math&gt;PA=3, PB=4&lt;/math&gt; and &lt;math&gt;PC=5.&lt;/math&gt; Then<br /> &lt;cmath&gt;\left (a+\dfrac{x}{2}\right )^2 +b^2=3^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;\left (a-\dfrac{x}{2}\right )^2 +b^2=4^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+\left (b-\dfrac{\sqrt{3}}{2}x\right )^2 =5^2.&lt;/cmath&gt;<br /> Subtraction and addition of the first two equations yield &lt;math&gt;2ax=-7, 2a^2+\dfrac{1}{2}x^2+2b^2=25.&lt;/math&gt; The third equation gives &lt;math&gt;2a^2+2b^2-2\sqrt{3}bx+\dfrac{3}{2}x^2=50.&lt;/math&gt; Then &lt;math&gt;x^2-2\sqrt{3}bx=25.&lt;/math&gt; We can then solve for &lt;math&gt;a, b&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt; and have a substitution. We have <br /> &lt;cmath&gt;2\left (\dfrac{-7}{2x}\right )^2 + \dfrac{1}{2}x^2 + 2\left (\dfrac{x^2-25}{2\sqrt{3}x}\right )^2 =25.&lt;/cmath&gt;<br /> Simplify it we have a quadratic equation for &lt;math&gt;x^2: x^4-50x^2+193=0.&lt;/math&gt; So &lt;math&gt;x^2=25\pm 12\sqrt{3}.&lt;/math&gt; The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.<br /> <br /> -JZ<br /> <br /> ==See also==<br /> {{AIME box|year=2012|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=117665 2001 AIME I Problems/Problem 12 2020-02-11T18:17:15Z <p>Jzhao000: </p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> == See also ==<br /> *&lt;url&gt;viewtopic.php?p=384205#384205 Discussion on AoPS&lt;/url&gt;<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_10&diff=116764 2000 AIME I Problems/Problem 10 2020-02-03T20:40:25Z <p>Jzhao000: </p> <hr /> <div>== Problem ==<br /> A [[sequence]] of numbers &lt;math&gt;x_{1},x_{2},x_{3},\ldots,x_{100}&lt;/math&gt; has the property that, for every [[integer]] &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,&lt;/math&gt; inclusive, the number &lt;math&gt;x_{k}&lt;/math&gt; is &lt;math&gt;k&lt;/math&gt; less than the sum of the other &lt;math&gt;99&lt;/math&gt; numbers. Given that &lt;math&gt;x_{50} = m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Let the sum of all of the terms in the sequence be &lt;math&gt;\mathbb{S}&lt;/math&gt;. Then for each integer &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k&lt;/math&gt;. Summing this up for all &lt;math&gt;k&lt;/math&gt; from &lt;math&gt;1, 2, \ldots, 100&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &amp;= 1 + 2 + \cdots + 100\\<br /> 100\mathbb{S} - 2\mathbb{S} &amp;= \frac{100 \cdot 101}{2} = 5050\\<br /> \mathbb{S}&amp;=\frac{2525}{49}\end{align*}&lt;/cmath&gt;<br /> <br /> Now, substituting for &lt;math&gt;x_{50}&lt;/math&gt;, we get &lt;math&gt;2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}&lt;/math&gt;, and the answer is &lt;math&gt;75+98=\boxed{173}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Consider &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;x_{k+1}&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the sum of the rest 98 terms. Then &lt;math&gt;x_k+k=S+x_{k+1}&lt;/math&gt; and &lt;math&gt;x_{k+1}+(k+1)=S+x_k.&lt;/math&gt; Eliminating &lt;math&gt;S&lt;/math&gt; we have &lt;math&gt;x_{k+1}-x_k=-\dfrac{1}{2}.&lt;/math&gt; So the sequence is arithmetic with common difference &lt;math&gt;-\dfrac{1}{2}.&lt;/math&gt; <br /> <br /> In terms of &lt;math&gt;x_{50},&lt;/math&gt; the sequence is &lt;math&gt;x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.&lt;/math&gt; Therefore &lt;math&gt;x_{50}+50=99x_{50}-\dfrac{50}{2}&lt;/math&gt;. We are done by solving for &lt;math&gt;x_{50}&lt;/math&gt;.<br /> <br /> -JZ<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1973_USAMO_Problems/Problem_4&diff=94661 1973 USAMO Problems/Problem 4 2018-05-21T19:16:19Z <p>Jzhao000: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Determine all the [[root]]s, [[real]] or [[complex]], of the system of simultaneous [[equation]]s<br /> &lt;center&gt;&lt;math&gt;x+y+z=3&lt;/math&gt;,<br /> &lt;math&gt;x^2+y^2+z^2=3&lt;/math&gt;,<br /> &lt;math&gt;x^3+y^3+z^3=3&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; be the [[root]]s of the [[cubic polynomial]] &lt;math&gt;t^3+at^2+bt+c&lt;/math&gt;. Let &lt;math&gt;S_1=x+y+z=3&lt;/math&gt;, &lt;math&gt;S_2=x^2+y^2+z^2=3&lt;/math&gt;, and &lt;math&gt;S_3=x^3+y^3+z^3=3&lt;/math&gt;. From this, &lt;math&gt;S_1+a=0&lt;/math&gt;, &lt;math&gt;S_2+aS_1+2b=0&lt;/math&gt;, and &lt;math&gt;S_3+aS_2+bS_1+3c=0&lt;/math&gt;. Solving each of these, &lt;math&gt;a=-3&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, and &lt;math&gt;c=-1&lt;/math&gt;. Thus &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are the roots of the polynomial &lt;math&gt;t^3-3t^2+3t-1=(t-1)^3&lt;/math&gt;. Thus &lt;math&gt;x=y=z=1&lt;/math&gt;, and there are no other solutions.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P(t)=t^3-at^2+bt-c&lt;/math&gt; have roots x, y, and z. Then &lt;cmath&gt;0=P(x)+P(y)+P(z)=3-3a+3b-3c&lt;/cmath&gt; using our system of equations, so &lt;math&gt;P(1)=0&lt;/math&gt;. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let &lt;math&gt;x=1&lt;/math&gt;. Then we can use the system of equations to find that &lt;math&gt;y=z=1&lt;/math&gt; as well, and so &lt;math&gt;\boxed{(1,1,1)}&lt;/math&gt; is the only solution to the system of equations.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a=x-1,&lt;/math&gt; &lt;math&gt;b=y-1&lt;/math&gt; and &lt;math&gt;c=z-1.&lt;/math&gt; Then <br /> &lt;cmath&gt;a+b+c=0,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+b^2+c^2=0,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^3+b^3+c^3=0.&lt;/cmath&gt;<br /> We have<br /> &lt;cmath&gt;\begin{align*}<br /> 0&amp;=(a+b+c)^3\\<br /> &amp;=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\<br /> &amp;=0-3a^3-3b^3-3c^3+6abc\\<br /> &amp;=6abc.<br /> \end{align*}&lt;/cmath&gt;<br /> Then one of &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; has to be 0, and easy to prove the other two are also 0. So &lt;math&gt;\boxed{(1,1,1)}&lt;/math&gt; is the only solution to the system of equations.<br /> <br /> J.Z.<br /> <br /> {{alternate solutions}}<br /> <br /> ==See Also==<br /> [[Newton's Sums]]<br /> {{USAMO box|year=1973|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1973_USAMO_Problems/Problem_4&diff=94619 1973 USAMO Problems/Problem 4 2018-05-18T21:48:18Z <p>Jzhao000: </p> <hr /> <div>==Problem==<br /> Determine all the [[root]]s, [[real]] or [[complex]], of the system of simultaneous [[equation]]s<br /> &lt;center&gt;&lt;math&gt;x+y+z=3&lt;/math&gt;,<br /> &lt;math&gt;x^2+y^2+z^2=3&lt;/math&gt;,<br /> &lt;math&gt;x^3+y^3+z^3=3&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; be the [[root]]s of the [[cubic polynomial]] &lt;math&gt;t^3+at^2+bt+c&lt;/math&gt;. Let &lt;math&gt;S_1=x+y+z=3&lt;/math&gt;, &lt;math&gt;S_2=x^2+y^2+z^2=3&lt;/math&gt;, and &lt;math&gt;S_3=x^3+y^3+z^3=3&lt;/math&gt;. From this, &lt;math&gt;S_1+a=0&lt;/math&gt;, &lt;math&gt;S_2+aS_1+2b=0&lt;/math&gt;, and &lt;math&gt;S_3+aS_2+bS_1+3c=0&lt;/math&gt;. Solving each of these, &lt;math&gt;a=-3&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, and &lt;math&gt;c=-1&lt;/math&gt;. Thus &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are the roots of the polynomial &lt;math&gt;t^3-3t^2+3t-1=(t-1)^3&lt;/math&gt;. Thus &lt;math&gt;x=y=z=1&lt;/math&gt;, and there are no other solutions.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P(t)=t^3-at^2+bt-c&lt;/math&gt; have roots x, y, and z. Then &lt;cmath&gt;0=P(x)+P(y)+P(z)=3-3a+3b-3c&lt;/cmath&gt; using our system of equations, so &lt;math&gt;P(1)=0&lt;/math&gt;. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let &lt;math&gt;x=1&lt;/math&gt;. Then we can use the system of equations to find that &lt;math&gt;y=z=1&lt;/math&gt; as well, and so &lt;math&gt;\boxed{(1,1,1)}&lt;/math&gt; is the only solution to the system of equations.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a=x-1,&lt;/math&gt; &lt;math&gt;b=y-1&lt;/math&gt; and &lt;math&gt;c=z-1.&lt;/math&gt; Then <br /> &lt;cmath&gt;a+b+c=0,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+b^2+c^2=0,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^3+b^3+c^3=0.&lt;/cmath&gt;<br /> We have<br /> &lt;cmath&gt;\begin{align*}<br /> 0&amp;=(a+b+c)^3\\<br /> &amp;=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\<br /> &amp;=0-3a^3-3b^3-3c^3+6abc\\<br /> &amp;=6abc.<br /> \end{align*}&lt;/cmath&gt;<br /> Then one of &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; has to be 0, and easy to prove the other two are also 0. So &lt;math&gt;\boxed{(1,1,1)}&lt;/math&gt; is the only solution to the system of equations.<br /> <br /> J.Z.<br /> <br /> {{alternate solutions}}<br /> <br /> ==See Also==<br /> [[Newton's Sums]]<br /> {{USAMO box|year=1973|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1987_USAMO_Problems/Problem_3&diff=94617 1987 USAMO Problems/Problem 3 2018-05-18T20:49:48Z <p>Jzhao000: /* Solution */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;X&lt;/math&gt; is the smallest set of polynomials &lt;math&gt;p(x)&lt;/math&gt; such that: <br /> <br /> : 1. &lt;math&gt;p(x) = x&lt;/math&gt; belongs to &lt;math&gt;X&lt;/math&gt;.<br /> : 2. If &lt;math&gt;r(x)&lt;/math&gt; belongs to &lt;math&gt;X&lt;/math&gt;, then &lt;math&gt;x\cdot r(x)&lt;/math&gt; and &lt;math&gt;(x + (1 - x) \cdot r(x) )&lt;/math&gt; both belong to &lt;math&gt;X&lt;/math&gt;. <br /> <br /> Show that if &lt;math&gt;r(x)&lt;/math&gt; and &lt;math&gt;s(x)&lt;/math&gt; are distinct elements of &lt;math&gt;X&lt;/math&gt;, then &lt;math&gt;r(x) \neq s(x)&lt;/math&gt; for any &lt;math&gt;0 &lt; x &lt; 1&lt;/math&gt;. <br /> <br /> ==Solution==<br /> Let &lt;math&gt;s(x)&lt;/math&gt; be an arbitrary polynomial in &lt;math&gt;X.&lt;/math&gt; Then &lt;math&gt;0&lt;s(x)&lt;1&lt;/math&gt; when &lt;math&gt;0&lt;x&lt;1.&lt;/math&gt; Define &lt;math&gt;X_1=\{s(x)\in X:s(x)=x\cdot s_1(x)&lt;/math&gt; for some &lt;math&gt;s_1(x)\in X\},&lt;/math&gt; and &lt;math&gt;X_2=\{t(x)\in X: t(x)=x+(1-x)t_1(x)&lt;/math&gt; for some &lt;math&gt;t_1(x)\in X\}.&lt;/math&gt;<br /> <br /> If &lt;math&gt;s(x) \in X_1&lt;/math&gt; and &lt;math&gt;t(x)\in X_2,&lt;/math&gt; we have &lt;math&gt;s(x) &lt;x&lt;t(x)&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; with &lt;math&gt;0&lt;x&lt;1.&lt;/math&gt; Therefore &lt;math&gt;s(x)\ne t(x)&lt;/math&gt; for any &lt;math&gt;0&lt;x&lt;1.&lt;/math&gt;<br /> <br /> For any &lt;math&gt;s(x), t(x) \in X_1&lt;/math&gt;, Let &lt;math&gt;s(x)=x\cdot s_1(x)&lt;/math&gt; and &lt;math&gt;t(x)=x\cdot t_1(x)&lt;/math&gt; for &lt;math&gt;s_1(x), t_1(x) \in X.&lt;/math&gt; If &lt;math&gt;s_1(x) \ne t_1(x)&lt;/math&gt; for &lt;math&gt;0&lt;x&lt;1,&lt;/math&gt; then<br /> &lt;math&gt;s(x)-t(x)=x(s_1(x)-t_1(x))\ne 0&lt;/math&gt; for &lt;math&gt;0&lt;x&lt;1.&lt;/math&gt;<br /> <br /> Similarly, for any &lt;math&gt;s(x), t(x) \in X_2&lt;/math&gt;, Let &lt;math&gt;s(x)=x+(1-x) s_1(x)&lt;/math&gt; and &lt;math&gt;t(x)=x+(1-x) t_1(x)&lt;/math&gt; for &lt;math&gt;s_1(x), t_1(x) \in X.&lt;/math&gt; If &lt;math&gt;s_1(x) \ne t_1(x)&lt;/math&gt; for &lt;math&gt;0&lt;x&lt;1,&lt;/math&gt; then<br /> &lt;math&gt;s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\ne 0&lt;/math&gt; for &lt;math&gt;0&lt;x&lt;1.&lt;/math&gt;<br /> <br /> The proof is done by an induction.<br /> <br /> J.Z.<br /> <br /> ==See Also==<br /> {{USAMO box|year=1987|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1985_USAMO_Problems/Problem_2&diff=94593 1985 USAMO Problems/Problem 2 2018-05-18T12:32:49Z <p>Jzhao000: /* Solution */</p> <hr /> <div>== Problem ==<br /> Determine each real root of<br /> <br /> &lt;math&gt;x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0&lt;/math&gt;<br /> <br /> correct to four decimal places.<br /> <br /> ==Solution==<br /> The equation can be re-written as <br /> &lt;cmath&gt;\begin{align}\label{eqn1}<br /> (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0.<br /> \end{align}&lt;/cmath&gt;<br /> <br /> We first prove that the equation has no negative roots.<br /> Let &lt;math&gt;x\le 0.&lt;/math&gt; The equation above can be further re-arranged as<br /> &lt;cmath&gt;\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*}&lt;/cmath&gt;<br /> The right hand side of the equation is negative. Therefore &lt;cmath&gt;[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]&lt;0,&lt;/cmath&gt; and we have<br /> &lt;math&gt;-1&lt;(x+10^5)(x-10^5) &lt;2.&lt;/math&gt; Then the left hand side of the equation is bounded by<br /> &lt;cmath&gt;|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3.&lt;/cmath&gt;<br /> However, since &lt;math&gt;|(x+10^5)(x-10^5)|\le 2&lt;/math&gt; and &lt;math&gt;x&lt;0,&lt;/math&gt; it follows that &lt;math&gt;|x+10^5| &lt;\frac{2}{|x-10^5|}&lt;2\times 10^{-5}&lt;/math&gt; for negative &lt;math&gt;x.&lt;/math&gt; Then &lt;math&gt;x&lt;2\times 10^{-5}-10^5.&lt;/math&gt; The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.<br /> <br /> Now let &lt;math&gt;x&gt;0.&lt;/math&gt; When &lt;math&gt;x=10^5,&lt;/math&gt; the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of &lt;math&gt;10^5&lt;/math&gt;, as its leading coefficient is positive. We will prove that &lt;math&gt;x=10^5&lt;/math&gt; is a good approximation of the roots (within &lt;math&gt;10^{-2}&lt;/math&gt;). In fact, we can solve the &quot;quadratic&quot; equation (1) for &lt;math&gt;(x+10^5)(x-10^5)&lt;/math&gt;:<br /> &lt;cmath&gt;(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}.&lt;/cmath&gt;<br /> Then &lt;cmath&gt;x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}.&lt;/cmath&gt;<br /> Easy to see that<br /> &lt;math&gt;|x-10^5| &lt;1&lt;/math&gt; for positve &lt;math&gt;x.&lt;/math&gt; Therefore, &lt;math&gt;10^5-1&lt;x&lt;10^5+1.&lt;/math&gt; Then<br /> &lt;cmath&gt;\begin{align*}<br /> |x-10^5|&amp;=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\<br /> &amp;\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\<br /> &amp;\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\<br /> &amp;&lt;10^{-2}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;x_1&lt;/math&gt; be a root of the equation with &lt;math&gt;x_1&lt;10^5.&lt;/math&gt; Then &lt;math&gt;0&lt;10^5-x_1&lt;10^{-2}&lt;/math&gt; and <br /> &lt;cmath&gt;x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.&lt;/cmath&gt;<br /> An aproximation of &lt;math&gt;x_1&lt;/math&gt; is defined as follows:<br /> &lt;cmath&gt;\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.&lt;/cmath&gt;<br /> We check the error of the estimate:<br /> &lt;cmath&gt;\begin{align*}<br /> |\tilde{x}_1-x_1|&amp;=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\<br /> &amp;\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> The first absolute value <br /> &lt;cmath&gt; \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}&lt;10^{-12}.&lt;/cmath&gt;<br /> <br /> The second absolute value <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}<br /> \right |\\<br /> &amp;\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\<br /> &amp;\le 10^{-7}+10^{-9},<br /> \end{align*}&lt;/cmath&gt;<br /> through a rationalized numerator.Therefore &lt;math&gt;|\tilde{x}_1-x_1|\le 10^{-6}.&lt;/math&gt;<br /> <br /> For a real root &lt;math&gt;x_2&lt;/math&gt; with &lt;math&gt;x_2&gt;10^5,&lt;/math&gt; we choose <br /> &lt;cmath&gt;\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.&lt;/cmath&gt;<br /> We can similarly prove it has the desired approximation.<br /> <br /> ==Notes==<br /> <br /> Another round of iteration can increase the accuracy to more than 10 decimal places:<br /> &lt;cmath&gt;\begin{align*}<br /> \tilde{x}_1^\prime=10^5+\frac{1-\sqrt{1+4(\tilde{x}_1+1)}}{2(\tilde{x}_1+10^5)},\\<br /> \tilde{x}_2^\prime=10^5+\frac{1+\sqrt{1+4(\tilde{x}_2+1)}}{2(\tilde{x}_2+10^5)}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> J.Z.<br /> <br /> == See Also ==<br /> {{USAMO box|year=1985|num-b=1|num-a=3}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1985_USAMO_Problems/Problem_2&diff=94592 1985 USAMO Problems/Problem 2 2018-05-18T11:38:50Z <p>Jzhao000: New solution</p> <hr /> <div>== Problem ==<br /> Determine each real root of<br /> <br /> &lt;math&gt;x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0&lt;/math&gt;<br /> <br /> correct to four decimal places.<br /> <br /> ==Solution==<br /> The equation can be re-written as <br /> &lt;cmath&gt;\begin{align}\label{eqn1}<br /> (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0.<br /> \end{align}&lt;/cmath&gt;<br /> <br /> We first prove that the equation has no negative roots.<br /> Let &lt;math&gt;x\le 0.&lt;/math&gt; The equation above can be further re-arranged as<br /> &lt;cmath&gt;\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*}&lt;/cmath&gt;<br /> The right hand side of the equation is negative. Therefore &lt;cmath&gt;[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]&lt;0,&lt;/cmath&gt; and we have<br /> &lt;math&gt;-1&lt;(x+10^5)(x-10^5) &lt;2.&lt;/math&gt; Then the left hand side of the equation is bounded by<br /> &lt;cmath&gt;|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3.&lt;/cmath&gt;<br /> However, since &lt;math&gt;|(x+10^5)(x-10^5)|\le 2&lt;/math&gt; and &lt;math&gt;x&lt;0,&lt;/math&gt; it follows that &lt;math&gt;|x+10^5| &lt;\frac{2}{|x-10^5|}&lt;2\times 10^{-5}&lt;/math&gt; for negative &lt;math&gt;x.&lt;/math&gt; Then &lt;math&gt;x&lt;2\times 10^{-5}-10^5.&lt;/math&gt; The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.<br /> <br /> Now let &lt;math&gt;x&gt;0.&lt;/math&gt; When &lt;math&gt;x=10^5,&lt;/math&gt; the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of &lt;math&gt;10^5&lt;/math&gt;, as its leading coefficient is positive. We will prove that &lt;math&gt;x=10^5&lt;/math&gt; is a good approximation of the roots (within &lt;math&gt;10^{-2}&lt;/math&gt;). In fact, we can solve the &quot;quadratic&quot; equation (1) for &lt;math&gt;(x+10^5)(x-10^5)&lt;/math&gt;:<br /> &lt;cmath&gt;(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}.&lt;/cmath&gt;<br /> Then &lt;cmath&gt;x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}.&lt;/cmath&gt;<br /> Easy to see that<br /> &lt;math&gt;|x-10^5| &lt;1&lt;/math&gt; for positve &lt;math&gt;x.&lt;/math&gt; Therefore, &lt;math&gt;10^5-1&lt;x&lt;10^5+1.&lt;/math&gt; Then<br /> &lt;cmath&gt;\begin{align*}<br /> |x-10^5|&amp;=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\<br /> &amp;\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\<br /> &amp;\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\<br /> &amp;&lt;10^{-2}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;x_1&lt;/math&gt; be a root of the equation with &lt;math&gt;x_1&lt;10^5.&lt;/math&gt; Then &lt;math&gt;0&lt;10^5-x_1&lt;10^{-2}&lt;/math&gt; and <br /> &lt;cmath&gt;x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.&lt;/cmath&gt;<br /> An aproximation of &lt;math&gt;x_1&lt;/math&gt; is defined as follows:<br /> &lt;cmath&gt;\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.&lt;/cmath&gt;<br /> We check the error of the estimate:<br /> &lt;cmath&gt;\begin{align*}<br /> |\tilde{x}_1-x_1|&amp;=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\<br /> &amp;\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> The first absolute value <br /> &lt;cmath&gt; \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}&lt;10^{-12}.&lt;/cmath&gt;<br /> <br /> The second absolute value <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}<br /> \right |\\<br /> &amp;\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\<br /> &amp;\le 10^{-7}+10^{-9},<br /> \end{align*}&lt;/cmath&gt;<br /> through a rationalized numerator.Therefore &lt;math&gt;|\tilde{x}_1-x_1|\le 10^{-6}.&lt;/math&gt;<br /> <br /> For a real root &lt;math&gt;x_2&lt;/math&gt; with &lt;math&gt;x_2&gt;10^5,&lt;/math&gt; we choose <br /> &lt;cmath&gt;\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.&lt;/cmath&gt;<br /> We can similarly prove it has the desired approximation.<br /> <br /> == See Also ==<br /> {{USAMO box|year=1985|num-b=1|num-a=3}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1978_USAMO_Problems/Problem_1&diff=94520 1978 USAMO Problems/Problem 1 2018-05-16T20:39:15Z <p>Jzhao000: </p> <hr /> <div>== Problem ==<br /> Given that &lt;math&gt;a,b,c,d,e&lt;/math&gt; are real numbers such that<br /> <br /> &lt;math&gt;a+b+c+d+e=8&lt;/math&gt;,<br /> <br /> &lt;math&gt;a^2+b^2+c^2+d^2+e^2=16&lt;/math&gt;.<br /> <br /> Determine the maximum value of &lt;math&gt;e&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> By Cauchy Schwarz, we can see that &lt;math&gt;(1+1+1+1)(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2&lt;/math&gt;<br /> thus &lt;math&gt;4(16-e^2)\geq (8-e)^2&lt;/math&gt; <br /> Finally, &lt;math&gt;e(5e-16) \geq 0&lt;/math&gt; which means &lt;math&gt;\frac{16}{5} \geq e \geq 0&lt;/math&gt;<br /> so the maximum value of &lt;math&gt;e&lt;/math&gt; is &lt;math&gt;\frac{16}{5}&lt;/math&gt;.<br /> <br /> '''from:''' [http://image.ohozaa.com/view2/vUGiXdRQdAPyw036 Image from Gon Mathcenter.net]<br /> <br /> == Solution 2==<br /> Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem.<br /> We get the following equations:<br /> <br /> &lt;math&gt;(1)\hspace*{0.5cm} a+b+c+d+e=8\\<br /> (2)\hspace*{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\\<br /> (3)\hspace*{0.5cm} 0=\lambda+2a\mu\\<br /> (4)\hspace*{0.5cm} 0=\lambda+2b\mu\\<br /> (5)\hspace*{0.5cm} 0=\lambda+2c\mu\\<br /> (6)\hspace*{0.5cm} 0=\lambda+2d\mu\\<br /> (7)\hspace*{0.5cm} 1=\lambda+2e\mu&lt;/math&gt;<br /> <br /> If &lt;math&gt;\mu=0&lt;/math&gt;, then &lt;math&gt;\lambda=0&lt;/math&gt; according to &lt;math&gt;(6)&lt;/math&gt; and &lt;math&gt;\lambda=1&lt;/math&gt; according to &lt;math&gt;(7)&lt;/math&gt;, so &lt;math&gt;\mu \neq 0&lt;/math&gt;. Setting the right sides of &lt;math&gt;(3)&lt;/math&gt; and &lt;math&gt;(4)&lt;/math&gt; equal yields &lt;math&gt;\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b&lt;/math&gt;. Similar steps yield that &lt;math&gt;a=b=c=d&lt;/math&gt;. Thus, &lt;math&gt;(1)&lt;/math&gt; becomes &lt;math&gt;4d+e=8&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt; becomes &lt;math&gt;4d^{2}+e^{2}=16&lt;/math&gt;. Solving the system yields &lt;math&gt;e=0,\frac{16}{5}&lt;/math&gt;, so the maximum possible value of &lt;math&gt;e&lt;/math&gt; is &lt;math&gt;\frac{16}{5}&lt;/math&gt;.<br /> <br /> == Solution 3==<br /> A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have<br /> &lt;cmath&gt;(a+b+c+d)^2=(8-e)^2,&lt;/cmath&gt; and <br /> &lt;cmath&gt;a^2+b^2+c^2+d^2=16-e^2.&lt;/cmath&gt;<br /> The second equation times 4, then minus the first equation, <br /> &lt;cmath&gt;(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=4(16-e^2)-(8-e)^2.&lt;/cmath&gt;<br /> The rest follows.<br /> <br /> J.Z.<br /> == See Also ==<br /> {{USAMO box|year=1978|before=First Question|num-a=2}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=1974_USAMO_Problems/Problem_2&diff=94519 1974 USAMO Problems/Problem 2 2018-05-16T18:57:21Z <p>Jzhao000: </p> <hr /> <div>==Problem==<br /> Prove that if &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive real numbers, then<br /> &lt;center&gt;&lt;math&gt;a^ab^bc^c\ge (abc)^{(a+b+c)/3}&lt;/math&gt;&lt;/center&gt;<br /> <br /> ==Solution 1==<br /> Consider the function &lt;math&gt;f(x)=x\ln{x}&lt;/math&gt;. &lt;math&gt;f''(x)=\frac{1}{x}&gt;0&lt;/math&gt; for &lt;math&gt;x&gt;0&lt;/math&gt;; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:<br /> &lt;center&gt;&lt;math&gt;\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)&lt;/math&gt;&lt;/center&gt;<br /> Apply [[AM-GM]] to get<br /> &lt;center&gt;&lt;math&gt;\frac{a+b+c}{3}\ge \sqrt{abc}&lt;/math&gt;&lt;/center&gt;<br /> which implies<br /> &lt;center&gt;&lt;math&gt;\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt{abc}\right)&lt;/math&gt;&lt;/center&gt;<br /> Rearranging,<br /> &lt;center&gt;&lt;math&gt;a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)&lt;/math&gt;&lt;/center&gt;<br /> Because &lt;math&gt;f(x) = e^x&lt;/math&gt; is an increasing function, we can conclude that:<br /> &lt;center&gt;&lt;math&gt;e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}&lt;/math&gt;&lt;/center&gt;<br /> which simplifies to the desired inequality.<br /> <br /> ==Solution 2==<br /> Note that &lt;math&gt;(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3&lt;/math&gt;.<br /> <br /> So if we can prove that &lt;math&gt;a^ab^bc^c\ge a^bb^cc^a&lt;/math&gt; and &lt;math&gt;a^ab^bc^c\ge a^cb^ac^b&lt;/math&gt;, then we are done.<br /> <br /> WLOG let &lt;math&gt;a\ge b\ge c&lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a&lt;/math&gt;. Since &lt;math&gt;\dfrac{c}{a} \le 1&lt;/math&gt;, &lt;math&gt;\dfrac{c}{b} \le 1&lt;/math&gt;, &lt;math&gt;a-b \ge 0&lt;/math&gt;, and &lt;math&gt;b-c \ge 0&lt;/math&gt;, it follows that &lt;math&gt;a^ab^bc^c \ge a^bb^cc^a&lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b&lt;/math&gt;. Since &lt;math&gt;\dfrac{b}{a} \le 1&lt;/math&gt;, &lt;math&gt;\dfrac{c}{a} \le 1&lt;/math&gt;, &lt;math&gt;a-b \ge 0&lt;/math&gt;, and &lt;math&gt;b-c \ge 0&lt;/math&gt;, it follows that &lt;math&gt;a^ab^bc^c \ge a^cb^ac^b&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3&lt;/math&gt;, and cube-rooting both sides gives &lt;math&gt;a^ab^bc^c\ge (abc)^{(a+b+c)/3}&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> <br /> WLOG let &lt;math&gt;a\ge b\ge c&lt;/math&gt;. Let &lt;math&gt;b = ax&lt;/math&gt; and &lt;math&gt;c = ay&lt;/math&gt;, where &lt;math&gt;x \ge 1&lt;/math&gt; and &lt;math&gt;y \ge 1&lt;/math&gt;.<br /> <br /> We want to prove that &lt;math&gt;(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}&lt;/math&gt;.<br /> <br /> Simplifying and combining terms on each side, we get &lt;math&gt;a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;a &gt; 0&lt;/math&gt;, we can divide out &lt;math&gt;a^{a + ax + ay}&lt;/math&gt; to get &lt;math&gt;x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}&lt;/math&gt;.<br /> <br /> Take the &lt;math&gt;a&lt;/math&gt;th root of each side and then cube both sides to get &lt;math&gt;x^{3x}y^{3y} \ge (xy)^{1 + x + y}&lt;/math&gt;.<br /> <br /> This simplifies to &lt;math&gt;x^{2x-1}y^{2y-1} \ge x^{y}y^{x}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;2x - 1 \ge x&lt;/math&gt; and &lt;math&gt;2y - 1 \ge y&lt;/math&gt;, we only need to prove &lt;math&gt;x^{x}y^{y} \ge x^{y}y^{x}&lt;/math&gt; for our given &lt;math&gt;x, y&lt;/math&gt;.<br /> <br /> WLOG, let &lt;math&gt;y \ge x&lt;/math&gt; and &lt;math&gt; y =kx&lt;/math&gt; for &lt;math&gt;k \ge 1&lt;/math&gt;. Then our expression becomes <br /> <br /> &lt;math&gt;x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}&lt;/math&gt;<br /> <br /> &lt;math&gt;k^{xk} \ge k^{x}&lt;/math&gt;<br /> <br /> &lt;math&gt;k^k \ge k&lt;/math&gt;<br /> <br /> This is clearly true for &lt;math&gt;k \ge 1&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> WLOG let &lt;math&gt;a\ge b\ge c&lt;/math&gt;. Then sequence &lt;math&gt;(a,b,c)&lt;/math&gt; majorizes &lt;math&gt;(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})&lt;/math&gt;. Thus by Muirhead's Inequality, we have &lt;math&gt;\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}&lt;/math&gt;, so &lt;math&gt;a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;x=\frac{a}{\sqrt{abc}},&lt;/math&gt; &lt;math&gt;y=\frac{b}{\sqrt{abc}}&lt;/math&gt; and &lt;math&gt;z=\frac{c}{\sqrt{abc}}.&lt;/math&gt; Then &lt;math&gt;xyz=1&lt;/math&gt; and a straightforward calculation reduces the problem to <br /> &lt;cmath&gt;x^xy^yz^z \ge 1.&lt;/cmath&gt;<br /> WLOG, assume &lt;math&gt;x\ge y\ge z.&lt;/math&gt; Then &lt;math&gt;x\ge 1,&lt;/math&gt; &lt;math&gt;z\le 1&lt;/math&gt; and &lt;math&gt;xy=\frac{1}{z} \ge 1.&lt;/math&gt; Therefore,<br /> &lt;cmath&gt; x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.&lt;/cmath&gt;<br /> <br /> J.Z.<br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> {{USAMO box|year=1974|num-b=1|num-a=3}}<br /> *[http://www.mathlinks.ro/viewtopic.php?t=102633 Simple Olympiad Inequality]<br /> *[http://www.mathlinks.ro/viewtopic.php?t=98846 Hard inequality]<br /> *[http://www.mathlinks.ro/viewtopic.php?t=85663 Inequality]<br /> *[http://www.mathlinks.ro/Forum/viewtopic.php?t=82706 Some q's on usamo write ups]<br /> *[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq]<br /> *[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)]<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Jzhao000 https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_4&diff=94514 2013 USAMO Problems/Problem 4 2018-05-16T15:55:43Z <p>Jzhao000: </p> <hr /> <div>Find all real numbers &lt;math&gt;x,y,z\geq 1&lt;/math&gt; satisfying &lt;cmath&gt;\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.&lt;/cmath&gt;<br /> <br /> <br /> == Solution (Cauchy or AM-GM) ==<br /> The key Lemma is:<br /> &lt;cmath&gt;\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}&lt;/cmath&gt; for all &lt;math&gt;a,b \ge 1&lt;/math&gt;. Equality holds when &lt;math&gt;(a-1)(b-1)=1&lt;/math&gt;.<br /> <br /> This is proven easily.<br /> &lt;cmath&gt;\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}&lt;/cmath&gt; by Cauchy.<br /> Equality then holds when &lt;math&gt;a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1&lt;/math&gt;.<br /> <br /> Now assume that &lt;math&gt;x = \min(x,y,z)&lt;/math&gt;. Now note that, by the Lemma,<br /> <br /> &lt;cmath&gt;\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}&lt;/cmath&gt;. So equality must hold.<br /> So &lt;math&gt;(y-1)(z-1) = 1&lt;/math&gt; and &lt;math&gt;(x-1)(yz) = 1&lt;/math&gt;. If we let &lt;math&gt;z = c&lt;/math&gt;, then we can easily compute that &lt;math&gt;y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}&lt;/math&gt;.<br /> Now it remains to check that &lt;math&gt;x \le y, z&lt;/math&gt;.<br /> <br /> But by easy computations, &lt;math&gt;x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0&lt;/math&gt;, which is obvious.<br /> Also &lt;math&gt;x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1&lt;/math&gt;, which is obvious, since &lt;math&gt;c \ge 1&lt;/math&gt;.<br /> <br /> So all solutions are of the form &lt;math&gt;\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}&lt;/math&gt;, and all permutations for &lt;math&gt;c &gt; 1&lt;/math&gt;.<br /> <br /> '''Remark:''' An alternative proof of the key Lemma is the following:<br /> By AM-GM, &lt;cmath&gt;(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}&lt;/cmath&gt;. Now taking the square root of both sides gives the desired. Equality holds when &lt;math&gt;(a-1)(b-1) = 1&lt;/math&gt;.<br /> ==Solution with Thought Process==<br /> Without loss of generality, let &lt;math&gt;1 \le x \le y \le z&lt;/math&gt;. Then &lt;math&gt;\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}&lt;/math&gt;.<br /> <br /> Suppose x = y = z. Then &lt;math&gt;\sqrt{x + x^3} = 3\sqrt{x-1}&lt;/math&gt;, so &lt;math&gt;x + x^3 = 9x - 9&lt;/math&gt;. It is easily verified that &lt;math&gt;x^3 - 8x + 9 = 0&lt;/math&gt; has no solution in positive numbers greater than 1. Thus, &lt;math&gt;\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}&lt;/math&gt; for x = y = z. We suspect if the inequality always holds.<br /> <br /> Let x = 1. Then we have &lt;math&gt;\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}&lt;/math&gt;, which simplifies to &lt;cmath&gt;1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}&lt;/cmath&gt; and hence &lt;cmath&gt;yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}&lt;/cmath&gt; Let us try a few examples: if y = z = 2, we have &lt;math&gt;3 &gt; 2&lt;/math&gt;; if y = z, we have &lt;math&gt;y^2 - 2y + 3 \ge 2(y-1)&lt;/math&gt;, which reduces to &lt;math&gt;y^2 - 4y + 5 \ge 0&lt;/math&gt;. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let &lt;math&gt;u = \sqrt{(y-1)(z-1)}&lt;/math&gt;! Thus, &lt;cmath&gt;u^2 - 2u + 2 = (u-1)^2 + 1 &gt; 0&lt;/cmath&gt; and the claim holds for x = 1.<br /> <br /> If x &gt; 1, we see the &lt;math&gt;\sqrt{x - 1}&lt;/math&gt; will provide a huge obstacle when squaring. But, using the identity &lt;math&gt;(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz&lt;/math&gt;:<br /> &lt;cmath&gt;x + xyz \ge x - 1 + y - 1 + z - 1 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(y-1)}&lt;/cmath&gt;<br /> which leads to<br /> &lt;cmath&gt;xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}&lt;/cmath&gt;<br /> Again, we experiment. If x = 2, y = 3, and z = 3, then &lt;math&gt;18 &gt; 7 + 4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, we see the finish: setting &lt;math&gt;u = \sqrt{x-1}&lt;/math&gt; gives &lt;math&gt;x = u^2 + 1&lt;/math&gt;. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations:<br /> &lt;cmath&gt;u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0&lt;/cmath&gt;<br /> <br /> Because the coefficient of &lt;math&gt;u^2&lt;/math&gt; is positive, all we need to do is to verify that the discriminant is nonpositive:<br /> &lt;cmath&gt;b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8\sqrt{(y-1)(z-1)})&lt;/cmath&gt;<br /> <br /> Let us try a few examples. If y = z, then the discriminant D = &lt;math&gt;8(y-1) - 8(y-1) - yz(12 + 4y^2 - 8y - 8(y-1)) = -yz(4y^2 - 16y + 20) = -4yz(y^2 - 4y + 5) &lt; 0&lt;/math&gt;.<br /> <br /> We are almost done, but we need to find the correct argument. (How frustrating!)<br /> Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.<br /> <br /> --Thinking Process by suli<br /> <br /> == Solution 2 ==<br /> WLOG, assume that &lt;math&gt;x = \min(x,y,z)&lt;/math&gt;. Let &lt;math&gt;a=\sqrt{x-1},&lt;/math&gt; &lt;math&gt;b=\sqrt{y-1}&lt;/math&gt; and &lt;math&gt;c=\sqrt{z-1}&lt;/math&gt;. Then &lt;math&gt;x=a^2+1&lt;/math&gt;, &lt;math&gt;y=b^2+1&lt;/math&gt; and &lt;math&gt;z=c^2+1&lt;/math&gt;. The equation becomes<br /> &lt;cmath&gt;(a^2+1)+(a^2+1)(b^2+1)(c^2+1)=(a+b+c)^2.&lt;/cmath&gt;<br /> Rearranging the terms, we have <br /> &lt;cmath&gt;(1+a^2)(bc-1)^2+[a(b+c)-1]^2=0.&lt;/cmath&gt;<br /> Therefore &lt;math&gt;bc=1&lt;/math&gt; and &lt;math&gt;a(b+c)=1.&lt;/math&gt; Express &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; in terms of &lt;math&gt;c&lt;/math&gt;, we have &lt;math&gt;a=\frac{c}{c^2+1}&lt;/math&gt; and &lt;math&gt;b=\frac{1}{c}.&lt;/math&gt; Easy to check that &lt;math&gt;a&lt;/math&gt; is the smallest among &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c.&lt;/math&gt; Then &lt;math&gt;x=\frac{c^4+3c^2+1}{(c^2+1)^2}&lt;/math&gt;, &lt;math&gt;y=\frac{c^2+1}{c^2}&lt;/math&gt; and &lt;math&gt;z=c^2+1.&lt;/math&gt;<br /> Let &lt;math&gt;c^2=t&lt;/math&gt;, we have the solutions for &lt;math&gt;(x,y,z)&lt;/math&gt; as follows:<br /> &lt;math&gt;(\frac{t^2+3t+1}{(t+1)^2}, \frac{t+1}{t}, t+1)&lt;/math&gt; and permutations for all &lt;math&gt;t&gt;0.&lt;/math&gt; <br /> <br /> --J.Z.<br /> <br /> {{MAA Notice}}</div> Jzhao000