https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Nli&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-22T11:13:40Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=113642 2012 AMC 12B Problems/Problem 23 2019-12-29T02:22:14Z <p>Nli: /* Solution 3 */</p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (dubious) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> == Solution 2 ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> First, notice that &lt;math&gt;z=1&lt;/math&gt; cannot be a root of the polynomial because &lt;math&gt;a,b,c,d \geq 0&lt;/math&gt;. Multiplying the polynomial by &lt;math&gt;(z-1)&lt;/math&gt; yields &lt;math&gt;P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d&lt;/math&gt;, so for &lt;math&gt;z \neq 1&lt;/math&gt; to be a root of &lt;math&gt;P(z)&lt;/math&gt;, &lt;math&gt;4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d&lt;/math&gt;. Now we consider the root &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1&lt;/math&gt;. &lt;math&gt;|4z_0^5| = 4&lt;/math&gt;, so the right hand side must have absolute value 4. By the triangle inequality, &lt;math&gt;|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|&lt;/math&gt; &lt;math&gt;\leq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d &lt;/math&gt; &lt;math&gt;= (4-a)+(a-b)+(b-c)+(c-d)+d=4&lt;/math&gt;, with equality if and only if each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, &lt;math&gt;(c-d)z_0&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; is either zero or in the same direction as all the others when looked at as vectors in the complex plane. <br /> <br /> We can now divide into two cases: &lt;math&gt;d \neq 0&lt;/math&gt; and &lt;math&gt;d=0&lt;/math&gt;. If &lt;math&gt;d \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^5&lt;/math&gt; must be real by the previous argument, so &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity. Also, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;(c-d)z_0&lt;/math&gt; must all be zero because if &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a&lt;/math&gt;, &lt;math&gt;c=b&lt;/math&gt;, and &lt;math&gt;d=c&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,4)&lt;/math&gt;. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.<br /> <br /> If &lt;math&gt;d=0&lt;/math&gt;, then each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;cz_0&lt;/math&gt; must either be zero or in the same direction as all the others, so each of &lt;math&gt;4z_0^4&lt;/math&gt;, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, &lt;math&gt;(b-c)z_0&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; must either be zero or in the same direction as all the others. We can divide this into two cases: &lt;math&gt;c \neq 0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;. If &lt;math&gt;c \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^4&lt;/math&gt; must be real. Then, &lt;math&gt;z_0&lt;/math&gt; is a fourth root of unity. If &lt;math&gt;z_0&lt;/math&gt; is not a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero, implying that &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a=4&lt;/math&gt;, and &lt;math&gt;c=b=4&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,0)&lt;/math&gt;. If &lt;math&gt;z_0&lt;/math&gt; is also a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt; and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero but &lt;math&gt;(a-b)z_0^2&lt;/math&gt; can be anything. This implies &lt;math&gt;a=4&lt;/math&gt; and &lt;math&gt;b=c&lt;/math&gt; with no other restrictions, leading to the new solutions &lt;math&gt;(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;c=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^3&lt;/math&gt;, &lt;math&gt;(4-a)z_0^2&lt;/math&gt;, &lt;math&gt;(a-b)z_0&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; must be zero or in the same direction as all the others. If &lt;math&gt;b \neq 0&lt;/math&gt;, then &lt;math&gt;z_0&lt;/math&gt; must be a third root of unity, so &lt;math&gt;(4-a)z_0^2&lt;/math&gt; and &lt;math&gt;(a-b)z_0&lt;/math&gt; must be zero, implying &lt;math&gt;a=b=4&lt;/math&gt;, leading to the new solution &lt;math&gt;(a,b,c,d)=(4,4,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;b=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^2&lt;/math&gt;, &lt;math&gt;(4-a)z_0&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; must be zero or in the same direction as the others. For &lt;math&gt;|z_0|=1&lt;/math&gt;, &lt;math&gt;a = 4&lt;/math&gt;, but we have already counted the solution &lt;math&gt;(a,b,c,d)=(4,0,0,0)&lt;/math&gt;. <br /> <br /> Then, the complete list of solutions is &lt;math&gt;(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)&lt;/math&gt;, leading to a sum of &lt;math&gt;\framebox{B}=92&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=113639 2012 AMC 12B Problems/Problem 23 2019-12-29T02:09:32Z <p>Nli: /* Solution 3 */</p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (dubious) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> == Solution 2 ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> First, notice that &lt;math&gt;z=1&lt;/math&gt; cannot be a root of the polynomial because &lt;math&gt;a,b,c,d \geq 0&lt;/math&gt;. Multiplying the polynomial by &lt;math&gt;(z-1)&lt;/math&gt; yields &lt;math&gt;P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d&lt;/math&gt;, so for &lt;math&gt;z \neq 1&lt;/math&gt; to be a root of &lt;math&gt;P(z)&lt;/math&gt;, &lt;math&gt;4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d&lt;/math&gt;. Now we consider the root &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1&lt;/math&gt;. &lt;math&gt;|4z_0^5| = 4&lt;/math&gt;, so the right hand side must have absolute value 4. By the triangle inequality, &lt;math&gt;|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|&lt;/math&gt; &lt;math&gt;\geq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d &lt;/math&gt; &lt;math&gt;= (4-a)+(a-b)+(b-c)+(c-d)+d=4&lt;/math&gt;, with equality if and only if each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, &lt;math&gt;(c-d)z_0&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; is either zero or in the same direction as all the others when looked at as vectors in the complex plane. <br /> <br /> We can now divide into two cases: &lt;math&gt;d \neq 0&lt;/math&gt; and &lt;math&gt;d=0&lt;/math&gt;. If &lt;math&gt;d \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^5&lt;/math&gt; must be real by the previous argument, so &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity. Also, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;(c-d)z_0&lt;/math&gt; must all be zero because if &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a&lt;/math&gt;, &lt;math&gt;c=b&lt;/math&gt;, and &lt;math&gt;d=c&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,4)&lt;/math&gt;. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.<br /> <br /> If &lt;math&gt;d=0&lt;/math&gt;, then each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;cz_0&lt;/math&gt; must either be zero or in the same direction as all the others, so each of &lt;math&gt;4z_0^4&lt;/math&gt;, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, &lt;math&gt;(b-c)z_0&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; must either be zero or in the same direction as all the others. We can divide this into two cases: &lt;math&gt;c \neq 0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;. If &lt;math&gt;c \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^4&lt;/math&gt; must be real. Then, &lt;math&gt;z_0&lt;/math&gt; is a fourth root of unity. If &lt;math&gt;z_0&lt;/math&gt; is not a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero, implying that &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a=4&lt;/math&gt;, and &lt;math&gt;c=b=4&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,0)&lt;/math&gt;. If &lt;math&gt;z_0&lt;/math&gt; is also a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt; and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero but &lt;math&gt;(a-b)z_0^2&lt;/math&gt; can be anything. This implies &lt;math&gt;a=4&lt;/math&gt; and &lt;math&gt;b=c&lt;/math&gt; with no other restrictions, leading to the new solutions &lt;math&gt;(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;c=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^3&lt;/math&gt;, &lt;math&gt;(4-a)z_0^2&lt;/math&gt;, &lt;math&gt;(a-b)z_0&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; must be zero or in the same direction as all the others. If &lt;math&gt;b \neq 0&lt;/math&gt;, then &lt;math&gt;z_0&lt;/math&gt; must be a third root of unity, so &lt;math&gt;(4-a)z_0^2&lt;/math&gt; and &lt;math&gt;(a-b)z_0&lt;/math&gt; must be zero, implying &lt;math&gt;a=b=4&lt;/math&gt;, leading to the new solution &lt;math&gt;(a,b,c,d)=(4,4,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;b=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^2&lt;/math&gt;, &lt;math&gt;(4-a)z_0&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; must be zero or in the same direction as the others. For &lt;math&gt;|z_0|=1&lt;/math&gt;, &lt;math&gt;a = 4&lt;/math&gt;, but we have already counted the solution &lt;math&gt;(a,b,c,d)=(4,0,0,0)&lt;/math&gt;. <br /> <br /> Then, the complete list of solutions is &lt;math&gt;(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)&lt;/math&gt;, leading to a sum of &lt;math&gt;\framebox{B}=92&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=113638 2012 AMC 12B Problems/Problem 23 2019-12-29T02:08:00Z <p>Nli: /* Solution 3 */</p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (dubious) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> == Solution 2 ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> First, notice that &lt;math&gt;z=1&lt;/math&gt; cannot be a root of the polynomial because &lt;math&gt;a,b,c,d \geq 0&lt;/math&gt;. Multiplying the polynomial by &lt;math&gt;(z-1)&lt;/math&gt; yields &lt;math&gt;P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d&lt;/math&gt;, so for &lt;math&gt;z \neq 1&lt;/math&gt; to be a root of &lt;math&gt;P(z)&lt;/math&gt;, &lt;math&gt;4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d&lt;/math&gt;. Now we consider the root &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1&lt;/math&gt;. &lt;math&gt;|4z_0^5| = 4&lt;/math&gt;, so the right hand side must have absolute value 4. By the triangle inequality, &lt;math&gt;|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|&lt;/math&gt; &lt;math&gt;\geq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d &lt;/math&gt; &lt;math&gt;= (4-a)+(a-b)+(b-c)+(c-d)+d=4&lt;/math&gt;, with equality if and only if each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, &lt;math&gt;(c-d)z_0&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; is either zero or in the same direction as all the others when looked at as vectors in the complex plane. <br /> <br /> We can now divide into two cases: &lt;math&gt;d \neq 0&lt;/math&gt; and &lt;math&gt;d=0&lt;/math&gt;. If &lt;math&gt;d \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^5&lt;/math&gt; must be real by the previous argument, so &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity. Also, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;(c-d)z_0&lt;/math&gt; must all be zero because if &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a&lt;/math&gt;, &lt;math&gt;c=b&lt;/math&gt;, and &lt;math&gt;d=c&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,4)&lt;/math&gt;. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.<br /> <br /> If &lt;math&gt;d=0&lt;/math&gt;, then each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;cz_0&lt;/math&gt; must either be zero or in the same direction as all the others, so each of &lt;math&gt;4z_0^4&lt;/math&gt;, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, &lt;math&gt;(b-c)z_0&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; must either be zero or in the same direction as all the others. We can divide this into two cases: &lt;math&gt;c \neq 0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;. If &lt;math&gt;c \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^4&lt;/math&gt; must be real. Then, &lt;math&gt;z_0&lt;/math&gt; is a fourth root of unity. If &lt;math&gt;z_0&lt;/math&gt; is not a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero, implying that &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a=4&lt;/math&gt;, and &lt;math&gt;c=b=4&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,0)&lt;/math&gt;. If &lt;math&gt;z_0&lt;/math&gt; is also a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt; and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero but &lt;math&gt;(a-b)z_0^2&lt;/math&gt; can be anything. This implies &lt;math&gt;a=4&lt;/math&gt; and &lt;math&gt;b=c&lt;/math&gt; with no other restrictions, leading to the new solutions &lt;math&gt;(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;c=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^3&lt;/math&gt;, &lt;math&gt;(4-a)z_0^2&lt;/math&gt;, &lt;math&gt;(a-b)z_0&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; must be zero or in the same direction as all the others. If &lt;math&gt;b \neq 0&lt;/math&gt;, then &lt;math&gt;z_0&lt;/math&gt; must be a third root of unity, so &lt;math&gt;(4-a)z_0^2&lt;/math&gt; and &lt;math&gt;(a-b)z_0&lt;/math&gt; must be zero, implying &lt;math&gt;a=b=4&lt;/math&gt;, leading to the new solution &lt;math&gt;(a,b,c,d)=(4,4,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;b=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^2&lt;/math&gt;, &lt;math&gt;(4-a)z_0&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; must be zero or in the same direction as the others. For &lt;math&gt;|z_0|=1&lt;/math&gt;, &lt;math&gt;a = 4&lt;/math&gt;, but we have already counted the solution &lt;math&gt;(a,b,c,d)=(4,0,0,0)&lt;/math&gt;. <br /> <br /> Then, the complete list of solutions is &lt;math&gt;(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)&lt;/math&gt;, leading to a sum of \framebox{(B)=92}.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=113637 2012 AMC 12B Problems/Problem 23 2019-12-29T02:05:56Z <p>Nli: Added another solution that uses a straightforward casework argument</p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (dubious) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> == Solution 2 ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> First, notice that &lt;math&gt;z=1&lt;/math&gt; cannot be a root of the polynomial because &lt;math&gt;a,b,c,d \geq 0&lt;/math&gt;. Multiplying the polynomial by &lt;math&gt;(z-1)&lt;/math&gt; yields &lt;math&gt;P(z)(z-1) = 4z^5-(4-a)z^4-(a-b)z^3-(b-c)z^2-(c-d)z-d&lt;/math&gt;, so for &lt;math&gt;z \neq 1&lt;/math&gt; to be a root of &lt;math&gt;P(z)&lt;/math&gt;, &lt;math&gt;4z^5 = (4-a)z^4+(a-b)z^3+(b-c)z^2+(c-d)z+d&lt;/math&gt;. Now we consider the root &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1&lt;/math&gt;. &lt;math&gt;|4z_0^5| = 4&lt;/math&gt;, so the right hand side must have absolute value 4. By the triangle inequality, &lt;math&gt;|(4-a)z_0^4+(a-b)z_0^3+(b-c)z_0^2+(c-d)z_0+d|&lt;/math&gt; &lt;math&gt;\geq |(4-a)z_0^4|+|(a-b)z_0^3| + |(b-c)z_0^2| + |(c-d)z_0|+d &lt;/math&gt; &lt;math&gt;= (4-a)+(a-b)+(b-c)+(c-d)+d=4&lt;/math&gt;, with equality if and only if each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, &lt;math&gt;(c-d)z_0&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; is either zero or in the same direction as all the others when looked at as vectors in the complex plane. <br /> <br /> We can now divide into two cases: &lt;math&gt;d \neq 0&lt;/math&gt; and &lt;math&gt;d=0&lt;/math&gt;. If &lt;math&gt;d \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^5&lt;/math&gt; must be real by the previous argument, so &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity. Also, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;(c-d)z_0&lt;/math&gt; must all be zero because if &lt;math&gt;z_0&lt;/math&gt; is a fifth root of unity, none of these can be real numbers with positive absolute value. Therefore, &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=a&lt;/math&gt;, &lt;math&gt;c=b&lt;/math&gt;, and &lt;math&gt;d=c&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,4)&lt;/math&gt;. Just to be sure, we can easily verify that this solution leads to the six complex numbers under question being in the same direction.<br /> <br /> If &lt;math&gt;d=0&lt;/math&gt;, then each of &lt;math&gt;4z_0^5&lt;/math&gt;, &lt;math&gt;(4-a)z_0^4&lt;/math&gt;, &lt;math&gt;(a-b)z_0^3&lt;/math&gt;, &lt;math&gt;(b-c)z_0^2&lt;/math&gt;, and &lt;math&gt;cz_0&lt;/math&gt; must either be zero or in the same direction as all the others, so each of &lt;math&gt;4z_0^4&lt;/math&gt;, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, &lt;math&gt;(b-c)z_0&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; must either be zero or in the same direction as all the others. We can divide this into two cases: &lt;math&gt;c \neq 0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;. If &lt;math&gt;c \neq 0&lt;/math&gt;, then &lt;math&gt;4z_0^4&lt;/math&gt; must be real. Then, &lt;math&gt;z_0&lt;/math&gt; is a fourth root of unity. If &lt;math&gt;z_0&lt;/math&gt; is not a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt;, &lt;math&gt;(a-b)z_0^2&lt;/math&gt;, and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero, implying that &lt;math&gt;a=4, b=a=4, and c=b=4&lt;/math&gt;, leading to the solution &lt;math&gt;(a,b,c,d)=(4,4,4,0)&lt;/math&gt;. If &lt;math&gt;z_0&lt;/math&gt; is also a second root of unity, &lt;math&gt;(4-a)z_0^3&lt;/math&gt; and &lt;math&gt;(b-c)z_0&lt;/math&gt; must be zero but &lt;math&gt;(a-b)z_0^2&lt;/math&gt; can be anything. This implies &lt;math&gt;a=4&lt;/math&gt; and &lt;math&gt;b=c&lt;/math&gt; with no other restrictions, leading to the new solutions &lt;math&gt;(a,b,c,d) = (4,3,3,0), (4,2,2,0), (4,1,1,0), (4,0,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;c=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^3&lt;/math&gt;, &lt;math&gt;(4-a)z_0^2&lt;/math&gt;, &lt;math&gt;(a-b)z_0&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; must be zero or in the same direction as all the others. If &lt;math&gt;b \neq 0&lt;/math&gt;, then &lt;math&gt;z_0&lt;/math&gt; must be a third root of unity, so &lt;math&gt;(4-a)z_0^2&lt;/math&gt; and &lt;math&gt;(a-b)z_0&lt;/math&gt; must be zero, implying &lt;math&gt;a=b=4&lt;/math&gt;, leading to the new solution &lt;math&gt;(a,b,c,d)=(4,4,0,0)&lt;/math&gt;. <br /> <br /> If &lt;math&gt;b=0&lt;/math&gt;, then we can similarly show that each of &lt;math&gt;4z_0^2&lt;/math&gt;, &lt;math&gt;(4-a)z_0&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; must be zero or in the same direction as the others. For &lt;math&gt;|z_0|=1&lt;/math&gt;, &lt;math&gt;a = 4&lt;/math&gt;, but we have already counted the solution &lt;math&gt;(a,b,c,d)=(4,0,0,0)&lt;/math&gt;. <br /> <br /> Then, the complete list of solutions is &lt;math&gt;(a,b,c,d)=(4,4,4,4),(4,4,4,0),(4,3,3,0),(4,2,2,0),(4,1,1,0),(4,0,0,0),(4,4,0,0)&lt;/math&gt;, leading to a sum of \framebox{(B)=92}. <br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=94248 2012 AIME II Problems/Problem 15 2018-04-30T04:05:24Z <p>Nli: /* Solution 4 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle BFC=\angle CFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=94247 2012 AIME II Problems/Problem 15 2018-04-30T04:00:32Z <p>Nli: /* Solution 4 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CFB=\angle BEC= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=94246 2012 AIME II Problems/Problem 15 2018-04-30T03:59:43Z <p>Nli: /* Solution 4 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CEB=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_13&diff=93158 2018 AIME I Problems/Problem 13 2018-03-11T22:56:31Z <p>Nli: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have side lengths &lt;math&gt;AB=30&lt;/math&gt;, &lt;math&gt;BC=32&lt;/math&gt;, and &lt;math&gt;AC=34&lt;/math&gt;. Point &lt;math&gt;X&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt;, and points &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; are the incenters of &lt;math&gt;\triangle ABX&lt;/math&gt; and &lt;math&gt;\triangle ACX&lt;/math&gt;, respectively. Find the minimum possible area of &lt;math&gt;\triangle AI_1I_2&lt;/math&gt; as &lt;math&gt;X&lt;/math&gt; varies along &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> ==Solution==<br /> First note that &lt;cmath&gt;\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2&lt;/cmath&gt; is a constant not depending on &lt;math&gt;X&lt;/math&gt;, so by &lt;math&gt;[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2&lt;/math&gt; it suffices to minimize &lt;math&gt;(AI_1)(AI_2)&lt;/math&gt;. Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = AC&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt;, and &lt;math&gt;\alpha = \angle AXB&lt;/math&gt;. Remark that &lt;cmath&gt;\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.&lt;/cmath&gt; Applying the Law of Sines to &lt;math&gt;\triangle ABI_1&lt;/math&gt; gives &lt;cmath&gt;\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.&lt;/cmath&gt; Analogously one can derive &lt;math&gt;AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}&lt;/math&gt;, and so &lt;cmath&gt;[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,&lt;/cmath&gt; with equality when &lt;math&gt;\alpha = 90^\circ&lt;/math&gt;, that is, when &lt;math&gt;X&lt;/math&gt; is the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;. In this case the desired area is &lt;math&gt;bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2&lt;/math&gt;. To make this feasible to compute, note that &lt;cmath&gt;\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.&lt;/cmath&gt; Applying similar logic to &lt;math&gt;\sin \tfrac B2&lt;/math&gt; and &lt;math&gt;\sin\tfrac C2&lt;/math&gt; and simplifying yields a final answer of &lt;cmath&gt;\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&amp;=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&amp;=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}&lt;/cmath&gt;<br /> Thank who help me to modify my solution - S.B.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Nli https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_14&diff=92378 2016 AIME II Problems/Problem 14 2018-02-26T04:29:16Z <p>Nli: /* Short Simple Solution */</p> <hr /> <div>Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;600&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie outside the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; and are on opposite sides of the plane. Furthermore, &lt;math&gt;PA=PB=PC&lt;/math&gt;, and &lt;math&gt;QA=QB=QC&lt;/math&gt;, and the planes of &lt;math&gt;\triangle PAB&lt;/math&gt; and &lt;math&gt;\triangle QAB&lt;/math&gt; form a &lt;math&gt;120^{\circ}&lt;/math&gt; dihedral angle (the angle between the two planes). There is a point &lt;math&gt;O&lt;/math&gt; whose distance from each of &lt;math&gt;A,B,C,P,&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt;. Find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> The inradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100\sqrt 3&lt;/math&gt; and the circumradius is &lt;math&gt;200 \sqrt 3&lt;/math&gt;. Now, consider the line perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; through the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. Note that &lt;math&gt;P,Q,O&lt;/math&gt; must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since &lt;math&gt;P, Q, O&lt;/math&gt; are collinear, and &lt;math&gt;OP=OQ&lt;/math&gt;, we must have &lt;math&gt;O&lt;/math&gt; is the midpoint of &lt;math&gt;PQ&lt;/math&gt;. Now, Let &lt;math&gt;K&lt;/math&gt; be the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, and &lt;math&gt;L&lt;/math&gt; be the foot of the altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;. We must have &lt;math&gt;\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})&lt;/math&gt;. Setting &lt;math&gt;KP=x&lt;/math&gt; and &lt;math&gt;KQ=y&lt;/math&gt;, assuming WLOG &lt;math&gt;x&gt;y&lt;/math&gt;, we must have &lt;math&gt;\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}&lt;/math&gt;. Therefore, we must have &lt;math&gt;100(x+y)=xy-30000&lt;/math&gt;. Also, we must have &lt;math&gt;\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000&lt;/math&gt; by the Pythagorean theorem, so we have &lt;math&gt;xy=120000&lt;/math&gt;, so substituting into the other equation we have &lt;math&gt;90000=100(x+y)&lt;/math&gt;, or &lt;math&gt;x+y=900&lt;/math&gt;. Since we want &lt;math&gt;\dfrac{x+y}{2}&lt;/math&gt;, the desired answer is &lt;math&gt;\boxed{450}&lt;/math&gt;.<br /> <br /> Solution by Shaddoll<br /> <br /> ==Short Simple Solution==<br /> Draw a good diagram. Draw &lt;math&gt;CH&lt;/math&gt; as an altitude of the triangle. Scale everything down by a factor of &lt;math&gt;100\sqrt{3}&lt;/math&gt;, so that &lt;math&gt;AB=2\sqrt{3}&lt;/math&gt;. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line &lt;math&gt;CH&lt;/math&gt;, which of course includes &lt;math&gt;P, Q&lt;/math&gt;. From there, we can call &lt;math&gt;OU=h&lt;/math&gt;. There are two crucial equations we can thus generate. WLOG set &lt;math&gt;PU&lt;QU&lt;/math&gt;, then we call &lt;math&gt;PU=d-h, QU=d+h&lt;/math&gt;. First equation: using the Pythagorean Theorem on &lt;math&gt;\triangle UOB&lt;/math&gt;, &lt;math&gt;h^2+2^2=d^2&lt;/math&gt;. Next, using the tangent addition formula on angles &lt;math&gt;\angle PHU, \angle UHQ&lt;/math&gt; we see that after simplifying &lt;math&gt;-d^2+h^2=-4, 2d=3\sqrt{3}&lt;/math&gt; in the numerator, so &lt;math&gt;d=\frac{3\sqrt{3}}{2}&lt;/math&gt;. Multiply back the scalar and you get &lt;math&gt;\boxed{450}&lt;/math&gt;. Not that hard, was it?<br /> <br /> ==Solution 3==<br /> To make numbers more feasible, we'll scale everything down by a factor of &lt;math&gt;100&lt;/math&gt; so that &lt;math&gt;\overline{AB}=\overline{BC}=\overline{AC}=6&lt;/math&gt;. We should also note that &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; must lie on the line that is perpendicular to the plane of &lt;math&gt;ABC&lt;/math&gt; and also passes through the circumcenter of &lt;math&gt;ABC&lt;/math&gt; (due to &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; being equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;), let &lt;math&gt;D&lt;/math&gt; be the altitude from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. We can draw a vertical cross-section of the figure then: &lt;asy&gt;pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label(&quot;$C$&quot;, C, E); label(&quot;$D$&quot;, D, W); label(&quot;$I$&quot;, I, NW); label(&quot;$P$&quot;, P, N); label(&quot;$Q$&quot;, Q, S); label(&quot;$O$&quot;, O, SW); dot(O); dot(I);&lt;/asy&gt; We let &lt;math&gt;\angle PDI=\alpha&lt;/math&gt; so &lt;math&gt;\angle QDI=\alpha&lt;/math&gt;, also note that &lt;math&gt;\overline{PO}=\overline{QO}=\overline{CO}=d&lt;/math&gt;. Because &lt;math&gt;I&lt;/math&gt; is the centroid of &lt;math&gt;ABC&lt;/math&gt;, we know that ratio of &lt;math&gt;\overline{CI}&lt;/math&gt; to &lt;math&gt;\overline{DI}&lt;/math&gt; is &lt;math&gt;2:1&lt;/math&gt;. Since we've scaled the figure down, the length of &lt;math&gt;CD&lt;/math&gt; is &lt;math&gt;3\sqrt{3}&lt;/math&gt;, from this it's easy to know that &lt;math&gt;\overline{CI}=2\sqrt{3}&lt;/math&gt; and &lt;math&gt;\overline{DI}=\sqrt{3}&lt;/math&gt;. The following two equations arise: &lt;cmath&gt;\begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&amp;=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &amp;= \sqrt{d^{2}-12} \end{align}&lt;/cmath&gt; Using trig identities for the tangent, we find that &lt;cmath&gt;\begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&amp;=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &amp;= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &amp;= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &amp;= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*}&lt;/cmath&gt; Okay, now we can plug this into &lt;math&gt;\left(1\right)&lt;/math&gt; to get: &lt;cmath&gt;\begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&amp;=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &amp;= \sqrt{d^{2}-12} \end{align}&lt;/cmath&gt; Notice that &lt;math&gt;\alpha&lt;/math&gt; only appears in the above system of equations in the form of &lt;math&gt;\sqrt{3}\tan{\left(\alpha\right)}&lt;/math&gt;, we can set &lt;math&gt;\sqrt{3}\tan{\left(\alpha\right)}=a&lt;/math&gt; for convenience since we really only care about &lt;math&gt;d&lt;/math&gt;. Now we have &lt;cmath&gt;\begin{align}a+\frac{a+3}{a-1}&amp;=2d \\ a - d &amp;= \sqrt{d^{2}-12} \end{align}&lt;/cmath&gt; Looking at &lt;math&gt;\left(2\right)&lt;/math&gt;, it's tempting to square it to get rid of the square-root so now we have: &lt;cmath&gt;\begin{align*}a^{2}-2ad+d^{2}&amp;=d^{2}-12 \\ a - 2ad &amp;= \text{-}12 \end{align*}&lt;/cmath&gt; See the sneaky &lt;math&gt;2d&lt;/math&gt; in the above equation? That we means we can substitute it for &lt;math&gt;a+\frac{a+3}{a-1}&lt;/math&gt;: &lt;cmath&gt;\begin{align*}a^{2}-2ad+d^{2}&amp;=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &amp;= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &amp;=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&amp;=\text{-}12 \\ \text{-}a^{2}-3a&amp;=\text{-}12a+12 \\ 0 &amp;= a^{2}-9a+12 \end{align*}&lt;/cmath&gt; Use the quadratic formula, we find that &lt;math&gt;a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}&lt;/math&gt; - the two solutions were expected because &lt;math&gt;a&lt;/math&gt; can be &lt;math&gt;\angle PDI&lt;/math&gt; or &lt;math&gt;\angle QDI&lt;/math&gt;. We can plug this into &lt;math&gt;\left(1\right)&lt;/math&gt;: &lt;cmath&gt;\begin{align*}a+\frac{a+3}{a-1}&amp;=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &amp;= 2d\end{align*}&lt;/cmath&gt; I'll use &lt;math&gt;a=\frac{9+\sqrt{33}}{2}&lt;/math&gt; because both values should give the same answer for &lt;math&gt;d&lt;/math&gt;. &lt;cmath&gt;\begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &amp;= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &amp;= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &amp;= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &amp;= 2d \\ 9 &amp;= 2d \\ \frac{9}{2} &amp;= d\end{align*}&lt;/cmath&gt; Wait! Before you get excited, remember that we scaled the entire figure by &lt;math&gt;100&lt;/math&gt;?? That means that the answer is &lt;math&gt;d=100\times\frac{9}{2}=\boxed{450}&lt;/math&gt;. An alternate way of proceeding after finding &lt;math&gt;a&lt;/math&gt; (credit to riemanntensor), was to average the two possible values, you can see for yourself why this would work.<br /> <br /> -fatant<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]</div> Nli