https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kkyyhe&feedformat=atom AoPS Wiki - User contributions [en] 2022-09-27T02:36:24Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_15&diff=148700 2002 AIME II Problems/Problem 15 2021-03-06T15:20:49Z <p>Kkyyhe: /* Solution 2 (Alcumus) */</p> <hr /> <div>== Problem ==<br /> Circles &lt;math&gt;\mathcal{C}_{1}&lt;/math&gt; and &lt;math&gt;\mathcal{C}_{2}&lt;/math&gt; intersect at two points, one of which is &lt;math&gt;(9,6)&lt;/math&gt;, and the product of the radii is &lt;math&gt;68&lt;/math&gt;. The x-axis and the line &lt;math&gt;y = mx&lt;/math&gt;, where &lt;math&gt;m &gt; 0&lt;/math&gt;, are tangent to both circles. It is given that &lt;math&gt;m&lt;/math&gt; can be written in the form &lt;math&gt;a\sqrt {b}/c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers, &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime, and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime. Find &lt;math&gt;a + b + c&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Let the smaller angle between the &lt;math&gt;x&lt;/math&gt;-axis and the line &lt;math&gt;y=mx&lt;/math&gt; be &lt;math&gt;\theta&lt;/math&gt;. Note that the centers of the two circles lie on the angle bisector of the angle between the &lt;math&gt;x&lt;/math&gt;-axis and the line &lt;math&gt;y=mx&lt;/math&gt;. Also note that if &lt;math&gt;(x,y)&lt;/math&gt; is on said angle bisector, we have that &lt;math&gt;\frac{y}{x}=\tan{\frac{\theta}{2}}&lt;/math&gt;. Let &lt;math&gt;\tan{\frac{\theta}{2}}=m_1&lt;/math&gt;, for convenience. Therefore if &lt;math&gt;(x,y)&lt;/math&gt; is on the angle bisector, then &lt;math&gt;x=\frac{y}{m_1}&lt;/math&gt;. Now let the centers of the two relevant circles be &lt;math&gt;(a/m_1 , a)&lt;/math&gt; and &lt;math&gt;(b/m_1 , b)&lt;/math&gt; for some positive reals &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. These two circles are tangent to the &lt;math&gt;x&lt;/math&gt;-axis, so the radii of the circles are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively. We know that the point &lt;math&gt;(9,6)&lt;/math&gt; is a point on both circles, so we have that<br /> <br /> &lt;cmath&gt;(9-\frac{a}{m_1})^2+(6-a)^2=a^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(9-\frac{b}{m_1})^2+(6-b)^2=b^2&lt;/cmath&gt;<br /> <br /> Expanding these and manipulating terms gives<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0&lt;/cmath&gt;<br /> <br /> It follows that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are the roots of the quadratic<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0&lt;/cmath&gt;<br /> <br /> It follows from Vieta's Formulas that the product of the roots of this quadratic is &lt;math&gt;117m_1^2&lt;/math&gt;, but we were also given that the product of the radii was 68. Therefore &lt;math&gt;68=117m_1^2&lt;/math&gt;, or &lt;math&gt;m_1^2=\frac{68}{117}&lt;/math&gt;. Note that the half-angle formula for tangents is<br /> <br /> &lt;cmath&gt;\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}&lt;/cmath&gt;<br /> <br /> Therefore<br /> <br /> &lt;cmath&gt;\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;\cos{\theta}&lt;/math&gt; gives that &lt;math&gt;\cos{\theta}=\frac{49}{185}&lt;/math&gt;. It then follows that &lt;math&gt;\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}&lt;/math&gt;.<br /> <br /> It then follows that &lt;math&gt;m=\tan{\theta}=\frac{12\sqrt{221}}{49}&lt;/math&gt;. Therefore &lt;math&gt;a=12&lt;/math&gt;, &lt;math&gt;b=221&lt;/math&gt;, and &lt;math&gt;c=49&lt;/math&gt;. The desired answer is then &lt;math&gt;12+221+49=\boxed{282}&lt;/math&gt;.<br /> <br /> == Solution 2 (Alcumus)==<br /> Let &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; be the radii of the circles. Then the centers of the circles are of the form &lt;math&gt;(kr_1,r_1)&lt;/math&gt; and &lt;math&gt;(kr_2,r_2)&lt;/math&gt; for the same constant &lt;math&gt;k,&lt;/math&gt; since the two centers are collinear with the origin. Since &lt;math&gt;(9,6)&lt;/math&gt; lies on both circles,<br /> &lt;cmath&gt;(kr - 9)^2 + (r - 6)^2 = r^2,&lt;/cmath&gt;where &lt;math&gt;r&lt;/math&gt; represents either radius. Expanding, we get<br /> &lt;cmath&gt;k^2 r^2 - (18k + 12) r + 117 = 0.&lt;/cmath&gt;We are told the product of the circles is 68, so by Vieta's formulas, &lt;math&gt;\frac{117}{k^2} = 68.&lt;/math&gt; Hence, &lt;math&gt;k^2 = \frac{117}{68},&lt;/math&gt; and &lt;math&gt;k = \sqrt{\frac{117}{68}}.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(0.25 cm);<br /> <br /> pair[] O;<br /> real[] r;<br /> pair P;<br /> <br /> r = 4.096;<br /> r = 16.6;<br /> O = (r/(2/3*sqrt(17/13)),r);<br /> O = (r/(2/3*sqrt(17/13)),r);<br /> P = reflect(O,O)*(9,6);<br /> <br /> draw(Circle(O,r));<br /> //draw(Circle(O,r));<br /> draw(arc(O,r,130,300));<br /> draw((0,0)--(8,12*sqrt(221)/49*8));<br /> draw((0,0)--(30,0));<br /> draw((0,0)--O--(O.x,0));<br /> draw(O--(O + reflect((0,0),(10,12*sqrt(221)/49*10))*(O))/2);<br /> <br /> label(&quot;$y = mx$&quot;, (8,12*sqrt(221)/49*8), N);<br /> <br /> dot(&quot;$(9,6)$&quot;, (9,6), NE);<br /> dot(&quot;$(kr,r)$&quot;, O, N);<br /> dot(P,red);<br /> &lt;/asy&gt;<br /> <br /> Since the circle is tangent to the line &lt;math&gt;y = mx,&lt;/math&gt; the distance from the center &lt;math&gt;(kr,r)&lt;/math&gt; to the line is &lt;math&gt;r.&lt;/math&gt; We can write &lt;math&gt;y = mx&lt;/math&gt; as &lt;math&gt;y - mx = 0,&lt;/math&gt; so from the distance formula,<br /> &lt;cmath&gt;\frac{|r - krm|}{\sqrt{1 + m^2}} = r.&lt;/cmath&gt;Squaring both sides, we get<br /> &lt;cmath&gt;\frac{(r - krm)^2}{1 + m^2} = r^2,&lt;/cmath&gt;so &lt;math&gt;(r - krm)^2 = r^2 (1 + m^2).&lt;/math&gt; Since &lt;math&gt;r \neq 0,&lt;/math&gt; we can divide both sides by r, to get<br /> &lt;cmath&gt;(1 - km)^2 = 1 + m^2.&lt;/cmath&gt;Then &lt;math&gt;1 - 2km + k^2 m^2 = 1 + m^2,&lt;/math&gt; so &lt;math&gt;m^2 (1 - k^2) + 2km = 0.&lt;/math&gt; Since &lt;math&gt;m \neq 0,&lt;/math&gt;<br /> &lt;cmath&gt;m(1 - k^2) + 2k = 0.&lt;/cmath&gt;Hence,<br /> &lt;cmath&gt;m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \boxed{\frac{12 \sqrt{221}}{49}}.&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_14&diff=147734 2004 AIME II Problems/Problem 14 2021-02-22T17:36:21Z <p>Kkyyhe: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Consider a string of &lt;math&gt; n &lt;/math&gt; &lt;math&gt; 7 &lt;/math&gt;'s, &lt;math&gt; 7777\cdots77, &lt;/math&gt; into which &lt;math&gt; + &lt;/math&gt; signs are inserted to produce an arithmetic [[expression]]. For example, &lt;math&gt; 7+77+777+7+7=875 &lt;/math&gt; could be obtained from eight &lt;math&gt; 7 &lt;/math&gt;'s in this way. For how many values of &lt;math&gt; n &lt;/math&gt; is it possible to insert &lt;math&gt; + &lt;/math&gt; signs so that the resulting expression has value &lt;math&gt; 7000 &lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> Suppose we require &lt;math&gt;a&lt;/math&gt; &lt;math&gt;7&lt;/math&gt;s, &lt;math&gt;b&lt;/math&gt; &lt;math&gt;77&lt;/math&gt;s, and &lt;math&gt;c&lt;/math&gt; &lt;math&gt;777&lt;/math&gt;s to sum up to &lt;math&gt;7000&lt;/math&gt; (&lt;math&gt;a,b,c \ge 0&lt;/math&gt;). Then &lt;math&gt;7a + 77b + 777c = 7000&lt;/math&gt;, or dividing by &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;a + 11b + 111c = 1000&lt;/math&gt;. Then the question is asking for the number of values of &lt;math&gt;n = a + 2b + 3c&lt;/math&gt;.<br /> <br /> Manipulating our equation, we have &lt;math&gt;a + 2b + 3c = n = 1000 - 9(b + 12c) \Longrightarrow 0 \le 9(b+12c) &lt; 1000&lt;/math&gt;. Thus the number of potential values of &lt;math&gt;n&lt;/math&gt; is the number of multiples of &lt;math&gt;9&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1000&lt;/math&gt;, or &lt;math&gt;112&lt;/math&gt;.<br /> <br /> <br /> However, we forgot to consider the condition that &lt;math&gt;a \ge 0&lt;/math&gt;. For a solution set &lt;math&gt;(b,c): n=1000-9(b+12c)&lt;/math&gt;, it is possible that &lt;math&gt;a = n-2b-3c &lt; 0&lt;/math&gt; (for example, suppose we counted the solution set &lt;math&gt;(b,c) = (1,9) \Longrightarrow n = 19&lt;/math&gt;, but substituting into our original equation we find that &lt;math&gt;a = -10&lt;/math&gt;, so it is invalid). In particular, this invalidates the values of &lt;math&gt;n&lt;/math&gt; for which their only expressions in terms of &lt;math&gt;(b,c)&lt;/math&gt; fall into the inequality &lt;math&gt;9b + 108c &lt; 1000 &lt; 11b + 111c&lt;/math&gt;. <br /> <br /> For &lt;math&gt;1000 - n = 9k \le 9(7 \cdot 12 + 11) = 855&lt;/math&gt;, we can express &lt;math&gt;k&lt;/math&gt; in terms of &lt;math&gt;(b,c): n \equiv b \pmod{12}, 0 \le b \le 11&lt;/math&gt; and &lt;math&gt;c = \frac{n-b}{12} \le 7&lt;/math&gt; (in other words, we take the greatest possible value of &lt;math&gt;c&lt;/math&gt;, and then &quot;fill in&quot; the remainder by incrementing &lt;math&gt;b&lt;/math&gt;). Then &lt;math&gt;11b + 111c \le 855 + 2b + 3c \le 855 + 2(11) + 3(7) = 898 &lt; 1000&lt;/math&gt;, so these values work. <br /> <br /> Similarily, for &lt;math&gt;855 \le 9k \le 9(8 \cdot 12 + 10) = 954&lt;/math&gt;, we can let &lt;math&gt;(b,c) = (k-8 \cdot 12,8)&lt;/math&gt;, and the inequality &lt;math&gt;11b + 111c \le 954 + 2b + 3c \le 954 + 2(10) + 3(8) = 998 &lt; 1000&lt;/math&gt;. However, for &lt;math&gt;9k \ge 963 \Longrightarrow n \le 37&lt;/math&gt;, we can no longer apply this approach. <br /> <br /> So we now have to examine the numbers on an individual basis. For &lt;math&gt;9k = 972&lt;/math&gt;, &lt;math&gt;(b,c) = (0,9)&lt;/math&gt; works. For &lt;math&gt;9k = 963, 981, 990, 999 \Longrightarrow n = 37, 19, 10, 1&lt;/math&gt;, we find (using that respectively, &lt;math&gt;b = 11,9,10,11 + 12p&lt;/math&gt; for integers &lt;math&gt;p&lt;/math&gt;) that their is no way to satisfy the inequality &lt;math&gt;11b + 111c &lt; 1000&lt;/math&gt;. <br /> <br /> Thus, the answer is &lt;math&gt;112 - 4 = \boxed{108}&lt;/math&gt;.<br /> <br /> <br /> ---- <br /> <br /> A note: Above, we formulated the solution in a forward manner (the last four paragraphs are devoted to showing that all the solutions we found worked except for the four cases pointed out; in a contest setting, we wouldn't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that &lt;math&gt;n \equiv 1 \pmod{9}&lt;/math&gt;, and noting that small values of &lt;math&gt;n&lt;/math&gt; would not work. <br /> <br /> Looking at the number &lt;math&gt;7000&lt;/math&gt;, we obviously see the maximum number of &lt;math&gt;7's&lt;/math&gt;: a string of &lt;math&gt;1000 \ 7's&lt;/math&gt;. Then, we see that the minimum is &lt;math&gt;28 \ 7's: \ 777*9 + 7 = 7000&lt;/math&gt;. The next step is to see by what interval the value of &lt;math&gt;n&lt;/math&gt; increases. Since &lt;math&gt;777&lt;/math&gt; is &lt;math&gt;3 \ 7's, \ 77*10 + 7&lt;/math&gt; is &lt;math&gt;21 \ 7's&lt;/math&gt;, we can convert a &lt;math&gt;777&lt;/math&gt; into &lt;math&gt;77's&lt;/math&gt; and &lt;math&gt;7's&lt;/math&gt; and add &lt;math&gt;18&lt;/math&gt; to the value of &lt;math&gt;n&lt;/math&gt;. Since we have &lt;math&gt;9 \ 777's&lt;/math&gt; to work with, this gives us &lt;math&gt;28,46,64,82,100,118,136,154,172,190 ( = 28 + 18n | 1\leq n\leq 9)&lt;/math&gt; as values for &lt;math&gt;n&lt;/math&gt;. Since &lt;math&gt;77&lt;/math&gt; can be converted into &lt;math&gt;7*11&lt;/math&gt;, we can add &lt;math&gt;9&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; by converting &lt;math&gt;77&lt;/math&gt; into &lt;math&gt;7's&lt;/math&gt;. Our &lt;math&gt;n = 190&lt;/math&gt;, which has &lt;math&gt;0 \ 777's \ 90 \ 77's \ 10 7's&lt;/math&gt;. We therefore can add &lt;math&gt;9&lt;/math&gt; to &lt;math&gt;n \ 90&lt;/math&gt; times by doing this. All values of &lt;math&gt;n&lt;/math&gt; not covered by this can be dealt with with the &lt;math&gt;n = 46 \ (8 \ 777's \ 10 \ 77's \ 2 \ 7's)&lt;/math&gt; up to &lt;math&gt;190&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> To simplify, replace all the &lt;math&gt;7&lt;/math&gt;’s with &lt;math&gt;1&lt;/math&gt;’s. <br /> Because the sum is congruent to &lt;math&gt;n \pmod 9&lt;/math&gt; and &lt;cmath&gt;1000 \equiv 1 \pmod 9 \implies n \equiv 1 \pmod 9&lt;/cmath&gt;<br /> Also, &lt;math&gt;n \leq 1000&lt;/math&gt;. There are &lt;cmath&gt; \left \lfloor \frac{1000}{9} \right \rfloor + 1 = 112 \textrm{ positive integers that satisfy both conditions i.e. } \{1, 10, 19, 28, 37, 46, . . . , 1000\}.&lt;/cmath&gt; <br /> <br /> For &lt;math&gt;n = 1, 10, 19&lt;/math&gt;, the greatest sum that is less than or equal to &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;6 \cdot 111 + 1 = 677 \implies 112-3 = 109&lt;/math&gt;.<br /> <br /> <br /> Thus &lt;math&gt;n \geq 28&lt;/math&gt; and let &lt;math&gt;S = \{28, 37, 46, . . . , 1000\}&lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;n=28&lt;/math&gt; is possible because &lt;math&gt;9 \cdot 111+1 \cdot 1 = 1000&lt;/math&gt;.<br /> <br /> When &lt;math&gt;n = 37&lt;/math&gt;, the greatest sum that is at most &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;8 \cdot 111+6\cdot 11+1 \cdot 1 = 955&lt;/math&gt;. <br /> <br /> <br /> All other elements of &lt;math&gt;S&lt;/math&gt; are possible because if any element &lt;math&gt;n&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt; between &lt;math&gt;46&lt;/math&gt; and &lt;math&gt;991&lt;/math&gt; is possible, then &lt;math&gt;(n+ 9)&lt;/math&gt; must be too. <br /> <br /> <br /> &lt;math&gt;\textrm{Case } 1:\text{ Sum has no } 11&lt;/math&gt;’s<br /> <br /> It must have at least one &lt;math&gt;1&lt;/math&gt;. <br /> If it has exactly one &lt;math&gt;1&lt;/math&gt;, there must be nine &lt;math&gt;111&lt;/math&gt;’s and &lt;math&gt;n = 28&lt;/math&gt;. <br /> Thus, for &lt;math&gt;n \geq 46&lt;/math&gt;, the sum has more than one &lt;math&gt;1&lt;/math&gt;, so it must have at least &lt;math&gt;1000 - 8 \cdot 111 = 112&lt;/math&gt; number of &lt;math&gt;1&lt;/math&gt;’s.<br /> For &lt;math&gt;n \leq 1000&lt;/math&gt;, at least one &lt;math&gt;111&lt;/math&gt;. <br /> To show that if &lt;math&gt;n&lt;/math&gt; is possible, then &lt;math&gt;(n + 9)&lt;/math&gt; is possible, replace a &lt;math&gt;111&lt;/math&gt; with &lt;math&gt;1 + 1 + 1&lt;/math&gt;, replace eleven &lt;math&gt;(1 + 1)&lt;/math&gt;’s with eleven &lt;math&gt;11&lt;/math&gt;’s, and include nine new &lt;math&gt;1&lt;/math&gt;’s as &lt;math&gt;+1&lt;/math&gt;’s. The sum remains &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;\textrm{Case } 2: \textrm{ Sum has at least one } 11&lt;/math&gt;.<br /> <br /> Replace an &lt;math&gt;11&lt;/math&gt; with &lt;math&gt;1 + 1&lt;/math&gt;, and include nine new &lt;math&gt;1&lt;/math&gt;’s as &lt;math&gt;+1&lt;/math&gt;’s.<br /> Now note that &lt;math&gt;46&lt;/math&gt; is possible because &lt;math&gt;8 \cdot 111 + 10 \cdot 11 + 2 \cdot 1 = 1000&lt;/math&gt;.<br /> Thus all elements of &lt;math&gt;S&lt;/math&gt; except &lt;math&gt;37&lt;/math&gt; are possible. <br /> <br /> <br /> Thus there are &lt;math&gt;\boxed{108}&lt;/math&gt; possible values for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> It's obvious that we cannot have any number &lt;math&gt;\ge 7777&lt;/math&gt; because &lt;math&gt;7777 &gt; 7000&lt;/math&gt; so the max number that an occur is &lt;math&gt;777&lt;/math&gt;<br /> <br /> Let's say we have &lt;math&gt;a&lt;/math&gt; 777's , &lt;math&gt;b&lt;/math&gt; 77's and &lt;math&gt;c&lt;/math&gt; 7's<br /> <br /> From here we get our required equation as &lt;math&gt;777a + 77b + 7c = 7000&lt;/math&gt;<br /> <br /> Now comes the main problem , one might think that if we find number of &lt;math&gt;(a,b,c)&lt;/math&gt; then we're done , but in reality we are over-counting our number of &lt;math&gt;n&lt;/math&gt;'s. This is because &lt;math&gt;n&lt;/math&gt; is the total number of 7's and from our equation we'll get &lt;math&gt;n&lt;/math&gt; as &lt;math&gt;3a + 2b + c&lt;/math&gt; (because there are three 7's , two 7's and one 7) <br /> <br /> The reason why we're over-counting is because , say &lt;math&gt;a_1 , b_1 , c_1&lt;/math&gt; be a solution of our original equation and &lt;math&gt;a_2 , b_2 , c_2&lt;/math&gt; be another solution of our original equation , then there can be a possibility that &lt;math&gt;3a_1 + 2b_1 + c_1 = 3a_2 + 2b_2 + c_2&lt;/math&gt; where &lt;math&gt;a_1 \neq a_2 , b_1 \neq b_2 , c_1 \neq c_2&lt;/math&gt; (example : &lt;math&gt; 2 + 3 + 4 = 1 + 5 + 3&lt;/math&gt; but &lt;math&gt;2 \neq 1 , 3 \neq 5 , 4 \neq 3&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;0 \le a \le 9&lt;/math&gt; , &lt;math&gt;0 \le b \le 90&lt;/math&gt; , &lt;math&gt;0 \le c \le 1000&lt;/math&gt;<br /> The bound on &lt;math&gt;a&lt;/math&gt; is easier to handle with , so lets start putting values on &lt;math&gt;a&lt;/math&gt; and calculate &lt;math&gt;b , c , n&lt;/math&gt; by making cases<br /> <br /> Reduced equation : &lt;math&gt;111a + 11b + c = 1000&lt;/math&gt;<br /> <br /> Case 1 : &lt;math&gt;a = 9&lt;/math&gt;<br /> <br /> We get &lt;math&gt;11b + c = 1 \implies b = 0 , c = 1&lt;/math&gt; is our only solution thus only &lt;math&gt;\boxed{1}&lt;/math&gt; value of &lt;math&gt;n&lt;/math&gt;<br /> <br /> Case 2 : &lt;math&gt;a = 8&lt;/math&gt;<br /> <br /> We get &lt;math&gt;11b + c = 112 \implies b = 0,1,2,\cdots ,10&lt;/math&gt; and &lt;math&gt;c = 112 , 101 , 90,\cdots ,2&lt;/math&gt; From here we get different &lt;math&gt;n&lt;/math&gt;'s as &lt;math&gt;24 + 112 , 24 + 103 , 24 + 94, \cdots ,24 + 22&lt;/math&gt; (remember &lt;math&gt;n = 3a + 2b + c&lt;/math&gt; and if you have difficulty in understanding how we got &lt;math&gt;n = 24 + (\cdots)&lt;/math&gt; then just put the values of &lt;math&gt;a,b,c&lt;/math&gt; i am sure you will get it :) )<br /> <br /> Let's write the sequences of &lt;math&gt;n&lt;/math&gt;'s in a compact form , &lt;math&gt;T_p = 24 + 22 + 9(p-1)&lt;/math&gt; (This will be helpful later on)<br /> There are &lt;math&gt;\boxed{11}&lt;/math&gt; values of &lt;math&gt;n&lt;/math&gt;<br /> <br /> Case 3 : &lt;math&gt;a = 7&lt;/math&gt;<br /> <br /> We get &lt;math&gt;11b + c = 223 \implies b=0,1,2, \cdots ,20&lt;/math&gt; and &lt;math&gt;c = 223,212,201, \cdots ,3&lt;/math&gt; From here we get different &lt;math&gt;n&lt;/math&gt;'s as &lt;math&gt;21 + 223, 21 + 214 , 21 + 205, \cdots ,21 + 43&lt;/math&gt;<br /> <br /> Let's also write the sequences of &lt;math&gt;n&lt;/math&gt;'s in a compact form , &lt;math&gt;T_q = 21 + 43 + 9(q-1)&lt;/math&gt;<br /> <br /> Now comes the major part , since we need to find distinct &lt;math&gt;n&lt;/math&gt;'s so let's subtract the cases where we find common values , from the total number of values.<br /> <br /> To do this we need to make &lt;math&gt;T_p = T_q \implies p - q = 2&lt;/math&gt; (after some calculations you'll get &lt;math&gt;p - q = 2&lt;/math&gt;) . Now we know that &lt;math&gt;0 \le p \le 10&lt;/math&gt; and &lt;math&gt;0 \le q \le 20&lt;/math&gt; so we get &lt;math&gt;p&lt;/math&gt; as &lt;math&gt;10,9,\cdots,2&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; as &lt;math&gt;8,7,\cdots,0&lt;/math&gt; so there are 9 common solutions out of 21(diff values of &lt;math&gt;q&lt;/math&gt;) total , so there are &lt;math&gt;\boxed{21 - 9}&lt;/math&gt; values of &lt;math&gt;n&lt;/math&gt;<br /> <br /> Case 4 : &lt;math&gt;a = 6&lt;/math&gt;<br /> <br /> We get &lt;math&gt;11b + c = 334 \implies b = 0,1,2, \cdots ,30&lt;/math&gt; and &lt;math&gt;c = 334,323,312, \cdots ,4&lt;/math&gt; From here we get the different &lt;math&gt;n&lt;/math&gt;'s as &lt;math&gt;18 + 334,18 + 325,18 + 316, \cdots ,18 + 64&lt;/math&gt;<br /> <br /> Compact form of terms is &lt;math&gt;T_r = 18 + 64 + 9(r-1)&lt;/math&gt;<br /> Now let's repeat the process of eliminating common solutions (one thing to notice is that we've removed common solutions of case 2 from case 3 so if we check case 4 with case 3 we'll remove common solutions from all previous cases and hence we do not have to handle common solutions of case 4 with case 2,1 and in general case X with case X-2,X-3,...2,1 , {basically means checking case X with case X-1 is enough})<br /> <br /> &lt;math&gt;T_q = T_r \implies 21 + 43 + 9(q-1) = 18 + 64 + 9(r-1) \implies q - r = 2&lt;/math&gt;<br /> &lt;math&gt;\implies q = 20,19,18, \cdots ,2&lt;/math&gt; and &lt;math&gt;r = 18,17,16, \cdots ,0&lt;/math&gt; so there are 19 common solutions out of 31 (diff values of r) so there are &lt;math&gt;\boxed{31 - 19}&lt;/math&gt; values of &lt;math&gt;n&lt;/math&gt;<br /> <br /> Now hopefully you've seen a pattern , if not try another case 5 with &lt;math&gt;a = 5&lt;/math&gt; you'll get &lt;math&gt;\boxed{41 - 29}&lt;/math&gt; values of &lt;math&gt;n&lt;/math&gt;<br /> <br /> Like this we get the value of &quot;distinct&quot; &lt;math&gt;n&lt;/math&gt; by summing all the sub-values from the cases that we've solved.<br /> <br /> &lt;math&gt;n = 1 + 11 + (21 - 9) + (31 - 19) + (41 - 29) + \cdots + (91 - 79)&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies n = (1 + 11 + 21 + 31 + 41 + \cdots + 91) - (0 + 0 + 9 + 19 + 29 + \cdots + 79)&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies \boxed{n = 108}&lt;/math&gt;<br /> <br /> ~MrOreoJuice<br /> <br /> == See also ==<br /> <br /> {{AIME box|year=2004|n=II|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_2&diff=147020 2005 AIME II Problems/Problem 2 2021-02-16T01:07:03Z <p>Kkyyhe: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the [[probability]] each guest got one roll of each type is &lt;math&gt; \frac mn, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are [[relatively prime]] [[integer]]s, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Use [[construction]]. We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined. <br /> <br /> *Person 1: &lt;math&gt;\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}&lt;/math&gt;<br /> <br /> *Person 2: &lt;math&gt;\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25&lt;/math&gt;<br /> *Person 3: One roll of each type is left, so the probability here is &lt;math&gt;1&lt;/math&gt;. <br /> <br /> Our answer is thus &lt;math&gt;\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}&lt;/math&gt;, and &lt;math&gt;m + n = \boxed{79}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in &lt;math&gt;\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216&lt;/math&gt; different manners. The total number of possible strings is &lt;math&gt;\frac{9!}{3!3!3!} = 1680&lt;/math&gt;. The solution is therefore &lt;math&gt;\frac{216}{1680} = \frac{9}{70}&lt;/math&gt;, and &lt;math&gt;m + n = \boxed{79}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to &lt;math&gt;{9 \choose 3}{6 \choose3}&lt;/math&gt; as the amount of ways to select three rolls out of 9 to give to the first person is &lt;math&gt;{9 \choose 3}&lt;/math&gt;, and three rolls out of 6 is &lt;math&gt;{6 \choose3}&lt;/math&gt;. After that, the three remaining rolls have no more configurations. <br /> <br /> The numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored. <br /> <br /> xxx, yyy, zzz<br /> <br /> So you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving &lt;math&gt;3^3&lt;/math&gt; combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently &lt;math&gt;2^3&lt;/math&gt; since there are two of each letter left. <br /> <br /> &lt;math&gt;(27*8)/{9 \choose 3}{6 \choose3}&lt;/math&gt; simplifies down to our fraction m/n, which is &lt;math&gt;9/70&lt;/math&gt;. Adding them up gives &lt;math&gt;9 + 70 = \boxed{79}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=128160 2016 AMC 12B Problems/Problem 16 2020-07-12T21:57:23Z <p>Kkyyhe: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.<br /> <br /> For the first case, we can cleverly choose the convenient form of our sequence to be<br /> &lt;cmath&gt;a-n,\cdots, a-1, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> <br /> because then our sum will just be &lt;math&gt;(2n+1)a&lt;/math&gt;. We now have <br /> &lt;cmath&gt;(2n+1)a = 345&lt;/cmath&gt;<br /> and &lt;math&gt;a&lt;/math&gt; will have a solution when &lt;math&gt;\frac{345}{2n+1}&lt;/math&gt; is an integer, namely when &lt;math&gt;2n+1&lt;/math&gt; is a divisor of 345. We check that <br /> &lt;cmath&gt;2n+1 = 3, 5, 15, 23&lt;/cmath&gt;<br /> work, and no more, because &lt;math&gt;2n+1=1&lt;/math&gt; does not satisfy the requirements of two or more consecutive integers, and when &lt;math&gt;2n+1&lt;/math&gt; equals the next biggest factor, &lt;math&gt;69&lt;/math&gt;, there must be negative integers in the sequence. Our solutions are &lt;math&gt;\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}&lt;/math&gt;.<br /> <br /> For the even cases, we choose our sequence to be of the form:<br /> &lt;cmath&gt;a-(n-1), \cdots, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> so the sum is &lt;math&gt;\frac{(2n)(2a+1)}{2} = n(2a+1)&lt;/math&gt;. In this case, we find our solutions to be &lt;math&gt;\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}&lt;/math&gt;.<br /> <br /> We have found all 7 solutions and our answer is &lt;math&gt;\boxed{\textbf{(E)} \, 7}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> <br /> The sum from &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and &lt;math&gt;a&gt;b &lt;/math&gt; is <br /> <br /> &lt;math&gt;S=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;345=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)&lt;/math&gt;<br /> <br /> Let &lt;math&gt;c=a-b+1&lt;/math&gt; and &lt;math&gt;d=a+b&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=c\cdot d&lt;/math&gt;<br /> <br /> If we factor &lt;math&gt;690&lt;/math&gt; into all of its factor groups &lt;math&gt;(\text{exg}~ (10,69) ~\text{or} ~(15,46))&lt;/math&gt; we will have several ordered pairs &lt;math&gt;(c,d)&lt;/math&gt; where &lt;math&gt;c&lt;d&lt;/math&gt;<br /> <br /> The number of possible values for &lt;math&gt;c&lt;/math&gt; is half the number of factors of &lt;math&gt;690&lt;/math&gt; which is &lt;math&gt;\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8&lt;/math&gt; <br /> <br /> However, we have one extraneous case of &lt;math&gt;(1,690)&lt;/math&gt; because here, &lt;math&gt;a=b&lt;/math&gt; and we have the sum of one consecutive number which is not allowed by the question.<br /> <br /> Thus the answer is &lt;math&gt;8-1=7&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Logic)==<br /> <br /> The consecutive sums can be written as &lt;math&gt;345=kn+\sum_{i=1}^{k-1}{i}&lt;/math&gt; where &lt;math&gt;k&lt;/math&gt; is the number of terms in a sequence, and &lt;math&gt;n&lt;/math&gt; is the first term. Then, &lt;math&gt;\{k,n\}\in \mathbb{N}&lt;/math&gt; and &lt;math&gt;k\geq2&lt;/math&gt;. Evaluating the sum and rearranging yields &lt;math&gt;n=\frac{345}{k}-\frac{k-1}{2}&lt;/math&gt;. <br /> <br /> <br /> The prime factorization of &lt;math&gt;345&lt;/math&gt; is &lt;math&gt;1\cdot3\cdot5\cdot23&lt;/math&gt;. Then, &lt;math&gt;3\cdot5&lt;/math&gt;, &lt;math&gt;3\cdot23&lt;/math&gt;, and &lt;math&gt;5\cdot23&lt;/math&gt; are also divisors. As odd divisors of &lt;math&gt;345&lt;/math&gt;, note that they all produce integer solutions to &lt;math&gt;n&lt;/math&gt; as &lt;math&gt;k&lt;/math&gt;. Only &lt;math&gt;k!=1&lt;/math&gt; is not valid, as &lt;math&gt;k\geq2&lt;/math&gt;. Similarly, quickly notice that &lt;math&gt;k=2&lt;/math&gt; is a solution. Multiplying an odd divisor by &lt;math&gt;2&lt;/math&gt; always yields an integer solution (see below). As such, the even integer solutions are &lt;math&gt;k=2, 2\cdot3, 2\cdot5, 2\cdot15 ... 2\cdot345&lt;/math&gt;.<br /> <br /> <br /> Note that the function is decreasing for increasing values of &lt;math&gt;k&lt;/math&gt; and that &lt;math&gt;\frac{345}{k}-\frac{k-1}{2}=4&lt;/math&gt; for &lt;math&gt;k=23&lt;/math&gt;. Thus, when &lt;math&gt;n&lt;/math&gt; is negative, &lt;math&gt;k&lt;/math&gt; is only slightly more than &lt;math&gt;k=23&lt;/math&gt;. Recall &lt;math&gt;\{k,n\}\in \mathbb{N}&lt;/math&gt;. Since the next highest solution, &lt;math&gt;k=2\cdot23=46&lt;/math&gt; is twice &lt;math&gt;k=23&lt;/math&gt;, &lt;math&gt;k\leq23&lt;/math&gt;. Thus, the remaining solutions are when &lt;math&gt;k=2,6,10,3,5,15,23&lt;/math&gt; &lt;math&gt;\implies \boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> <br /> <br /> :''This is proven by the fact that an odd number (&lt;math&gt;345&lt;/math&gt; in this case) divided by an divisor always yields an odd number. Dividing this by &lt;math&gt;2&lt;/math&gt; produces a number in the form &lt;math&gt;z+\frac{k-1}{2}&lt;/math&gt;, where z is an integer. Since an even number multiplied by an odd number also yields an even number, &lt;math&gt;\frac{k-1}{2}&lt;/math&gt; similarly is in the form &lt;math&gt;z+\frac{k-1}{2}&lt;/math&gt;, producing an integer.''<br /> <br /> (Solution by BJHHar)<br /> <br /> <br /> <br /> ==Solution 5==<br /> <br /> We're dealing with an increasing arithmetic progression of common difference 1. Let &lt;math&gt;x&lt;/math&gt; be the number of terms in a summation. Let &lt;math&gt;y&lt;/math&gt; be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x \cdot \frac{y+y+[(x-1)1]}{2}&amp;=345 \\<br /> 2xy+x^2-x&amp;=690<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that &lt;math&gt;x&lt;/math&gt;... voilà!<br /> <br /> &lt;cmath&gt;(x)(x+2y-1)=690&lt;/cmath&gt;<br /> <br /> There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is &lt;math&gt;2y-1&lt;/math&gt; more than the expression in the parenthesis on the left. &lt;math&gt;y&lt;/math&gt; is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the &quot;increasing sequence&quot; has to have &quot;two or more&quot; terms, so &lt;math&gt;x \geqslant 2&lt;/math&gt;. This eliminates the factor pair &lt;math&gt;1 \cdot 690&lt;/math&gt;. With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found &lt;math&gt;\boxed{\textbf{(E)}\ 7}&lt;/math&gt; ways to write 345 in the silly way outlined by the problem.<br /> <br /> <br /> ==Solution 6==<br /> <br /> By the sum of an arithmetic sequence... this ultimately comes to<br /> &lt;math&gt;n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\cdot3\cdot5\cdot2&lt;/math&gt;.<br /> <br /> Quick testing (would take you roughly a minute)<br /> <br /> <br /> We see that the first 7 values of &lt;math&gt;p&lt;/math&gt; that work are<br /> <br /> &lt;math&gt;p=1,2,4,5,9,14,22&lt;/math&gt;.<br /> <br /> <br /> We see that each one of them works.<br /> Hence, the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_12&diff=122247 2020 AMC 12B Problems/Problem 12 2020-05-10T20:40:28Z <p>Kkyyhe: /* Solution 2 (Power of a Point) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\overline{AB}&lt;/math&gt; be a diameter in a circle of radius &lt;math&gt;5\sqrt2.&lt;/math&gt; Let &lt;math&gt;\overline{CD}&lt;/math&gt; be a chord in the circle that intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at a point &lt;math&gt;E&lt;/math&gt; such that &lt;math&gt;BE=2\sqrt5&lt;/math&gt; and &lt;math&gt;\angle AEC = 45^{\circ}.&lt;/math&gt; What is &lt;math&gt;CE^2+DE^2?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle, and &lt;math&gt;X&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{CD}&lt;/math&gt;. Let &lt;math&gt;CX=a&lt;/math&gt; and &lt;math&gt;EX=b&lt;/math&gt;. This implies that &lt;math&gt;DE = a - b&lt;/math&gt;. Since &lt;math&gt;CE = CX + EX = a + b&lt;/math&gt;, we now want to find &lt;math&gt;(a+b)^2+(a-b)^2=2(a^2+b^2)&lt;/math&gt;. Since &lt;math&gt;\angle CXO&lt;/math&gt; is a right angle, by Pythagorean theorem &lt;math&gt;a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50&lt;/math&gt;. Thus, our answer is &lt;math&gt;2(50)=\boxed{\textbf{(E) } 100}&lt;/math&gt;. <br /> <br /> ~JHawk0224<br /> <br /> ==Solution 2 (Power of a Point)==<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle, and &lt;math&gt;X&lt;/math&gt; be the midpoint of &lt;math&gt;CD&lt;/math&gt;. Draw triangle &lt;math&gt;OCD&lt;/math&gt;, and median &lt;math&gt;OX&lt;/math&gt;. Because &lt;math&gt;OC = OD&lt;/math&gt;, &lt;math&gt;OCD&lt;/math&gt; is isosceles, so &lt;math&gt;OX&lt;/math&gt; is also an altitude of &lt;math&gt;OCD&lt;/math&gt;. &lt;math&gt;OD = 5\sqrt2 - 2\sqrt5&lt;/math&gt;, and because angle &lt;math&gt;OEC&lt;/math&gt; is &lt;math&gt;45&lt;/math&gt; degrees and triangle &lt;math&gt;OXE&lt;/math&gt; is right, &lt;math&gt;OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}&lt;/math&gt;. Because triangle &lt;math&gt;OXC&lt;/math&gt; is right, &lt;math&gt;CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}&lt;/math&gt;. Thus, &lt;math&gt;CD = 2\sqrt{15 + 10\sqrt{10}}&lt;/math&gt;. We are looking for &lt;math&gt;CE^2&lt;/math&gt; + &lt;math&gt;DE^2&lt;/math&gt; which is also &lt;math&gt;(CE + DE)^2 - 2 \cdot CE \cdot DE&lt;/math&gt;. Because &lt;math&gt;CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}, (CE + DE)^2 = 4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}&lt;/math&gt;. By power of a point, &lt;math&gt;CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20&lt;/math&gt; so &lt;math&gt;2 \cdot CE \cdot DE = 40\sqrt{10} - 40&lt;/math&gt;. Finally, &lt;math&gt;CE^2 + DE^2 = 60 + 40\sqrt{10} - (40\sqrt{10} - 40) = \boxed{(E) 100}&lt;/math&gt;.<br /> <br /> ~CT17<br /> <br /> ==Solution 3 (Law of Cosines)==<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Notice how &lt;math&gt;OC = OD = r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of the circle. By applying the law of cosines on triangle &lt;math&gt;OCE&lt;/math&gt;, &lt;math&gt;r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}&lt;/math&gt;. Similarly, by applying the law of cosines on triangle &lt;math&gt;ODE&lt;/math&gt;, &lt;math&gt;r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}&lt;/math&gt;. By subtracting these two equations, we get &lt;math&gt;CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0&lt;/math&gt;. We can rearrange it to get &lt;math&gt;CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2})&lt;/math&gt;. Because both &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; are both positive, we can safely divide both sides by &lt;math&gt;(CE+DE)&lt;/math&gt; to obtain &lt;math&gt;CE-DE=OE\sqrt{2}&lt;/math&gt;. Because &lt;math&gt;OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}&lt;/math&gt;, &lt;math&gt;(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}&lt;/math&gt;. Through power of a point, we can find out that &lt;math&gt;(CE)(DE)=20\sqrt{10} - 20&lt;/math&gt;, so &lt;math&gt;CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{\textbf{(E) } 100}&lt;/math&gt;. <br /> <br /> ~Math_Wiz_3.14<br /> <br /> ==Solution 4 (Reflections)==<br /> &lt;asy&gt;<br /> draw(circle((0,0),7.07));<br /> dot((-7.07,0));<br /> label(&quot;A&quot;,(-7.07,0),W);<br /> dot((7.07,0));<br /> label(&quot;B&quot;,(7.07,0),E);<br /> dot((0,0));<br /> label(&quot;O&quot;,(0,0),N);<br /> dot((-2.24,-6.71));<br /> label(&quot;C&quot;,(-2.24,-6.71),SSW);<br /> dot((6.71,2.24));<br /> label(&quot;D&quot;,(6.71,2.24),NE);<br /> draw((-2.24,-6.71)--(6.71,2.24));<br /> dot((6.71,-2.24));<br /> label(&quot;D'&quot;,(6.71,-2.24),SE);<br /> draw((4.51,0)--(6.71,-2.24));<br /> dot((4.51,0));<br /> label(&quot;E&quot;,(4.51,0),NNW);<br /> draw((-7.07,0)--(7.07,0));<br /> draw((0,0)--(-2.24,-6.71));<br /> draw((0,0)--(6.71,-2.24));<br /> draw((-2.24,-6.71)--(6.71,-2.24));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. By reflecting &lt;math&gt;D&lt;/math&gt; across the line &lt;math&gt;AB&lt;/math&gt; to produce &lt;math&gt;D'&lt;/math&gt;, we have that &lt;math&gt;\angle BED'=45&lt;/math&gt;. Since &lt;math&gt;\angle AEC=45&lt;/math&gt;, &lt;math&gt;\angle CED'=90&lt;/math&gt;. Since &lt;math&gt;DE=ED'&lt;/math&gt;, by the Pythagorean Theorem, our desired solution is just &lt;math&gt;CD'^2&lt;/math&gt;.<br /> Looking next to circle arcs, we know that &lt;math&gt;\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45&lt;/math&gt;, so &lt;math&gt;\overarc{AC}+\overarc{BD}=90&lt;/math&gt;. Since &lt;math&gt;\overarc{BD'}=\overarc{BD}&lt;/math&gt;, and &lt;math&gt;\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180&lt;/math&gt;, &lt;math&gt;\overarc{CD'}=90&lt;/math&gt;. Thus, &lt;math&gt;\angle COD'=90&lt;/math&gt;.<br /> Since &lt;math&gt;OC=OD'=5\sqrt{2}&lt;/math&gt;, by the Pythagorean Theorem, the desired &lt;math&gt;CD'^2= \boxed{\textbf{(E) } 100}&lt;/math&gt;.<br /> <br /> ~sofas103<br /> <br /> ==Video Solution==<br /> On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/0xgTR3UEqbQ<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=108081 2019 AMC 10B Problems/Problem 21 2019-07-31T00:50:45Z <p>Kkyyhe: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We firstly want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the second head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is &lt;math&gt;THTHH&lt;/math&gt;, which has a probability of &lt;math&gt;\frac{1}{2^5} = \frac{1}{32}&lt;/math&gt;. Furthermore, she can prolong her coin flipping by adding an extra &lt;math&gt;TH&lt;/math&gt;, which itself has a probability of &lt;math&gt;\frac{1}{2^2} = \frac{1}{4}&lt;/math&gt;. Since she can do this indefinitely, this gives an infinite geometric series, which means the answer (by the geometric series sum formula) is &lt;math&gt;\frac{\frac{1}{32}}{1-\frac{1}{4}} = \boxed{\textbf{(B) }\frac{1}{24}}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Easier)==<br /> Note that the sequence must start in THT, which happens with &lt;math&gt;\frac{1}{8}&lt;/math&gt; probability. Now, let &lt;math&gt;P&lt;/math&gt; be the probability that Debra will get two heads in a row after flipping THT. Either Debra flips two heads in a row immediately (probability &lt;math&gt;\frac{1}{4}&lt;/math&gt;), or flips a head and then a tail and reverts back to the &quot;original position&quot; (probability &lt;math&gt;\frac{1}{4}P&lt;/math&gt;). Therefore, &lt;math&gt;P=\frac{1}{4}+\frac{1}{4}P&lt;/math&gt;, so &lt;math&gt;P=\frac{1}{3}&lt;/math&gt;, so our final answer is &lt;math&gt;\frac{1}{8}\times\frac{1}{3}=\boxed{\textbf{(B) }\frac{1}{24}}&lt;/math&gt;. -Stormersyle get rect<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=2f1zEvfUe9o<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Kkyyhe https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=108079 2019 AMC 10B Problems/Problem 20 2019-07-31T00:44:01Z <p>Kkyyhe: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br /> ==Problem==<br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;<br /> where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> ==Solution==<br /> Divide the circle into four parts: the top semicircle (&lt;math&gt;A&lt;/math&gt;); the bottom sector (&lt;math&gt;B&lt;/math&gt;), whose arc angle is &lt;math&gt;120^{\circ}&lt;/math&gt; because the large circle's radius is &lt;math&gt;2&lt;/math&gt; and the short length (the radius of the smaller semicircles) is &lt;math&gt;1&lt;/math&gt;, giving a &lt;math&gt;30^{\circ}-60^{\circ}-90^{\circ}&lt;/math&gt; triangle; the triangle formed by the radii of &lt;math&gt;A&lt;/math&gt; and the chord (&lt;math&gt;C&lt;/math&gt;), and the four parts which are the corners of a circle inscribed in a square (&lt;math&gt;D&lt;/math&gt;). Then the area is &lt;math&gt;A + B - C + D&lt;/math&gt; (in &lt;math&gt;B-C&lt;/math&gt;, we find the area of the shaded region above the semicircles but below the diameter, and in &lt;math&gt;D&lt;/math&gt; we find the area of the bottom shaded region).<br /> <br /> The area of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \pi \cdot 2^2 = 2\pi&lt;/math&gt;.<br /> <br /> The area of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}&lt;/math&gt;.<br /> <br /> For the area of &lt;math&gt;C&lt;/math&gt;, the radius of &lt;math&gt;2&lt;/math&gt;, and the distance of &lt;math&gt;1&lt;/math&gt; (the smaller semicircles' radius) to &lt;math&gt;BC&lt;/math&gt;, creates two &lt;math&gt;30^{\circ}-60^{\circ}-90^{\circ}&lt;/math&gt; triangles, so &lt;math&gt;C&lt;/math&gt;'s area is &lt;math&gt;2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}&lt;/math&gt;.<br /> <br /> The area of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi&lt;/math&gt;.<br /> <br /> Hence, finding &lt;math&gt;A+B-C+D&lt;/math&gt;, the desired area is &lt;math&gt;\frac{7\pi}{3}-\sqrt{3}+4&lt;/math&gt;, so the answer is &lt;math&gt;7+3+3+4=\boxed{\textbf{(E) } 17}&lt;/math&gt;.<br /> <br /> ~scrabbler94<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Kkyyhe