https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Burunduchok&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-11T10:51:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User:Burunduchok&diff=93843 User:Burunduchok 2018-04-08T13:40:03Z <p>Burunduchok: </p> <hr /> <div>Hello. &lt;math&gt;e^{i \theta} = \cos \theta + i \sin \theta&lt;/math&gt;</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=User:Burunduchok&diff=93842 User:Burunduchok 2018-04-08T13:38:43Z <p>Burunduchok: Created page with &quot;Hello.&quot;</p> <hr /> <div>Hello.</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_11&diff=93841 2018 AIME II Problems/Problem 11 2018-04-08T13:38:07Z <p>Burunduchok: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of permutations of &lt;math&gt;1, 2, 3, 4, 5, 6&lt;/math&gt; such that for each &lt;math&gt;k&lt;/math&gt; with &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\leq&lt;/math&gt; &lt;math&gt;k&lt;/math&gt; &lt;math&gt;\leq&lt;/math&gt; &lt;math&gt;5&lt;/math&gt;, at least one of the first &lt;math&gt;k&lt;/math&gt; terms of the permutation is greater than &lt;math&gt;k&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> If the first number is &lt;math&gt;6&lt;/math&gt;, then there are no restrictions. There are &lt;math&gt;5!&lt;/math&gt;, or &lt;math&gt;120&lt;/math&gt; ways to place the other &lt;math&gt;5&lt;/math&gt; numbers.<br /> <br /> <br /> If the first number is &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt; can go in four places, and there are &lt;math&gt;4!&lt;/math&gt; ways to place the other &lt;math&gt;4&lt;/math&gt; numbers. &lt;math&gt;4 \cdot 4! = 96&lt;/math&gt; ways.<br /> <br /> <br /> If the first number is &lt;math&gt;4&lt;/math&gt;, ....<br /> <br /> 4 6 _ _ _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 4 _ 6 _ _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 4 _ _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 4 _ _ _ 6 _ &lt;math&gt;\implies&lt;/math&gt; 5 must go between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, so there are &lt;math&gt;3 \cdot 3! = 18&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;24 + 24 + 24 + 18 = 90&lt;/math&gt; ways if 4 is first.<br /> <br /> <br /> If the first number is &lt;math&gt;3&lt;/math&gt;, ....<br /> <br /> 3 6 _ _ _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 3 _ 6 _ _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 3 1 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 4 ways <br /> <br /> 3 2 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 4 ways<br /> <br /> 3 4 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 3 5 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 3 5 _ _ 6 _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 3 _ 5 _ 6 _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 3 _ _ 5 6 _ &lt;math&gt;\implies&lt;/math&gt; 4 ways<br /> <br /> &lt;math&gt;24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84&lt;/math&gt; ways<br /> <br /> <br /> If the first number is &lt;math&gt;2&lt;/math&gt;, ....<br /> <br /> 2 6 _ _ _ _ &lt;math&gt;\implies&lt;/math&gt; 24 ways<br /> <br /> 2 _ 6 _ _ _ &lt;math&gt;\implies&lt;/math&gt; 18 ways <br /> <br /> 2 3 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 4 ways<br /> <br /> 2 4 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 4 ways<br /> <br /> 2 4 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 2 5 _ 6 _ _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 2 5 _ _ 6 _ &lt;math&gt;\implies&lt;/math&gt; 6 ways<br /> <br /> 2 _ 5 _ 6 _ &lt;math&gt;\implies&lt;/math&gt; 4 ways<br /> <br /> 2 4 _ 5 6 _ &lt;math&gt;\implies&lt;/math&gt; 2 ways<br /> <br /> 2 3 4 5 6 1 &lt;math&gt;\implies&lt;/math&gt; 1 way<br /> <br /> <br /> &lt;math&gt;24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71&lt;/math&gt; ways<br /> <br /> <br /> Grand Total : &lt;math&gt;120 + 96 + 90 + 84 + 71 = &lt;/math&gt;&lt;math&gt;\boxed{461}&lt;/math&gt;<br /> <br /> {{AIME box|year=2018|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_6&diff=89957 2015 AIME II Problems/Problem 6 2018-01-24T20:33:31Z <p>Burunduchok: /* Solution 2 (Algebra+ Brute Force) */</p> <hr /> <div>==Problem==<br /> <br /> Steve says to Jon, &quot;I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form &lt;math&gt;P(x) = 2x^3-2ax^2+(a^2-81)x-c&lt;/math&gt; for some positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. Can you tell me the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;?&quot;<br /> <br /> After some calculations, Jon says, &quot;There is more than one such polynomial.&quot;<br /> <br /> Steve says, &quot;You're right. Here is the value of &lt;math&gt;a&lt;/math&gt;.&quot; He writes down a positive integer and asks, &quot;Can you tell me the value of &lt;math&gt;c&lt;/math&gt;?&quot;<br /> <br /> Jon says, &quot;There are still two possible values of &lt;math&gt;c&lt;/math&gt;.&quot;<br /> <br /> Find the sum of the two possible values of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> ==Solution 1 (Algebra)==<br /> <br /> We call the three roots (some may be equal to one another) &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, and &lt;math&gt;x_3&lt;/math&gt;. Using Vieta's formulas, we get &lt;math&gt;x_1+x_2+x_3 = a&lt;/math&gt;, &lt;math&gt;x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}&lt;/math&gt;, and &lt;math&gt;x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}&lt;/math&gt;. <br /> <br /> Squaring our first equation we get &lt;math&gt;x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2&lt;/math&gt;.<br /> <br /> We can then subtract twice our second equation to get &lt;math&gt;x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}&lt;/math&gt;.<br /> <br /> Simplifying the right side:<br /> <br /> &lt;math&gt;a^2-2 \cdot \frac{a^2-81}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;a^2-a^2+81&lt;/math&gt;<br /> <br /> &lt;math&gt;81&lt;/math&gt;<br /> <br /> So, we know &lt;math&gt;x_1^2+x_2^2+x_3^2 = 81&lt;/math&gt;.<br /> <br /> We can then list out all the triples of positive integers whose squares sum to &lt;math&gt;81&lt;/math&gt;:<br /> <br /> We get &lt;math&gt;(1, 4, 8)&lt;/math&gt;, &lt;math&gt;(3, 6, 6)&lt;/math&gt;, and &lt;math&gt;(4, 4, 7)&lt;/math&gt;. <br /> <br /> These triples give &lt;math&gt;a&lt;/math&gt; values of &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, and &lt;math&gt;15&lt;/math&gt;, respectively, and &lt;math&gt;c&lt;/math&gt; values of &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;224&lt;/math&gt;, respectively. <br /> <br /> We know that Jon still found two possible values of &lt;math&gt;c&lt;/math&gt; when Steve told him the &lt;math&gt;a&lt;/math&gt; value, so the &lt;math&gt;a&lt;/math&gt; value must be &lt;math&gt;15&lt;/math&gt;. Thus, the two &lt;math&gt;c&lt;/math&gt; values are &lt;math&gt;216&lt;/math&gt; and &lt;math&gt;224&lt;/math&gt;, which sum to &lt;math&gt;\boxed{\text{440}}&lt;/math&gt;.<br /> <br /> ~BealsConjecture~<br /> <br /> ==Solution 2 (Algebra+ Brute Force)==<br /> First things first. Vietas gives us the following:<br /> <br /> &lt;math&gt;x_1+x_2+x_3 = a&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}&lt;/math&gt; (2)<br /> <br /> &lt;math&gt;x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}&lt;/math&gt; (3)<br /> <br /> From (2), &lt;math&gt;a&lt;/math&gt; must have odd parity, meaning &lt;math&gt;a^2-81&lt;/math&gt; must be a multiple of 4, which implies that both sides of (2) are even. Then, from (1), we see that an odd number of &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, and &lt;math&gt;x_3&lt;/math&gt; must be odd, because we have already deduced that &lt;math&gt;a&lt;/math&gt; is odd. In order for both sides of (2) to be even, there must only be one odd number and 2 even numbers.<br /> <br /> Now, the theoretical maximum value of the left side of (2) is &lt;math&gt;3 \cdot \frac{a}{3}^2=\frac{a^2}{3}&lt;/math&gt;. That means that the maximum bound of &lt;math&gt;a&lt;/math&gt; is where<br /> <br /> &lt;math&gt;\frac{a^2}{3}&gt; \frac{a^2-81}{2}&lt;/math&gt; (4)<br /> <br /> which simplifies to<br /> <br /> &lt;math&gt;\sqrt{243}&gt;a&lt;/math&gt;<br /> <br /> meaning<br /> <br /> &lt;math&gt;16&gt;a&lt;/math&gt; (5)<br /> <br /> So now we have that &lt;math&gt;9&lt;a&lt;/math&gt; from (2), &lt;math&gt;a&lt;16&lt;/math&gt; from (5), and &lt;math&gt;a&lt;/math&gt; is odd from (2). This means that &lt;math&gt;a&lt;/math&gt; could equal &lt;math&gt;11&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, or &lt;math&gt;15&lt;/math&gt;. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of &lt;math&gt;(1, 4, 8)&lt;/math&gt;, &lt;math&gt;(3, 6, 6)&lt;/math&gt;, and &lt;math&gt;(4, 4, 7)&lt;/math&gt;, of which the last two return equal &lt;math&gt;a&lt;/math&gt; values. Then, &lt;math&gt;2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}&lt;/math&gt; AWD.<br /> <br /> Thank you, Rowechen Zhong.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_8&diff=89902 2017 AIME I Problems/Problem 8 2018-01-21T00:22:18Z <p>Burunduchok: /* Solution 3 (Quicker Trig) */</p> <hr /> <div>==Problem 8==<br /> Two real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0, 75)&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; be two points on the plane with &lt;math&gt;OP = 200&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be on the same side of line &lt;math&gt;OP&lt;/math&gt; such that the degree measures of &lt;math&gt;\angle POQ&lt;/math&gt; and &lt;math&gt;\angle POR&lt;/math&gt; are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, and &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are both right angles. The probability that &lt;math&gt;QR \leq 100&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Noting that &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are right angles, we realize that we can draw a semicircle with diameter &lt;math&gt;\overline{OP}&lt;/math&gt; and points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; on the semicircle. Since the radius of the semicircle is &lt;math&gt;100&lt;/math&gt;, if &lt;math&gt;\overline{QR} \leq 100&lt;/math&gt;, then &lt;math&gt;\overarc{QR}&lt;/math&gt; must be less than or equal to &lt;math&gt;60^{\circ}&lt;/math&gt;.<br /> <br /> This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:<br /> <br /> Given &lt;math&gt;a, b&lt;/math&gt; such that &lt;math&gt;0&lt;a, b&lt;75&lt;/math&gt;, what is the probability that &lt;math&gt;|a-b| \leq 30&lt;/math&gt;? <br /> Through simple geometric probability, we get that &lt;math&gt;P = \frac{16}{25}&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;16+25=\boxed{041}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==Solution 2 (Trig Bash)==<br /> Put &lt;math&gt;\triangle POQ&lt;/math&gt; and &lt;math&gt;\triangle POR&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt; on the origin and the triangles on the &lt;math&gt;1^{st}&lt;/math&gt; quadrant.<br /> The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(200 \cos^{2}a,200 \cos a\sin a )&lt;/math&gt;, &lt;math&gt;(200\cos^{2}b,200\cos(b)\sin b)&lt;/math&gt;. So &lt;math&gt;PQ^{2}&lt;/math&gt; = &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}&lt;/math&gt;, which we want to be less then &lt;math&gt;100^{2}&lt;/math&gt;.<br /> So &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} &lt;= 100^{2} &lt;/math&gt;<br /> &lt;cmath&gt;(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt; \sin^{2} (b-a) \le \frac{1}{4} &lt;/cmath&gt;<br /> So we want &lt;math&gt; -\frac{1}{2} \le \sin (b-a) \le \frac{1}{2} &lt;/math&gt;, which is equivalent to &lt;math&gt; -30 \le b-a \le 30&lt;/math&gt; or &lt;math&gt; 150 \le b-a \le 210&lt;/math&gt;. The second inequality is impossible so we only consider what the first inequality does to our &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box in the &lt;math&gt;ab&lt;/math&gt; plane. This cuts off two isosceles right triangles from opposite corners with side lengths &lt;math&gt;45&lt;/math&gt; from the &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box. Hence the probability is &lt;math&gt;1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}&lt;/math&gt; and the answer is &lt;math&gt;16+25 = \boxed{41}&lt;/math&gt;<br /> <br /> Solution by Leesisi<br /> <br /> ==Solution 3 (Quicker Trig)==<br /> &lt;asy&gt;<br /> pair O, P, Q, R;<br /> draw(circle(O, 10));<br /> O = (10, 0);<br /> P = (-10, 0);<br /> Q = (10*cos(pi/3), 10*sin(pi/3));<br /> R = (10*cos(5*pi/6), 10*sin(5*pi/6));<br /> dot(Q);<br /> dot(O);<br /> dot(P);<br /> dot(R);<br /> draw(P--O--Q--P--R--O);<br /> draw(Q--R, red);<br /> label(&quot;$O$&quot;, O, 2*E);<br /> label(&quot;$P$&quot;, P, 2*W);<br /> label(&quot;$Q$&quot;, Q, NE);<br /> label(&quot;$R$&quot;, R, NW);<br /> label(&quot;$200$&quot;, (0,0), 2*S);<br /> label(&quot;$x$&quot;, (Q+R)/2, N);<br /> draw(rightanglemark(O, Q, P, 38));<br /> draw(rightanglemark(O, R, P, 38));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;QR=x.&lt;/math&gt; Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: &lt;math&gt;OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.&lt;/math&gt; Now observe that quadrilateral &lt;math&gt;OQRP&lt;/math&gt; is a [[cyclic quadrilateral]]. Thus, we are able to apply [[Ptolemy's Theorem]] to it:<br /> &lt;cmath&gt;200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x + 200 (\cos a \sin b) = 200 (\sin a \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200(\sin a \cos b - \sin b \cos a),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200 \sin(a-b).&lt;/cmath&gt;<br /> We want &lt;math&gt;|x| \le 100&lt;/math&gt; (the absolute value comes from the fact that &lt;math&gt;a&lt;/math&gt; is not necessarily greater than &lt;math&gt;b,&lt;/math&gt; so we cannot assume that &lt;math&gt;Q&lt;/math&gt; is to the right of &lt;math&gt;R&lt;/math&gt; as in the diagram), so we substitute:<br /> &lt;cmath&gt;|200 \sin(a-b)| \le 100,&lt;/cmath&gt;<br /> &lt;cmath&gt;|\sin(a-b)| \le \frac{1}{2},&lt;/cmath&gt;<br /> &lt;cmath&gt;|a-b| \le 30 ^\circ,&lt;/cmath&gt;<br /> &lt;cmath&gt;-30 \le a-b \le 30.&lt;/cmath&gt;<br /> By simple geometric probability (see Solution 2 for complete explanation), &lt;math&gt;\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},&lt;/math&gt; so &lt;math&gt;m+n = \boxed{041}.&lt;/math&gt;<br /> <br /> ~burunduchok<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_8&diff=89901 2017 AIME I Problems/Problem 8 2018-01-21T00:22:02Z <p>Burunduchok: /* Solution 3 (Quicker Trig) */</p> <hr /> <div>==Problem 8==<br /> Two real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0, 75)&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; be two points on the plane with &lt;math&gt;OP = 200&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be on the same side of line &lt;math&gt;OP&lt;/math&gt; such that the degree measures of &lt;math&gt;\angle POQ&lt;/math&gt; and &lt;math&gt;\angle POR&lt;/math&gt; are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, and &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are both right angles. The probability that &lt;math&gt;QR \leq 100&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Noting that &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are right angles, we realize that we can draw a semicircle with diameter &lt;math&gt;\overline{OP}&lt;/math&gt; and points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; on the semicircle. Since the radius of the semicircle is &lt;math&gt;100&lt;/math&gt;, if &lt;math&gt;\overline{QR} \leq 100&lt;/math&gt;, then &lt;math&gt;\overarc{QR}&lt;/math&gt; must be less than or equal to &lt;math&gt;60^{\circ}&lt;/math&gt;.<br /> <br /> This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:<br /> <br /> Given &lt;math&gt;a, b&lt;/math&gt; such that &lt;math&gt;0&lt;a, b&lt;75&lt;/math&gt;, what is the probability that &lt;math&gt;|a-b| \leq 30&lt;/math&gt;? <br /> Through simple geometric probability, we get that &lt;math&gt;P = \frac{16}{25}&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;16+25=\boxed{041}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==Solution 2 (Trig Bash)==<br /> Put &lt;math&gt;\triangle POQ&lt;/math&gt; and &lt;math&gt;\triangle POR&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt; on the origin and the triangles on the &lt;math&gt;1^{st}&lt;/math&gt; quadrant.<br /> The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(200 \cos^{2}a,200 \cos a\sin a )&lt;/math&gt;, &lt;math&gt;(200\cos^{2}b,200\cos(b)\sin b)&lt;/math&gt;. So &lt;math&gt;PQ^{2}&lt;/math&gt; = &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}&lt;/math&gt;, which we want to be less then &lt;math&gt;100^{2}&lt;/math&gt;.<br /> So &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} &lt;= 100^{2} &lt;/math&gt;<br /> &lt;cmath&gt;(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt; \sin^{2} (b-a) \le \frac{1}{4} &lt;/cmath&gt;<br /> So we want &lt;math&gt; -\frac{1}{2} \le \sin (b-a) \le \frac{1}{2} &lt;/math&gt;, which is equivalent to &lt;math&gt; -30 \le b-a \le 30&lt;/math&gt; or &lt;math&gt; 150 \le b-a \le 210&lt;/math&gt;. The second inequality is impossible so we only consider what the first inequality does to our &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box in the &lt;math&gt;ab&lt;/math&gt; plane. This cuts off two isosceles right triangles from opposite corners with side lengths &lt;math&gt;45&lt;/math&gt; from the &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box. Hence the probability is &lt;math&gt;1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}&lt;/math&gt; and the answer is &lt;math&gt;16+25 = \boxed{41}&lt;/math&gt;<br /> <br /> Solution by Leesisi<br /> <br /> ==Solution 3 (Quicker Trig)==<br /> &lt;asy&gt;<br /> pair O, P, Q, R;<br /> draw(circle(O, 10));<br /> O = (10, 0);<br /> P = (-10, 0);<br /> Q = (10*cos(pi/3), 10*sin(pi/3));<br /> R = (10*cos(5*pi/6), 10*sin(5*pi/6));<br /> dot(Q);<br /> dot(O);<br /> dot(P);<br /> dot(R);<br /> draw(P--O--Q--P--R--O);<br /> draw(Q--R, red);<br /> label(&quot;$O$&quot;, O, 2*E);<br /> label(&quot;$P$&quot;, P, 2*W);<br /> label(&quot;$Q$&quot;, Q, NE);<br /> label(&quot;$R$&quot;, R, NW);<br /> label(&quot;$200$&quot;, (0,0), 2*S);<br /> label(&quot;$x$&quot;, (Q+R)/2, N);<br /> draw(rightanglemark(O, Q, P, 38));<br /> draw(rightanglemark(O, R, P, 38));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;QR=x.&lt;/math&gt; Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: &lt;math&gt;OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.&lt;/math&gt; Now observe that quadrilateral &lt;math&gt;OQRP&lt;/math&gt; is a [[cyclic quadrilateral]]. Thus, we are able to apply [[Ptolemy's Theorem] to it:<br /> &lt;cmath&gt;200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x + 200 (\cos a \sin b) = 200 (\sin a \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200(\sin a \cos b - \sin b \cos a),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200 \sin(a-b).&lt;/cmath&gt;<br /> We want &lt;math&gt;|x| \le 100&lt;/math&gt; (the absolute value comes from the fact that &lt;math&gt;a&lt;/math&gt; is not necessarily greater than &lt;math&gt;b,&lt;/math&gt; so we cannot assume that &lt;math&gt;Q&lt;/math&gt; is to the right of &lt;math&gt;R&lt;/math&gt; as in the diagram), so we substitute:<br /> &lt;cmath&gt;|200 \sin(a-b)| \le 100,&lt;/cmath&gt;<br /> &lt;cmath&gt;|\sin(a-b)| \le \frac{1}{2},&lt;/cmath&gt;<br /> &lt;cmath&gt;|a-b| \le 30 ^\circ,&lt;/cmath&gt;<br /> &lt;cmath&gt;-30 \le a-b \le 30.&lt;/cmath&gt;<br /> By simple geometric probability (see Solution 2 for complete explanation), &lt;math&gt;\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},&lt;/math&gt; so &lt;math&gt;m+n = \boxed{041}.&lt;/math&gt;<br /> <br /> ~burunduchok<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_8&diff=89900 2017 AIME I Problems/Problem 8 2018-01-21T00:16:51Z <p>Burunduchok: /* Solution 3 (Quicker Trig) */</p> <hr /> <div>==Problem 8==<br /> Two real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0, 75)&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; be two points on the plane with &lt;math&gt;OP = 200&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be on the same side of line &lt;math&gt;OP&lt;/math&gt; such that the degree measures of &lt;math&gt;\angle POQ&lt;/math&gt; and &lt;math&gt;\angle POR&lt;/math&gt; are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, and &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are both right angles. The probability that &lt;math&gt;QR \leq 100&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Noting that &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are right angles, we realize that we can draw a semicircle with diameter &lt;math&gt;\overline{OP}&lt;/math&gt; and points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; on the semicircle. Since the radius of the semicircle is &lt;math&gt;100&lt;/math&gt;, if &lt;math&gt;\overline{QR} \leq 100&lt;/math&gt;, then &lt;math&gt;\overarc{QR}&lt;/math&gt; must be less than or equal to &lt;math&gt;60^{\circ}&lt;/math&gt;.<br /> <br /> This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:<br /> <br /> Given &lt;math&gt;a, b&lt;/math&gt; such that &lt;math&gt;0&lt;a, b&lt;75&lt;/math&gt;, what is the probability that &lt;math&gt;|a-b| \leq 30&lt;/math&gt;? <br /> Through simple geometric probability, we get that &lt;math&gt;P = \frac{16}{25}&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;16+25=\boxed{041}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==Solution 2 (Trig Bash)==<br /> Put &lt;math&gt;\triangle POQ&lt;/math&gt; and &lt;math&gt;\triangle POR&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt; on the origin and the triangles on the &lt;math&gt;1^{st}&lt;/math&gt; quadrant.<br /> The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(200 \cos^{2}a,200 \cos a\sin a )&lt;/math&gt;, &lt;math&gt;(200\cos^{2}b,200\cos(b)\sin b)&lt;/math&gt;. So &lt;math&gt;PQ^{2}&lt;/math&gt; = &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}&lt;/math&gt;, which we want to be less then &lt;math&gt;100^{2}&lt;/math&gt;.<br /> So &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} &lt;= 100^{2} &lt;/math&gt;<br /> &lt;cmath&gt;(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt; \sin^{2} (b-a) \le \frac{1}{4} &lt;/cmath&gt;<br /> So we want &lt;math&gt; -\frac{1}{2} \le \sin (b-a) \le \frac{1}{2} &lt;/math&gt;, which is equivalent to &lt;math&gt; -30 \le b-a \le 30&lt;/math&gt; or &lt;math&gt; 150 \le b-a \le 210&lt;/math&gt;. The second inequality is impossible so we only consider what the first inequality does to our &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box in the &lt;math&gt;ab&lt;/math&gt; plane. This cuts off two isosceles right triangles from opposite corners with side lengths &lt;math&gt;45&lt;/math&gt; from the &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box. Hence the probability is &lt;math&gt;1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}&lt;/math&gt; and the answer is &lt;math&gt;16+25 = \boxed{41}&lt;/math&gt;<br /> <br /> Solution by Leesisi<br /> <br /> ==Solution 3 (Quicker Trig)==<br /> &lt;asy&gt;<br /> pair O, P, Q, R;<br /> draw(circle(O, 10));<br /> O = (10, 0);<br /> P = (-10, 0);<br /> Q = (10*cos(pi/3), 10*sin(pi/3));<br /> R = (10*cos(5*pi/6), 10*sin(5*pi/6));<br /> dot(Q);<br /> dot(O);<br /> dot(P);<br /> dot(R);<br /> draw(P--O--Q--P--R--O);<br /> draw(Q--R, red);<br /> label(&quot;$O$&quot;, O, 2*E);<br /> label(&quot;$P$&quot;, P, 2*W);<br /> label(&quot;$Q$&quot;, Q, NE);<br /> label(&quot;$R$&quot;, R, NW);<br /> label(&quot;$200$&quot;, (0,0), 2*S);<br /> label(&quot;$x$&quot;, (Q+R)/2, N);<br /> draw(rightanglemark(O, Q, P, 38));<br /> draw(rightanglemark(O, R, P, 38));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;QR=x.&lt;/math&gt; Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: &lt;math&gt;OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.&lt;/math&gt; Now observe that quadrilateral &lt;math&gt;OQRP&lt;/math&gt; is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it:<br /> &lt;cmath&gt;200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x + 200 (\cos a \sin b) = 200 (\sin a \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200(\sin a \cos b - \sin b \cos a),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200 \sin(a-b).&lt;/cmath&gt;<br /> We want &lt;math&gt;|x| \le 100&lt;/math&gt; (the absolute value comes from the fact that &lt;math&gt;a&lt;/math&gt; is not necessarily greater than &lt;math&gt;b,&lt;/math&gt; so we cannot assume that &lt;math&gt;Q&lt;/math&gt; is to the right of &lt;math&gt;R&lt;/math&gt; as in the diagram), so we substitute:<br /> &lt;cmath&gt;|200 \sin(a-b)| \le 100,&lt;/cmath&gt;<br /> &lt;cmath&gt;|\sin(a-b)| \le \frac{1}{2},&lt;/cmath&gt;<br /> &lt;cmath&gt;|a-b| \le 30 ^\circ,&lt;/cmath&gt;<br /> &lt;cmath&gt;-30 \le a-b \le 30.&lt;/cmath&gt;<br /> By simple geometric probability (see Solution 2 for complete explanation), &lt;math&gt;\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},&lt;/math&gt; so &lt;math&gt;m+n = \boxed{041}.&lt;/math&gt;<br /> <br /> ~burunduchok<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_8&diff=89899 2017 AIME I Problems/Problem 8 2018-01-21T00:16:21Z <p>Burunduchok: </p> <hr /> <div>==Problem 8==<br /> Two real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0, 75)&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; be two points on the plane with &lt;math&gt;OP = 200&lt;/math&gt;. Let &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be on the same side of line &lt;math&gt;OP&lt;/math&gt; such that the degree measures of &lt;math&gt;\angle POQ&lt;/math&gt; and &lt;math&gt;\angle POR&lt;/math&gt; are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, and &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are both right angles. The probability that &lt;math&gt;QR \leq 100&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Noting that &lt;math&gt;\angle OQP&lt;/math&gt; and &lt;math&gt;\angle ORP&lt;/math&gt; are right angles, we realize that we can draw a semicircle with diameter &lt;math&gt;\overline{OP}&lt;/math&gt; and points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; on the semicircle. Since the radius of the semicircle is &lt;math&gt;100&lt;/math&gt;, if &lt;math&gt;\overline{QR} \leq 100&lt;/math&gt;, then &lt;math&gt;\overarc{QR}&lt;/math&gt; must be less than or equal to &lt;math&gt;60^{\circ}&lt;/math&gt;.<br /> <br /> This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:<br /> <br /> Given &lt;math&gt;a, b&lt;/math&gt; such that &lt;math&gt;0&lt;a, b&lt;75&lt;/math&gt;, what is the probability that &lt;math&gt;|a-b| \leq 30&lt;/math&gt;? <br /> Through simple geometric probability, we get that &lt;math&gt;P = \frac{16}{25}&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;16+25=\boxed{041}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==Solution 2 (Trig Bash)==<br /> Put &lt;math&gt;\triangle POQ&lt;/math&gt; and &lt;math&gt;\triangle POR&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt; on the origin and the triangles on the &lt;math&gt;1^{st}&lt;/math&gt; quadrant.<br /> The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(200 \cos^{2}a,200 \cos a\sin a )&lt;/math&gt;, &lt;math&gt;(200\cos^{2}b,200\cos(b)\sin b)&lt;/math&gt;. So &lt;math&gt;PQ^{2}&lt;/math&gt; = &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}&lt;/math&gt;, which we want to be less then &lt;math&gt;100^{2}&lt;/math&gt;.<br /> So &lt;math&gt;(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} &lt;= 100^{2} &lt;/math&gt;<br /> &lt;cmath&gt;(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt;(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} &lt;/cmath&gt;<br /> &lt;cmath&gt; \sin^{2} (b-a) \le \frac{1}{4} &lt;/cmath&gt;<br /> So we want &lt;math&gt; -\frac{1}{2} \le \sin (b-a) \le \frac{1}{2} &lt;/math&gt;, which is equivalent to &lt;math&gt; -30 \le b-a \le 30&lt;/math&gt; or &lt;math&gt; 150 \le b-a \le 210&lt;/math&gt;. The second inequality is impossible so we only consider what the first inequality does to our &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box in the &lt;math&gt;ab&lt;/math&gt; plane. This cuts off two isosceles right triangles from opposite corners with side lengths &lt;math&gt;45&lt;/math&gt; from the &lt;math&gt;75&lt;/math&gt; by &lt;math&gt;75&lt;/math&gt; box. Hence the probability is &lt;math&gt;1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}&lt;/math&gt; and the answer is &lt;math&gt;16+25 = \boxed{41}&lt;/math&gt;<br /> <br /> Solution by Leesisi<br /> <br /> ==Solution 3 (Quicker Trig)==<br /> &lt;asy&gt;<br /> pair O, P, Q, R;<br /> draw(circle(O, 10));<br /> O = (10, 0);<br /> P = (-10, 0);<br /> Q = (10*cos(pi/3), 10*sin(pi/3));<br /> R = (10*cos(5*pi/6), 10*sin(5*pi/6));<br /> dot(Q);<br /> dot(O);<br /> dot(P);<br /> dot(R);<br /> draw(P--O--Q--P--R--O);<br /> draw(Q--R, red);<br /> label(&quot;$O$&quot;, O, 2*E);<br /> label(&quot;$P$&quot;, P, 2*W);<br /> label(&quot;$Q$&quot;, Q, NE);<br /> label(&quot;$R$&quot;, R, NW);<br /> label(&quot;$200$&quot;, (0,0), 2*S);<br /> label(&quot;$x$&quot;, (Q+R)/2, N);<br /> draw(rightanglemark(O, Q, P, 38));<br /> draw(rightanglemark(O, R, P, 38));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;QR=x.&lt;/math&gt; Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: &lt;math&gt;OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.&lt;/math&gt; Now observe that quadrilateral &lt;math&gt;OQRP&lt;/math&gt; is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it:<br /> &lt;cmath&gt;200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x + 200 (\cos a \sin b) = 200 (\sin a \cos b),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200(\sin a \cos b - \sin b \cos a),&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 200 \sin(a-b).&lt;/cmath&gt;<br /> We want &lt;math&gt;|x| \le 100&lt;/math&gt; (the absolute value comes from the fact that &lt;math&gt;a&lt;/math&gt; is not necessarily greater than &lt;math&gt;b,&lt;/math&gt; so we cannot assume that &lt;math&gt;Q&lt;/math&gt; is to the right of &lt;math&gt;R&lt;/math&gt; as in the diagram), so we substitute:<br /> &lt;cmath&gt;|200 \sin(a-b)| \le 100,&lt;/cmath&gt;<br /> &lt;cmath&gt;|\sin(a-b)| \le \frac{1}{2},&lt;/cmath&gt;<br /> &lt;cmath&gt;|a-b| \le 30 ^\circ,&lt;/cmath&gt;<br /> &lt;cmath&gt;-30 \le a-b \le 30.&lt;/cmath&gt;<br /> By simple geometric probability (see Solution 2 for complete explanation), &lt;math&gt;\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},&lt;/math&gt; so &lt;math&gt;m+n = \boxed{041}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=Characteristic_polynomial&diff=87814 Characteristic polynomial 2017-10-15T12:49:10Z <p>Burunduchok: /* Intermediate */</p> <hr /> <div>The '''characteristic polynomial''' of a linear [[operator]] refers to the [[polynomial]] whose roots are the [[eigenvalue]]s of the operator. It carries much information about the operator. <br /> <br /> In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of [[#Linear recurrences|linear recurrences]].<br /> <br /> == Definition ==<br /> Suppose &lt;math&gt;A&lt;/math&gt; is a &lt;math&gt;n \times n&lt;/math&gt; [[matrix]] (over a field &lt;math&gt;K&lt;/math&gt;). Then the characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is defined as &lt;math&gt;P_A(t) = \det(tI - A)&lt;/math&gt;, which is a &lt;math&gt;n&lt;/math&gt;th degree polynomial in &lt;math&gt;t&lt;/math&gt;. Here, &lt;math&gt;I&lt;/math&gt; refers to the &lt;math&gt;n\times n&lt;/math&gt; [[identity matrix]].<br /> <br /> Written out, the characteristic polynomial is the [[determinant]]<br /> <br /> &lt;center&gt;&lt;cmath&gt;P_A(t) = \begin{vmatrix}t-a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; t-a_{22} &amp; \cdots &amp; a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; t-a_{nn}\end{vmatrix}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> == Properties ==<br /> An [[eigenvector]] &lt;math&gt;\bold{v} \in K^n&lt;/math&gt; is a non-zero vector that satisfies the relation &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, for some scalar &lt;math&gt;\lambda \in K&lt;/math&gt;. In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number &lt;math&gt;\lambda&lt;/math&gt; is called the [[eigenvalue]]. <br /> <br /> There are at most &lt;math&gt;n&lt;/math&gt; distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is &lt;math&gt;0&lt;/math&gt; [[iff]] the column vectors of the matrix are [[linearly dependent]]. Observe that if &lt;math&gt;\lambda&lt;/math&gt; satisfies &lt;math&gt;\det(\lambda I-A) = 0&lt;/math&gt;, then the column vectors of the &lt;math&gt;n\times n&lt;/math&gt; matrix &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent. Indeed, if we define &lt;math&gt;T = \lambda I - A&lt;/math&gt; and let &lt;math&gt;\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n&lt;/math&gt; denote the column vectors of &lt;math&gt;T&lt;/math&gt;, this is equivalent to saying that there exists not all zero scalars &lt;math&gt;v_i \in K&lt;/math&gt; such that <br /> <br /> &lt;cmath&gt;v_1 \cdot \bold{T}^1 + v_2 \cdot \bold{T}^2 + \cdots + v_n \cdot \bold{T}^n = \bold{O}.&lt;/cmath&gt;<br /> <br /> Hence, there exists a non-zero vector &lt;math&gt;\bold{v} = (v_1, v_2, \ldots, v_n) \in K^n&lt;/math&gt; such that &lt;math&gt;(\lambda I - A) \bold{v} = \bold{O}&lt;/math&gt;. Distributing and re-arranging, &lt;math&gt;A\bold{v} = \lambda\bold{v}&lt;/math&gt;, as desired. In the other direction, if &lt;math&gt;A \bold{v} = \lambda \bold{v}&lt;/math&gt;, then &lt;math&gt;\lambda I \bold{v} - A \bold{v} = \bold{O}&lt;/math&gt;. But then, the column vectors of &lt;math&gt;\lambda I - A&lt;/math&gt; are linearly dependent, so it follows that &lt;math&gt;\det(\lambda I - A) = 0&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Note that if &lt;math&gt;t = 0&lt;/math&gt;, then &lt;math&gt;P_A(0) = \det (tI - A) = \det (-A) = (-1)^n \det (A)&lt;/math&gt;. Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the [[coefficient]] of the &lt;math&gt;t^{n-1}&lt;/math&gt; term of &lt;math&gt;P_A(t)&lt;/math&gt; gives the negative of the [[trace]] of the matrix (which follows from [[Vieta's formulas]]). <br /> <br /> By the [[Hamilton-Cayley Theorem]], the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is &lt;math&gt;P_A(A) = 0&lt;/math&gt;. The [[minimal polynomial]] of &lt;math&gt;A&lt;/math&gt; thus divides the characteristic polynomial &lt;math&gt;p_A&lt;/math&gt;. <br /> <br /> == Linear recurrences ==<br /> Let &lt;math&gt;x_1, x_2, \ldots, &lt;/math&gt; be a sequence of real numbers. Consider a monic [[homogenous]] [[linear recurrence]] of the form <br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n} = c_1x_{n-1} + c_2x_{n-2} + \cdots + c_kx_{n-k}, \quad (*)&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> where &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt; are real constants. The characteristic polynomial of this recurrence is defined as the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(x) = x^k - c_1x^{k-1} - c_2x^{k-2} - \cdots -c_{k-1}x - c_k.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> For example, let &lt;math&gt;F_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] defined by &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt;, and <br /> <br /> &lt;center&gt;&lt;cmath&gt;F_n = F_{n-1} + F_{n-2} \Longleftrightarrow F_n - F_{n-1} - F_{n-2} = 0.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, its characteristic polynomial is &lt;math&gt;x^2 - x - 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial &lt;math&gt;P&lt;/math&gt; are distinct, then suppose the roots are &lt;math&gt;r_1,r_2, \cdots, r_k&lt;/math&gt;. Then, there exists real constants &lt;math&gt;a_1, a_2, \cdots, a_k&lt;/math&gt; such that<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_n = a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> If we evaluate &lt;math&gt;k&lt;/math&gt; different values of &lt;math&gt;x_i&lt;/math&gt; (typically &lt;math&gt;x_0, x_1, \cdots, x_{k-1}&lt;/math&gt;), we can find a linear system in the &lt;math&gt;a_i&lt;/math&gt;s that can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields [[Binet's formula]]. &lt;br&gt;&lt;br&gt;<br /> <br /> If there are roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, suppose &lt;math&gt;r_1 = r_2 = \cdots = r_{k_1}&lt;/math&gt;. Then we would replace the term &lt;math&gt;a_1 \cdot r^n + a_2 \cdot r_2^n + \cdots + a_{k_1} \cdot r_{k_1}^n&lt;/math&gt; with the expression <br /> <br /> &lt;center&gt;&lt;cmath&gt;(a_1 \cdot n^{k_1-1} + a_2 \cdot n^{k_1 - 2} + \cdots + a_{k_1-1} \cdot n + a_{k_1}) \cdot r^n.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> That is, there is now a polynomial in &lt;math&gt;n&lt;/math&gt; multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation &lt;math&gt;x_n = -2x_{n-1} -x_{n-2} \Longleftrightarrow x_n + 2x_{n-1} + x_{n-2} = 0&lt;/math&gt;. It’s characteristic polynomial, &lt;math&gt;x^2 + 2x + 1&lt;/math&gt;, has a &lt;math&gt;-1&lt;/math&gt; double root. Then, its closed form solution is of the type &lt;math&gt;x_n = (-1)^n(an + b)&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Given a linear recurrence of the form &lt;math&gt;x_n - a_1x_{n-1} - \cdots - a_kx_{n-k} = f(n)&lt;/math&gt;, we often try to find a new sequence &lt;math&gt;y_n = x_n - g(n)&lt;/math&gt; such that &lt;math&gt;y_n&lt;/math&gt; is a homogenous linear recurrence. Then, we can find a closed form for &lt;math&gt;y_n&lt;/math&gt;, and then the answer is given by &lt;math&gt;x_n = y_n + g(n)&lt;/math&gt;. <br /> <br /> Of the following proofs, the second and third are more approachable. <br /> <br /> === Proof 1 (Linear Algebra) ===<br /> Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of [[linear algebra]] (upto the [[Spectral Theorem]]).<br /> <br /> Let &lt;math&gt;\bold{y}_{n-1} = \begin{pmatrix} x_{n-1} \\ x_{n-2} \\ \vdots \\ x_{n-k} \end{pmatrix}&lt;/math&gt;. Then, we can express our linear recurrence as the matrix <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = \begin{pmatrix}a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_{k-1} &amp; a_{k} \\ 1 &amp; 0 &amp; 0 &amp; \cdots &amp;0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1 &amp; 0 \end{pmatrix},&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> so that &lt;math&gt;\bold{y}_n = A\bold{y}_{n-1}&lt;/math&gt; (try to verify this). The characteristic polynomial of &lt;math&gt;A&lt;/math&gt; is precisely &lt;math&gt;P_A(t) = \det (tI - A) = a_1 \cdot t^{k-1} + a_2 \cdot t^{k-2} \cdots + a_k = 0&lt;/math&gt;. This is not difficult to show via [[Laplace's expansion]] and induction, and is left as an exercise to the reader. <br /> <br /> If the roots of &lt;math&gt;P_A&lt;/math&gt; are distinct, then there exists a [[basis]] (of &lt;math&gt;\mathbb{R}^{k-1}&lt;/math&gt;) consisting of [[eigenvector]]s of &lt;math&gt;A&lt;/math&gt; (since eigenvectors of different eigenvalues are [[linearly independent]]). That is, applying a change of bases, we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;A = U^{-1} D U&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for a matrix &lt;math&gt;U&lt;/math&gt; and a diagonal matrix &lt;math&gt;D&lt;/math&gt;. Then, &lt;math&gt;A^2 = U^{-1} D U U^{-1} D U = U^{-1} D^2 U&lt;/math&gt;, and in general, &lt;math&gt;A^n = U^{-1} D^n U&lt;/math&gt;. Thus, &lt;math&gt;\bold{y}_{n} = A^{n}\bold{y}_0 = U^{-1}D^{n+k}U\bold{y}_0&lt;/math&gt;. Here, &lt;math&gt;U^{-1}, U,&lt;/math&gt; and &lt;math&gt;\bold{y}_0&lt;/math&gt; are fixed (note that to find the values of &lt;math&gt;\bold{y}_0&lt;/math&gt;, we may need to trace the recurrence backwards. We only take the &lt;math&gt;0&lt;/math&gt;th index for simplicity). It follows that &lt;math&gt;y_{n}&lt;/math&gt; is a linear combination of the diagonal elements of &lt;math&gt;D&lt;/math&gt;, namely &lt;math&gt;r_1^n, r_2^n, \ldots, r_k^n&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> Suppose now that that the roots of &lt;math&gt;P_A&lt;/math&gt; are not distinct. Then, we can write the matrix in the [[Jordan normal form]]. For simplicity, first consider just a single root &lt;math&gt;r&lt;/math&gt; repeated &lt;math&gt;k&lt;/math&gt; times. The corresponding Jordan form of the matrix is given by (that is, the matrix is [[similar]] to the following):<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r &amp; 1 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; r &amp; 1 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r \end{pmatrix}&lt;/cmath&gt;.&lt;/center&gt;<br /> <br /> Exponentiating this matrix to the &lt;math&gt;n&lt;/math&gt;th power will yield [[binomial coefficient]]s as follows<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{pmatrix} r^n &amp; {n \choose 1}r^{n-1} &amp; {n \choose 2}r^{n-2} &amp; \cdots &amp; {n \choose k}r^{n-k} \\ 0 &amp; r &amp; {n \choose 1}r^{n-1} &amp; \cdots &amp; {n \choose {k-1}}r^{n-k+1} \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; r^n \end{pmatrix}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> We can treat the binomial coefficient as a polynomial in &lt;math&gt;n&lt;/math&gt;. Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows. &lt;br&gt;&lt;br&gt;<br /> <br /> See also a &lt;url&gt;viewtopic.php?t=290351 graph-theoretic&lt;/url&gt; approach.<br /> <br /> === Proof 2 (Induction) ===<br /> <br /> There are a couple of lower-level ways to prove this. One is by [[induction]], though the proof is not very revealing; we can explicitly check that a sequence &lt;math&gt;\{a_1 \cdot r_1^n + a_2 \cdot r_2^n + \cdots + a_k \cdot r_k^n\}&lt;/math&gt;, for real numbers &lt;math&gt;a_1, a_2, \ldots, a_k&lt;/math&gt;, satisfies the linear recurrence relation &lt;math&gt;(*)&lt;/math&gt;. If the two sequences are the same for the first &lt;math&gt;k&lt;/math&gt; values of the sequence, it follows by induction that the two sequences must be exactly the same.<br /> <br /> In particular, for &lt;math&gt;k=2&lt;/math&gt;, we can check this by using the identity <br /> <br /> &lt;center&gt;&lt;cmath&gt;ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - xy(ax^{n-1} + by^{n-1}).&lt;/cmath&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> It is also possible to reduce the recurrence to a [[telescoping sum]]. However, the details are slightly messy. <br /> <br /> === Proof 3 (Partial fractions) ===<br /> <br /> Another method uses [[partial fractions]] and [[generating function]]s, though not much knowledge of each is required for this proof. Let &lt;math&gt;G_n = a_1G_{n-1} + \cdots + a_kG_{n-k}&lt;/math&gt; be a linear recurrence. Consider the [[generating function]] given by <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = G_0 + G_1x + G_2x^2 + \cdots&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Then, writing the following expressing out and carefully comparing coefficients (try it), <br /> <br /> &lt;center&gt;&lt;cmath&gt;a_kG(x) \cdot x^{k-1} + a_{k-1}G(x) \cdot x^{k-2} + \cdots + a_{2}G(x) \cdot x + a_{1}G(x) = \frac{G(x) + R(x)}x,&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;R(x)&lt;/math&gt; is a remainder polynomial with degree &lt;math&gt;\le k-1&lt;/math&gt;. Re-arranging, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{R(x)}{a_k\cdot x^{k} + a_{k-1}\cdot x^{k-1} + \cdots + a_{1}x - 1}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are &lt;math&gt;1/r_1, 1/r_2, \ldots, 1/r_k&lt;/math&gt;, assuming that the roots are distinct. Using [[partial fraction]] decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = \frac{c_1}{1 - r_1x} + \frac{c_2}{1-r_2x} + \cdots + \frac{c_k}{1-r_kx } &lt;/cmath&gt;&lt;/center&gt;<br /> <br /> for some constants &lt;math&gt;c_1, \ldots, c_k&lt;/math&gt;. Using the [[geometric series]] formula, we have &lt;math&gt;\frac{c}{1-y} = c + cy + cy^2 + \ldots&lt;/math&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;cmath&gt;G(x) = (c_1 + \cdots + c_k) + (c_1r_1 + \cdots + c_kr_k)x + (c_1r_1^2 + \cdots + c_kr_k^2)x^2 + \cdots.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> Comparing coefficients with our original definition of &lt;math&gt;G(x)&lt;/math&gt; gives &lt;math&gt;G_n = c_1r_1^n + c_2r_2^n + \cdots + c_kr_k^n&lt;/math&gt;, as desired. &lt;br&gt;&lt;br&gt;<br /> <br /> The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form &lt;math&gt;\frac{c_i}{(1-r_ix)^\alpha}&lt;/math&gt; for &lt;math&gt;\alpha &gt; 1&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=250866 blog][http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833 posts] for further ideas.<br /> <br /> == Differential equations ==<br /> Given a monic linear [[homogenous]] [[differential equation]] of the form &lt;math&gt;D^ny +c_{n-1}D^{n-1}y + \cdots + c_1Dy + c_0y = 0&lt;/math&gt;, then the characteristic polynomial of the equation is the polynomial <br /> <br /> &lt;center&gt;&lt;cmath&gt;P(t) = t^n + c_{n-1}t^{n-1} + c_{n-2}t^{n-2} + \cdots + c_0.&lt;/cmath&gt;&lt;/center&gt; <br /> <br /> Here, &lt;math&gt;D = \frac{d}{dx}&lt;/math&gt; is short-hand for the differential operator. &lt;br&gt;&lt;br&gt;<br /> <br /> If the roots of the polynomial are distinct, say &lt;math&gt;r_1, r_2, \cdots, r_n&lt;/math&gt;, then the solutions of this differential equation are precisely the linear combinations &lt;math&gt;y(x) = a_1e^{r_1x} + a_2e^{r_2x} + \cdots + a_ne^{r_nx}&lt;/math&gt;. Similarly, if there is a set of roots with multiplicity greater than &lt;math&gt;1&lt;/math&gt;, say &lt;math&gt;r_1, r_2, \cdots, r_k&lt;/math&gt;, then we would replace the term &lt;math&gt;a_1e^{r_1x} + \cdots + a_ke^{r_kx}&lt;/math&gt; with the expression &lt;math&gt;(a_1x^{k-1} + a_2x^{k-2} + \cdots + a_{k-1}x + a_k)e^{r_1x}&lt;/math&gt;. &lt;br&gt;&lt;br&gt;<br /> <br /> In general, given a linear differential equation of the form &lt;math&gt;Ly = f&lt;/math&gt;, where &lt;math&gt;L = c_nD^n + c_{n-1}D^{n-1} + \cdots + c_0&lt;/math&gt; is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation &lt;math&gt;Ly = 0&lt;/math&gt; and a specific solution to &lt;math&gt;Ly = f&lt;/math&gt;.<br /> <br /> === Proof ===<br /> We can apply an induction technique similar to the section on linear recurrences above. <br /> <br /> From linear algebra, we can use the following [[vector space]] decomposition theorem. Let &lt;math&gt;A: V \to V&lt;/math&gt; be a linear operator for a [[vector space]]s &lt;math&gt;V&lt;/math&gt; over a field &lt;math&gt;K&lt;/math&gt;. Suppose that there exists a polynomial &lt;math&gt;f(x) \in K[x]&lt;/math&gt; such that &lt;math&gt;f = gh&lt;/math&gt;, where &lt;math&gt;g&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are non-zero polynomials such that &lt;math&gt;\text{gcd}\,(g,h) = 1&lt;/math&gt;, and such that &lt;math&gt;f(A) = O&lt;/math&gt;. Then &lt;math&gt;V = \ker g(A) \oplus \ker h(A)&lt;/math&gt;. This allows us to reduce the differential equation into finding the solutions to the equation &lt;math&gt;(D - \lambda I)^my = 0&lt;/math&gt;, which has a basis of functions &lt;math&gt;e^{\lambda t}, te^{\lambda t}, \ldots, t^{m-1}e^{\lambda t}&lt;/math&gt;.<br /> <br /> == Problems ==<br /> === Introductory ===<br /> *Prove [[Binet's formula]]. Find a similar closed form equation for the [[Lucas sequence]], defined with the starting terms &lt;math&gt;L_1 = 2, L_2 = 1&lt;/math&gt;, and satisfying the recursion &lt;math&gt;L_n = L_{n-1} + L_{n-2}&lt;/math&gt;. <br /> *Let &lt;math&gt;\{x_n\}&lt;/math&gt; denote the sequence defined by the recursion &lt;math&gt;x_0 = 3, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_n = 2x_{n-1} + 3x_{n-2}&lt;/math&gt;. Find a closed form expression for &lt;math&gt;x_n&lt;/math&gt;.<br /> *Given &lt;math&gt;a_0 = 1&lt;/math&gt;, &lt;math&gt;a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt;a_n^2 - a_{n - 1}a_{n + 1} = ( - 1)^n&lt;/math&gt; for &lt;math&gt;n \ge 1&lt;/math&gt;. Find a linear recurrence for &lt;math&gt;a_n&lt;/math&gt;. (AHSME 1958, Problem 40)<br /> <br /> === Intermediate ===<br /> *Let &lt;math&gt;S_n&lt;/math&gt; denote the number of ternary sequences (consisting of &lt;math&gt;0&lt;/math&gt;,&lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt;s) of length &lt;math&gt;n&lt;/math&gt;, such that they do not contain a substring of &quot;10&quot;, &quot;01&quot;, or &quot;11&quot;. Find a closed form expression for &lt;math&gt;S_n&lt;/math&gt;. <br /> *Let &lt;math&gt;\{X_n\}&lt;/math&gt; and &lt;math&gt;\{Y_n\}&lt;/math&gt; be sequences defined as follows: <br /> <br /> &lt;center&gt;&lt;cmath&gt;X_0 = Y_0 = X_1 = Y_1 = 1, X_{n+1} = X_n + 2X_{n-1}, Y_{n+1} = 3Y_n + 4Y_{n-1}.&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :Let &lt;math&gt;k&lt;/math&gt; be the largest integer that satisfies all of the following conditions:<br /> <br /> ::(i) &lt;math&gt;|X_i - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;i&lt;/math&gt;;<br /> ::(ii) &lt;math&gt;|Y_j - k| \le 2007&lt;/math&gt;, for some positive integer &lt;math&gt;j&lt;/math&gt;;<br /> ::(iii) &lt;math&gt;k &lt; 10^{2007}&lt;/math&gt;. <br /> <br /> :Find the remainder when &lt;math&gt;k&lt;/math&gt; is divided by &lt;math&gt;2007&lt;/math&gt;. (2007 [[iTest]], #47)<br /> <br /> *Let &lt;math&gt;a_{n}&lt;/math&gt;, &lt;math&gt;b_{n}&lt;/math&gt;, and &lt;math&gt;c_{n}&lt;/math&gt; be geometric sequences with different common ratios and let &lt;math&gt;a_{n}+b_{n}+c_{n}=d_{n}&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt;. If &lt;math&gt;d_{1}=1&lt;/math&gt;, &lt;math&gt;d_{2}=2&lt;/math&gt;, &lt;math&gt;d_{3}=3&lt;/math&gt;, &lt;math&gt;d_{4}=-7&lt;/math&gt;, &lt;math&gt;d_{5}=13&lt;/math&gt;, and &lt;math&gt;d_{6}=-16&lt;/math&gt;, find &lt;math&gt;d_{7}&lt;/math&gt;. ([[Mock AIME 1 2006-2007/Problem 13|Mock AIME 1 2006-2007, Problem 13]])<br /> <br /> *Find all possible values of &lt;math&gt;x_0&lt;/math&gt; and &lt;math&gt;x_1&lt;/math&gt; such that the sequence defined by:<br /> <br /> &lt;center&gt;&lt;cmath&gt;x_{n + 1} = \frac {x_{n - 1} x_n}{3x_{n - 1} - 2x_n}, \quad n \ge 1&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> :contains infinitely many natural numbers.<br /> <br /> *Show that &lt;math&gt;a^n + b^n + c^n&lt;/math&gt; can be written as a linear combination of the [[elementary symmetric polynomials]] &lt;math&gt;abc, ab+bc+ca, a+b+c&lt;/math&gt;. In general, prove [[Newton's sums]]. <br /> <br /> === Olympiad ===<br /> *Let &lt;math&gt;r&lt;/math&gt; be a real number, and let &lt;math&gt;x_n&lt;/math&gt; be a sequence such that &lt;math&gt;x_0 = 0, x_1 = 1&lt;/math&gt;, and &lt;math&gt;x_{n+2} = rx_{n+1} - x_n&lt;/math&gt; for &lt;math&gt;n \ge 0&lt;/math&gt;. For which values of &lt;math&gt;r&lt;/math&gt; does &lt;math&gt;x_1 + x_3 + \cdots + x_{2m-1} = x_m^2&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt;? ([[WOOT]])<br /> * Let &lt;math&gt;a(x,y)&lt;/math&gt; be the polynomial &lt;math&gt;x^2y + xy^2&lt;/math&gt;, and &lt;math&gt;b(x,y)&lt;/math&gt; the polynomial &lt;math&gt;x^2 + xy + y^2&lt;/math&gt;. Prove that we can find a polynomial &lt;math&gt;p_n(a, b)&lt;/math&gt; which is identically equal to &lt;math&gt;(x + y)^n + (-1)^n (x^n + y^n)&lt;/math&gt;. For example, &lt;math&gt;p_4(a, b) = 2b^2&lt;/math&gt;. ([[1976 Putnam Problems/Problem A2|1976 Putnam, Problem A2]]).<br /> *&lt;math&gt;a_1,a_2,a_3,b_1,b_2,b_3&lt;/math&gt; are distinct positive integers such that &lt;math&gt;(n + 1)a_1^n + na_2^n + (n - 1)a_3^n|(n + 1)b_1^n + nb_2^n + (n - 1)b_3^n&lt;/math&gt; holds for all positive integer &lt;math&gt;n&lt;/math&gt;. Prove that there exists &lt;math&gt;k\in N&lt;/math&gt; such that &lt;math&gt;b_i = ka_i&lt;/math&gt; for &lt;math&gt;i = 1,2,3&lt;/math&gt;. (2010 Chinese MO, Problem 6)<br /> <br /> == Hints/Solutions ==<br /> === Introductory ===<br /> * A proof of [[Binet's formula]] may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients. <br /> * We will work out this problem in full detail. The recurrence relation is &lt;math&gt;x_n -2x_{n-1} -3x_{n-2}&lt;/math&gt;, and its characteristic polynomial is given by &lt;math&gt;x^2-2x-3&lt;/math&gt;. The roots of this polynomial are &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. Thus, a closed form solution is given by &lt;math&gt;x_n = a \cdot (-1)^n + b \cdot 3^n&lt;/math&gt;. For &lt;math&gt;n = 0&lt;/math&gt;, we get &lt;math&gt;x_0 = a + b = 3&lt;/math&gt;, and for &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;x_1 = -a + 3b = 1&lt;/math&gt;. Solving gives &lt;math&gt;a = 2, b = 1&lt;/math&gt;. Thus, our answer is &lt;math&gt;x_n = 2 \cdot (-1)^n + 3^n&lt;/math&gt;. <br /> *&lt;url&gt;viewtopic.php?t=282744 Discussion&lt;/url&gt; (1958 AHSME, 40)<br /> <br /> === Intermediate ===<br /> *&lt;url&gt;viewtopic.php?t=335138 Discussion&lt;/url&gt;<br /> *Answer (2007 iTest 47): The answer is &lt;math&gt;k = 7468&lt;/math&gt; (before taking &lt;math&gt;\mod{2007}&lt;/math&gt;).<br /> *&lt;url&gt;viewtopic.php?t=287441 Discussion&lt;/url&gt; <br /> *Hint (Newton’s Sum): Let us work backwards. Suppose &lt;math&gt;a,b,c&lt;/math&gt; are the roots of the characteristic polynomial of a linear recurrence. Then apply [[Vieta's formulas]].<br /> <br /> === Olympiad ===<br /> *Hint (WOOT): substitute the closed form solution. It is true for all &lt;math&gt;r&lt;/math&gt;. <br /> *Hint (1976 Putnam A2): do the even and odd cases separately.<br /> *&lt;url&gt;viewtopic.php?search_id=804457492&amp;t=327474 Discussion&lt;/url&gt; (2010 Chinese MO, 6)<br /> <br /> [[Category:Linear algebra]]</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_7&diff=85412 1988 AIME Problems/Problem 7 2017-04-23T22:40:01Z <p>Burunduchok: /* Solution */</p> <hr /> <div>== Problem ==<br /> In [[triangle]] &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;\tan \angle CAB = 22/7&lt;/math&gt;, and the [[altitude]] from &lt;math&gt;A&lt;/math&gt; divides &lt;math&gt;BC&lt;/math&gt; into [[segment]]s of length 3 and 17. What is the area of triangle &lt;math&gt;ABC&lt;/math&gt;?<br /> <br /> == Solution ==<br /> &lt;center&gt;[[Image:AIME_1988_Solution_07.png]]&lt;/center&gt;<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the intersection of the [[altitude]] with &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt; be the length of the altitude. [[Without loss of generality]], let &lt;math&gt;BD = 17&lt;/math&gt; and &lt;math&gt;CD = 3&lt;/math&gt;. Then &lt;math&gt;\tan \angle DAB = \frac{17}{h}&lt;/math&gt; and &lt;math&gt;\tan \angle CAD = \frac{3}{h}&lt;/math&gt;. Using the [[Trigonometric_identities#Angle_Addition.2FSubtraction_Identities|tangent sum formula]],<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \tan CAB &amp;= \tan (DAB + CAD)\\<br /> \frac{22}{7} &amp;= \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\<br /> &amp;=\frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\<br /> \frac{22}{7} &amp;= \frac{20h}{h^2 - 51}\\<br /> 0 &amp;= 22h^2 - 140h - 22 \cdot 51\\<br /> 0 &amp;= (11h + 51)(h - 11)<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> The postive value of &lt;math&gt;h = 11&lt;/math&gt;, so the area is &lt;math&gt;\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_17&diff=81440 2016 AMC 8 Problems/Problem 17 2016-11-23T16:25:08Z <p>Burunduchok: </p> <hr /> <div>17. An ATM password at Fred's Bank is composed of four digits from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, with repeated digits allowable. If no password may begin with the sequence &lt;math&gt;9,1,1,&lt;/math&gt; then how many passwords are possible?<br /> &lt;math&gt;(A)\mbox{ }30\mbox{ }(B)\mbox{ }7290\mbox{ }(C)\mbox{ }9000\mbox{ }(D)\mbox{ }9990\mbox{ }(E)\mbox{ }9999\mbox{ }&lt;/math&gt;<br /> <br /> {{AMC8 box|year=2016|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=81432 2016 AMC 8 Problems/Problem 6 2016-11-23T16:08:49Z <p>Burunduchok: </p> <hr /> <div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br /> <br /> <br /> ==Solution==</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_4&diff=81334 2016 AMC 8 Problems/Problem 4 2016-11-23T14:07:12Z <p>Burunduchok: /* Solution */</p> <hr /> <div>4. When Cheenu was a boy he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;3&lt;/math&gt; hours and &lt;math&gt;30&lt;/math&gt; minutes. As an old man he can now walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;4&lt;/math&gt; hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30&lt;/math&gt;<br /> <br /> ==Solution==<br /> When Cheenu was a boy, he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;210&lt;/math&gt; minutes, thus running &lt;math&gt;14&lt;/math&gt; minutes per mile. When he is an old man, he can walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;240&lt;/math&gt; minutes, thus walking &lt;math&gt;24&lt;/math&gt; minutes per mile. Therefore it takes him &lt;math&gt;\textbf{(B) }10&lt;/math&gt; minutes longer to walk a mile now compared to when he was a boy.<br /> <br /> <br /> {{AMC8 box|year=2016|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_4&diff=81333 2016 AMC 8 Problems/Problem 4 2016-11-23T14:06:57Z <p>Burunduchok: </p> <hr /> <div>4. When Cheenu was a boy he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;3&lt;/math&gt; hours and &lt;math&gt;30&lt;/math&gt; minutes. As an old man he can now walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;4&lt;/math&gt; hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30&lt;/math&gt;<br /> <br /> ==Solution==<br /> When Cheenu was a boy, he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;210&lt;/math&gt; minutes, thus running &lt;math&gt;14&lt;/math&gt; minutes per mile. When he is an old man, he can walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;240&lt;/math&gt; minutes, thus walking &lt;math&gt;24&lt;/math&gt; minutes per mile. Therefore it takes him &lt;math&gt;\textbf{(B) } 10&lt;/math&gt; minutes longer to walk a mile now compared to when he was a boy.<br /> <br /> <br /> {{AMC8 box|year=2016|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_3&diff=81324 2016 AMC 8 Problems/Problem 3 2016-11-23T14:03:38Z <p>Burunduchok: </p> <hr /> <div>Four students take an exam. Three of their scores are &lt;math&gt;70, 80,&lt;/math&gt; and &lt;math&gt;90&lt;/math&gt;. If the average of their four scores is &lt;math&gt;70&lt;/math&gt;, then what is the remaining score?<br /> <br /> &lt;math&gt;\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that &lt;math&gt;80-70=10&lt;/math&gt; and &lt;math&gt;90-70=20&lt;/math&gt;. We then find that &lt;math&gt;10+20=30&lt;/math&gt;. We want our average to be &lt;math&gt;70&lt;/math&gt;, so we find &lt;math&gt;70-30=40&lt;/math&gt;. So our final answer is &lt;math&gt;\boxed{\textbf{(A) }40}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Answer_Key&diff=81322 2016 AMC 8 Answer Key 2016-11-23T14:02:27Z <p>Burunduchok: </p> <hr /> <div>#C<br /> #A<br /> #A<br /> #B<br /> #E<br /> #<br /> #B<br /> #C<br /> #B<br /> #D<br /> #B<br /> #B<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #E<br /> #A<br /> #<br /> #C<br /> #<br /> #A<br /> #B</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=Trigonometric_identities&diff=81198 Trigonometric identities 2016-11-21T13:31:38Z <p>Burunduchok: /* Other Identities */</p> <hr /> <div>'''Trigonometric identities''' are used to manipulate [[trigonometry]] [[equation]]s in certain ways. Here is a list of them:<br /> <br /> == Basic Definitions ==<br /> The six basic trigonometric functions can be defined using a right triangle:<br /> &lt;center&gt;[[Image:righttriangle.png]]&lt;/center&gt;<br /> <br /> The six trig functions are sine, cosine, tangent, cosecant, secant, and cotangent. They are abbreviated by using the first three letters of their name (except for cosecant which uses &lt;math&gt;\csc&lt;/math&gt;). They are defined as follows:<br /> {| class=&quot;wikitable&quot;<br /> |+ Basic Definitions<br /> |- &lt;!-- Start of a new row --&gt;<br /> | &lt;math&gt;\sin A = \frac ac&lt;/math&gt; || &lt;math&gt;\csc A = \frac ca&lt;/math&gt; || &lt;math&gt; \cos A = \frac bc&lt;/math&gt; || &lt;math&gt;\sec A = \frac cb&lt;/math&gt; || &lt;math&gt; \tan A = \frac ab&lt;/math&gt; || &lt;math&gt; \cot A = \frac ba&lt;/math&gt;<br /> |}<br /> <br /> == Even-Odd Identities ==<br /> *&lt;math&gt;\sin (-\theta) = -\sin (\theta) &lt;/math&gt;<br /> <br /> *&lt;math&gt;\cos (-\theta) = \cos (\theta) &lt;/math&gt;<br /> <br /> *&lt;math&gt;\tan (-\theta) = -\tan (\theta) &lt;/math&gt;<br /> <br /> *&lt;math&gt;\sec (-\theta) = \sec (\theta) &lt;/math&gt;<br /> <br /> *&lt;math&gt;\csc (-\theta) = -\csc (\theta) &lt;/math&gt;<br /> <br /> *&lt;math&gt;\cot (-\theta) = -\cot (\theta) &lt;/math&gt;<br /> <br /> ===Further Conclusions===<br /> <br /> Based on the above identities, we can also claim that<br /> <br /> *&lt;math&gt;\sin(\cos(-\theta)) = \sin(\cos(\theta))&lt;/math&gt;<br /> <br /> *&lt;math&gt;\cos(\sin(-\theta)) = \cos(-\sin(\theta)) = \cos(\sin(\theta))&lt;/math&gt;<br /> <br /> This is only true when &lt;math&gt;\sin(\theta)&lt;/math&gt; is in the domain of &lt;math&gt;\cos(\theta)&lt;/math&gt;.<br /> <br /> == Reciprocal Relations ==<br /> From the first section, it is easy to see that the following hold:<br /> <br /> *&lt;math&gt; \sin A = \frac 1{\csc A}&lt;/math&gt; <br /> <br /> *&lt;math&gt; \cos A = \frac 1{\sec A}&lt;/math&gt;<br /> <br /> *&lt;math&gt; \tan A = \frac 1{\cot A}&lt;/math&gt;<br /> <br /> Another useful identity that isn't a reciprocal relation is that &lt;math&gt; \tan A =\frac{\sin A}{\cos A} &lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;\sin^{-1} A \neq \csc A&lt;/math&gt;; the former refers to the [[inverse trigonometric function]]s.<br /> <br /> == Pythagorean Identities ==<br /> Using the [[Pythagorean Theorem]] on our triangle above, we know that &lt;math&gt;a^2 + b^2 = c^2 &lt;/math&gt;. If we divide by &lt;math&gt;c^2 &lt;/math&gt; we get &lt;math&gt;\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1 &lt;/math&gt;, which is just &lt;math&gt;\sin^2 A + \cos^2 A =1 &lt;/math&gt;. Dividing by &lt;math&gt; a^2 &lt;/math&gt; or &lt;math&gt; b^2 &lt;/math&gt; instead produces two other similar identities. The Pythagorean Identities are listed below:<br /> <br /> *&lt;math&gt;\sin^2x + \cos^2x = 1&lt;/math&gt;<br /> *&lt;math&gt;1 + \cot^2x = \csc^2x&lt;/math&gt;<br /> *&lt;math&gt;\tan^2x + 1 = \sec^2x&lt;/math&gt;<br /> <br /> (Note that the last two are easily derived by dividing the first by &lt;math&gt;\sin^2x&lt;/math&gt; and &lt;math&gt;\cos^2x&lt;/math&gt;, respectively.)<br /> <br /> == Angle Addition/Subtraction Identities ==<br /> Once we have formulas for angle addition, angle subtraction is rather easy to derive. For example, we just look at &lt;math&gt; \sin(\alpha+(-\beta))&lt;/math&gt; and we can derive the sine angle subtraction formula using the sine angle addition formula.<br /> <br /> *&lt;math&gt; \sin(\alpha \pm \beta) = \sin \alpha\cos \beta \pm\sin \beta \cos \alpha&lt;/math&gt;<br /> *&lt;math&gt; \cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta &lt;/math&gt;<br /> *&lt;math&gt;\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1\mp\tan \alpha \tan \beta} &lt;/math&gt;<br /> <br /> We can prove &lt;math&gt; \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta &lt;/math&gt; easily by using &lt;math&gt; \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha&lt;/math&gt; and &lt;math&gt;\sin(x)=\cos(90-x)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\cos (\alpha + \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt; = \sin((90 -\alpha) - \beta) &lt;/math&gt;&lt;math&gt;= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha) &lt;/math&gt;<br /> <br /> &lt;math&gt;=\cos \alpha \cos \beta - \sin \beta \sin \alpha &lt;/math&gt;<br /> <br /> == Double Angle Identities ==<br /> Double angle identities are easily derived from the angle addition formulas by just letting &lt;math&gt; \alpha = \beta &lt;/math&gt;. Doing so yields:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sin 2\alpha &amp;=&amp; 2\sin \alpha \cos \alpha\\<br /> \cos 2\alpha &amp;=&amp; \cos^2 \alpha - \sin^2 \alpha\\<br /> &amp;=&amp; 2\cos^2 \alpha - 1\\<br /> &amp;=&amp; 1-2\sin^2 \alpha\\<br /> \tan 2\alpha &amp;=&amp; \frac{2\tan \alpha}{1-\tan^2\alpha} \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> =Further Conclusions=<br /> <br /> We can see from the above that<br /> <br /> *&lt;math&gt;\csc(2a) = \frac{\csc(a)\sec(a)}{2}&lt;/math&gt;<br /> *&lt;math&gt;\sec(2a) = \frac{1}{2\cos^2(a) - 1} = \frac{1}{\cos^2(a) - \sin^2(a)} = \frac{1}{1 - 2\sin^2(a)}&lt;/math&gt;<br /> *&lt;math&gt;\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}&lt;/math&gt;<br /> <br /> == Half Angle Identities ==<br /> Using the double angle identities, we can derive half angle identities. The double angle formula for cosine tells us &lt;math&gt;\cos 2\alpha = 2\cos^2 \alpha - 1 &lt;/math&gt;. Solving for &lt;math&gt;\cos \alpha &lt;/math&gt; we get &lt;math&gt;\cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}&lt;/math&gt; where we look at the quadrant of &lt;math&gt;\alpha &lt;/math&gt; to decide if it's positive or negative. Likewise, we can use the fact that &lt;math&gt;\cos 2\alpha = 1 - 2\sin^2 \alpha &lt;/math&gt; to find a half angle identity for sine. Then, to find a half angle identity for tangent, we just use the fact that &lt;math&gt;\tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2} &lt;/math&gt; and plug in the half angle identities for sine and cosine.<br /> <br /> To summarize:<br /> <br /> *&lt;math&gt; \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2} &lt;/math&gt;<br /> *&lt;math&gt; \cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2} &lt;/math&gt;<br /> *&lt;math&gt; \tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} &lt;/math&gt;<br /> <br /> == Prosthaphaeresis Identities ==<br /> (Otherwise known as sum-to-product identities)<br /> <br /> * &lt;math&gt;\sin \theta \pm \sin \gamma = 2 \sin \frac{\theta\pm \gamma}2 \cos \frac{\theta\mp \gamma}2&lt;/math&gt;<br /> * &lt;math&gt;\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2&lt;/math&gt;<br /> * &lt;math&gt;\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2&lt;/math&gt;<br /> <br /> == Law of Sines ==<br /> {{main|Law of Sines}}<br /> The extended [[Law of Sines]] states<br /> <br /> *&lt;math&gt;\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.&lt;/math&gt;<br /> <br /> == Law of Cosines ==<br /> {{main|Law of Cosines}}<br /> The [[Law of Cosines]] states <br /> <br /> *&lt;math&gt;a^2 = b^2 + c^2 - 2bc\cos A. &lt;/math&gt;<br /> <br /> == Law of Tangents ==<br /> {{main|Law of Tangents}}<br /> The [[Law of Tangents]] states that if &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> <br /> &lt;math&gt; \frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} . &lt;/math&gt;<br /> <br /> A further extension of the [[Law of Tangents]] states that if &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are angles in a triangle, then<br /> &lt;math&gt;\tan(A)\cdot\tan(B)\cdot\tan(C)=\tan(A)+\tan(B)+\tan(C)&lt;/math&gt;<br /> <br /> == Other Identities ==<br /> *&lt;math&gt;\sin(90-\theta) = \cos(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\cos(90-\theta)=\sin(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\tan(90-\theta)=\cot(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\sin(180-\theta) = \sin(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\cos(180-\theta) = -\cos(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\tan(180-\theta) = -\tan(\theta)&lt;/math&gt;<br /> *&lt;math&gt;e^{i\theta} = \cos \theta + i\sin \theta&lt;/math&gt; (This is also written as &lt;math&gt;\text{cis }\theta&lt;/math&gt;)<br /> *&lt;math&gt;|1-e^{i\theta}|=2\sin\frac{\theta}{2}&lt;/math&gt;<br /> *&lt;math&gt;\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}&lt;/math&gt;<br /> *&lt;math&gt;\sin(\theta) = \cos(\theta)\tan(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}&lt;/math&gt;<br /> *&lt;math&gt;\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}&lt;/math&gt;<br /> *&lt;math&gt;\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)&lt;/math&gt;<br /> *&lt;math&gt;\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)&lt;/math&gt;<br /> <br /> The two identities right above here were based on identites others posted on this site with a substitution.<br /> <br /> *&lt;math&gt;\cos(2\theta) = (\cos(\theta) + \sin(\theta))(\cos(\theta) - \sin(\theta))&lt;/math&gt;<br /> <br /> ==See also==<br /> * [[Trigonometry]]<br /> * [[Trigonometric substitution]]<br /> * [http://www.sosmath.com/trig/Trig5/trig5/trig5.html Trigonometric Identities]<br /> <br /> [[Category:Trigonometry]]</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=Proofs&diff=80264 Proofs 2016-09-10T22:56:35Z <p>Burunduchok: /* Pythagorean Theorem */</p> <hr /> <div>==Quadratic Formula==<br /> <br /> Let &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;. Then<br /> &lt;cmath&gt;x^2+\frac{b}{a}x+\frac{c}{a}=0&lt;/cmath&gt;<br /> Completing the square, we get<br /> &lt;cmath&gt;\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}&lt;/cmath&gt;<br /> Simplifying, we see<br /> &lt;cmath&gt;\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}&lt;/cmath&gt;<br /> <br /> ==Pythagorean Theorem==<br /> <br /> http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif<br /> <br /> Since area of green square is &lt;math&gt;a^2&lt;/math&gt;<br /> <br /> Since are of blue square is &lt;math&gt;b^2&lt;/math&gt;<br /> <br /> Since red square is &lt;math&gt;c^2&lt;/math&gt;<br /> <br /> We have the following relationship<br /> <br /> Based on this, we get we get &lt;math&gt;\boxed {a^2+b^2=c^2}&lt;/math&gt; (here we get the Pythagorean Theorem)</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=Proofs&diff=80263 Proofs 2016-09-10T22:55:39Z <p>Burunduchok: /* Pythagorean Theorem */</p> <hr /> <div>==Quadratic Formula==<br /> <br /> Let &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;. Then<br /> &lt;cmath&gt;x^2+\frac{b}{a}x+\frac{c}{a}=0&lt;/cmath&gt;<br /> Completing the square, we get<br /> &lt;cmath&gt;\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}&lt;/cmath&gt;<br /> Simplifying, we see<br /> &lt;cmath&gt;\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}&lt;/cmath&gt;<br /> <br /> ==Pythagorean Theorem==<br /> <br /> http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif<br /> <br /> Since area of green square is &lt;math&gt;a^2&lt;/math&gt;<br /> <br /> Since are of blue square is &lt;math&gt;b^2&lt;/math&gt;<br /> <br /> Since red square is &lt;math&gt;c^2&lt;/math&gt;<br /> <br /> We have the following relationship<br /> <br /> Based on this, we get we get &lt;math&gt;\boxed {a^2+b^2=c^2}&lt;/math&gt; (here we get the pythagorean theorem)</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_6&diff=79771 2016 AIME I Problems/Problem 6 2016-07-31T01:08:06Z <p>Burunduchok: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; let &lt;math&gt;I&lt;/math&gt; be the center of the inscribed circle, and let the bisector of &lt;math&gt;\angle ACB&lt;/math&gt; intersect &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;L&lt;/math&gt;. The line through &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; intersects the circumscribed circle of &lt;math&gt;\triangle ABC&lt;/math&gt; at the two points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. If &lt;math&gt;LI=2&lt;/math&gt; and &lt;math&gt;LD=3&lt;/math&gt;, then &lt;math&gt;IC=\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> =Solution=<br /> ==Solution 1==<br /> Suppose we label the angles as shown below.<br /> &lt;asy&gt;<br /> size(150);<br /> import olympiad;<br /> real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;<br /> pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));<br /> pair C=intersectionpoints(circle(A,b),circle(B,a));<br /> pair I=incenter(A,B,C);<br /> pair L=extension(C,D,A,B);<br /> dot(I^^A^^B^^C^^D);<br /> draw(C--D);<br /> path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}<br /> draw(A--B--D--cycle);<br /> draw(circumcircle(A,B,D));<br /> draw(A--C--B);<br /> draw(A--I--B^^C--I);<br /> draw(incircle(A,B,C));<br /> label(&quot;$A$&quot;,A,SW,fontsize(8));<br /> label(&quot;$B$&quot;,B,SE,fontsize(8));<br /> label(&quot;$C$&quot;,C,N,fontsize(8));<br /> label(&quot;$D$&quot;,D,S,fontsize(8));<br /> label(&quot;$I$&quot;,I,NE,fontsize(8));<br /> label(&quot;$L$&quot;,L,SW,fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(C,A,I)),fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(I,A,B)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(A,B,I)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(I,B,C)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(A,C,I)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(I,C,B)),fontsize(8));<br /> &lt;/asy&gt;<br /> As &lt;math&gt;\angle BCD&lt;/math&gt; and &lt;math&gt;\angle BAD&lt;/math&gt; intercept the same arc, we know that &lt;math&gt;\angle BAD=\gamma&lt;/math&gt;. Similarly, &lt;math&gt;\angle ABD=\gamma&lt;/math&gt;. Also, using &lt;math&gt;\triangle ICA&lt;/math&gt;, we find &lt;math&gt;\angle CIA=180-\alpha-\gamma&lt;/math&gt;. Therefore, &lt;math&gt;\angle AID=\alpha+\gamma&lt;/math&gt;. Therefore, &lt;math&gt;\angle DAI=\angle AID=\alpha+\gamma&lt;/math&gt;, so &lt;math&gt;\triangle AID&lt;/math&gt; must be isosceles with &lt;math&gt;AD=ID=5&lt;/math&gt;. Similarly, &lt;math&gt;BD=ID=5&lt;/math&gt;. Then &lt;math&gt;\triangle DLB \sim \triangle ALC&lt;/math&gt;, hence &lt;math&gt;\frac{AL}{AC} = \frac{3}{5}&lt;/math&gt;. Also, &lt;math&gt;AI&lt;/math&gt; bisects &lt;math&gt;\angle LAC&lt;/math&gt;, so by the Angle Bisector Theorem &lt;math&gt;\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}&lt;/math&gt;. Thus &lt;math&gt;CI = \frac{10}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> WLOG assume &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles. Then, &lt;math&gt;L&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt;, and &lt;math&gt;\angle CLB=\angle CLA=90^\circ&lt;/math&gt;. Draw the perpendicular from &lt;math&gt;I&lt;/math&gt; to &lt;math&gt;CB&lt;/math&gt;, and let it meet &lt;math&gt;CB&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. Since &lt;math&gt;IL=2&lt;/math&gt;, &lt;math&gt;IE&lt;/math&gt; is also &lt;math&gt;2&lt;/math&gt; (they are both inradii). Set &lt;math&gt;BD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;. Then, triangles &lt;math&gt;BLD&lt;/math&gt; and &lt;math&gt;CEI&lt;/math&gt; are similar, and &lt;math&gt;\tfrac{2}{3}=\tfrac{CI}{x}&lt;/math&gt;. Thus, &lt;math&gt;CI=\tfrac{2x}{3}&lt;/math&gt;. &lt;math&gt;\triangle CBD \sim \triangle CEI&lt;/math&gt;, so &lt;math&gt;\tfrac{IE}{DB}=\tfrac{CI}{CD}&lt;/math&gt;. Thus &lt;math&gt;\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt;, we have:<br /> &lt;math&gt;x^2-2x-15=0&lt;/math&gt;, or &lt;math&gt;x=5, -3&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is positive, so &lt;math&gt;x=5&lt;/math&gt;. As a result, &lt;math&gt;CI=\tfrac{2x}{3}=\tfrac{10}{3}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{013}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> WLOG assume &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles (with vertex &lt;math&gt;C&lt;/math&gt;). Let &lt;math&gt;O&lt;/math&gt; be the center of the circumcircle, &lt;math&gt;R&lt;/math&gt; the circumradius, and &lt;math&gt;r&lt;/math&gt; the inradius. A simple sketch will reveal that &lt;math&gt;\triangle ABC&lt;/math&gt; must be obtuse (as an acute triangle will result in &lt;math&gt;LI&lt;/math&gt; being greater than &lt;math&gt;DL&lt;/math&gt;) and that &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; are collinear. Next, if &lt;math&gt;OI=d&lt;/math&gt;, &lt;math&gt;DO+OI=R+d&lt;/math&gt; and &lt;math&gt;R+d=DL+LI=5&lt;/math&gt;. Euler gives us that &lt;math&gt;d^{2}=R(R-2r)&lt;/math&gt;, and in this case, &lt;math&gt;r=LI=2&lt;/math&gt;. Thus, &lt;math&gt;d=\sqrt{R^{2}-4R}&lt;/math&gt;. Solving for &lt;math&gt;d&lt;/math&gt;, we have &lt;math&gt;R+\sqrt{R^{2}-4R}=5&lt;/math&gt;, then &lt;math&gt;R^{2}-4R=25-10R+R^{2}&lt;/math&gt;, yielding &lt;math&gt;R=\frac{25}{6}&lt;/math&gt;. Next, &lt;math&gt;R+d=5&lt;/math&gt; so &lt;math&gt;d=\frac{5}{6}&lt;/math&gt;. Finally, &lt;math&gt;OC=OI+IC&lt;/math&gt; gives us &lt;math&gt;R=d+IC&lt;/math&gt;, and &lt;math&gt;IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}&lt;/math&gt;. Our answer is then &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Since &lt;math&gt;\angle{LAD} = \angle{BDC}&lt;/math&gt; and &lt;math&gt;\angle{DLA}=\angle{DBC}&lt;/math&gt;, &lt;math&gt;\triangle{DLA}\sim\triangle{DBC}&lt;/math&gt;. Also, &lt;math&gt;\angle{DAC}=\angle{BLC}&lt;/math&gt; and &lt;math&gt;\angle{ACD}=\angle{LCB}&lt;/math&gt; so &lt;math&gt;\triangle{DAC}\sim\triangle{BLC}&lt;/math&gt;. Now we can call &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. By angle bisector theorem, &lt;math&gt;\frac{AD}{DB}=\frac{AC}{BC}&lt;/math&gt;. So let &lt;math&gt;AD=bk&lt;/math&gt; and &lt;math&gt;DB=ak&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt;. Now call &lt;math&gt;IC=x&lt;/math&gt;. By the similar triangles we found earlier, &lt;math&gt;\frac{3}{ak}=\frac{bk}{x+2}&lt;/math&gt; and &lt;math&gt;\frac{b}{x+5}=\frac{x+2}{a}&lt;/math&gt;. We can simplify this to &lt;math&gt;abk^2=3x+6&lt;/math&gt; and &lt;math&gt;ab=(x+5)(x+2)&lt;/math&gt;. So we can plug the &lt;math&gt;ab&lt;/math&gt; into the first equation and get &lt;math&gt;(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3&lt;/math&gt;. We can now draw a line through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; that intersects &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. By mass points, we can assign a mass of &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;a+b&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. We can also assign a mass of &lt;math&gt;(a+b)k&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; by angle bisector theorem. So the ratio of &lt;math&gt;\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}&lt;/math&gt;. So since &lt;math&gt;k=\frac{2}{x}&lt;/math&gt;, we can plug this back into the original equation to get &lt;math&gt;\left(\frac{2}{x}\right)^2(x+5)=3&lt;/math&gt;. This means that &lt;math&gt;\frac{3x^2}{4}-x-5=0&lt;/math&gt; which has roots -2 and &lt;math&gt;\frac{10}{3}&lt;/math&gt; which means our &lt;math&gt;CI=\frac{10}{3}&lt;/math&gt; and our answer is &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_7&diff=74854 2014 AMC 10B Problems/Problem 7 2016-01-26T22:18:58Z <p>Burunduchok: /* Solution */</p> <hr /> <div><br /> ==Problem==<br /> <br /> Suppose &lt;math&gt;A&gt;B&gt;0&lt;/math&gt; and A is &lt;math&gt;x&lt;/math&gt;% greater than &lt;math&gt;B&lt;/math&gt;. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf {(A) } 100(\frac{A-B}{B}) \qquad \textbf {(B) } 100(\frac{A+B}{B}) \qquad \textbf {(C) } 100(\frac{A+B}{A})\qquad \textbf {(D) } 100(\frac{A-B}{A}) \qquad \textbf {(E) } 100(\frac{A}{B})&lt;/math&gt;<br /> <br /> ==Solution==<br /> We have that A is &lt;math&gt;x\%&lt;/math&gt; greater than B, so &lt;math&gt;A=\frac{100+x}{100}(B)&lt;/math&gt;. We solve for &lt;math&gt;x&lt;/math&gt;. We get <br /> <br /> &lt;math&gt;\frac{A}{B}=\frac{100+x}{100}&lt;/math&gt;<br /> <br /> &lt;math&gt;100\frac{A}{B}=100+x&lt;/math&gt;<br /> <br /> &lt;math&gt;100(\frac{A}{B}-1)=x&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{100(\frac{A-B}{B}) (\textbf{A})}=x&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_4&diff=74853 2014 AMC 10B Problems/Problem 4 2016-01-26T22:10:02Z <p>Burunduchok: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Susie pays for &lt;math&gt;4&lt;/math&gt; muffins and &lt;math&gt;3&lt;/math&gt; bananas. Calvin spends twice as much paying for &lt;math&gt;2&lt;/math&gt; muffins and &lt;math&gt;16&lt;/math&gt; bananas. A muffin is how many times as expensive as a banana?<br /> <br /> &lt;math&gt; \textbf {(A) } \frac{3}{2} \qquad \textbf {(B) } \frac{5}{3} \qquad \textbf {(C) } \frac{7}{4} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \frac{13}{4}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; be the cost of a muffin and &lt;math&gt;b&lt;/math&gt; be the cost of a banana. From the given information, &lt;cmath&gt;2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\text{(B) } \boxed{\frac{5}{3}b}&lt;/cmath&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_4&diff=74852 2014 AMC 10B Problems/Problem 4 2016-01-26T22:09:44Z <p>Burunduchok: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Susie pays for &lt;math&gt;4&lt;/math&gt; muffins and &lt;math&gt;3&lt;/math&gt; bananas. Calvin spends twice as much paying for &lt;math&gt;2&lt;/math&gt; muffins and &lt;math&gt;16&lt;/math&gt; bananas. A muffin is how many times as expensive as a banana?<br /> <br /> &lt;math&gt; \textbf {(A) } \frac{3}{2} \qquad \textbf {(B) } \frac{5}{3} \qquad \textbf {(C) } \frac{7}{4} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \frac{13}{4}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; be the cost of a muffin and &lt;math&gt;b&lt;/math&gt; be the cost of a banana. From the given information, &lt;cmath&gt;2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\text{(B)} \boxed{\frac{5}{3}b}&lt;/cmath&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_20&diff=74849 2014 AMC 10A Problems/Problem 20 2016-01-26T20:08:20Z <p>Burunduchok: /* Solution */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #16]] and [[2014 AMC 10A Problems|2014 AMC 10A #20]]}}<br /> ==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;k&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that for &lt;math&gt;k\ge{2}&lt;/math&gt;, &lt;math&gt;8 \cdot \underbrace{888...8}_{k \text{ digits}}=7\underbrace{111...1}_{k-2 \text{ ones}}04&lt;/math&gt;, which has a digit sum of &lt;math&gt;7+k-2+0+4=9+k&lt;/math&gt;. Since we are given that said number has a digit sum of &lt;math&gt;1000&lt;/math&gt;, we have &lt;math&gt;9+k=1000 \Rightarrow k=\boxed{\textbf{(D) }991}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2014|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Number Theory Problems]]</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_21&diff=74831 2015 AMC 10A Problems/Problem 21 2016-01-25T22:17:20Z <p>Burunduchok: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #16]] and [[2015 AMC 10A Problems|2015 AMC 10A #21]]}}<br /> ==Problem==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;AC=3&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;BD=4&lt;/math&gt;, &lt;math&gt;AD=3&lt;/math&gt;, and &lt;math&gt;CD=\tfrac{12}5\sqrt2&lt;/math&gt;. What is the volume of the tetrahedron?<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the midpoint of &lt;math&gt;CD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt;. We have &lt;math&gt;CE = \dfrac{6}{5} \sqrt{2}&lt;/math&gt;, and so by the Pythagorean Theorem &lt;math&gt;AE = \dfrac{\sqrt{153}}{5}&lt;/math&gt; and &lt;math&gt;BE = \dfrac{\sqrt{328}}{5}&lt;/math&gt;. Because the altitude from &lt;math&gt;A&lt;/math&gt; of tetrahedron &lt;math&gt;ABCD&lt;/math&gt; passes touches plane &lt;math&gt;BCD&lt;/math&gt; on &lt;math&gt;BE&lt;/math&gt;, it is also an altitude of triangle &lt;math&gt;ABE&lt;/math&gt;. The area &lt;math&gt;A&lt;/math&gt; of triangle &lt;math&gt;ABE&lt;/math&gt; is, by Heron's Formula, given by<br /> <br /> &lt;cmath&gt;16A^2 = 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 = -(a^2 + b^2 - c^2)^2 + 4a^2 b^2.&lt;/cmath&gt;<br /> Substituting &lt;math&gt;a = AE, b = BE, c = 5&lt;/math&gt; and performing huge (but manageable) computations yield &lt;math&gt;A^2 = 18&lt;/math&gt;, so &lt;math&gt;A = 3\sqrt{2}&lt;/math&gt;. Thus, if &lt;math&gt;h&lt;/math&gt; is the length of the altitude from &lt;math&gt;A&lt;/math&gt; of the tetrahedron, &lt;math&gt;BE \cdot h = 2A = 6\sqrt{2}&lt;/math&gt;. Our answer is thus<br /> &lt;cmath&gt;V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},&lt;/cmath&gt;<br /> and so our answer is &lt;math&gt;\boxed{\textbf{(C) } \dfrac{24}{5}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Drop altitudes of triangle &lt;math&gt;ABC&lt;/math&gt; and triangle &lt;math&gt;ABD&lt;/math&gt; down from &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, respectively. Both will hit the same point; let this point be &lt;math&gt;T&lt;/math&gt;. Because both triangle &lt;math&gt;ABC&lt;/math&gt; and triangle &lt;math&gt;ABD&lt;/math&gt; are 3-4-5 triangles, &lt;math&gt;CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}&lt;/math&gt;. Because &lt;math&gt;CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}&lt;/math&gt;, it follows that the &lt;math&gt;CTD&lt;/math&gt; is a right triangle, meaning that &lt;math&gt;\angle CTD = 90^\circ&lt;/math&gt;, and it follows that planes &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;ABD&lt;/math&gt; are perpendicular to each other. Now, we can treat &lt;math&gt;ABC&lt;/math&gt; as the base of the tetrahedron and &lt;math&gt;TD&lt;/math&gt; as the height. Thus, the desired volume is &lt;cmath&gt;V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}&lt;/cmath&gt; which is answer &lt;math&gt;\boxed{\textbf{(C) } \dfrac{24}{5}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2015|ab=A|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=AA_similarity&diff=72620 AA similarity 2015-10-27T18:56:58Z <p>Burunduchok: Created page with &quot;Theorem: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar. Proof: Let ABC and DEF be two triangles such that &lt;math&gt;\angle ...&quot;</p> <hr /> <div>Theorem:<br /> In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.<br /> <br /> Proof:<br /> Let ABC and DEF be two triangles such that &lt;math&gt;\angle A = \angle D&lt;/math&gt; and &lt;math&gt;\angle B = \angle E&lt;/math&gt;.<br /> &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt; (Sum of all angles in a triangle is &lt;math&gt;180&lt;/math&gt;)<br /> &lt;math&gt;\angle D + \angle E + \angle F = 180&lt;/math&gt; (Sum of all angles in a triangle is &lt;math&gt;180&lt;/math&gt;)<br /> &lt;math&gt;\angle A + \angle B + \angle C=\angle D + \angle E + \angle F&lt;/math&gt;<br /> &lt;math&gt;\angle D + \angle E + \angle C = \angle D + \angle E + \angle F&lt;/math&gt; (since &lt;math&gt;\angle A = \angle D&lt;/math&gt; and &lt;math&gt;\angle B = \angle E&lt;/math&gt;)<br /> &lt;math&gt;\angle C = \angle F&lt;/math&gt;.</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_2&diff=68609 2009 AIME II Problems/Problem 2 2015-03-09T22:19:24Z <p>Burunduchok: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive real numbers such that &lt;math&gt;a^{\log_3 7} = 27&lt;/math&gt;, &lt;math&gt;b^{\log_7 11} = 49&lt;/math&gt;, and &lt;math&gt;c^{\log_{11}25} = \sqrt{11}&lt;/math&gt;. Find<br /> &lt;cmath&gt; a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. &lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> <br /> First, we have:<br /> &lt;cmath&gt;<br /> x^{(\log_y z)^2}<br /> = x^{\left( (\log_y z)^2 \right) }<br /> = x^{(\log_y z) \cdot (\log_y z) }<br /> = \left( x^{\log_y z} \right)^{\log_y z}<br /> &lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;x=y^w&lt;/math&gt;, then we have:<br /> &lt;cmath&gt;<br /> x^{\log_y z} <br /> = \left( y^w \right)^{\log_y z} <br /> = y^{w\log_y z} <br /> = y^{\log_y (z^w)} <br /> = z^w<br /> &lt;/cmath&gt;<br /> <br /> This is all we need to evaluate the given formula. Note that in our case we have &lt;math&gt;27=3^3&lt;/math&gt;, &lt;math&gt;49=7^2&lt;/math&gt;, and &lt;math&gt;\sqrt{11}=11^{1/2}&lt;/math&gt;. We can now compute:<br /> <br /> &lt;cmath&gt;<br /> a^{(\log_3 7)^2}<br /> = \left( a^{\log_3 7} \right)^{\log_3 7}<br /> = 27^{\log_3 7}<br /> = (3^3)^{\log_3 7}<br /> = 7^3<br /> = 343<br /> &lt;/cmath&gt;<br /> <br /> Similarly, we get<br /> &lt;cmath&gt;<br /> b^{(\log_7 11)^2} <br /> = (7^2)^{\log_7 11}<br /> = 11^2 <br /> = 121<br /> &lt;/cmath&gt;<br /> <br /> and<br /> &lt;cmath&gt;<br /> c^{(\log_{11} 25)^2}<br /> = (11^{1/2})^{\log_{11} 25}<br /> = 25^{1/2}<br /> = 5<br /> &lt;/cmath&gt;<br /> <br /> and therefore the answer is &lt;math&gt;343+121+5 = \boxed{469}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> We know from the first three equations that &lt;math&gt;\log_a27&lt;/math&gt; = &lt;math&gt;\log_37&lt;/math&gt;, &lt;math&gt;\log_b49&lt;/math&gt; = &lt;math&gt;\log_711&lt;/math&gt;, and &lt;math&gt;\log_c\sqrt{11}&lt;/math&gt; = &lt;math&gt;\log_{11}25&lt;/math&gt;. Substituting, we get <br /> <br /> &lt;math&gt;a^{(\log_a27)(\log_37)}&lt;/math&gt; + &lt;math&gt;b^{(\log_b49)(\log_711)}&lt;/math&gt; + &lt;math&gt;c^{(\log_c\sqrt {11})(\log_{11}25)}&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;x^{\log_xy}&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;, so we get<br /> <br /> &lt;math&gt;27^{\log_37}&lt;/math&gt; + &lt;math&gt;49^{\log_711}&lt;/math&gt; + &lt;math&gt;\sqrt {11}^{\log_{11}25}&lt;/math&gt;<br /> <br /> &lt;math&gt;(3^{\log_37})^3&lt;/math&gt; + &lt;math&gt;(7^{\log_711})^2&lt;/math&gt; + &lt;math&gt;({11^{\log_{11}25}})^{1/2}&lt;/math&gt;<br /> <br /> The &lt;math&gt;3&lt;/math&gt; and the &lt;math&gt;\log_37&lt;/math&gt; cancel out to make &lt;math&gt;7&lt;/math&gt;, and we can do this for the other two terms. We obtain <br /> <br /> &lt;math&gt;7^3&lt;/math&gt; + &lt;math&gt;11^2&lt;/math&gt; + &lt;math&gt;25^{1/2}&lt;/math&gt;<br /> <br /> = &lt;math&gt;343&lt;/math&gt; + &lt;math&gt;121&lt;/math&gt; + &lt;math&gt;5&lt;/math&gt;<br /> = &lt;math&gt;\boxed {469}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AIME box|year=2009|n=II|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=68278 2015 AMC 12A Problems/Problem 9 2015-03-03T23:36:02Z <p>Burunduchok: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=68277 2015 AMC 12A Problems/Problem 9 2015-03-03T23:35:32Z <p>Burunduchok: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=68275 2015 AMC 12A Problems/Problem 9 2015-03-03T23:34:02Z <p>Burunduchok: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac12 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> Burunduchok https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=68274 2015 AMC 12A Problems/Problem 9 2015-03-03T23:31:49Z <p>Burunduchok: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac16 \qquad\textbf{(C)}\ \frac15 \qquad\textbf{(D)}}\ \frac13 \qquad\textbf{(E)}\ \frac12 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> Burunduchok