https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ketans73&feedformat=atom AoPS Wiki - User contributions [en] 2022-08-15T19:22:39Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_15&diff=169120 2007 AIME II Problems/Problem 15 2022-01-02T10:38:10Z <p>Ketans73: /* Solution 1 (homothety) */</p> <hr /> <div>== Problem ==<br /> Four [[circle]]s &lt;math&gt;\omega,&lt;/math&gt; &lt;math&gt;\omega_{A},&lt;/math&gt; &lt;math&gt;\omega_{B},&lt;/math&gt; and &lt;math&gt;\omega_{C}&lt;/math&gt; with the same [[radius]] are drawn in the interior of [[triangle]] &lt;math&gt;ABC&lt;/math&gt; such that &lt;math&gt;\omega_{A}&lt;/math&gt; is [[tangent]] to sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;\omega_{B}&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;BA&lt;/math&gt;, &lt;math&gt;\omega_{C}&lt;/math&gt; to &lt;math&gt;CA&lt;/math&gt; and &lt;math&gt;CB&lt;/math&gt;, and &lt;math&gt;\omega&lt;/math&gt; is [[externally tangent]] to &lt;math&gt;\omega_{A},&lt;/math&gt; &lt;math&gt;\omega_{B},&lt;/math&gt; and &lt;math&gt;\omega_{C}&lt;/math&gt;. If the sides of triangle &lt;math&gt;ABC&lt;/math&gt; are &lt;math&gt;13,&lt;/math&gt; &lt;math&gt;14,&lt;/math&gt; and &lt;math&gt;15,&lt;/math&gt; the radius of &lt;math&gt;\omega&lt;/math&gt; can be represented in the form &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 /> __TOC__<br /> == Solution ==<br /> [[Image:2007 AIME II-15.png]]<br /> === Solution 1 (homothety)===<br /> First, apply [[Heron's formula]] to find that &lt;math&gt;[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84&lt;/math&gt;. The semiperimeter is &lt;math&gt;21&lt;/math&gt;, so the [[inradius]] is &lt;math&gt;\frac{A}{s} = \frac{84}{21} = 4&lt;/math&gt;. <br /> <br /> Now consider the [[incenter]] &lt;math&gt;I&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the [[radius]] of one of the small circles be &lt;math&gt;r&lt;/math&gt;. Let the centers of the three little circles tangent to the sides of &lt;math&gt;\triangle ABC&lt;/math&gt; be &lt;math&gt;O_A&lt;/math&gt;, &lt;math&gt;O_B&lt;/math&gt;, and &lt;math&gt;O_C&lt;/math&gt;. Let the center of the circle tangent to those three circles be &lt;math&gt;O&lt;/math&gt;. The [[homothety]] &lt;math&gt;\mathcal{H}\left(I, \frac{4-r}{4}\right)&lt;/math&gt; maps &lt;math&gt;\triangle ABC&lt;/math&gt; to &lt;math&gt;\triangle XYZ&lt;/math&gt;; since &lt;math&gt;OO_A = OO_B = OO_C = 2r&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle XYZ&lt;/math&gt; and &lt;math&gt;\mathcal{H}&lt;/math&gt; therefore maps the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt; to &lt;math&gt;O&lt;/math&gt;. Thus, &lt;math&gt;2r = R \cdot \frac{4 - r}{4}&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the [[circumradius]] of &lt;math&gt;\triangle ABC&lt;/math&gt;. Substituting &lt;math&gt;R = \frac{abc}{4[ABC]} = \frac{65}{8}&lt;/math&gt;, &lt;math&gt;r = \frac{260}{129}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{389}&lt;/math&gt;.<br /> <br /> <br /> https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png<br /> <br /> === Solution 2 ===<br /> [[Image:2007 AIME II-15b.gif]]<br /> <br /> Consider a 13-14-15 triangle. &lt;math&gt;A=84.&lt;/math&gt; [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]<br /> <br /> The inradius is &lt;math&gt;r=\frac{A}{s}=\frac{84}{21}=4&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, &lt;math&gt;u.&lt;/math&gt;<br /> <br /> The circumradius is &lt;math&gt;R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},&lt;/math&gt; where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are the side-lengths. Scale the triangle with the circumradius by a [[line]]ar scale factor, &lt;math&gt;v&lt;/math&gt;.<br /> <br /> Cut and combine the triangles, as shown. Then solve for &lt;math&gt;4u&lt;/math&gt;:<br /> :&lt;math&gt;\frac{65}{8}v=8u&lt;/math&gt;<br /> <br /> :&lt;math&gt;v=\frac{64}{65}u&lt;/math&gt;<br /> <br /> :&lt;math&gt;u+v=1&lt;/math&gt;<br /> <br /> :&lt;math&gt;u+\frac{64}{65}u=1&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{129}{65}u=1&lt;/math&gt;<br /> <br /> :&lt;math&gt;4u=\frac{260}{129}&lt;/math&gt;<br /> <br /> The solution is &lt;math&gt;260+129=\boxed{389}&lt;/math&gt;.<br /> <br /> === Solution 3 (elementary)===<br /> <br /> Let &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, &lt;math&gt;C'&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt; be the centers of circles &lt;math&gt;\omega_{A}&lt;/math&gt;, &lt;math&gt;\omega_{B}&lt;/math&gt;, &lt;math&gt;\omega_{C}&lt;/math&gt;, &lt;math&gt;\omega&lt;/math&gt;, respectively, and let &lt;math&gt;x&lt;/math&gt; be their radius.<br /> <br /> Now, triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'&lt;/math&gt; are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for &lt;math&gt;x&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;OA'=OB'=OC'=2x&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt; is the circumcenter of triangle &lt;math&gt;A'B'C'&lt;/math&gt; and its circumradius is &lt;math&gt;2x&lt;/math&gt;. Let &lt;math&gt;I&lt;/math&gt; denote the incenter of triangle &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; the inradius of &lt;math&gt;ABC&lt;/math&gt;. Then the inradius of &lt;math&gt;A'B'C'=r-x&lt;/math&gt;, so now we compute r. Computing the inradius by &lt;math&gt;A=rs&lt;/math&gt;, we find that the inradius of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Additionally, using the circumradius formula &lt;math&gt;R=\frac{abc}{4K}&lt;/math&gt; where &lt;math&gt;K&lt;/math&gt; is the area of &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; is the circumradius, we find &lt;math&gt;R=\frac{65}{8}&lt;/math&gt;. Now we can equate the ratio of circumradius to inradius in triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}&lt;/cmath&gt;<br /> <br /> Solving, we get &lt;math&gt;x=\frac{260}{129}&lt;/math&gt;, so our answer is &lt;math&gt;260+129=\boxed{389}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is&lt;math&gt;2r&lt;/math&gt;. Now denoting &lt;math&gt;AB=13;BC=14;AC=15&lt;/math&gt;, and centers of circles tangent to &lt;math&gt;AB,AC;AC,BC;AB,BC&lt;/math&gt; are relatively &lt;math&gt;M,N,O&lt;/math&gt; with &lt;math&gt;OJ,NK&lt;/math&gt; both perpendicular to &lt;math&gt;BC&lt;/math&gt;. It is easy to know that &lt;math&gt;tanB=\frac{12}{5}&lt;/math&gt;, so &lt;math&gt;tan\angle OBJ=\frac{2}{3}&lt;/math&gt; according to half angle formula. Similarly, we can find &lt;math&gt;tan\angle NCK=\frac{1}{2}&lt;/math&gt;. So we can see that &lt;math&gt;JK=ON=14-\frac{7x}{2}&lt;/math&gt;. Obviously, &lt;math&gt;\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}&lt;/math&gt; . After solving, we get<br /> &lt;math&gt;x=\frac{260}{129}&lt;/math&gt;, so our answer is &lt;math&gt;260+129=\boxed{389}&lt;/math&gt;. ~bluesoul<br /> <br /> == Diagram for Solution 1 ==<br /> <br /> Here is a diagram illustrating solution 1. Note that unlike in the solution &lt;math&gt;O&lt;/math&gt; refers to the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. Instead, &lt;math&gt;O_\omega&lt;/math&gt; is used for the center of the third circle, &lt;math&gt;\omega&lt;/math&gt;.<br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> pair A, B, C, Oa, Ob, Oc, Od, O, I;<br /> path circ1, circ2;<br /> <br /> // Homotethy factor - backplugged from solution<br /> real k = 64/129;<br /> real r = 260/129;<br /> <br /> B = (0, 0);<br /> C = (14, 0);<br /> <br /> circ1 = circle(B, 13);<br /> circ2 = circle(C, 15);<br /> <br /> A = intersectionpoints(circ1, circ2);<br /> I = incenter(A, B, C);<br /> <br /> Oa = (65*A + 64*I)/129;<br /> Ob = (65*B + 64*I)/129;<br /> Oc = (65*C + 64*I)/129;<br /> <br /> Od = circumcenter(Oa, Ob, Oc);<br /> O = circumcenter(A, B, C);<br /> <br /> draw(circle(Oa, r));<br /> draw(circle(Ob, r));<br /> draw(circle(Oc, r));<br /> draw(circle(Od, r));<br /> <br /> draw(incircle(Oa, Ob, Oc)^^incircle(A, B, C)^^I--foot(I, A, C), green);<br /> draw(A--B--C--cycle);<br /> draw(Oa--Ob--Oc--cycle, blue);<br /> draw(A--I--B^^I--C, blue);<br /> draw(Oa--foot(Oa, A, C)^^Oc--foot(Oc, A, C), blue);<br /> draw(rightanglemark(Oa, foot(Oa, A, C), C)^^rightanglemark(Oc, foot(Oc, A, C), A));<br /> dot(I);<br /> dot(Oa);<br /> dot(Ob);<br /> dot(Oc);<br /> dot(Od);<br /> dot(O, red);<br /> <br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, S);<br /> label(&quot;$C$&quot;, C, S);<br /> label(&quot;$I$&quot;, I, S);<br /> label(&quot;$O_A$&quot;, Oa, NW);<br /> label(&quot;$O_B$&quot;, Ob, SW);<br /> label(&quot;$O_C$&quot;, Oc, SE);<br /> label(&quot;$O_\omega$&quot;, Od, N);<br /> label(&quot;$O$&quot;, O, SE, red);<br /> <br /> &lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ketans73 https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems&diff=169114 2000 AIME I Problems 2022-01-02T08:02:38Z <p>Ketans73: /* Problem 1 */</p> <hr /> <div>{{AIME Problems|year=2000|n=I}}<br /> <br /> == Problem 1 ==<br /> Find the least positive integer(s) &lt;math&gt;n&lt;/math&gt; such that no matter how &lt;math&gt;10^{n}&lt;/math&gt; is expressed as the product of any two positive integers, at least one of these two integers contains the digit &lt;math&gt;0&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;u&lt;/math&gt; and &lt;math&gt;v&lt;/math&gt; be integers satisfying &lt;math&gt;0 &lt; v &lt; u&lt;/math&gt;. Let &lt;math&gt;A = (u,v)&lt;/math&gt;, let &lt;math&gt;B&lt;/math&gt; be the reflection of &lt;math&gt;A&lt;/math&gt; across the line &lt;math&gt;y = x&lt;/math&gt;, let &lt;math&gt;C&lt;/math&gt; be the reflection of &lt;math&gt;B&lt;/math&gt; across the y-axis, let &lt;math&gt;D&lt;/math&gt; be the reflection of &lt;math&gt;C&lt;/math&gt; across the x-axis, and let &lt;math&gt;E&lt;/math&gt; be the reflection of &lt;math&gt;D&lt;/math&gt; across the y-axis. The area of pentagon &lt;math&gt;ABCDE&lt;/math&gt; is &lt;math&gt;451&lt;/math&gt;. Find &lt;math&gt;u + v&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> In the expansion of &lt;math&gt;(ax + b)^{2000},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers, the coefficients of &lt;math&gt;x^{2}&lt;/math&gt; and &lt;math&gt;x^{3}&lt;/math&gt; are equal. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.<br /> <br /> &lt;center&gt;&lt;asy&gt;defaultpen(linewidth(0.7));<br /> draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));<br /> draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61));<br /> draw((34,36)--(34,45)--(25,45));<br /> draw((36,36)--(36,38)--(34,38));<br /> draw((36,38)--(41,38));<br /> draw((34,45)--(41,45));&lt;/asy&gt;&lt;/center&gt;<br /> <br /> [[2000 AIME I Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is &lt;math&gt;25.&lt;/math&gt; One marble is taken out of each box randomly. The probability that both marbles are black is &lt;math&gt;\frac{27}{50},&lt;/math&gt; and the probability that both marbles are white is &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. What is &lt;math&gt;m + n&lt;/math&gt;?<br /> <br /> [[2000 AIME I Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> For how many ordered pairs &lt;math&gt;(x,y)&lt;/math&gt; of integers is it true that &lt;math&gt;0 &lt; x &lt; y &lt; 10^{6}&lt;/math&gt; and that the arithmetic mean of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is exactly &lt;math&gt;2&lt;/math&gt; more than the geometric mean of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;?<br /> <br /> [[2000 AIME I Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Suppose that &lt;math&gt;x,&lt;/math&gt; &lt;math&gt;y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are three positive numbers that satisfy the equations &lt;math&gt;xyz = 1,&lt;/math&gt; &lt;math&gt;x + \frac {1}{z} = 5,&lt;/math&gt; and &lt;math&gt;y + \frac {1}{x} = 29.&lt;/math&gt; Then &lt;math&gt;z + \frac {1}{y} = \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 /> [[2000 AIME I Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is &lt;math&gt;m - n\sqrt {p},&lt;/math&gt; where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the cube of any prime number. Find &lt;math&gt;m + n + p&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> A sequence of numbers &lt;math&gt;x_{1},x_{2},x_{3},\ldots,x_{100}&lt;/math&gt; has the property that, for every integer &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,&lt;/math&gt; inclusive, the number &lt;math&gt;x_{k}&lt;/math&gt; is &lt;math&gt;k&lt;/math&gt; less than the sum of the other &lt;math&gt;99&lt;/math&gt; numbers. Given that &lt;math&gt;x_{50} = \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 /> [[2000 AIME I Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the sum of all numbers of the form &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive divisors of &lt;math&gt;1000.&lt;/math&gt; What is the greatest integer that does not exceed &lt;math&gt;\frac{S}{10}&lt;/math&gt;?<br /> <br /> [[2000 AIME I Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Given a function &lt;math&gt;f&lt;/math&gt; for which<br /> &lt;center&gt;&lt;math&gt;f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)&lt;/math&gt;&lt;/center&gt;<br /> holds for all real &lt;math&gt;x,&lt;/math&gt; what is the largest number of different values that can appear in the list &lt;math&gt;f(0),f(1),f(2),\ldots,f(999)&lt;/math&gt;?<br /> <br /> [[2000 AIME I Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at &lt;math&gt;50&lt;/math&gt; miles per hour along the highways and at &lt;math&gt;14&lt;/math&gt; miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is &lt;math&gt;\frac{m}{n}&lt;/math&gt; square miles, 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 /> [[2000 AIME I Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; it is given that angles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are congruent. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB},&lt;/math&gt; respectively, so that &lt;math&gt;AP = PQ = QB = BC.&lt;/math&gt; Angle &lt;math&gt;ACB&lt;/math&gt; is &lt;math&gt;r&lt;/math&gt; times as large as angle &lt;math&gt;APQ,&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is a positive real number. Find &lt;math&gt;\lfloor 1000r \rfloor&lt;/math&gt;.<br /> <br /> [[2000 AIME I Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> A stack of &lt;math&gt;2000&lt;/math&gt; cards is labelled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2000,&lt;/math&gt; with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: &lt;math&gt;1,2,3,\ldots,1999,2000.&lt;/math&gt; In the original stack of cards, how many cards were above the card labeled &lt;math&gt;1999&lt;/math&gt;?<br /> <br /> [[2000 AIME I Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> <br /> {{AIME box|year = 2000|n=I|before=[[1999 AIME Problems]]|after=[[2000 AIME II Problems]]}}<br /> <br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ketans73 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_15&diff=158407 1989 AIME Problems/Problem 15 2021-07-14T16:26:18Z <p>Ketans73: /* Solution */</p> <hr /> <div>== Problem ==<br /> Point &lt;math&gt;P&lt;/math&gt; is inside &lt;math&gt;\triangle ABC&lt;/math&gt;. Line segments &lt;math&gt;APD&lt;/math&gt;, &lt;math&gt;BPE&lt;/math&gt;, and &lt;math&gt;CPF&lt;/math&gt; are drawn with &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AB&lt;/math&gt; (see the figure below). Given that &lt;math&gt;AP=6&lt;/math&gt;, &lt;math&gt;BP=9&lt;/math&gt;, &lt;math&gt;PD=6&lt;/math&gt;, &lt;math&gt;PE=3&lt;/math&gt;, and &lt;math&gt;CF=20&lt;/math&gt;, find the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> [[Image:AIME_1989_Problem_15.png|center]]<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Because we're given three concurrent [[cevian]]s and their lengths, it seems very tempting to apply [[Mass points]]. We immediately see that &lt;math&gt;w_E = 3&lt;/math&gt;, &lt;math&gt;w_B = 1&lt;/math&gt;, and &lt;math&gt;w_A = w_D = 2&lt;/math&gt;. Now, we recall that the masses on the three sides of the triangle must be balanced out, so &lt;math&gt;w_C = 1&lt;/math&gt; and &lt;math&gt;w_F = 3&lt;/math&gt;. Thus, &lt;math&gt;CP = 15&lt;/math&gt; and &lt;math&gt;PF = 5&lt;/math&gt;. <br /> <br /> Recalling that &lt;math&gt;w_C = w_B = 1&lt;/math&gt;, we see that &lt;math&gt;DC = DB&lt;/math&gt; and &lt;math&gt;DP&lt;/math&gt; is a [[median]] to &lt;math&gt;BC&lt;/math&gt; in &lt;math&gt;\triangle BCP&lt;/math&gt;. Applying [[Stewart's Theorem]], &lt;math&gt;BC^2 + 12^2 = 2(15^2 + 9^2)&lt;/math&gt;, and &lt;math&gt;BC = 6\sqrt {13}&lt;/math&gt;. Now notice that &lt;math&gt;2[BCP] = [ABC]&lt;/math&gt;, because both triangles share the same base and the &lt;math&gt;h_{\triangle ABC} = 2h_{\triangle BCP}&lt;/math&gt;. Applying [[Heron's formula]] on triangle &lt;math&gt;BCP&lt;/math&gt; with sides &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, and &lt;math&gt;6\sqrt{13}&lt;/math&gt;, &lt;math&gt;[BCP] = 54&lt;/math&gt; and &lt;math&gt;[ABC] = \boxed{108}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Using a different form of [[Ceva's Theorem]], we have &lt;math&gt;\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}&lt;/math&gt;<br /> <br /> Solving &lt;math&gt;4y = x + y&lt;/math&gt; and &lt;math&gt;x + y = 20&lt;/math&gt;, we obtain &lt;math&gt;x = CP = 15&lt;/math&gt; and &lt;math&gt;y = FP = 5&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the point on &lt;math&gt;AB&lt;/math&gt; such that &lt;math&gt;FC \parallel QD&lt;/math&gt;.<br /> Since &lt;math&gt;AP = PD&lt;/math&gt; and &lt;math&gt;FP\parallel QD&lt;/math&gt;, &lt;math&gt;QD = 2FP = 10&lt;/math&gt;. (Stewart's Theorem)<br /> <br /> Also, since &lt;math&gt;FC\parallel QD&lt;/math&gt; and &lt;math&gt;QD = \frac{FC}{2}&lt;/math&gt;, we see that &lt;math&gt;FQ = QB&lt;/math&gt;, &lt;math&gt;BD = DC&lt;/math&gt;, etc. ([[Stewart's Theorem]])<br /> Similarly, we have &lt;math&gt;PR = RB&lt;/math&gt; (&lt;math&gt;= \frac12PB = 7.5&lt;/math&gt;) and thus &lt;math&gt;RD = \frac12PC = 4.5&lt;/math&gt;.<br /> <br /> &lt;math&gt;PDR&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; [[right triangle]], so &lt;math&gt;\angle PDR&lt;/math&gt; (&lt;math&gt;\angle ADQ&lt;/math&gt;) is &lt;math&gt;90^\circ&lt;/math&gt;.<br /> Therefore, the area of &lt;math&gt;\triangle ADQ = \frac12\cdot 12\cdot 6 = 36&lt;/math&gt;.<br /> Using area ratio, &lt;math&gt;\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Because the length of cevian &lt;math&gt;BE&lt;/math&gt; is unknown, we can examine what happens when we extend it or decrease its length and see that it simply changes the angles between the cevians. Wouldn't it be great if it the length of &lt;math&gt;BE&lt;/math&gt; was such that &lt;math&gt;\angle APC = 90^\circ&lt;/math&gt;? Let's first assume it's a right angle and hope that everything works out. <br /> <br /> Extend &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; so that &lt;math&gt;PD = DQ = 6&lt;/math&gt;. The result is that &lt;math&gt;BQ = 9&lt;/math&gt;, &lt;math&gt;PQ = 12&lt;/math&gt;, and &lt;math&gt;BP = 15&lt;/math&gt; because &lt;math&gt;\triangle CDP\cong \triangle BDQ&lt;/math&gt;. Now we see that if we are able to show that &lt;math&gt;BE = 20&lt;/math&gt;, that is &lt;math&gt;PE = 5&lt;/math&gt;, then our right angle assumption will be true.<br /> <br /> Apply the [[Pythagorean Theorem]] on &lt;math&gt;\triangle APC&lt;/math&gt; to get &lt;math&gt;AC = 3\sqrt {13}&lt;/math&gt;, so &lt;math&gt;AE = \sqrt {13}&lt;/math&gt; and &lt;math&gt;CE = 2\sqrt {13}&lt;/math&gt;. Now, we apply the [[Law of Cosines]] on triangles &lt;math&gt;CEP&lt;/math&gt; and &lt;math&gt;AEP&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;PE = x&lt;/math&gt;. Notice that &lt;math&gt;\angle CEB = 180^\circ - \angle AEB&lt;/math&gt; and &lt;math&gt;\cos CEB = - \cos AEB&lt;/math&gt;, so we get two nice equations.<br /> <br /> &lt;math&gt;81 = 52 + y^2 - 2y \sqrt {13}\cos CEF&lt;/math&gt;<br /> &lt;math&gt;36 = 13 + y^2 + y \sqrt {13} \cos CEF&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt;y = 5&lt;/math&gt; (yay!). <br /> <br /> Now, the area is easy to find. &lt;math&gt;[ABC] = [AQB] + [APC] = \frac12(9)(18) + \frac12(6)(9) = \boxed{108}&lt;/math&gt;.<br /> <br /> [however, I think this solution is wrong, the A, B, and Cs do not match with the picture]<br /> <br /> === Solution 4 ===<br /> <br /> First, let &lt;math&gt;[AEP]=a, [AFP]=b,&lt;/math&gt; and &lt;math&gt;[ECP]=c.&lt;/math&gt; Thus, we can easily find that &lt;math&gt;\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.&lt;/math&gt; Now, &lt;math&gt;\frac{[ABP]}{BPD}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.&lt;/math&gt; In the same manner, we find that &lt;math&gt;[CPD]=a+c.&lt;/math&gt; Now, we can find that &lt;math&gt;\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.&lt;/math&gt; We can now use this to find that &lt;math&gt;\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.&lt;/math&gt; Plugging this value in, we find that &lt;math&gt;\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.&lt;/math&gt; Now, since &lt;math&gt;\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},&lt;/math&gt; we can find that &lt;math&gt;2AE=EC.&lt;/math&gt; Setting &lt;math&gt;AC=b,&lt;/math&gt; we can apply Stewart's Theorem on triangle &lt;math&gt;APC&lt;/math&gt; to find that &lt;math&gt;(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).&lt;/math&gt; Solving, we find that &lt;math&gt;b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.&lt;/math&gt; But, &lt;math&gt;3^2+6^2=45,&lt;/math&gt; meaning that &lt;math&gt;\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.&lt;/math&gt; Since &lt;math&gt;[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,&lt;/math&gt; we conclude that the answer is &lt;math&gt;\boxed{108.00}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=14|after=Final Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ketans73