https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Drunkenninja&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-16T06:10:01Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_17&diff=116643 2020 AMC 12A Problems/Problem 17 2020-02-02T19:44:11Z <p>Drunkenninja: /* Solution 1 */</p> <hr /> <div>==Problem 17==<br /> The vertices of a quadrilateral lie on the graph of &lt;math&gt;y=\ln{x}&lt;/math&gt;, and the &lt;math&gt;x&lt;/math&gt;-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is &lt;math&gt;\ln{\frac{91}{90}}&lt;/math&gt;. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of the leftmost vertex?<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let the left-most &lt;math&gt;x&lt;/math&gt;-coordinate be &lt;math&gt;n.&lt;/math&gt;<br /> <br /> Recall that, by the shoelace formula, the area of the triangle must be &lt;math&gt;-\ln{(n)}+\ln{(n+1)}+\ln{(n+2)}-\ln{(n+3)}.&lt;/math&gt; That equals to &lt;math&gt;\ln\frac{(n+1)(n+2)}{n(n+3)}.&lt;/math&gt; <br /> <br /> &lt;math&gt;\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}&lt;/math&gt; <br /> <br /> &lt;math&gt;\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{91}{90}&lt;/math&gt;<br /> <br /> &lt;math&gt;\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \frac{182}{180}&lt;/math&gt;<br /> <br /> &lt;math&gt;n^{2}+3n = 180&lt;/math&gt;<br /> <br /> &lt;math&gt;n^{2}+3n-180 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(n-12)(n+15) = 0&lt;/math&gt;<br /> <br /> The &lt;math&gt;x&lt;/math&gt;-coordinate is, therefore, &lt;math&gt;\boxed{\textbf{(D) } 12.}&lt;/math&gt;~lopkiloinm.<br /> <br /> ==Solution 2==<br /> Like above, use the shoelace formula to find that the area of the triangle is equal to &lt;math&gt;\ln\frac{(n+1)(n+2)}{n(n+3)}&lt;/math&gt;. Because the final area we are looking for is &lt;math&gt;\ln\frac{91}{90}&lt;/math&gt;, the numerator factors into &lt;math&gt;13&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;, which one of &lt;math&gt;n+1&lt;/math&gt; and &lt;math&gt;n+2&lt;/math&gt; has to be a multiple of &lt;math&gt;13&lt;/math&gt; and the other has to be a multiple of &lt;math&gt;7&lt;/math&gt;. Clearly, the only choice for that is &lt;math&gt;\boxed{12}&lt;/math&gt;<br /> <br /> ~Solution by IronicNinja<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_22&diff=107710 2018 AMC 12B Problems/Problem 22 2019-07-18T18:46:41Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Consider polynomials &lt;math&gt;P(x)&lt;/math&gt; of degree at most &lt;math&gt;3&lt;/math&gt;, each of whose coefficients is an element of &lt;math&gt;\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}&lt;/math&gt;. How many such polynomials satisfy &lt;math&gt;P(-1) = -9&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Suppose our polynomial is equal to<br /> &lt;cmath&gt;ax^3+bx^2+cx+d&lt;/cmath&gt;Then we are given that<br /> &lt;cmath&gt;-9=b+d-a-c.&lt;/cmath&gt;If we let &lt;math&gt;-a=a'-9, -c=c'-9&lt;/math&gt; then we have<br /> &lt;cmath&gt;9=a'+c'+b+d.&lt;/cmath&gt; This way all four variables are within 0 and 9. The number of solutions to this equation is simply &lt;math&gt;\binom{12}{3}=220&lt;/math&gt; by stars and bars, so our answer is &lt;math&gt;\boxed{\textbf{D}.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Suppose our polynomial is equal to<br /> &lt;cmath&gt;ax^3+bx^2+cx+d&lt;/cmath&gt;Then we are given that<br /> &lt;cmath&gt;9=b+d-a-c.&lt;/cmath&gt;Then the polynomials &lt;cmath&gt;cx^3+bx^2+ax+d&lt;/cmath&gt;, &lt;cmath&gt;ax^3+dx^2+cx+b,&lt;/cmath&gt; &lt;cmath&gt;cx^3+dx^2+ax+b&lt;/cmath&gt;also have &lt;cmath&gt;b+d-a-c=-9&lt;/cmath&gt; when &lt;cmath&gt;x=-1.&lt;/cmath&gt; So the number of solutions must be divisible by 4. So the answer must be &lt;math&gt;\boxed{\textbf{D}.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Algebra Problems]]</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_18&diff=90188 2016 AMC 10A Problems/Problem 18 2018-02-03T23:35:33Z <p>Drunkenninja: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First of all, the adjacent faces have the same sum &lt;math&gt;(18&lt;/math&gt;, because &lt;math&gt;1+2+3+4+5+6+7+8=36&lt;/math&gt;, &lt;math&gt;36/2=18)&lt;/math&gt;, <br /> so now consider the &lt;math&gt;opposite \text{ } sides&lt;/math&gt; (the two sides which are parallel but not in same face of the cube);<br /> they must have the same sum value too.<br /> Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of &lt;math&gt;opposite \text{ } sides&lt;/math&gt;,<br /> we should have &lt;math&gt;1+X=8+Y&lt;/math&gt;, but no solution for &lt;math&gt;[2,7]&lt;/math&gt;, contradiction. <br /> <br /> Now we know &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; must share the same side, which sum is &lt;math&gt;9&lt;/math&gt;, the &lt;math&gt;opposite \text{ } side&lt;/math&gt; also must have sum of &lt;math&gt;9&lt;/math&gt;, same thing for the other two parallel sides.<br /> <br /> Now we have &lt;math&gt;4&lt;/math&gt; parallel sides &lt;math&gt;1-8, 2-7, 3-6, 4-5&lt;/math&gt;.<br /> thinking about &lt;math&gt;4&lt;/math&gt; end points number need to have sum of &lt;math&gt;18&lt;/math&gt;.<br /> It is easy to notice only &lt;math&gt;1-7-6-4&lt;/math&gt; vs &lt;math&gt;8-2-3-5&lt;/math&gt; would work.<br /> <br /> So if we fix one direction &lt;math&gt;1-8 (&lt;/math&gt;or &lt;math&gt;8-1)&lt;/math&gt; all other &lt;math&gt;3&lt;/math&gt; parallel sides must lay in one particular direction. &lt;math&gt;(1-8,7-2,6-3,4-5)&lt;/math&gt; or &lt;math&gt;(8-1,2-7,3-6,5-4)&lt;/math&gt;<br /> <br /> Now, the problem is same as the problem to arrange &lt;math&gt;4&lt;/math&gt; points in a two-dimensional square. which is &lt;math&gt;4!/4&lt;/math&gt;=&lt;math&gt;\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Again, all faces sum to &lt;math&gt;18.&lt;/math&gt; If &lt;math&gt;x,y,z&lt;/math&gt; are the vertices next to one, then the remaining vertices are &lt;math&gt;17-x-y, 17-y-z, 17-x-z, x+y+z-16.&lt;/math&gt; Now it remains to test possibilities. Note that we must have &lt;math&gt;x+y+z&gt;17.&lt;/math&gt; Without loss of generality, let &lt;math&gt;x&lt;y&lt;z.&lt;/math&gt;<br /> <br /> &lt;math&gt;3,7,8:&lt;/math&gt; Does not work.<br /> &lt;math&gt;4,6,8:&lt;/math&gt; Works.<br /> &lt;math&gt;5,6,7:&lt;/math&gt; Does not work.<br /> &lt;math&gt;5,6,8:&lt;/math&gt; Works.<br /> &lt;math&gt;5,7,8:&lt;/math&gt; Does not work.<br /> &lt;math&gt;6,7,8:&lt;/math&gt; Works.<br /> <br /> So our answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> We know the sum of each face is &lt;math&gt;18.&lt;/math&gt; If we look at an edge of the cube whose numbers sum to &lt;math&gt;x&lt;/math&gt;, it must be possible to achieve the sum &lt;math&gt;18-x&lt;/math&gt; in two distinct ways, looking at the two faces which contain the edge. If &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; were on the same face, it is possible to achieve the desired sum only with the numbers &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; since the values must be distinct. Similarly, if &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; were on the same face, the only way to get the sum is with &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. This means that &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are not on the same edge as &lt;math&gt;8&lt;/math&gt;, or in other words they are diagonally across from it on the same face, or on the other end of the cube.<br /> <br /> Now we look at three cases, each yielding two solutions which are reflections of each other:<br /> <br /> 1) &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are diagonally opposite &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 2) &lt;math&gt;6&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;7&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 3) &lt;math&gt;7&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;6&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> <br /> This means the answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_19&diff=89597 2011 AMC 12A Problems/Problem 19 2018-01-08T17:22:05Z <p>Drunkenninja: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> At a competition with &lt;math&gt;N&lt;/math&gt; players, the number of players given elite status is equal to &lt;math&gt;2^{1+\lfloor \log_{2} (N-1) \rfloor}-N&lt;/math&gt;. Suppose that &lt;math&gt;19&lt;/math&gt; players are given elite status. What is the sum of the two smallest possible values of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 38 \qquad<br /> \textbf{(B)}\ 90 \qquad<br /> \textbf{(C)}\ 154 \qquad<br /> \textbf{(D)}\ 406 \qquad<br /> \textbf{(E)}\ 1024 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> We start with &lt;math&gt; 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19&lt;/math&gt;. After rearranging, we get &lt;math&gt;\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt; \lfloor\log_{2}(N-1)\rfloor &lt;/math&gt; is a positive integer, &lt;math&gt; \frac{N+19}{2}&lt;/math&gt; must be in the form of &lt;math&gt;2^{m} &lt;/math&gt; for some positive integer &lt;math&gt; m &lt;/math&gt;. From this fact, we get &lt;math&gt;N=2^{m+1}-19&lt;/math&gt;.<br /> <br /> If we now check integer values of N that satisfy this condition, starting from &lt;math&gt;N=19&lt;/math&gt;, we quickly see that the first values that work for &lt;math&gt;N&lt;/math&gt; are &lt;math&gt;2^6 -19&lt;/math&gt; and &lt;math&gt;2^7 -19&lt;/math&gt;, giving values of &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; for &lt;math&gt;m&lt;/math&gt;, respectively. Adding up these two values for &lt;math&gt;N&lt;/math&gt;, we get &lt;math&gt;45 + 109 = 154 \rightarrow \boxed{\textbf{C}}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> We examine the value that &lt;math&gt; 2^{1+\lfloor\log_{2}(N-1)\rfloor}&lt;/math&gt; takes over various intervals. The &lt;math&gt;\lfloor\log_{2}(N-1)\rfloor&lt;/math&gt; means it changes on each multiple of 2, like so:<br /> <br /> 2 --&gt; 1<br /> <br /> 3 - 4 --&gt; 2<br /> <br /> 5 - 8 --&gt; 3<br /> <br /> 9 - 16 --&gt; 4<br /> <br /> From this, we see that &lt;math&gt; 2^{1+\lfloor\log_{2}(N-1)\rfloor} - N&lt;/math&gt; is the difference between the next power of 2 above &lt;math&gt; 2^{\lfloor\log_{2}(N-1)\rfloor}&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;. We are looking for &lt;math&gt;N&lt;/math&gt; such that this difference is 19. The first two &lt;math&gt;N&lt;/math&gt; that satisfy this are &lt;math&gt;45 = 64-19&lt;/math&gt; and &lt;math&gt;109=128-19&lt;/math&gt; for a final answer of &lt;math&gt;45 + 109 = 154 \rightarrow \boxed{\textbf{C}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_18&diff=89473 2011 AMC 12B Problems/Problem 18 2018-01-06T06:07:34Z <p>Drunkenninja: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}&lt;/math&gt;<br /> <br /> ==Solution== <br /> <br /> ===Solution 1===<br /> We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths &lt;math&gt;\frac{1}{2}, 1&lt;/math&gt; and &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base.<br /> <br /> This triangle is isosceles with a base of 1 and two sides of length &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides &lt;math&gt;\frac{1}{2}, \frac{\sqrt2}{2}&lt;/math&gt; and &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> The cube, touching all four triangular faces, will form a similar pyramid which sits on top of the cube. If the cube has side length &lt;math&gt;x&lt;/math&gt;, the small pyramid has height &lt;math&gt;\frac{x\sqrt{2}}{2}&lt;/math&gt; (because the pyramids are similar).<br /> <br /> Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.<br /> <br /> &lt;math&gt;x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}&lt;/math&gt;.<br /> <br /> &lt;math&gt;x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x\left(2+\sqrt{2}\right) = \sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 = &lt;/math&gt;side length of cube.<br /> <br /> &lt;math&gt;\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2011|num-b=17|num-a=19|ab=B}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_18&diff=89472 2011 AMC 12B Problems/Problem 18 2018-01-06T06:07:17Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>==Problem==<br /> A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}&lt;/math&gt;<br /> <br /> ==Solution== <br /> <br /> ===Solution 1===<br /> We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths &lt;math&gt;\frac{1}{2}, 1&lt;/math&gt; and &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base.<br /> <br /> This triangle is isosceles with a base of 1 and two sides of length &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides &lt;math&gt;\frac{1}{2}, \frac{\sqrt2}{2}&lt;/math&gt; and &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> The cube, touching all four triangular faces, will form a similar pyramid which sits on top of the cube. If the cube has side length &lt;math&gt;x&lt;/math&gt;, the small pyramid has height &lt;math&gt;\frac{x\sqrt{2}}{2}&lt;/math&gt; (because the pyramids are similar).<br /> <br /> Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.<br /> <br /> &lt;math&gt;x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}&lt;/math&gt;.<br /> <br /> &lt;math&gt;x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x\left(2+\sqrt{2}\right) = \sqrt{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 = &lt;/math&gt;side length of cube.<br /> <br /> &lt;math&gt;\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2011|num-b=17|num-a=19|ab=B}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=89211 2010 AMC 12B Problems/Problem 18 2017-12-27T06:34:27Z <p>Drunkenninja: /* Solution 2 (Geometric) */</p> <hr /> <div>== Problem ==<br /> A frog makes &lt;math&gt;3&lt;/math&gt; jumps, each exactly &lt;math&gt;1&lt;/math&gt; meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than &lt;math&gt;1&lt;/math&gt; meter from its starting position?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1 (Complex Numbers)===<br /> We will let the moves be complex numbers &lt;math&gt; a&lt;/math&gt;, &lt;math&gt; b&lt;/math&gt;, and &lt;math&gt; c&lt;/math&gt;, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br /> &lt;cmath&gt; \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}&lt;/cmath&gt;<br /> has magnitude less than or equal to &lt;math&gt; 1&lt;/math&gt;. Hence, the probability is &lt;math&gt; \boxed{\text{(C)} \frac {1}{4}}&lt;/math&gt;.<br /> <br /> ===Solution 2 (Geometric)===<br /> <br /> The first frog hop doesn't matter because no matter where the frog hops, it lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.<br /> <br /> <br /> No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how &lt;math&gt;\frac{1}{2}x^2&lt;/math&gt; and &lt;math&gt;x^2&lt;/math&gt; are tangent. The area ratio of the two circles is &lt;cmath&gt;\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}&lt;/cmath&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_13&diff=89180 2010 AMC 12A Problems/Problem 13 2017-12-26T01:34:21Z <p>Drunkenninja: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> For how many integer values of &lt;math&gt;k&lt;/math&gt; do the graphs of &lt;math&gt;x^2+y^2=k^2&lt;/math&gt; and &lt;math&gt;xy = k&lt;/math&gt; not intersect?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> The image below shows the two curves for &lt;math&gt;k=4&lt;/math&gt;. The blue curve is &lt;math&gt;x^2+y^2=k^2&lt;/math&gt;, which is clearly a circle with radius &lt;math&gt;k&lt;/math&gt;, and the red curve is a part of the curve &lt;math&gt;xy=k&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(200);<br /> <br /> real f(real x) {return 4/x;};<br /> real g1(real x) {return sqrt(4*4-x*x);};<br /> real g2(real x) {return -sqrt(4*4-x*x);};<br /> draw(graph(f,-20./3,-0.6),red);<br /> draw(graph(f,0.6,20./3),red);<br /> draw(graph(g1,-4,4),blue);<br /> draw(graph(g2,-4,4),blue);<br /> axes(&quot;$x$&quot;,&quot;$y$&quot;);<br /> &lt;/asy&gt;<br /> <br /> In the special case &lt;math&gt;k=0&lt;/math&gt; the blue curve is just the point &lt;math&gt;(0,0)&lt;/math&gt;, and as &lt;math&gt;0\cdot 0=0&lt;/math&gt;, this point is on the red curve as well, hence they intersect. <br /> <br /> The case &lt;math&gt;k&lt;0&lt;/math&gt; is symmetric to &lt;math&gt;k&gt;0&lt;/math&gt;: the blue curve remains the same and the red curve is flipped according to the &lt;math&gt;x&lt;/math&gt; axis. Hence we just need to focus on &lt;math&gt;k&gt;0&lt;/math&gt;.<br /> <br /> Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as &lt;math&gt;x&lt;/math&gt; approaches 0, &lt;math&gt;y&lt;/math&gt; approaches &lt;math&gt;\infty&lt;/math&gt;. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most &lt;math&gt;k&lt;/math&gt;.<br /> <br /> <br /> At this point we can guess that on the red curve the point where &lt;math&gt;x=y&lt;/math&gt; is always closest to the origin, and skip the rest of this solution.<br /> <br /> <br /> For an exact solution, fix &lt;math&gt;k&lt;/math&gt; and consider any point &lt;math&gt;(x,y)&lt;/math&gt; on the red curve. Its distance from the origin is &lt;math&gt;\sqrt{ x^2 + (k/x)^2 }&lt;/math&gt;. To minimize this distance, it is enough to minimize &lt;math&gt;x^2 + (k/x)^2&lt;/math&gt;. By the [[Arithmetic Mean-Geometric Mean Inequality]] we get that this value is at least &lt;math&gt;2k&lt;/math&gt;, and that equality holds whenever &lt;math&gt;x^2 = (k/x)^2&lt;/math&gt;, i.e., &lt;math&gt;x=\pm\sqrt k&lt;/math&gt;.<br /> <br /> <br /> Now recall that the red curve intersects the blue one if and only if its closest point is at most &lt;math&gt;k&lt;/math&gt; from the origin. We just computed that the distance between the origin and the closest point on the red curve is &lt;math&gt;\sqrt{2k}&lt;/math&gt;. Therefore, we want to find all positive integers &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;\sqrt{2k} &gt; k&lt;/math&gt;.<br /> <br /> Clearly the only such integer is &lt;math&gt;k=1&lt;/math&gt;, hence the two curves are only disjoint for &lt;math&gt;k=1&lt;/math&gt; and &lt;math&gt;k=-1&lt;/math&gt;. <br /> This is a total of &lt;math&gt;\boxed{2\ \textbf{(C)}}&lt;/math&gt; values.<br /> <br /> == Solution 2 ==<br /> <br /> <br /> From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle. <br /> <br /> Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates &lt;math&gt; (x, k/x) &lt;/math&gt; derived from the equation. Then, all coordinates that satisfy &lt;math&gt;\sqrt{ x^2+ (k/x)^2 } \leq k &lt;/math&gt; intersect the circle.<br /> Squaring, we find &lt;math&gt; x^2+(k/x)^2 \leq k^2. &lt;/math&gt;<br /> After multiplying through by &lt;math&gt; x^2 &lt;/math&gt; and rearranging, we find &lt;math&gt; x^4-x^2k^2+k^2 \leq 0 &lt;/math&gt;.<br /> We see this is a quadratic in &lt;math&gt; x^2 &lt;/math&gt; and consider taking the determinant, which tells us that solutions are real when, after factoring:<br /> &lt;math&gt; k^2(k^2-4) \geq 0 &lt;/math&gt;<br /> We plot this inequality on the number line to find it is satisfied for all values except: &lt;math&gt; (-1, 0, 1) &lt;/math&gt;.<br /> We then eliminate 0 because it is extraneous as both &lt;math&gt; xy=0 &lt;/math&gt; and &lt;math&gt; x^2+y^2=0 &lt;/math&gt; are points which coincide.<br /> Therefore, there are a total of &lt;math&gt;\boxed{2\ \textbf{(C)}}&lt;/math&gt; values.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=12|num-a=14|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_6&diff=89104 2010 AMC 12A Problems/Problem 6 2017-12-23T18:13:20Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>== Problem ==<br /> A &lt;math&gt;\text{palindrome}&lt;/math&gt;, such as 83438, is a number that remains the same when its digits are reversed. The numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;x+32&lt;/math&gt; are three-digit and four-digit palindromes, respectively. What is the sum of the digits of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> &lt;math&gt;x&lt;/math&gt; is at most &lt;math&gt;999&lt;/math&gt;, so &lt;math&gt;x+32&lt;/math&gt; is at most &lt;math&gt;1031&lt;/math&gt;. The minimum value of &lt;math&gt;x+32&lt;/math&gt; is &lt;math&gt;1000&lt;/math&gt;. However, the only palindrome between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;1032&lt;/math&gt; is &lt;math&gt;1001&lt;/math&gt;, which means that &lt;math&gt;x+32&lt;/math&gt; must be &lt;math&gt;1001&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;969&lt;/math&gt;, so the sum of the digits is &lt;math&gt;\boxed{\textbf{(E)}\ 24}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> For &lt;math&gt;x+32&lt;/math&gt; to be a four-digit number, &lt;math&gt;x&lt;/math&gt; is in between &lt;math&gt;968&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;. The palindromes in this range are &lt;math&gt;969&lt;/math&gt;, &lt;math&gt;979&lt;/math&gt;, &lt;math&gt;989&lt;/math&gt;, and &lt;math&gt;999&lt;/math&gt;, so the sum of digits of &lt;math&gt;x&lt;/math&gt; can be &lt;math&gt;24&lt;/math&gt;, &lt;math&gt;25&lt;/math&gt;, &lt;math&gt;26&lt;/math&gt;, or &lt;math&gt;27&lt;/math&gt;. Only &lt;math&gt;\boxed{\textbf{(E)}\ 24}&lt;/math&gt; is an option, and upon checking, &lt;math&gt;x+32=1001&lt;/math&gt; is indeed a palindrome.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_12&diff=88957 2009 AMC 12A Problems/Problem 12 2017-12-18T00:06:46Z <p>Drunkenninja: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> How many positive integers less than &lt;math&gt;1000&lt;/math&gt; are &lt;math&gt;6&lt;/math&gt; times the sum of their digits?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> The sum of the digits is at most &lt;math&gt;9+9+9=27&lt;/math&gt;. Therefore the number is at most &lt;math&gt;6\cdot 27 = 162&lt;/math&gt;. Out of the numbers &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;162&lt;/math&gt; the one with the largest sum of digits is &lt;math&gt;99&lt;/math&gt;, and the sum is &lt;math&gt;9+9=18&lt;/math&gt;. Hence the sum of digits will be at most &lt;math&gt;18&lt;/math&gt;.<br /> <br /> Also, each number with this property is divisible by &lt;math&gt;6&lt;/math&gt;, therefore it is divisible by &lt;math&gt;3&lt;/math&gt;, and thus also its sum of digits is divisible by &lt;math&gt;3&lt;/math&gt;. Thus, the number is divisible by &lt;math&gt;18&lt;/math&gt;.<br /> <br /> We only have six possibilities left for the sum of the digits: &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, and &lt;math&gt;18&lt;/math&gt;, but since the number is divisible by &lt;math&gt;18&lt;/math&gt;, the digits can only add to &lt;math&gt;9&lt;/math&gt; or &lt;math&gt;18&lt;/math&gt;. This leads to the integers &lt;math&gt;18&lt;/math&gt;, &lt;math&gt;36&lt;/math&gt;, &lt;math&gt;54&lt;/math&gt;, &lt;math&gt;72&lt;/math&gt;, &lt;math&gt;90&lt;/math&gt;, and &lt;math&gt;108&lt;/math&gt; being possibilities. We can check to see that &lt;math&gt;\boxed{1}&lt;/math&gt; solution: the number &lt;math&gt;54&lt;/math&gt; is the only solution that satisfies the conditions in the problem.<br /> <br /> === Solution 2 ===<br /> <br /> We can write each integer between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt; inclusive as &lt;math&gt;\overline{abc}=100a+10b+c&lt;/math&gt; where &lt;math&gt;a,b,c\in\{0,1,\dots,9\}&lt;/math&gt; and &lt;math&gt;a+b+c&gt;0&lt;/math&gt;.<br /> The sum of digits of this number is &lt;math&gt;a+b+c&lt;/math&gt;, hence we get the equation &lt;math&gt;100a+10b+c = 6(a+b+c)&lt;/math&gt;. This simplifies to &lt;math&gt;94a + 4b - 5c = 0&lt;/math&gt;. Clearly for &lt;math&gt;a&gt;0&lt;/math&gt; there are no solutions, hence &lt;math&gt;a=0&lt;/math&gt; and we get the equation &lt;math&gt;4b=5c&lt;/math&gt;. This obviously has only one valid solution &lt;math&gt;(b,c)=(5,4)&lt;/math&gt;, hence the only solution is the number &lt;math&gt;54&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> The sum of the digits is at most &lt;math&gt;9+9+9=27&lt;/math&gt;. Therefore the number is at most &lt;math&gt;6\cdot 27 = 162&lt;/math&gt;. Since the number is &lt;math&gt;6&lt;/math&gt; times the sum of its digits, it must be divisible by &lt;math&gt;6&lt;/math&gt;, therefore also by &lt;math&gt;3&lt;/math&gt;, therefore the sum of its digits must be divisible by &lt;math&gt;3&lt;/math&gt;. With this in mind we can conclude that the number must be divisible by &lt;math&gt;18&lt;/math&gt;, not just by &lt;math&gt;6&lt;/math&gt;. Since the number is divisible by &lt;math&gt;18&lt;/math&gt;, it is also divisible by &lt;math&gt;9&lt;/math&gt;, therefore the sum of its digits is divisible by &lt;math&gt;9&lt;/math&gt;, therefore the number is divisible by &lt;math&gt;54&lt;/math&gt;, which leaves us with &lt;math&gt;54&lt;/math&gt;, &lt;math&gt;108&lt;/math&gt; and &lt;math&gt;162&lt;/math&gt;. Only &lt;math&gt;54&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt; times its digits, hence the answer is &lt;math&gt;\boxed{1}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=85052 2016 AMC 10B Problems/Problem 25 2017-03-28T03:24:55Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Since &lt;math&gt;x = \lfloor x \rfloor + \{ x \}&lt;/math&gt;, we have <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> The function can then be simplified into <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> which becomes<br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor&lt;/cmath&gt;<br /> <br /> We can see that for each value of k, &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; can equal integers from 0 to k-1. <br /> <br /> Clearly, the value of &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; changes only when x is equal to any of the fractions &lt;math&gt;\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}&lt;/math&gt;.<br /> <br /> So we want to count how many distinct fractions have the form &lt;math&gt;\frac{m}{n}&lt;/math&gt; where &lt;math&gt;n \le 10&lt;/math&gt;. We can find this easily by computing<br /> &lt;cmath&gt;\sum_{k=2}^{10} \phi(k)&lt;/cmath&gt;<br /> where &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function. Basically &lt;math&gt;\phi(k)&lt;/math&gt; counts the number of fractions with &lt;math&gt;k&lt;/math&gt; as its denominator (after simplification). This comes out to be &lt;math&gt;31&lt;/math&gt;.<br /> <br /> Because the value of &lt;math&gt;f(x)&lt;/math&gt; is at least 0 and can increase 31 times, there are a total of &lt;math&gt;\fbox{\textbf{(A)}\ 32}&lt;/math&gt; different possible values of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=85051 2016 AMC 10B Problems/Problem 25 2017-03-28T03:23:58Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Since &lt;math&gt;x = \lfloor x \rfloor + \{ x \}&lt;/math&gt;, we have <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> The function can then be simplified into <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> which becomes<br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor&lt;/cmath&gt;<br /> <br /> We can see that for each value of k, &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; can equal integers from 0 to k-1. <br /> <br /> Clearly, the value of &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; changes only when x is equal to any of the fractions &lt;math&gt;\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}&lt;/math&gt;.<br /> <br /> So we want to count how many distinct fractions have the form &lt;math&gt;\frac{m}{n}&lt;/math&gt; where &lt;math&gt;n \le 10&lt;/math&gt;. We can find this easily by computing<br /> &lt;cmath&gt;\sum_{k=2}^{10} \phi(k)&lt;/cmath&gt;<br /> where &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function. Basically &lt;math&gt;\phi(k)&lt;/math&gt; counts the number of fractions with &lt;math&gt;k&lt;/math&gt; as its denominator (after simplification). This comes out to be &lt;math&gt;31&lt;/math&gt;.<br /> <br /> Because the value of &lt;math&gt;f(x)&lt;/math&gt; is at least 0 and can increase 31 times, there are a total of &lt;math&gt;\fbox{\textbf{(A)}\ 32}&lt;/math&gt; different possible values of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_25&diff=85050 2016 AMC 10B Problems/Problem 25 2017-03-28T03:23:24Z <p>Drunkenninja: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Since &lt;math&gt;x = \lfloor x \rfloor + \{ x \}&lt;/math&gt;, we have <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} (\lfloor k \lfloor x \rfloor +\rfloor k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> The function can then be simplified into <br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} ( k \lfloor x \rfloor + \lfloor k \{ x \} \rfloor - k \lfloor x \rfloor)&lt;/cmath&gt;<br /> <br /> which becomes<br /> <br /> &lt;cmath&gt;f(x) = \sum_{k=2}^{10} \lfloor k \{ x \} \rfloor&lt;/cmath&gt;<br /> <br /> We can see that for each value of k, &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; can equal integers from 0 to k-1. <br /> <br /> Clearly, the value of &lt;math&gt;\lfloor k \{ x \} \rfloor&lt;/math&gt; changes only when x is equal to any of the fractions &lt;math&gt;\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}&lt;/math&gt;.<br /> <br /> So we want to count how many distinct fractions have the form &lt;math&gt;\frac{m}{n}&lt;/math&gt; where &lt;math&gt;n \le 10&lt;/math&gt;. We can find this easily by computing<br /> &lt;cmath&gt;\sum_{k=2}^{10} \phi(k)&lt;/cmath&gt;<br /> where &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function. Basically &lt;math&gt;\phi(k)&lt;/math&gt; counts the number of fractions with &lt;math&gt;k&lt;/math&gt; as its denominator (after simplification). This comes out to be &lt;math&gt;31&lt;/math&gt;.<br /> <br /> Because the value of &lt;math&gt;f(x)&lt;/math&gt; is at least 0 and can increase 31 times, there are a total of &lt;math&gt;\fbox{\textbf{(A)}\ 32}&lt;/math&gt; different possible values of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&diff=74647 2009 AIME I Problems/Problem 4 2016-01-18T01:00:01Z <p>Drunkenninja: /* Solution 4 */</p> <hr /> <div>== Problem 4 ==<br /> In parallelogram &lt;math&gt;ABCD&lt;/math&gt;, point &lt;math&gt;M&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\frac {AM}{AB} = \frac {17}{1000}&lt;/math&gt; and point &lt;math&gt;N&lt;/math&gt; is on &lt;math&gt;\overline{AD}&lt;/math&gt; so that &lt;math&gt;\frac {AN}{AD} = \frac {17}{2009}&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of intersection of &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Find &lt;math&gt;\frac {AC}{AP}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> One of the ways to solve this problem is to make this parallelogram a straight line.<br /> So the whole length of the line is &lt;math&gt;APC&lt;/math&gt;(&lt;math&gt;AMC&lt;/math&gt; or &lt;math&gt;ANC&lt;/math&gt;), and &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1000x+2009x=3009x.&lt;/math&gt;<br /> <br /> &lt;math&gt;AP&lt;/math&gt;(&lt;math&gt;AM&lt;/math&gt; or &lt;math&gt;AN&lt;/math&gt;) is &lt;math&gt;17x.&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;3009x/17x = \boxed{177}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Draw a diagram with all the given points and lines involved. Construct parallel lines &lt;math&gt;\overline{DF_2F_1}&lt;/math&gt; and &lt;math&gt;\overline{BB_1B_2}&lt;/math&gt; to &lt;math&gt;\overline{MN}&lt;/math&gt;, where for the lines the endpoints are on &lt;math&gt;\overline{AM}&lt;/math&gt; and &lt;math&gt;\overline{AN}&lt;/math&gt;, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral &lt;math&gt;BB_2DF_1&lt;/math&gt; &lt;math&gt;\overline{E_1E_2E_3}&lt;/math&gt; where the points are in order from top to bottom. Clearly, by similar triangles, &lt;math&gt;BB_2 = \frac {1000}{17}MN&lt;/math&gt; and &lt;math&gt;DF_1 = \frac {2009}{17}MN&lt;/math&gt;. It is not difficult to see that &lt;math&gt;E_2&lt;/math&gt; is the center of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; and thus the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt; as well as the midpoint of &lt;math&gt;\overline{B_1}{F_2}&lt;/math&gt; (all of this is easily proven with symmetry). From more triangle similarity, &lt;math&gt;E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP&lt;/math&gt;<br /> &lt;math&gt;= \boxed{177}AP&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Using vectors, note that &lt;math&gt;\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}&lt;/math&gt;. Note that &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; for some positive x and y, but at the same time is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. So, writing the equation &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; in terms of &lt;math&gt;\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AD}&lt;/math&gt;, we have &lt;math&gt;\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}&lt;/math&gt;. But the coefficients of the two vectors must be equal because, as already stated, &lt;math&gt;\overrightarrow{AP}&lt;/math&gt; is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. We then see that &lt;math&gt;\frac{x}{x+y}=\frac{1000}{3009}&lt;/math&gt; and &lt;math&gt;\frac{y}{x+y}=\frac{2009}{3009}&lt;/math&gt;. Finally, we have &lt;math&gt;\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})&lt;/math&gt; and, simplifying, &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}=177\overrightarrow{AP}&lt;/math&gt; and the desired quantity is &lt;math&gt;177&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We approach the problem using mass points on triangle &lt;math&gt;ABD&lt;/math&gt; as displayed below.<br /> &lt;asy&gt;<br /> <br /> pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);<br /> draw(A--B--C--D--cycle);<br /> draw(B--D^^A--C^^M--NN);<br /> pair O=extension(A,C,B,D);<br /> pair P=extension(A,C,M,NN);<br /> dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$N$&quot;,NN,NW);<br /> label(&quot;$P$&quot;,P,NNE);<br /> label(&quot;$O$&quot;,O,N);<br /> &lt;/asy&gt;<br /> <br /> But as &lt;math&gt;MN&lt;/math&gt; does not protrude from a vertex, we will have to &quot;split the mass&quot; at point &lt;math&gt;A&lt;/math&gt;. First, we know that &lt;math&gt;DO&lt;/math&gt; is congruent to &lt;math&gt;BO&lt;/math&gt; because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. In this case, we assign &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; a mass of 17 each. Now we split the mass at &lt;math&gt;A&lt;/math&gt;, so we balance segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; separately, and then the mass of &lt;math&gt;A&lt;/math&gt; is the sum of those masses. A mass of 983 is required to balance segment &lt;math&gt;AB&lt;/math&gt;, while a mass of 1992 is required to balance segment &lt;math&gt;AD&lt;/math&gt;. Therefore, &lt;math&gt;A&lt;/math&gt; has a mass of &lt;math&gt;1992+983=2975&lt;/math&gt;. Also, &lt;math&gt;O&lt;/math&gt; has a mass of 34. Therefore, &lt;math&gt;\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}&lt;/math&gt;, so &lt;math&gt;\frac{AC}{AP}=\frac{2 (3009)}{34}=177&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&diff=74623 2009 AIME I Problems/Problem 4 2016-01-18T00:24:26Z <p>Drunkenninja: /* Solution 4 */</p> <hr /> <div>== Problem 4 ==<br /> In parallelogram &lt;math&gt;ABCD&lt;/math&gt;, point &lt;math&gt;M&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\frac {AM}{AB} = \frac {17}{1000}&lt;/math&gt; and point &lt;math&gt;N&lt;/math&gt; is on &lt;math&gt;\overline{AD}&lt;/math&gt; so that &lt;math&gt;\frac {AN}{AD} = \frac {17}{2009}&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of intersection of &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Find &lt;math&gt;\frac {AC}{AP}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> One of the ways to solve this problem is to make this parallelogram a straight line.<br /> So the whole length of the line is &lt;math&gt;APC&lt;/math&gt;(&lt;math&gt;AMC&lt;/math&gt; or &lt;math&gt;ANC&lt;/math&gt;), and &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1000x+2009x=3009x.&lt;/math&gt;<br /> <br /> &lt;math&gt;AP&lt;/math&gt;(&lt;math&gt;AM&lt;/math&gt; or &lt;math&gt;AN&lt;/math&gt;) is &lt;math&gt;17x.&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;3009x/17x = \boxed{177}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Draw a diagram with all the given points and lines involved. Construct parallel lines &lt;math&gt;\overline{DF_2F_1}&lt;/math&gt; and &lt;math&gt;\overline{BB_1B_2}&lt;/math&gt; to &lt;math&gt;\overline{MN}&lt;/math&gt;, where for the lines the endpoints are on &lt;math&gt;\overline{AM}&lt;/math&gt; and &lt;math&gt;\overline{AN}&lt;/math&gt;, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral &lt;math&gt;BB_2DF_1&lt;/math&gt; &lt;math&gt;\overline{E_1E_2E_3}&lt;/math&gt; where the points are in order from top to bottom. Clearly, by similar triangles, &lt;math&gt;BB_2 = \frac {1000}{17}MN&lt;/math&gt; and &lt;math&gt;DF_1 = \frac {2009}{17}MN&lt;/math&gt;. It is not difficult to see that &lt;math&gt;E_2&lt;/math&gt; is the center of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; and thus the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt; as well as the midpoint of &lt;math&gt;\overline{B_1}{F_2}&lt;/math&gt; (all of this is easily proven with symmetry). From more triangle similarity, &lt;math&gt;E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP&lt;/math&gt;<br /> &lt;math&gt;= \boxed{177}AP&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Using vectors, note that &lt;math&gt;\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}&lt;/math&gt;. Note that &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; for some positive x and y, but at the same time is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. So, writing the equation &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; in terms of &lt;math&gt;\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AD}&lt;/math&gt;, we have &lt;math&gt;\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}&lt;/math&gt;. But the coefficients of the two vectors must be equal because, as already stated, &lt;math&gt;\overrightarrow{AP}&lt;/math&gt; is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. We then see that &lt;math&gt;\frac{x}{x+y}=\frac{1000}{3009}&lt;/math&gt; and &lt;math&gt;\frac{y}{x+y}=\frac{2009}{3009}&lt;/math&gt;. Finally, we have &lt;math&gt;\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})&lt;/math&gt; and, simplifying, &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}=177\overrightarrow{AP}&lt;/math&gt; and the desired quantity is &lt;math&gt;177&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We approach the problem using mass points on triangle &lt;math&gt;ABD&lt;/math&gt; as displayed below.<br /> &lt;asy&gt;<br /> <br /> pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);<br /> draw(A--B--C--D--cycle);<br /> draw(B--D^^A--C^^M--NN);<br /> pair O=extension(A,C,B,D);<br /> pair P=extension(A,C,M,NN);<br /> dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$N$&quot;,NN,NW);<br /> label(&quot;$P$&quot;,P,NNE);<br /> label(&quot;$O$&quot;,O,N);<br /> &lt;/asy&gt;<br /> <br /> But as &lt;math&gt;MN&lt;/math&gt; does not protrude from a vertex, we will have to &quot;split the mass&quot; at point &lt;math&gt;A&lt;/math&gt;. First, we know that &lt;math&gt;DO&lt;/math&gt; is congruent to &lt;math&gt;BO&lt;/math&gt; because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. In this case, we assign &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; a mass of 17 each. Now we split the mass at &lt;math&gt;A&lt;/math&gt;, so we balance segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; separately, and then the mass of &lt;math&gt;A&lt;/math&gt; is the sum of those masses. A mass of 983 is required to balance segment &lt;math&gt;AB&lt;/math&gt;, while a mass of 1992 is required to balance segment &lt;math&gt;AD&lt;/math&gt;. Therefore, &lt;math&gt;A&lt;/math&gt; has a mass of &lt;math&gt;1992+983=2975&lt;/math&gt;. Also, &lt;math&gt;O&lt;/math&gt; has a mass of 34. Therefore, &lt;math&gt;\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}&lt;/math&gt;, so &lt;math&gt;\frac{AC}{AP}=\frac{3009}{17}=177&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Drunkenninja https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&diff=74577 2009 AIME I Problems/Problem 4 2016-01-16T21:23:14Z <p>Drunkenninja: /* Solution 4 */</p> <hr /> <div>== Problem 4 ==<br /> In parallelogram &lt;math&gt;ABCD&lt;/math&gt;, point &lt;math&gt;M&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\frac {AM}{AB} = \frac {17}{1000}&lt;/math&gt; and point &lt;math&gt;N&lt;/math&gt; is on &lt;math&gt;\overline{AD}&lt;/math&gt; so that &lt;math&gt;\frac {AN}{AD} = \frac {17}{2009}&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of intersection of &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Find &lt;math&gt;\frac {AC}{AP}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> One of the ways to solve this problem is to make this parallelogram a straight line.<br /> So the whole length of the line is &lt;math&gt;APC&lt;/math&gt;(&lt;math&gt;AMC&lt;/math&gt; or &lt;math&gt;ANC&lt;/math&gt;), and &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1000x+2009x=3009x.&lt;/math&gt;<br /> <br /> &lt;math&gt;AP&lt;/math&gt;(&lt;math&gt;AM&lt;/math&gt; or &lt;math&gt;AN&lt;/math&gt;) is &lt;math&gt;17x.&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;3009x/17x = \boxed{177}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Draw a diagram with all the given points and lines involved. Construct parallel lines &lt;math&gt;\overline{DF_2F_1}&lt;/math&gt; and &lt;math&gt;\overline{BB_1B_2}&lt;/math&gt; to &lt;math&gt;\overline{MN}&lt;/math&gt;, where for the lines the endpoints are on &lt;math&gt;\overline{AM}&lt;/math&gt; and &lt;math&gt;\overline{AN}&lt;/math&gt;, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral &lt;math&gt;BB_2DF_1&lt;/math&gt; &lt;math&gt;\overline{E_1E_2E_3}&lt;/math&gt; where the points are in order from top to bottom. Clearly, by similar triangles, &lt;math&gt;BB_2 = \frac {1000}{17}MN&lt;/math&gt; and &lt;math&gt;DF_1 = \frac {2009}{17}MN&lt;/math&gt;. It is not difficult to see that &lt;math&gt;E_2&lt;/math&gt; is the center of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; and thus the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt; as well as the midpoint of &lt;math&gt;\overline{B_1}{F_2}&lt;/math&gt; (all of this is easily proven with symmetry). From more triangle similarity, &lt;math&gt;E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP&lt;/math&gt;<br /> &lt;math&gt;= \boxed{177}AP&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Using vectors, note that &lt;math&gt;\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}&lt;/math&gt;. Note that &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; for some positive x and y, but at the same time is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. So, writing the equation &lt;math&gt;\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}&lt;/math&gt; in terms of &lt;math&gt;\overrightarrow{AB}&lt;/math&gt; and &lt;math&gt;\overrightarrow{AD}&lt;/math&gt;, we have &lt;math&gt;\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}&lt;/math&gt;. But the coefficients of the two vectors must be equal because, as already stated, &lt;math&gt;\overrightarrow{AP}&lt;/math&gt; is a scalar multiple of &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}&lt;/math&gt;. We then see that &lt;math&gt;\frac{x}{x+y}=\frac{1000}{3009}&lt;/math&gt; and &lt;math&gt;\frac{y}{x+y}=\frac{2009}{3009}&lt;/math&gt;. Finally, we have &lt;math&gt;\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})&lt;/math&gt; and, simplifying, &lt;math&gt;\overrightarrow{AB}+\overrightarrow{AD}=177\overrightarrow{AP}&lt;/math&gt; and the desired quantity is &lt;math&gt;177&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We approach the problem using mass points on triangle &lt;math&gt;ABD&lt;/math&gt; as displayed below.<br /> &lt;asy&gt;<br /> <br /> pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);<br /> draw(A--B--C--D--cycle);<br /> draw(B--D^^A--C^^M--NN);<br /> pair O=extension(A,C,B,D);<br /> pair P=extension(A,C,M,NN);<br /> dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$N$&quot;,NN,NW);<br /> label(&quot;$P$&quot;,P,NNE);<br /> label(&quot;$O$&quot;,O,N);<br /> &lt;/asy&gt;<br /> <br /> But as &lt;math&gt;MN&lt;/math&gt; does not protrude from a vertex, we will have to &quot;split the mass&quot; at point &lt;math&gt;A&lt;/math&gt;. First, we know that &lt;math&gt;DO&lt;/math&gt; is congruent to &lt;math&gt;BO&lt;/math&gt; because diagonals of parallelograms bisect each other. Therefore, we assign equal masses to points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. In this case, we assign &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; a mass of 17 each. Now we split the mass at &lt;math&gt;A&lt;/math&gt;, so we balance segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; separately, and then the mass of &lt;math&gt;A&lt;/math&gt; is the sum of those masses. A mass of 983 is required to balance segment &lt;math&gt;AB&lt;/math&gt;, while a mass of 1992 is required to balance segment &lt;math&gt;AD&lt;/math&gt;. Therefore, &lt;math&gt;A&lt;/math&gt; has a mass of &lt;math&gt;1992+983=2975&lt;/math&gt;. Also, &lt;math&gt;O&lt;/math&gt; has a mass of 34. Therefore, &lt;math&gt;\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}&lt;/math&gt;, so &lt;math&gt;\frac{AC}{AP}=\frac{3009}{17}=177&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Drunkenninja