https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ye&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-21T10:47:49Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_15&diff=85007 2017 AIME II Problems/Problem 15 2017-03-25T04:48:57Z <p>Ye: /* Solution */</p> <hr /> <div>==Problem==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AD=BC=28&lt;/math&gt;, &lt;math&gt;AC=BD=44&lt;/math&gt;, and &lt;math&gt;AB=CD=52&lt;/math&gt;. For any point &lt;math&gt;X&lt;/math&gt; in space, define &lt;math&gt;f(X)=AX+BX+CX+DX&lt;/math&gt;. The least possible value of &lt;math&gt;f(X)&lt;/math&gt; can be expressed as &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be midpoints of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;. The given conditions imply that &lt;math&gt;\triangle ABD\cong\triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA\cong\triangle DCB&lt;/math&gt;, and therefore &lt;math&gt;MC=MD&lt;/math&gt; and &lt;math&gt;NA=NB&lt;/math&gt;. It follows that &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; both lie on the common perpendicular bisector of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, and thus line &lt;math&gt;MN&lt;/math&gt; is that common perpendicular bisector. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are symmetric to &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;. If &lt;math&gt;X&lt;/math&gt; is a point in space and &lt;math&gt;X'&lt;/math&gt; is the point symmetric to &lt;math&gt;X&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;, then &lt;math&gt;BX=AX'&lt;/math&gt; and &lt;math&gt;CX=DX'&lt;/math&gt;, so &lt;math&gt;f(X) = AX+AX'+DX+DX'&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection of &lt;math&gt;\overline{XX'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Then &lt;math&gt;AX+AX'\geq 2AQ&lt;/math&gt;, from which it follows that &lt;math&gt;f(X) \geq 2(AQ+DQ) = f(Q)&lt;/math&gt;. It remains to minimize &lt;math&gt;f(Q)&lt;/math&gt; as &lt;math&gt;Q&lt;/math&gt; moves along &lt;math&gt;\overline{MN}&lt;/math&gt;.<br /> <br /> Allow &lt;math&gt;D&lt;/math&gt; to rotate about &lt;math&gt;\overline{MN}&lt;/math&gt; to point &lt;math&gt;D'&lt;/math&gt; in the plane &lt;math&gt;AMN&lt;/math&gt; on the side of &lt;math&gt;\overline{MN}&lt;/math&gt; opposite &lt;math&gt;A&lt;/math&gt;. Because &lt;math&gt;\angle DNM&lt;/math&gt; is a right angle, &lt;math&gt;D'N=DN&lt;/math&gt;. It then follows that &lt;math&gt;f(Q) = 2(AQ+D'Q)\geq 2AD'&lt;/math&gt;, and equality occurs when &lt;math&gt;Q&lt;/math&gt; is the intersection of &lt;math&gt;\overline{AD'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Thus &lt;math&gt;\min f(Q) = 2AD'&lt;/math&gt;. Because &lt;math&gt;\overline{MD}&lt;/math&gt; is the median of &lt;math&gt;\triangle ADB&lt;/math&gt;, the Length of Median Formula shows that &lt;math&gt;4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2&lt;/math&gt; and &lt;math&gt;MD^2 = 684&lt;/math&gt;. By the Pythagorean Theorem &lt;math&gt;MN^2 = MD^2 - ND^2 = 8&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle AMN&lt;/math&gt; and &lt;math&gt;\angle D'NM&lt;/math&gt; are right angles, &lt;cmath&gt;(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.&lt;/cmath&gt;It follows that &lt;math&gt;\min f(Q) = 2AD' = 4\sqrt{678}&lt;/math&gt;. The requested sum is &lt;math&gt;4+678=\boxed{682}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Set &lt;math&gt;a=BC=28&lt;/math&gt;, &lt;math&gt;b=CA=44&lt;/math&gt;, &lt;math&gt;c=AB=52&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the point which minimizes &lt;math&gt;f(X)&lt;/math&gt;.<br /> <br /> Claim: &lt;math&gt;O&lt;/math&gt; is the gravity center &lt;math&gt;\tfrac14(\vec A + \vec B + \vec C + \vec D)&lt;/math&gt;.<br /> Proof. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote the midpoints of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. From &lt;math&gt;\triangle ABD \cong \triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA \cong \triangle DCB&lt;/math&gt;, we have &lt;math&gt;MC=MD&lt;/math&gt;, &lt;math&gt;NA=NB&lt;/math&gt; an hence &lt;math&gt;MN&lt;/math&gt; is a perpendicular bisector of both segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. Then if &lt;math&gt;X&lt;/math&gt; is any point inside tetrahedron &lt;math&gt;ABCD&lt;/math&gt;, its orthogonal projection onto line &lt;math&gt;MN&lt;/math&gt; will have smaller &lt;math&gt;f&lt;/math&gt;-value; hence we conclude that &lt;math&gt;O&lt;/math&gt; must lie on &lt;math&gt;MN&lt;/math&gt;. Similarly, &lt;math&gt;O&lt;/math&gt; must lie on the line joining the midpoints of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Claim: The gravity center &lt;math&gt;O&lt;/math&gt; coincides with the circumcenter.<br /> Proof. Let &lt;math&gt;G_D&lt;/math&gt; be the centroid of triangle &lt;math&gt;ABC&lt;/math&gt;; then &lt;math&gt;DO = \tfrac 34 DG_D&lt;/math&gt; (by vectors). If we define &lt;math&gt;G_A&lt;/math&gt;, &lt;math&gt;G_B&lt;/math&gt;, &lt;math&gt;G_C&lt;/math&gt; similarly, we get &lt;math&gt;AO = \tfrac 34 AG_A&lt;/math&gt; and so on. But from symmetry we have &lt;math&gt;AG_A = BG_B = CG_C = DG_D&lt;/math&gt;, hence &lt;math&gt;AO = BO = CO = DO&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Now we use the fact that an isosceles tetrahedron has circumradius &lt;math&gt;R = \sqrt{\frac18(a^2+b^2+c^2)}&lt;/math&gt;. Here &lt;math&gt;R = \sqrt{678}&lt;/math&gt; so &lt;math&gt;f(O) = 4R = 4\sqrt{678}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;4 + 678 = \boxed{682}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum F(X) occurs at the center of gravity, and = 2d, where d is the length of the spatial diagonal of the rectangular box.<br /> <br /> Let the three dimensions of the box be a, b, c.<br /> a^2+b^2=28^2; <br /> c^2+b^2=52^2; <br /> a^2+c^2=44^2.<br /> <br /> Add three equations, d^2=(28^2+52^2+44^2)/2.<br /> Hence f(X)=4sqrt(678).<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Ye https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=84096 2017 AMC 12B Problems/Problem 18 2017-02-19T18:43:47Z <p>Ye: /* Solution 2: Similar triangles */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AAA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{4}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{CB}{DE} = \frac{AB}{AE} \Longrightarrow \frac{CB}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;CB = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw BF// ED with D on AE. AE=5×(4/7)=20/7.<br /> [ABF]=2×20/7=40/7.<br /> AC:CB:CF=49:35:25. (7/5 ration applied twice)<br /> [ABC]=49/(49+25)[ABF]=140/37.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ye https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_24&diff=84034 2017 AMC 12B Problems/Problem 24 2017-02-17T16:48:21Z <p>Ye: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;\triangle ABC&lt;/math&gt; is similar to &lt;math&gt;\triangle BCD&lt;/math&gt;, and &lt;math&gt;AB &gt; BC&lt;/math&gt;. There exists a point &lt;math&gt;E&lt;/math&gt; in the interior of &lt;math&gt;ABCD&lt;/math&gt; such that &lt;math&gt;\triangle ABC&lt;/math&gt; is similar to &lt;math&gt;\triangle CEB&lt;/math&gt; and the area of Triangle &lt;math&gt;AED&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt; times the area of Triangle &lt;math&gt;CEB&lt;/math&gt;. Find &lt;math&gt;AB/BC&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;CD=1&lt;/math&gt;, &lt;math&gt;BC=x&lt;/math&gt;, and &lt;math&gt;AB=x^2&lt;/math&gt;. Note that &lt;math&gt;AB/BC=x&lt;/math&gt;. By the Pythagorean Theorem, &lt;math&gt;BD=\sqrt{x^2+1}&lt;/math&gt;. Since &lt;math&gt;\triangle BCD \sim \triangle ABC \sim \triangle CEB&lt;/math&gt;, the ratios of side lengths must be equal. Since &lt;math&gt;BC=x&lt;/math&gt;, &lt;math&gt;CE=\frac{x^2}{\sqrt{x^2+1}}&lt;/math&gt; and &lt;math&gt;BE=\frac{x}{\sqrt{x^2+1}}&lt;/math&gt;. Let F be a point on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{EF}&lt;/math&gt; is an altitude of triangle &lt;math&gt;CEB&lt;/math&gt;. Note that &lt;math&gt;\triangle CEB \sim \triangle CFE \sim \triangle EFB&lt;/math&gt;. Therefore, &lt;math&gt;BF=\frac{x}{x^2+1}&lt;/math&gt; and &lt;math&gt;CF=\frac{x^3}{x^2+1}&lt;/math&gt;. Since &lt;math&gt;\overline{CF}&lt;/math&gt; and &lt;math&gt;\overline{BF}&lt;/math&gt; form altitudes of triangles &lt;math&gt;CED&lt;/math&gt; and &lt;math&gt;BEA&lt;/math&gt;, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle &lt;math&gt;BEC&lt;/math&gt; can be calculated, as it is a right triangle. Solving for each of these yields:<br /> &lt;cmath&gt;[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))&lt;/cmath&gt;<br /> &lt;cmath&gt;[ABCD]=[AED]+[DEC]+[CEB]+[BEA]&lt;/cmath&gt;<br /> &lt;cmath&gt;(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]&lt;/cmath&gt;<br /> &lt;cmath&gt;(x^3+x)/2=(20x^3)/(2(x^2+1))&lt;/cmath&gt;<br /> &lt;cmath&gt;(x)(x^2+1)=20x^3/(x^2+1)&lt;/cmath&gt;<br /> &lt;cmath&gt;(x^2+1)^2=20x^2&lt;/cmath&gt;<br /> &lt;cmath&gt;x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5&lt;/cmath&gt;<br /> Therefore, the answer is &lt;math&gt;\boxed{\textbf{(D) } 2+\sqrt{5}}&lt;/math&gt;<br /> <br /> <br /> [SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2). <br /> <br /> Meanwhile, FE:EG=[CBE]:[ADE]=1:17. <br /> FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), <br /> Namely x^4-18x^2+1=0.<br /> <br /> The rest is the same with solution 1.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ye https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_22&diff=84032 2017 AMC 12B Problems/Problem 22 2017-02-17T16:06:53Z <p>Ye: /* Solution */</p> <hr /> <div>==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}&lt;/math&gt;<br /> <br /> ==Solution==<br /> It amounts to filling in a 4x4 matrix M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let &lt;RA&gt;=number of non-0 elements in RA.<br /> <br /> WOLG, let C1=[1,-1,0,0]. Parity demands that &lt;RA&gt; and&lt;RB&gt;= 1 or 3. <br /> <br /> CASE 1. &lt;RA&gt;=3, and &lt;RB&gt;=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here.<br /> <br /> Case 2. &lt;RA&gt;=1, and &lt;RB&gt;=3. 3 Choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of &lt;RA&gt;=3 and &lt;RB&gt;=1. Total 24 subcases.<br /> <br /> Case 3. &lt;RA&gt;=1, and &lt;RB&gt;=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.<br /> <br /> In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ye https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_22&diff=84031 2017 AMC 12B Problems/Problem 22 2017-02-17T16:03:25Z <p>Ye: /* Solution */</p> <hr /> <div>==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}&lt;/math&gt;<br /> <br /> ==Solution==<br /> It amounts to filling in a 4x4 matrux M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let &lt;RA&gt;=number of non-0 elements in RA.<br /> <br /> WOLG, let C1=[1,-1,0,0]. Parity demands that &lt;RA&gt; and &lt;RB&gt; = 1 or 3. <br /> <br /> CASE 1. &lt;RA&gt;=3, and &lt;RB&gt;=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here.<br /> <br /> Case 2. &lt;RA&gt;=1, and &lt;RB&gt;=3. 3 Choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of &lt;RA&gt;=3 and &lt;RB&gt;=1. Total 24 subcases.<br /> <br /> Case 3. &lt;RA&gt;=1, and &lt;RB&gt;=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.<br /> <br /> In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ye https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_22&diff=84030 2017 AMC 12B Problems/Problem 22 2017-02-17T15:55:27Z <p>Ye: /* Solution */</p> <hr /> <div>==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}&lt;/math&gt;<br /> <br /> ==Solution==<br /> It amountsI choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of &lt;RA&gt;=3 and &lt;RB&gt;=1. Total 24 subcases.<br /> <br /> Case 3. &lt;RA&gt;=1, and &lt;RB&gt;=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.<br /> <br /> In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ye