https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Binderclips1&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-27T11:37:59Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_14&diff=125509 2005 AMC 10B Problems/Problem 14 2020-06-15T19:02:25Z <p>Binderclips1: added note.</p> <hr /> <div>== Problem ==<br /> Equilateral &lt;math&gt; \triangle ABC&lt;/math&gt; has side length &lt;math&gt;2&lt;/math&gt;, &lt;math&gt; M&lt;/math&gt; is the midpoint of &lt;math&gt; \overline{AC}&lt;/math&gt;, and &lt;math&gt; C&lt;/math&gt; is the midpoint of &lt;math&gt; \overline{BD}&lt;/math&gt;. What is the area of &lt;math&gt; \triangle CDM&lt;/math&gt;?<br /> &lt;asy&gt;defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> pair B = (0,0);<br /> pair A = 2*dir(60);<br /> pair C = (2,0);<br /> pair D = (4,0);<br /> pair M = midpoint(A--C);<br /> <br /> label(&quot;$A$&quot;,A,NW);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C, SE);label(&quot;$M$&quot;,M,NE);label(&quot;$D$&quot;,D,SE);<br /> <br /> draw(A--B--C--cycle);<br /> draw(C--D--M--cycle);&lt;/asy&gt;&lt;math&gt; \textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}&lt;/math&gt;<br /> == Solution ==<br /> ===Solution 1 (simplest) ===<br /> The area of a triangle can be given by &lt;math&gt;\frac12 ab \sin C&lt;/math&gt;. &lt;math&gt;MC=1&lt;/math&gt; because it is the midpoint of a side, and &lt;math&gt;CD=2&lt;/math&gt; because it is the same length as &lt;math&gt;BC&lt;/math&gt;. Each angle of an equilateral triangle is &lt;math&gt;60^\circ&lt;/math&gt; so &lt;math&gt;\angle MCD = 120^\circ&lt;/math&gt;. The area is &lt;math&gt;\frac12 (1)(2) \sin<br /> 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}&lt;/math&gt;.<br /> Note: Even if you don't know the value of &lt;math&gt;\sin 120^\circ&lt;/math&gt;, you can use the fact that &lt;math&gt;\sin x = \sin 180^\circ - x&lt;/math&gt;, so &lt;math&gt;\sin 120^\circ = \sin 60^\circ&lt;/math&gt;.<br /> You can easily calculate &lt;math&gt;\sin 60^\circ&lt;/math&gt; to be &lt;math&gt;\frac{\sqrt3}{2}&lt;/math&gt; using equilateral triangles.<br /> ===Solution 2===<br /> In order to calculate the area of &lt;math&gt;\triangle CDM&lt;/math&gt;, we can use the formula &lt;math&gt;A=\dfrac{1}{2}bh&lt;/math&gt;, where &lt;math&gt;\overline{CD}&lt;/math&gt; is the base. We already know that &lt;math&gt;\overline{CD}=2&lt;/math&gt;, so the formula now becomes &lt;math&gt;A=h&lt;/math&gt;. We can drop verticals down from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; to points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, respectively. We can see that &lt;math&gt;\triangle AEC \sim \triangle MFC&lt;/math&gt;. Now, we establish the relationship that &lt;math&gt;\dfrac{AE}{MF}=\dfrac{AC}{MC}&lt;/math&gt;. We are given that &lt;math&gt;\overline{AC}=2&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, so &lt;math&gt;\overline{MC}=1&lt;/math&gt;. Because &lt;math&gt;\triangle AEB&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle and the ratio of the sides opposite the angles are &lt;math&gt;1-\sqrt{3}-2&lt;/math&gt; &lt;math&gt;\overline{AE}&lt;/math&gt; is &lt;math&gt;\sqrt{3}&lt;/math&gt;. Plugging those numbers in, we have &lt;math&gt;\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}&lt;/math&gt;. Cross-multiplying, we see that &lt;math&gt;2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}&lt;/math&gt; Since &lt;math&gt;\overline{MF}&lt;/math&gt; is the height &lt;math&gt;\triangle CDM&lt;/math&gt;, the area is &lt;math&gt;\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Draw a line from &lt;math&gt;M&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{BC}&lt;/math&gt;. Call the midpoint of &lt;math&gt;\overline{BC}&lt;/math&gt; &lt;math&gt;P&lt;/math&gt;. This is an equilateral triangle, since the two segments &lt;math&gt;\overline{PC}&lt;/math&gt; and &lt;math&gt;\overline{MC}&lt;/math&gt; are identical, and &lt;math&gt;\angle C&lt;/math&gt; is 60°. Using the Pythagorean Theorem, point &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\dfrac{\sqrt{3}}{2}&lt;/math&gt;. Also, the length of &lt;math&gt;\overline{CD}&lt;/math&gt; is 2, since &lt;math&gt;C&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;. So, our final equation is &lt;math&gt;\dfrac{\sqrt{3}}{2}\times2\over2&lt;/math&gt;, which just leaves us with &lt;math&gt;\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}&lt;/math&gt;.<br /> <br /> ===Solution 4 ===<br /> Drop a vertical down from &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;. WLOG, let us call the point of intersection &lt;math&gt;X&lt;/math&gt; and the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;. We can observe that &lt;math&gt;\triangle AYC&lt;/math&gt; and &lt;math&gt;\triangle MXC&lt;/math&gt; are similar. By the Pythagorean theorem, &lt;math&gt;AY&lt;/math&gt; is &lt;math&gt;\sqrt3&lt;/math&gt;. Since &lt;math&gt;AC:MC=2:1,&lt;/math&gt; we find &lt;math&gt;MX=\frac{\sqrt3}{2}.&lt;/math&gt; Because &lt;math&gt;C&lt;/math&gt; is the midpoint of &lt;math&gt;BD,&lt;/math&gt; and &lt;math&gt;BC=2,&lt;/math&gt; &lt;math&gt;CD=2.&lt;/math&gt; Using the area formula, &lt;math&gt;\frac{CD*MX}{2}=\frac{\sqrt3}{2},&lt;/math&gt; &lt;math&gt;\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}.&lt;/math&gt;<br /> <br /> sdk652<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_19&diff=125361 2006 AMC 10A Problems/Problem 19 2020-06-14T00:19:48Z <p>Binderclips1: </p> <hr /> <div>== Problem ==<br /> How many non-[[similar]] [[triangle]]s have [[angle]]s whose [[degree]] measures are distinct positive integers in [[arithmetic progression]]? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad&lt;/math&gt;<br /> <br /> == Solution ==<br /> The sum of the angles of a triangle is &lt;math&gt;180&lt;/math&gt; degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it &lt;math&gt;\frac{180}{3} = 60&lt;/math&gt; degrees. The minimum possible value for the smallest angle is &lt;math&gt;1&lt;/math&gt; and the highest possible is &lt;math&gt;59&lt;/math&gt; (since the numbers are distinct), so there are &lt;math&gt;59&lt;/math&gt; possibilities &lt;math&gt;\Longrightarrow \mathrm{C}&lt;/math&gt;.<br /> <br /> ==Solution 2(Stars and Bars)==<br /> Let the first angle be &lt;math&gt;x&lt;/math&gt;, and the common difference be &lt;math&gt;d&lt;/math&gt;. The arithmetic progression can now be expressed as &lt;math&gt;x + (x + d) + (x + 2d) = 180&lt;/math&gt;. Simplifiying, &lt;math&gt;x + d = 60&lt;/math&gt;. Now, using stars and bars, we have &lt;math&gt;61_[c1] = 61&lt;/math&gt;. <br /> However, we must subtract the two cases in which either &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;d&lt;/math&gt; equal &lt;math&gt;0&lt;/math&gt;, so we have &lt;math&gt;61 - 2&lt;/math&gt; = C.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_19&diff=125358 2006 AMC 10A Problems/Problem 19 2020-06-13T22:57:40Z <p>Binderclips1: Added solution</p> <hr /> <div>== Problem ==<br /> How many non-[[similar]] [[triangle]]s have [[angle]]s whose [[degree]] measures are distinct positive integers in [[arithmetic progression]]? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad&lt;/math&gt;<br /> <br /> == Solution ==<br /> The sum of the angles of a triangle is &lt;math&gt;180&lt;/math&gt; degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it &lt;math&gt;\frac{180}{3} = 60&lt;/math&gt; degrees. The minimum possible value for the smallest angle is &lt;math&gt;1&lt;/math&gt; and the highest possible is &lt;math&gt;59&lt;/math&gt; (since the numbers are distinct), so there are &lt;math&gt;59&lt;/math&gt; possibilities &lt;math&gt;\Longrightarrow \mathrm{C}&lt;/math&gt;.<br /> <br /> ==Solution 2(Stars and Bars)==<br /> Let the first angle be &lt;math&gt;x&lt;/math&gt;, and the common difference be &lt;math&gt;d&lt;/math&gt;. The arithmetic progression can now be expressed as &lt;math&gt;x + (x + d) + (x + 2d) = 180&lt;/math&gt;. Simplifiying, &lt;math&gt;x + d = 60&lt;/math&gt;. Now, using stars and bars, we have &lt;math&gt;61_(c1) = 61&lt;/math&gt;. <br /> However, we must subtract the two cases in which either &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;d&lt;/math&gt; equal &lt;math&gt;0&lt;/math&gt;, so we have &lt;math&gt;61 - 2&lt;/math&gt; = C.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_17&diff=124801 2009 AMC 10A Problems/Problem 17 2020-06-10T05:04:14Z <p>Binderclips1: </p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=4&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Segment &lt;math&gt;EF&lt;/math&gt; is constructed through &lt;math&gt;B&lt;/math&gt; so that &lt;math&gt;EF&lt;/math&gt; is perpendicular to &lt;math&gt;DB&lt;/math&gt;, and &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt;, respectively. What is &lt;math&gt;EF&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ \frac {125}{12}<br /> \qquad<br /> \mathrm{(D)}\ \frac {103}{9}<br /> \qquad<br /> \mathrm{(E)}\ 12<br /> &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> <br /> The situation is shown in the picture below.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(4,0), C=(4,3), D=(0,3);<br /> pair EF=rotate(90)*(D-B);<br /> pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) );<br /> pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) );<br /> draw(A--B--C--D--cycle);<br /> draw(B--D, dashed);<br /> draw(E--F);<br /> draw(A--E, dashed);<br /> draw(C--F, dashed);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,NE);<br /> label(&quot;$3$&quot;,A--D,W);<br /> label(&quot;$4$&quot;,C--D,N);<br /> &lt;/asy&gt;<br /> <br /> From the [[Pythagorean theorem]] we have &lt;math&gt;BD=5&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;EAB&lt;/math&gt; is similar to &lt;math&gt;DAB&lt;/math&gt;, as they have the same angles. Segment &lt;math&gt;BA&lt;/math&gt; is perpendicular to &lt;math&gt;DA&lt;/math&gt;, meaning that angle &lt;math&gt;DAB&lt;/math&gt; and &lt;math&gt;BAE&lt;/math&gt; are right angles and congruent. Also, angle &lt;math&gt;DBE&lt;/math&gt; is a right angle. Because it is a rectangle, angle &lt;math&gt;BDC&lt;/math&gt; is congruent to &lt;math&gt;DBA&lt;/math&gt; and angle &lt;math&gt;ADC&lt;/math&gt; is also a right angle. By the transitive property:<br /> <br /> &lt;math&gt;mADB + mBDC = mDBA + mABE&lt;/math&gt;<br /> <br /> &lt;math&gt;mBDC = mDBA&lt;/math&gt;<br /> <br /> &lt;math&gt;mADB + mBDC = mBDC + mABE&lt;/math&gt;<br /> <br /> &lt;math&gt;mADB = mABE&lt;/math&gt;<br /> <br /> Next, because every triangle has a degree measure of 180, angle &lt;math&gt;E&lt;/math&gt; and angle &lt;math&gt;DBA&lt;/math&gt; are congruent. <br /> <br /> <br /> Hence &lt;math&gt;BE/AB = DB/AD&lt;/math&gt;, and therefore &lt;math&gt;BE = AB\cdot DB/AD = 20/3&lt;/math&gt;.<br /> <br /> Also triangle &lt;math&gt;CBF&lt;/math&gt; is similar to &lt;math&gt;ABD&lt;/math&gt;. Hence &lt;math&gt;BF/BC = DB/AB&lt;/math&gt;, and therefore &lt;math&gt;BF=BC\cdot DB / AB = 15/4&lt;/math&gt;.<br /> <br /> We then have &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since &lt;math&gt;BD&lt;/math&gt; is the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;, we can use the equation &lt;math&gt;BD^2 = EB\cdot BF&lt;/math&gt;. <br /> <br /> Looking at the angles, we see that triangle &lt;math&gt;BDE&lt;/math&gt; is similar to &lt;math&gt;DCB&lt;/math&gt;. Because of this, &lt;math&gt;\frac{AB}{CB} = \frac{EB}{DB}&lt;/math&gt;. From the given information and the [[Pythagorean theorem]], &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;CB=3&lt;/math&gt;, and &lt;math&gt;DB=5&lt;/math&gt;. Solving gives &lt;math&gt;EB=20/3&lt;/math&gt;. <br /> <br /> We can use the above formula to solve for &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;BD^2 = 20/3\cdot BF&lt;/math&gt;. Solve to obtain &lt;math&gt;BF=15/4&lt;/math&gt;. <br /> <br /> We now know &lt;math&gt;EB&lt;/math&gt; and &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> Solution 3<br /> There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar.<br /> <br /> ==Solution 4(Coordinate Bash)==<br /> To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the &lt;math&gt;x&lt;/math&gt;-axis.It is also worth noting the &lt;math&gt;F&lt;/math&gt; will lie on the &lt;math&gt;x&lt;/math&gt; axis and &lt;math&gt;E&lt;/math&gt; on the &lt;math&gt;y&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the origin, &lt;math&gt;A(3,0)&lt;/math&gt;, &lt;math&gt;C(4,0)&lt;/math&gt;, and &lt;math&gt;B(4,3)&lt;/math&gt;. We can express segment &lt;math&gt;DB&lt;/math&gt; as the line &lt;math&gt;y=\frac{3x}{4}&lt;/math&gt;. <br /> Since &lt;math&gt;EF&lt;/math&gt; is perpendicular to &lt;math&gt;DB&lt;/math&gt;, and we know that &lt;math&gt;(4,3)&lt;/math&gt; lies on it, we can use this information to find that segment &lt;math&gt;EF&lt;/math&gt;<br /> is on the line &lt;math&gt;y=\frac{-4x}{3}+\frac{25}{3}&lt;/math&gt;. Since &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on the &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; axis, respectively, we plug in &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;<br /> and &lt;math&gt;y&lt;/math&gt;, we find that point &lt;math&gt;E&lt;/math&gt; is at &lt;math&gt;(0,\frac{25}{3})&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; is at &lt;math&gt;(\frac{25}{4},0)&lt;/math&gt;. Applying the distance formula,<br /> we obtain that &lt;math&gt;EF&lt;/math&gt;= &lt;math&gt;\boxed{\frac{125}{12}}&lt;/math&gt;.<br /> Binderclips1 01:04, 10 June 2020 (EDT)<br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_17&diff=124800 2009 AMC 10A Problems/Problem 17 2020-06-10T05:02:45Z <p>Binderclips1: Added solution</p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=4&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Segment &lt;math&gt;EF&lt;/math&gt; is constructed through &lt;math&gt;B&lt;/math&gt; so that &lt;math&gt;EF&lt;/math&gt; is perpendicular to &lt;math&gt;DB&lt;/math&gt;, and &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt;, respectively. What is &lt;math&gt;EF&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ \frac {125}{12}<br /> \qquad<br /> \mathrm{(D)}\ \frac {103}{9}<br /> \qquad<br /> \mathrm{(E)}\ 12<br /> &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> <br /> The situation is shown in the picture below.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(4,0), C=(4,3), D=(0,3);<br /> pair EF=rotate(90)*(D-B);<br /> pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) );<br /> pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) );<br /> draw(A--B--C--D--cycle);<br /> draw(B--D, dashed);<br /> draw(E--F);<br /> draw(A--E, dashed);<br /> draw(C--F, dashed);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,NE);<br /> label(&quot;$3$&quot;,A--D,W);<br /> label(&quot;$4$&quot;,C--D,N);<br /> &lt;/asy&gt;<br /> <br /> From the [[Pythagorean theorem]] we have &lt;math&gt;BD=5&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;EAB&lt;/math&gt; is similar to &lt;math&gt;DAB&lt;/math&gt;, as they have the same angles. Segment &lt;math&gt;BA&lt;/math&gt; is perpendicular to &lt;math&gt;DA&lt;/math&gt;, meaning that angle &lt;math&gt;DAB&lt;/math&gt; and &lt;math&gt;BAE&lt;/math&gt; are right angles and congruent. Also, angle &lt;math&gt;DBE&lt;/math&gt; is a right angle. Because it is a rectangle, angle &lt;math&gt;BDC&lt;/math&gt; is congruent to &lt;math&gt;DBA&lt;/math&gt; and angle &lt;math&gt;ADC&lt;/math&gt; is also a right angle. By the transitive property:<br /> <br /> &lt;math&gt;mADB + mBDC = mDBA + mABE&lt;/math&gt;<br /> <br /> &lt;math&gt;mBDC = mDBA&lt;/math&gt;<br /> <br /> &lt;math&gt;mADB + mBDC = mBDC + mABE&lt;/math&gt;<br /> <br /> &lt;math&gt;mADB = mABE&lt;/math&gt;<br /> <br /> Next, because every triangle has a degree measure of 180, angle &lt;math&gt;E&lt;/math&gt; and angle &lt;math&gt;DBA&lt;/math&gt; are congruent. <br /> <br /> <br /> Hence &lt;math&gt;BE/AB = DB/AD&lt;/math&gt;, and therefore &lt;math&gt;BE = AB\cdot DB/AD = 20/3&lt;/math&gt;.<br /> <br /> Also triangle &lt;math&gt;CBF&lt;/math&gt; is similar to &lt;math&gt;ABD&lt;/math&gt;. Hence &lt;math&gt;BF/BC = DB/AB&lt;/math&gt;, and therefore &lt;math&gt;BF=BC\cdot DB / AB = 15/4&lt;/math&gt;.<br /> <br /> We then have &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since &lt;math&gt;BD&lt;/math&gt; is the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;, we can use the equation &lt;math&gt;BD^2 = EB\cdot BF&lt;/math&gt;. <br /> <br /> Looking at the angles, we see that triangle &lt;math&gt;BDE&lt;/math&gt; is similar to &lt;math&gt;DCB&lt;/math&gt;. Because of this, &lt;math&gt;\frac{AB}{CB} = \frac{EB}{DB}&lt;/math&gt;. From the given information and the [[Pythagorean theorem]], &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;CB=3&lt;/math&gt;, and &lt;math&gt;DB=5&lt;/math&gt;. Solving gives &lt;math&gt;EB=20/3&lt;/math&gt;. <br /> <br /> We can use the above formula to solve for &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;BD^2 = 20/3\cdot BF&lt;/math&gt;. Solve to obtain &lt;math&gt;BF=15/4&lt;/math&gt;. <br /> <br /> We now know &lt;math&gt;EB&lt;/math&gt; and &lt;math&gt;BF&lt;/math&gt;. &lt;math&gt;EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}&lt;/math&gt;.<br /> <br /> Solution 3<br /> There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar.<br /> <br /> ==Solution 4(Coordinate Bash)==<br /> To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the &lt;math&gt;x&lt;/math&gt;-axis.It is also worth noting the &lt;math&gt;F&lt;/math&gt; will lie on the &lt;math&gt;x&lt;/math&gt; axis and &lt;math&gt;E&lt;/math&gt; on the &lt;math&gt;y&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the origin, &lt;math&gt;A(3,0)&lt;/math&gt;, &lt;math&gt;C(4,0)&lt;/math&gt;, and &lt;math&gt;B(4,3)&lt;/math&gt;. We can express segment &lt;math&gt;DB&lt;/math&gt; as the line &lt;math&gt;y=\frac{3x}{4}&lt;/math&gt;. <br /> Since &lt;math&gt;EF&lt;/math&gt; is perpendicular to &lt;math&gt;DB&lt;/math&gt;, and we know that &lt;math&gt;(4,3)&lt;/math&gt; lies on it, we can use this information to find that segment &lt;math&gt;EF&lt;/math&gt;<br /> is on the line &lt;math&gt;y=\frac{-4x}{3}+\frac{25}{3}&lt;/math&gt;. Since &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on the &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; axis, respectively, we plug in &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;<br /> and &lt;math&gt;y&lt;/math&gt;, we find that point &lt;math&gt;E&lt;/math&gt; is at &lt;math&gt;(0,\frac{25}{3})&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; is at &lt;math&gt;(\frac{25}{4},0)&lt;/math&gt;. Applying the distance formula,<br /> we obtain that &lt;math&gt;EF&lt;/math&gt;= &lt;math&gt;\boxed{\frac{125}{12}}&lt;/math&gt;.<br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=123307 2019 AMC 10B Problems/Problem 11 2020-05-30T04:56:51Z <p>Binderclips1: Added solution</p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar &lt;math&gt;1&lt;/math&gt; the ratio of blue to green marbles is &lt;math&gt;9:1&lt;/math&gt;, and the ratio of blue to green marbles in Jar &lt;math&gt;2&lt;/math&gt; is &lt;math&gt;8:1&lt;/math&gt;. There are &lt;math&gt;95&lt;/math&gt; green marbles in all. How many more blue marbles are in Jar &lt;math&gt;1&lt;/math&gt; than in Jar &lt;math&gt;2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the number of marbles in each jar &lt;math&gt;x&lt;/math&gt; (because the problem specifies that they each contain the same number). Thus, &lt;math&gt;\frac{x}{10}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{x}{9}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;2&lt;/math&gt;. Since &lt;math&gt;\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}&lt;/math&gt;, we have &lt;math&gt;\frac{19x}{90}=95&lt;/math&gt;, so there are &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. <br /> <br /> Because &lt;math&gt;\frac{9x}{10}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{8x}{9}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5&lt;/math&gt; more marbles in Jar &lt;math&gt;1&lt;/math&gt; than Jar &lt;math&gt;2&lt;/math&gt;. This means the answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> <br /> ==Solution 2(Completely Solve)==<br /> Let &lt;math&gt;b_1&lt;/math&gt;, &lt;math&gt;g_1&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;g_2&lt;/math&gt;, represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the <br /> the amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations, <br /> &lt;math&gt;\frac{&lt;/math&gt;b_1&lt;math&gt;}{&lt;/math&gt;g_1&lt;math&gt;} = \frac{9}{1}&lt;/math&gt;, &lt;math&gt;\frac{&lt;/math&gt;b_2&lt;math&gt;}{&lt;/math&gt;g_2&lt;math&gt;} = \frac{8}{1}&lt;/math&gt;, &lt;math&gt;g_1 + g_2 =95&lt;/math&gt;, and &lt;math&gt;b_1 + g_1 = b_2 + g_2&lt;/math&gt;.<br /> Since &lt;math&gt;b_1 = 9g_1&lt;/math&gt; and &lt;math&gt;b_2 = 8g_2&lt;/math&gt;, we substitue that in to obtain &lt;math&gt;10g_1 = 9g_2&lt;/math&gt;. <br /> Coupled with our third equation, we find that &lt;math&gt;g_1 = 45&lt;/math&gt;, and that &lt;math&gt;g_2 = 50&lt;/math&gt;. We now use this information to find &lt;math&gt;b_1 = 405&lt;/math&gt;<br /> and &lt;math&gt;b_2 = 400&lt;/math&gt;. Therefore, &lt;math&gt;b_1 - b_2 = 5&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> ~Binderclips1<br /> <br /> ==Video Solution==<br /> https://youtu.be/mXvetCMMzpU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_3&diff=123224 2020 AMC 10B Problems/Problem 3 2020-05-28T23:54:13Z <p>Binderclips1: fixed typo</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}}<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> WLOG, let &lt;math&gt;w=4&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;. <br /> <br /> Since the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;, we can substitute in the value of &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, so &lt;math&gt;\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y&lt;/math&gt; is then &lt;math&gt;\frac{4}{\frac{3}{4}}=\frac{16}{3}&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. <br /> <br /> &lt;math&gt;z:x=1:6=2:12&lt;/math&gt;, and since &lt;math&gt;y:z=3:2&lt;/math&gt;, we can link them together to get &lt;math&gt;y:z:x=3:2:12&lt;/math&gt;.<br /> <br /> Finally, since &lt;math&gt;x:w=3:4=12:16&lt;/math&gt;, we can link this again to get: &lt;math&gt;y:z:x:w=3:2:12:16&lt;/math&gt;, so &lt;math&gt;w:y = \boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 3==<br /> <br /> We have the equations &lt;math&gt;\frac{w}{x}=\frac{4}{3}&lt;/math&gt;, &lt;math&gt;\frac{y}{z}=\frac{3}{2}&lt;/math&gt;, and &lt;math&gt;\frac{z}{x}=\frac{1}{6}&lt;/math&gt;.<br /> Clearing denominators, we have &lt;math&gt;3w = 4x&lt;/math&gt;, &lt;math&gt;2y = 3z&lt;/math&gt;, and &lt;math&gt;6z = x&lt;/math&gt;. <br /> Since we want &lt;math&gt;\frac{w}{y}&lt;/math&gt;, we look to find &lt;math&gt;y&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt; since we know the relationship between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> We begin by multiplying both sides of &lt;math&gt;2y = 3z&lt;/math&gt; by two, obtaining &lt;math&gt;4y = 6z&lt;/math&gt;. We then substitute that into &lt;math&gt;6z = x&lt;/math&gt; <br /> to get &lt;math&gt;4y = x&lt;/math&gt; . Now, to be able to substitute this into out first equation, we need to have &lt;math&gt;4x&lt;/math&gt; on the RHS. <br /> Multiplying both sides by &lt;math&gt;4&lt;/math&gt;, we have &lt;math&gt;16y = 4x&lt;/math&gt;. <br /> Substituting this into our first equation, we have &lt;math&gt;3w = 16y&lt;/math&gt;, or &lt;math&gt;\frac{w}{y}=\frac{16}{3}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU (for AMC 10)<br /> https://youtu.be/WfTty8Fe5Fo (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{AMC12 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_3&diff=123223 2020 AMC 10B Problems/Problem 3 2020-05-28T23:52:59Z <p>Binderclips1: fixed typo</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}}<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> WLOG, let &lt;math&gt;w=4&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;. <br /> <br /> Since the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;, we can substitute in the value of &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, so &lt;math&gt;\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y&lt;/math&gt; is then &lt;math&gt;\frac{4}{\frac{3}{4}}=\frac{16}{3}&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. <br /> <br /> &lt;math&gt;z:x=1:6=2:12&lt;/math&gt;, and since &lt;math&gt;y:z=3:2&lt;/math&gt;, we can link them together to get &lt;math&gt;y:z:x=3:2:12&lt;/math&gt;.<br /> <br /> Finally, since &lt;math&gt;x:w=3:4=12:16&lt;/math&gt;, we can link this again to get: &lt;math&gt;y:z:x:w=3:2:12:16&lt;/math&gt;, so &lt;math&gt;w:y = \boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 3==<br /> <br /> We have the equations &lt;math&gt;\frac{w}{x}=\frac{4}{3}&lt;/math&gt;, &lt;math&gt;\frac{y}{z}=\frac{3}{2}&lt;/math&gt;, and &lt;math&gt;\frac{z}{x}=\frac{1}{6}&lt;/math&gt;.<br /> Clearing denominators, we have &lt;math&gt;3w = 4x&lt;/math&gt;, &lt;math&gt;2y = 3z&lt;/math&gt;, and &lt;math&gt;6z = x&lt;/math&gt;. <br /> Since we want &lt;math&gt;\frac{w}{y}&lt;/math&gt;, we look to find &lt;math&gt;y&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt; since we know the relationship between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> We begin by multiplying both sides of &lt;math&gt;2y = 3z&lt;/math&gt; by two, obtaining &lt;math&gt;4y = 6z&lt;/math&gt;. We then substitute that into &lt;math&gt;6z = x&lt;/math&gt; <br /> to get &lt;math&gt;4y = x&lt;/math&gt; . Now, to be able to substitute this into out first equation, we need to have &lt;math&gt;4x&lt;/math&gt; on the RHS. <br /> Multiplying both sides by &lt;math&gt;4&lt;/math&gt;, we have &lt;math&gt;16y = 4x&lt;/math&gt;. <br /> Substituting this into our first equation, we have &lt;math&gt;3w = 16y&lt;/math&gt;, or &lt;math&gt;\frac{w}{y}=\frac{16}{3}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU (for AMC 10)<br /> https://youtu.be/WfTty8Fe5Fo (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{AMC12 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_3&diff=123222 2020 AMC 10B Problems/Problem 3 2020-05-28T23:52:21Z <p>Binderclips1: fixed typo</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}}<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> WLOG, let &lt;math&gt;w=4&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;. <br /> <br /> Since the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;, we can substitute in the value of &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, so &lt;math&gt;\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y&lt;/math&gt; is then &lt;math&gt;\frac{4}{\frac{3}{4}}=\frac{16}{3}&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. <br /> <br /> &lt;math&gt;z:x=1:6=2:12&lt;/math&gt;, and since &lt;math&gt;y:z=3:2&lt;/math&gt;, we can link them together to get &lt;math&gt;y:z:x=3:2:12&lt;/math&gt;.<br /> <br /> Finally, since &lt;math&gt;x:w=3:4=12:16&lt;/math&gt;, we can link this again to get: &lt;math&gt;y:z:x:w=3:2:12:16&lt;/math&gt;, so &lt;math&gt;w:y = \boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 3==<br /> <br /> We have the equations &lt;math&gt;\frac{w}{x}=\frac{4}{3}&lt;/math&gt;, &lt;math&gt;\frac{y}{z}=\frac{3}{2}&lt;/math&gt;, and &lt;math&gt;\frac{z}{x}=\frac{1}{6}&lt;/math&gt;.<br /> Clearing denominators, we have &lt;math&gt;3w = 4x&lt;/math&gt;, &lt;math&gt;2y = 3z&lt;/math&gt;, and &lt;math&gt;6z = x&lt;/math&gt;. <br /> Since we want &lt;math&gt;\frac{w}{y}&lt;/math&gt;, we look to find &lt;math&gt;y&lt;/math&gt; in terms of x since we know the relationship between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> We begin by multiplying both sides of &lt;math&gt;2y = 3z&lt;/math&gt; by two, obtaining &lt;math&gt;4y = 6z&lt;/math&gt;. We then substitute that into &lt;math&gt;6z = x&lt;/math&gt; <br /> to get &lt;math&gt;4y = x&lt;/math&gt; . Now, to be able to substitute this into out first equation, we need to have &lt;math&gt;4x&lt;/math&gt; on the RHS. <br /> Multiplying both sides by &lt;math&gt;4&lt;/math&gt;, we have &lt;math&gt;16y = 4x&lt;/math&gt;. <br /> Substituting this into our first equation, we have &lt;math&gt;3w = 16y&lt;/math&gt;, or &lt;math&gt;\frac{w}{y}=\frac{16}{3}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU (for AMC 10)<br /> https://youtu.be/WfTty8Fe5Fo (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{AMC12 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_3&diff=123215 2020 AMC 10B Problems/Problem 3 2020-05-28T23:35:17Z <p>Binderclips1: Added solution</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #3]] and [[2020 AMC 12B Problems|2020 AMC 12B #3]]}}<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> WLOG, let &lt;math&gt;w=4&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;. <br /> <br /> Since the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;, we can substitute in the value of &lt;math&gt;x&lt;/math&gt; to get &lt;math&gt;\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, so &lt;math&gt;\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}&lt;/math&gt;. <br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y&lt;/math&gt; is then &lt;math&gt;\frac{4}{\frac{3}{4}}=\frac{16}{3}&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. <br /> <br /> &lt;math&gt;z:x=1:6=2:12&lt;/math&gt;, and since &lt;math&gt;y:z=3:2&lt;/math&gt;, we can link them together to get &lt;math&gt;y:z:x=3:2:12&lt;/math&gt;.<br /> <br /> Finally, since &lt;math&gt;x:w=3:4=12:16&lt;/math&gt;, we can link this again to get: &lt;math&gt;y:z:x:w=3:2:12:16&lt;/math&gt;, so &lt;math&gt;w:y = \boxed{\textbf{(E)}\ 16:3}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 3==<br /> <br /> We have the equations &lt;math&gt;\frac{w}{x}=\frac{4}{3}&lt;/math&gt;, &lt;math&gt;\frac{y}{z}=\frac{3}{2}&lt;/math&gt;, and &lt;math&gt;\frac{z}{x}=\frac{1}{6}&lt;/math&gt;.<br /> Clearing denominators, we have &lt;math&gt;3w = 4x&lt;/math&gt;, &lt;math&gt;2y = 3z&lt;/math&gt;, and &lt;math&gt;6z = x&lt;/math&gt;. <br /> Since we want &lt;math&gt;\frac{w}{y}&lt;/math&gt;, we look to find &lt;math&gt;y&lt;/math&gt; in terms of x since we know the relationship between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> We begin by multiplying both sides of &lt;math&gt;2y = 3z&lt;/math&gt; by two, obtaining &lt;math&gt;4y = 6z&lt;/math&gt;. We then substitute that into &lt;math&gt;6z = x&lt;/math&gt; <br /> to get &lt;math&gt;4y = x&lt;/math&gt; . Now, to be able to substitute this into out first equation, we need to have &lt;math&gt;4x&lt;/math&gt; on the RHS. <br /> Multiplying both sides by &lt;math&gt;4&lt;/math&gt;, we have &lt;math&gt;16y = 4x&lt;/math&gt;. <br /> Substituting this into our first equation, we have &lt;math&gt;3w = 16y&lt;/math&gt;, or &lt;math&gt;\frac{w}{y}=\frac{16}{3}&lt;/math&gt;, so our answer is \boxed{\textbf{(E)}\ 16:3}$<br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU (for AMC 10)<br /> https://youtu.be/WfTty8Fe5Fo (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{AMC12 box|year=2020|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123186 2020 AMC 10A Problems/Problem 14 2020-05-28T18:38:56Z <p>Binderclips1: fixed typo</p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Solution 7==<br /> <br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x + y&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744&lt;/math&gt;, so we have &lt;cmath&gt; 4 + \frac{1744}{4}.&lt;/cmath&gt;. This equals 440, so our answer is &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123148 2020 AMC 10A Problems/Problem 14 2020-05-28T05:27:02Z <p>Binderclips1: </p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Solution 7==<br /> <br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x^2 + y^2&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744&lt;/math&gt;, so we have &lt;cmath&gt; 4 + \frac{1744}{4}.&lt;/cmath&gt;. This equals 440, so our answer is &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123147 2020 AMC 10A Problems/Problem 14 2020-05-28T05:21:58Z <p>Binderclips1: </p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Solution 7==<br /> <br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x^2 + y^2&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744. Putting this value into our original equation, we have, &lt;cmath&gt; 4 + \frac{1744}{4}.&lt;/cmath&gt; This value sums to 440, so our answer is &lt;/math&gt; \boxed{\textbf{(D)}\ 440}$ <br /> <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123146 2020 AMC 10A Problems/Problem 14 2020-05-28T05:15:26Z <p>Binderclips1: </p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Solution 7==<br /> <br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x^2 + y^2&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744. Plugging this back into our original equation, we find &lt;cmath&gt; 4 + \frac{1744}{4} = 4 + 436&lt;/cmath&gt;, which equals &lt;/math&gt;\boxed{\textbf{(D)} 440}$. <br /> <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123145 2020 AMC 10A Problems/Problem 14 2020-05-28T05:12:34Z <p>Binderclips1: </p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Solution 7==<br /> <br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x^2 + y^2&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744. Plugging ''this '' back into our original equation, we find &lt;cmath&gt; 4 + \frac{1744}{4} = 4 + 436&lt;/cmath&gt;, which equals &lt;/math&gt;\boxed{\textbf{(D)} 440}$. <br /> <br /> ~Binderclips1<br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=123144 2020 AMC 10A Problems/Problem 14 2020-05-28T05:10:01Z <p>Binderclips1: </p> <hr /> <div>==Problem==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> == Solution 2 ==<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> <br /> ==Solution 4 (Bashing)==<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> <br /> ~Baolan<br /> <br /> <br /> ==Solution 5 (Bashing Part 2)==<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> ==Solution 6 (Complete Binomial Theorem)==<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> ~ fidgetboss_4000<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> ==Solution 7==<br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x^2 + y^2&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744. Plugging ''this '' back into our original equation, we find &lt;cmath&gt; 4 + \frac{1744}{4} = 4 + 436&lt;/cmath&gt;, which equals &lt;/math&gt;\boxed{\textbf{(D)} 440}\$. <br /> ~binderclips1<br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=123113 2019 AMC 8 Problems/Problem 18 2020-05-28T00:31:21Z <p>Binderclips1: </p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;<br /> <br /> - ViratKohli2018 (VK18)<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; evens.<br /> <br /> Totally, we have &lt;math&gt;6*6=36&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; dies.<br /> <br /> Therefore the answer is &lt;math&gt;\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;. - juliankuang<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~heeeeeeeeheeeee<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of rolling the desired number is 1/6, and the probability of rolling a second odd number is constant, being 2/3. Since there are four cases that involve odd numbers, the probability of getting an even sum with two odd numbers is (4)(1/9)=4/9. Using this same logic, realize that the probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~binderclips1 (feel free to edit as I am new to editing on this site)<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=123112 2019 AMC 8 Problems/Problem 18 2020-05-28T00:30:54Z <p>Binderclips1: </p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;~binderclips1 (feel free to edit as I am new to editing on this site)<br /> <br /> - ViratKohli2018 (VK18)<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; evens.<br /> <br /> Totally, we have &lt;math&gt;6*6=36&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; dies.<br /> <br /> Therefore the answer is &lt;math&gt;\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;. - juliankuang<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~heeeeeeeeheeeee<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of rolling the desired number is 1/6, and the probability of rolling a second odd number is constant, being 2/3. Since there are four cases that involve odd numbers, the probability of getting an even sum with two odd numbers is (4)(1/9)=4/9. Using this same logic, realize that the probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~binderclips1 (feel free to edit as I am new to editing on this site)<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=123111 2019 AMC 8 Problems/Problem 18 2020-05-28T00:30:21Z <p>Binderclips1: </p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;~binderclips1 (feel free to edit as I am new to editing on this site)<br /> <br /> - ViratKohli2018 (VK18)<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; evens.<br /> <br /> Totally, we have &lt;math&gt;6*6=36&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; dies.<br /> <br /> Therefore the answer is &lt;math&gt;\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;. - juliankuang<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~heeeeeeeeheeeee<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of rolling the desired number is 1/6, and the probability of rolling a second odd number is constant, being 2/3. Since there are four cases that involve odd numbers, the probability of getting an even sum with two odd numbers is (4)(1/9)=4/9. Using this same logic, realize that the probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Binderclips1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=123110 2019 AMC 8 Problems/Problem 18 2020-05-28T00:29:04Z <p>Binderclips1: </p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;<br /> <br /> - ViratKohli2018 (VK18)<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; evens.<br /> <br /> Totally, we have &lt;math&gt;6*6=36&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; dies.<br /> <br /> Therefore the answer is &lt;math&gt;\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;. - juliankuang<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~heeeeeeeeheeeee<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of rolling the desired number is 1/6, and the probability of rolling a second odd number is constant, being 2/3. Since there are four cases that involve odd numbers, the probability of getting an even sum with two odd numbers is (4)(1/9)=4/9. Using this same logic, realize that the probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Binderclips1