https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Edonahey5&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T08:45:49ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_23&diff=1646242017 AMC 12B Problems/Problem 232021-11-05T03:25:36Z<p>Edonahey5: /* Solution 2 */ Fixed Parenthesis</p>
<hr />
<div>==Problem==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<br />
<math>\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8</math><br />
<br />
==Solution==<br />
First, we can define <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>, which contains points <math>A</math>, <math>B</math>, and <math>C</math>. Now we find that lines <math>AB</math>, <math>AC</math>, and <math>BC</math> are defined by the equations <math>y = 5x - 6</math>, <math>y= 6x-8</math>, and <math>y=7x-12</math> respectively. Since we want to find the <math>x</math>-coordinates of the intersections of these lines and <math>f(x)</math>, we set each of them to <math>f(x)</math>, and synthetically divide by the solutions we already know exist (eg. if we were looking at line <math>AB</math>, we would synthetically divide by the solutions <math>x=2</math> and <math>x=3</math>, because we already know <math>AB</math> intersects the graph at <math>A</math> and <math>B</math>, which have <math>x</math>-coordinates of <math>2</math> and <math>3</math>). After completing this process on all three lines, we get that the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> are <math>\frac{4a-1}{a}</math>, <math>\frac{3a-1}{a}</math>, and <math>\frac{2a-1}{a}</math> respectively. Adding these together, we get <math>\frac{9a-3}{a} = 24</math> which gives us <math>a = -\frac{1}{5}</math>. Substituting this back into the original equation, we get <math>f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2</math>, and <math>f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}</math><br />
<br />
Solution by vedadehhc<br />
<br />
==Solution 2==<br />
<math>\boxed{\textbf{No need to find the equations for the lines, really.}}</math> First of all, <math>f(x) = a(x-2)(x-3)(x-4) +x^2</math>. Let's say the line <math>AB</math> is <math>y=bx+c</math>, and <math>x_1</math> is the <math>x</math> coordinate of the third intersection, then <math>2</math>, <math>3</math>, and <math>x_1</math> are the three roots of <math>f(x) - bx-c</math>. The values of <math>b</math> and <math>c</math> have no effect on the sum of the 3 roots, because the coefficient of the <math>x^2</math> term is always <math>-9a+1</math>. So we have <br />
<cmath> \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3</cmath><br />
Adding all three equations up, we get <br />
<cmath> 3\left(\frac{9a-1}{a}\right) = 18 + x_1 + x_2 + x_3 = 18 + 24</cmath><br />
Solving this equation, we get <math>a = -\frac{1}{5}</math>. We finish as Solution 1 does.<br />
<math>\boxed{\textbf{(D)}\frac{24}{5}}</math>.<br />
<br />
- Mathdummy<br />
<br />
Cleaned up by SSding<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_22&diff=1056702003 AMC 10B Problems/Problem 222019-05-04T21:50:42Z<p>Edonahey5: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A clock chimes once at <math>30</math> minutes past each hour and chimes on the hour according to the hour. For example, at <math>1 \text{PM}</math> there is one chime and at noon and midnight there are twelve chimes. Starting at <math>11:15 \text{AM}</math> on <math>\text{February 26, 2003},</math> on what date will the <math>2003^{\text{rd}}</math> chime occur?<br />
<br />
<math>\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}</math><br />
<br />
==Solution==<br />
First, find how many chimes will have already happened before midnight (the beginning of the day) of <math>\text{February 27, 2003}.</math> <math>13</math> half-hours have passed, and the number of chimes according to the hour is <math>1+2+3+\cdots+12.</math> The total number of chimes is <math>13+78=91</math><br />
<br />
Every day, there will be <math>24</math> half-hours and <math>2(1+2+3+\cdots+12)</math> chimes according to the arrow, resulting in <math>24+156=180</math> total chimes.<br />
<br />
On <math>\text{February 26},</math> the number of chimes that still need to occur is <math>2003-91=1912.</math> <math>1912 \div 180=10 \text{R}112.</math> Rounding up, it is <math>11</math> days past <math>\text{February 26},</math> which is <math>\boxed{\textbf{(B) \ } \text{March 9}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_33&diff=942611953 AHSME Problems/Problem 332018-05-01T00:06:13Z<p>Edonahey5: Created page with "The perimeter of an isosceles right triangle is <math>2p</math>. Its area is: <math>\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\s..."</p>
<hr />
<div>The perimeter of an isosceles right triangle is <math>2p</math>. Its area is:<br />
<br />
<math>\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\\ \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2</math><br />
<br />
Given leg length <math>x</math>, we can write the perimeter of this triangle to be <math>2x+x\sqrt{2}=2p</math>. Thus, <math>x(2+\sqrt{2})=2p</math>. Divide to get <math>x=\frac{2p}{2+\sqrt{2}}</math>. Multiply by the conjugate and simplify to get <math>x=p(2-\sqrt{2})</math>. Square and divide by two to get the area of the triangle, or <math>\frac{p^2*(2-\sqrt{2}^2)}{2}</math>, or <math>p^2(3-2\sqrt{2})</math>. <math>\boxed{C}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_27&diff=942601953 AHSME Problems/Problem 272018-04-30T23:47:00Z<p>Edonahey5: Created page with "The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely..."</p>
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<div>The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely. The sum of the areas of the circles is:<br />
<br />
<math>\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}</math><br />
<br />
Note the areas of these triangles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>. <math>\boxed{D}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_20&diff=942591953 AHSME Problems/Problem 202018-04-30T23:11:58Z<p>Edonahey5: Created page with "==Problem 20== If <math>y=x+\frac{1}{x}</math>, then <math>x^4+x^3-4x^2+x+1=0</math> becomes: <math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\\ \tex..."</p>
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<div>==Problem 20==<br />
<br />
If <math>y=x+\frac{1}{x}</math>, then <math>x^4+x^3-4x^2+x+1=0</math> becomes: <br />
<br />
<math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad<br />
\textbf{(B)}\ x^2(y^2+y-3)=0\\ <br />
\textbf{(C)}\ x^2(y^2+y-4)=0 \qquad<br />
\textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these} </math><br />
<br />
We multiply each of the answers to get: <math>x^2(y^2)+x^2(y)+nx^2</math>, where <math>n</math> is either <math>-2,-3,-4,</math> or <math>-6</math>. Looking at the first term, we have to square <math>y</math>, or <math>x+\frac{1}{x}</math>, doing so, we get the equation <math>x^2+\frac{1}{x^2}-2</math>. Plugging that into <math>x^2</math>, we get <math>x^4+2x^2+1</math>. Multiplying <math>y</math> by <math>x^2</math>, we get the expression <math>x^3+x</math>. Adding these two equations together, we get <math>x^4+x^3+2x^2+x+1+nx=0</math>. To get the term <math>-4x^2</math>, which was in the original equation, <math>n</math> must be <math>-6</math>, giving an answer of <math>\boxed{D}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_25&diff=942581953 AHSME Problems/Problem 252018-04-30T22:58:14Z<p>Edonahey5: Created page with "In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is: <math>\textbf{(A)}\ 1 \qquad \tex..."</p>
<hr />
<div>In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{2}{\sqrt{5}}</math><br />
<br />
Given first term <math>a</math> and common ratio <math>r</math>, we have <math>a=a*r+a*r^2</math>, and. We divide by <math>a</math> in the first equation to get <math>1=r+r^2</math>. Rewriting, we have <math>r^2+r-1=0</math>. We use the quadratic formula to get <math>r = \frac{-1+-\sqrt{1^2-4(1)(-1)}}{2(1)}</math>. Because the terms all have to be positive, we must add the discriminant, getting an answer of <math>\frac{\sqrt{5}-1}{2}</math> <math>\boxed{C}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_24&diff=942571953 AHSME Problems/Problem 242018-04-30T22:49:33Z<p>Edonahey5: Created page with "If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if: <math>\textbf{(A)}</math> <math>b+c=10</mat..."</p>
<hr />
<div>If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if:<br />
<br />
<math>\textbf{(A)}</math> <math>b+c=10</math><br />
<math>\textbf{(B)}</math> <math>b=c</math><br />
<math>\textbf{(C)}</math> <math>a+b=10</math><br />
<math>\textbf {(D)}</math> <math>a=b</math><br />
<math>\textbf{(E)}</math> <math>a+b+c=10</math><br />
<br />
<br />
Multiply out the LHS to get <math>100a^2+10ac+10ab+bc=100a(a+1)+bc</math>. Subtract <math>bc</math> and factor to get <math>10a(10a+b+c)=10a(10a+10)</math>. Divide both sides by <math>10a</math> and then subtract <math>10a</math> to get <math>b+c=10</math>, giving an answer of <math>\boxed{A}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_23&diff=942561953 AHSME Problems/Problem 232018-04-30T22:41:44Z<p>Edonahey5: </p>
<hr />
<div>The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:<br />
<br />
<math>\textbf{(A)}</math> an extraneous root between <math>-5</math> and <math>-1</math><br />
<br />
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math> <br />
<br />
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math><br />
<br />
<math>\textbf{(D)}</math> two true roots<br />
<br />
<math>\textbf{(E)}</math> two extraneous roots<br />
<br />
We multiply both sides by <math>\sqrt{x+10}</math> to get the equation <math>x+4=5\sqrt{x+10}</math>. We square both sides to get <math>x^2+8x+16=25x+250</math>, or <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. We plug these values in to find that <math>-9</math> is an extraneous root and that <math>26</math> is a true root, giving an answer of <math>\boxed{B}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_23&diff=942551953 AHSME Problems/Problem 232018-04-30T22:41:13Z<p>Edonahey5: </p>
<hr />
<div>The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:<br />
<br />
<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math><br />
<br />
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math> <br />
<br />
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math><br />
<br />
<math>\textbf{(D)}</math> two true roots<br />
<br />
<math>\textbf{(E)}</math> two extraneous roots<br />
<br />
We multiply both sides by <math>\sqrt{x+10}</math> to get the equation <math>x+4=5\sqrt{x+10}</math>. We square both sides to get <math>x^2+8x+16=25x+250</math>, or <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. We plug these values in to find that <math>-9</math> is an extraneous root and that <math>26</math> is a true root, giving an answer of <math>\boxed{B}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_23&diff=942541953 AHSME Problems/Problem 232018-04-30T22:40:41Z<p>Edonahey5: </p>
<hr />
<div>The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:<br />
<br />
<math>\textbf{A}</math> an extraneous root between <math>-5</math> and <math>-1</math> <math>//</math><br />
<math>\textbf{(B)}</math> an extraneous root between <math>-10</math> and <math>-6</math> //<br />
<math>\textbf{(C)}</math> a true root between <math>20</math> and <math>25</math> //<br />
<math>\textbf{(D)}</math> two true roots //<br />
<math>\textbf{(E)}</math> two extraneous roots //<br />
<br />
We multiply both sides by <math>\sqrt{x+10}</math> to get the equation <math>x+4=5\sqrt{x+10}</math>. We square both sides to get <math>x^2+8x+16=25x+250</math>, or <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. We plug these values in to find that <math>-9</math> is an extraneous root and that <math>26</math> is a true root, giving an answer of <math>\boxed{B}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_23&diff=942531953 AHSME Problems/Problem 232018-04-30T22:35:32Z<p>Edonahey5: Created page with "The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has: https://latex.artofproblemsolving.com/4/1/8/4187aeb2c71a4fedaeec246af6e6cc3cc2222ac6.png We mult..."</p>
<hr />
<div>The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:<br />
https://latex.artofproblemsolving.com/4/1/8/4187aeb2c71a4fedaeec246af6e6cc3cc2222ac6.png<br />
<br />
We multiply both sides by <math>\sqrt{x+10}</math> to get the equation <math>x+4=5\sqrt{x+10}</math>. We square both sides to get <math>x^2+8x+16=25x+250</math>, or <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. We plug these values in to find that <math>-9</math> is an extraneous root and that <math>26</math> is a true root, giving an answer of <math>\boxed{B}</math>.</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=Asymptote_(Vector_Graphics_Language)&diff=85894Asymptote (Vector Graphics Language)2017-05-29T21:31:03Z<p>Edonahey5: /* See also */</p>
<hr />
<div>{{Asymptote}}<br />
<br />
'''Asymptote''' is a powerful vector graphics language designed for creating mathematical diagrams and figures. It can output images in either eps or pdf format, and is compatible with the standard mathematics typesetting language, [[LaTeX]]. It is also a complete programming language, and has cleaner syntax than its predecessor, [http://netlib.bell-labs.com/who/hobby/MetaPost.html MetaPost], which was a language used only for two-dimensional graphics.<br />
<br />
Here is an example of an image that can be produced using Asymptote:<br />
<br />
<center>[[Image:Figure1.jpg]]</center><br />
<br />
In a sense, Asymptote is the ruler and compass of typesetting.<br />
<br />
<br />
You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within <tt><nowiki><asy>...</asy></nowiki></tt> tags. For example, the following code<br />
<pre><nowiki><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></nowiki></pre><br />
created the picture <br />
<center><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></center><br />
And on the AoPS forums you can use <tt><nowiki>[asy]..[/asy]</nowiki></tt><br />
<br />
Another example:<br />
<br />
<pre><nowiki>[asy]<br />
pair A,B,C,X,Y,Z; <br />
A = (0,0);<br />
B = (1,0);<br />
C = (0.3,0.8);<br />
draw(A--B--C--A);<br />
X = (B+C)/2;<br />
Y = (A+C)/2;<br />
Z = (A+B)/2;<br />
draw(A--X, red);<br />
draw(B--Y,red);<br />
draw(C--Z,red);<br />
[/asy]</nowiki></pre><br />
<br />
<asy><br />
pair A,B,C,X,Y,Z;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (0.3,0.8);<br />
draw(A--B--C--A);<br />
X = (B+C)/2;<br />
Y = (A+C)/2;<br />
Z = (A+B)/2;<br />
draw(A--X, red);<br />
draw(B--Y,red);<br />
draw(C--Z,red);</asy><br />
=== See also ===<br />
*[[LaTeX]]<br />
* [http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php AoPS's Getting Started with LaTeX guide.]<br />
* [http://www.artofproblemsolving.com/community/c68_latex_and_asymptote Asymptote Forum on AoPS]<br />
[[Asymptote: Getting Started | Next: Getting Started]]</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_31,_2011&diff=85893AoPS Wiki talk:Problem of the Day/July 31, 20112017-05-29T21:05:32Z<p>Edonahey5: /* Problem */</p>
<hr />
<div>==Problem==<br />
{{:AoPSWiki:Problem of the Day/July 31, 2011}}<br />
<br />
==Solution==<br />
{{potd_solution}}<br />
<br />
<math> 2^{100} * \frac{153}{103}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_31,_2011&diff=85892AoPS Wiki talk:Problem of the Day/July 31, 20112017-05-29T21:05:14Z<p>Edonahey5: /* Problem */</p>
<hr />
<div>==Problem==<br />
{{:AoPSWiki:Problem of the Day/July 31, 2011!}}<br />
<br />
==Solution==<br />
{{potd_solution}}<br />
<br />
<math> 2^{100} * \frac{153}{103}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_31,_2011&diff=85891AoPS Wiki talk:Problem of the Day/July 31, 20112017-05-29T21:04:56Z<p>Edonahey5: /* Solution */</p>
<hr />
<div>==Problem==<br />
{{:AoPSWiki:Problem of the Day/July 31, 2011}}<br />
==Solution==<br />
{{potd_solution}}<br />
<br />
<math> 2^{100} * \frac{153}{103}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_31,_2011&diff=85890AoPS Wiki talk:Problem of the Day/July 31, 20112017-05-29T21:04:32Z<p>Edonahey5: </p>
<hr />
<div>==Problem==<br />
{{:AoPSWiki:Problem of the Day/July 31, 2011}}<br />
==Solution==<br />
{{potd_solution!}}<br />
<br />
<math> 2^{100} * \frac{153}{103}</math></div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_24&diff=826392003 AMC 10B Problems/Problem 242017-01-31T00:58:52Z<p>Edonahey5: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The first four terms in an arithmetic sequence are <math>x+y</math>, <math>x-y</math>, <math>xy</math>, and <math>\frac{x}{y}</math>, in that order. What is the fifth term? <br />
<br />
<math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math><br />
<br />
==Solution==<br />
<br />
The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing.<br />
<br />
<cmath>\begin{align*}<br />
xy&=x-3y\\<br />
xy-x&=-3y\\<br />
x(y-1)&=-3y\\<br />
x&=\frac{-3y}{y-1}<br />
\end{align*}</cmath><br />
<br />
Substitute into our other equation.<br />
<br />
<cmath><br />
\frac{x}{y}=x-5y</cmath><br />
<cmath>\frac{-3}{y-1}=\frac{-3y}{y-1}-5y</cmath><br />
<cmath>-3=-3y-5y(y-1)</cmath><br />
<cmath>0=5y^2-2y-3</cmath><br />
<cmath>0=(5y+3)(y-1)</cmath><br />
<cmath>y=-\frac35, 1</cmath><br />
<br />
But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally,<br />
<br />
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath><br />
<br />
==See Also==<br />
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Edonahey5https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_24&diff=826382003 AMC 10B Problems/Problem 242017-01-31T00:58:31Z<p>Edonahey5: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The first four terms in an arithmetic sequence are <math>x+y</math>, <math>x-y</math>, <math>xy</math>, and <math>\frac{x}{y}</math>, in that order. What is the fifth term? <br />
<br />
<math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math><br />
<br />
==Solution==<br />
<br />
The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them eual to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing.<br />
<br />
<cmath>\begin{align*}<br />
xy&=x-3y\\<br />
xy-x&=-3y\\<br />
x(y-1)&=-3y\\<br />
x&=\frac{-3y}{y-1}<br />
\end{align*}</cmath><br />
<br />
Substitute into our other equation.<br />
<br />
<cmath><br />
\frac{x}{y}=x-5y</cmath><br />
<cmath>\frac{-3}{y-1}=\frac{-3y}{y-1}-5y</cmath><br />
<cmath>-3=-3y-5y(y-1)</cmath><br />
<cmath>0=5y^2-2y-3</cmath><br />
<cmath>0=(5y+3)(y-1)</cmath><br />
<cmath>y=-\frac35, 1</cmath><br />
<br />
But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally,<br />
<br />
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath><br />
<br />
==See Also==<br />
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Edonahey5