https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Wiggle+Wam&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T18:11:24ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_14&diff=817462010 AMC 12B Problems/Problem 142016-12-01T19:17:12Z<p>Wiggle Wam: /* Solution */</p>
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<div>== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?<br />
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<math>\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804</math><br />
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== Solution ==<br />
We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest.<br />
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Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671 \Rightarrow \boxed{B}</math>.<br />
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== Solution 2 == <br />
First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to <math>\frac{2010}{5}=402,</math> so none of C, D, or E can be the answer. Thus, the answer is A or B. We will show that A is unattainable, leaving us with B as the only possible answer. <br />
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Assume WLOG that <math>d+e</math> is the largest sum. So <math>d+e=670,</math> meaning <math>a+b+c=2010-670=1340.</math> Because we let <math>d+e=M,</math> we must have <math>a+b \leq M=670</math> and <math>b+c \leq M=670.</math> Adding these inequalities gives <math>a+2b+c \leq 1340.</math> But we just showed that <math>a+b+c=1340,</math> which means that <math>b=0,</math> a contradiction because we are told that all the variables are positive. <br />
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Therefore, the answer is <math>\boxed{B}.</math><br />
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== See also ==<br />
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}}<br />
{{MAA Notice}}</div>Wiggle Wamhttps://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_18&diff=789021976 AHSME Problems/Problem 182016-06-11T17:35:17Z<p>Wiggle Wam: Created page with "Extend <math>\overline{BD}</math> until it touches the opposite side of the circle, say at point <math>E.</math> By power of a point, we have <math>AB^2=(BC)(BE),</math> so <m..."</p>
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<div>Extend <math>\overline{BD}</math> until it touches the opposite side of the circle, say at point <math>E.</math> By power of a point, we have <math>AB^2=(BC)(BE),</math> so <math>BE=\frac{6^2}{3}=12.</math> Therefore, <math>DE=12-3-3=6.</math> <br />
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Now extend <math>\overline{OD}</math> in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains <math>OD</math>). Let the two endpoints of this diameter be <math>P</math> and <math>Q,</math> where <math>Q</math> is closer to <math>C.</math> Again use power of a point. We have <math>(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.</math> But if the radius of the circle is <math>r,</math> we see that <math>PD=r+2</math> and <math>DQ=r-2,</math> so we have the equation <math>(r+2)(r-2)=18.</math> Solving gives <math>r=\sqrt{22}.</math></div>Wiggle Wamhttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_29&diff=788321983 AHSME Problems/Problem 292016-06-07T16:05:29Z<p>Wiggle Wam: Created page with "Place the square on the coordinate plane with <math>A</math> as the origin. (This means that <math>B=(1,0), C=(1,1),</math> and <math>D=(0,1).</math>We are given that <math>PA..."</p>
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<div>Place the square on the coordinate plane with <math>A</math> as the origin. (This means that <math>B=(1,0), C=(1,1),</math> and <math>D=(0,1).</math>We are given that <math>PA^2+PB^2=PC^2,</math> so <br />
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\begin{align*}&(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\ &2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ &x^2+y^2=-2y+1\\ &x^2+y^2+2y-1=0\\ &x^2+(y+1)^2=2\end{align*}<br />
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Thus, we see that <math>P</math> is on a circle centered at <math>(0,-1)</math> with radius <math>\sqrt{2}.</math> The farthest point from <math>D</math> on this circle is at the bottom of the circle, at <math>(0, -1-\sqrt{2}),</math> so <math>PD</math> is <math>\boxed{2+\sqrt{2}}.</math></div>Wiggle Wam