https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Wiggle+Wam&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-18T08:16:10Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_14&diff=81746 2010 AMC 12B Problems/Problem 14 2016-12-01T19:17:12Z <p>Wiggle Wam: /* Solution */</p> <hr /> <div>== Problem 14 ==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; be positive integers with &lt;math&gt;a+b+c+d+e=2010&lt;/math&gt; and let &lt;math&gt;M&lt;/math&gt; be the largest of the sum &lt;math&gt;a+b&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;c+d&lt;/math&gt; and &lt;math&gt;d+e&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;M&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804&lt;/math&gt;<br /> <br /> == Solution ==<br /> We want to try make &lt;math&gt;a+b&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, &lt;math&gt;c+d&lt;/math&gt;, and &lt;math&gt;d+e&lt;/math&gt; as close as possible so that &lt;math&gt;M&lt;/math&gt;, the maximum of these, if smallest.<br /> <br /> Notice that &lt;math&gt;2010=670+670+670&lt;/math&gt;. In order to express &lt;math&gt;2010&lt;/math&gt; as a sum of &lt;math&gt;5&lt;/math&gt; 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): &lt;math&gt;2010=670+1+670+1+668&lt;/math&gt; or &lt;math&gt;2010=670+1+669+1+669&lt;/math&gt;. We see that in both cases, the value of &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;671&lt;/math&gt;, so the answer is &lt;math&gt;671 \Rightarrow \boxed{B}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to &lt;math&gt;\frac{2010}{5}=402,&lt;/math&gt; 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 /> <br /> Assume WLOG that &lt;math&gt;d+e&lt;/math&gt; is the largest sum. So &lt;math&gt;d+e=670,&lt;/math&gt; meaning &lt;math&gt;a+b+c=2010-670=1340.&lt;/math&gt; Because we let &lt;math&gt;d+e=M,&lt;/math&gt; we must have &lt;math&gt;a+b \leq M=670&lt;/math&gt; and &lt;math&gt;b+c \leq M=670.&lt;/math&gt; Adding these inequalities gives &lt;math&gt;a+2b+c \leq 1340.&lt;/math&gt; But we just showed that &lt;math&gt;a+b+c=1340,&lt;/math&gt; which means that &lt;math&gt;b=0,&lt;/math&gt; a contradiction because we are told that all the variables are positive. <br /> <br /> Therefore, the answer is &lt;math&gt;\boxed{B}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}}<br /> {{MAA Notice}}</div> Wiggle Wam https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_18&diff=78902 1976 AHSME Problems/Problem 18 2016-06-11T17:35:17Z <p>Wiggle Wam: Created page with &quot;Extend &lt;math&gt;\overline{BD}&lt;/math&gt; until it touches the opposite side of the circle, say at point &lt;math&gt;E.&lt;/math&gt; By power of a point, we have &lt;math&gt;AB^2=(BC)(BE),&lt;/math&gt; so &lt;m...&quot;</p> <hr /> <div>Extend &lt;math&gt;\overline{BD}&lt;/math&gt; until it touches the opposite side of the circle, say at point &lt;math&gt;E.&lt;/math&gt; By power of a point, we have &lt;math&gt;AB^2=(BC)(BE),&lt;/math&gt; so &lt;math&gt;BE=\frac{6^2}{3}=12.&lt;/math&gt; Therefore, &lt;math&gt;DE=12-3-3=6.&lt;/math&gt; <br /> <br /> Now extend &lt;math&gt;\overline{OD}&lt;/math&gt; in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains &lt;math&gt;OD&lt;/math&gt;). Let the two endpoints of this diameter be &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q,&lt;/math&gt; where &lt;math&gt;Q&lt;/math&gt; is closer to &lt;math&gt;C.&lt;/math&gt; Again use power of a point. We have &lt;math&gt;(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.&lt;/math&gt; But if the radius of the circle is &lt;math&gt;r,&lt;/math&gt; we see that &lt;math&gt;PD=r+2&lt;/math&gt; and &lt;math&gt;DQ=r-2,&lt;/math&gt; so we have the equation &lt;math&gt;(r+2)(r-2)=18.&lt;/math&gt; Solving gives &lt;math&gt;r=\sqrt{22}.&lt;/math&gt;</div> Wiggle Wam https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_29&diff=78832 1983 AHSME Problems/Problem 29 2016-06-07T16:05:29Z <p>Wiggle Wam: Created page with &quot;Place the square on the coordinate plane with &lt;math&gt;A&lt;/math&gt; as the origin. (This means that &lt;math&gt;B=(1,0), C=(1,1),&lt;/math&gt; and &lt;math&gt;D=(0,1).&lt;/math&gt;We are given that &lt;math&gt;PA...&quot;</p> <hr /> <div>Place the square on the coordinate plane with &lt;math&gt;A&lt;/math&gt; as the origin. (This means that &lt;math&gt;B=(1,0), C=(1,1),&lt;/math&gt; and &lt;math&gt;D=(0,1).&lt;/math&gt;We are given that &lt;math&gt;PA^2+PB^2=PC^2,&lt;/math&gt; so <br /> <br /> \begin{align*}&amp;(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\ &amp;2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ &amp;x^2+y^2=-2y+1\\ &amp;x^2+y^2+2y-1=0\\ &amp;x^2+(y+1)^2=2\end{align*}<br /> <br /> Thus, we see that &lt;math&gt;P&lt;/math&gt; is on a circle centered at &lt;math&gt;(0,-1)&lt;/math&gt; with radius &lt;math&gt;\sqrt{2}.&lt;/math&gt; The farthest point from &lt;math&gt;D&lt;/math&gt; on this circle is at the bottom of the circle, at &lt;math&gt;(0, -1-\sqrt{2}),&lt;/math&gt; so &lt;math&gt;PD&lt;/math&gt; is &lt;math&gt;\boxed{2+\sqrt{2}}.&lt;/math&gt;</div> Wiggle Wam