https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Xsrvmy&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:23:32ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_25&diff=683362015 AMC 10B Problems/Problem 252015-03-04T04:44:36Z<p>Xsrvmy: Created page with "==Problem== A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c<..."</p>
<hr />
<div>==Problem==<br />
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?<br />
<br />
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math><br />
<br />
==Solution==<br />
The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.<br />
<br />
Divide both sides by <math>2abc</math>, we get that <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath><br />
<br />
First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.<br />
<br />
Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.<br />
Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>.<br />
<br />
So we have <math>a=3, 4, 5</math> or <math>6</math>.<br />
<br />
Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>.<br />
<br />
From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math><br />
<br />
When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.<br />
<br />
When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.<br />
<br />
When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.<br />
<br />
When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.<br />
<br />
Thus, there are <math>\boxed{\textbf{(B)}\;10}</math> solutions<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|after=Last Problem|num-b=24}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&diff=683332015 AMC 12B Problems/Problem 232015-03-04T04:40:14Z<p>Xsrvmy: /* Solution */</p>
<hr />
<div>==Problem==<br />
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?<br />
<br />
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math><br />
<br />
==Solution==<br />
The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.<br />
<br />
Divide both sides by <math>2abc</math>, we get that <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath><br />
<br />
First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.<br />
<br />
Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.<br />
Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>.<br />
<br />
So we have <math>a=3, 4, 5</math> or <math>6</math>.<br />
<br />
Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>.<br />
<br />
From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math><br />
<br />
When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.<br />
<br />
When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.<br />
<br />
When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.<br />
<br />
When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.<br />
<br />
Thus, there are <math>\boxed{\textbf{(B)}\;10}</math> solutions<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&diff=683322015 AMC 12B Problems/Problem 232015-03-04T04:39:29Z<p>Xsrvmy: /* Solution */</p>
<hr />
<div>==Problem==<br />
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?<br />
<br />
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math><br />
<br />
==Solution==<br />
The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.<br />
<br />
Divide both sides by <math>2abc</math>, we get that <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath><br />
<br />
First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.<br />
<br />
Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.<br />
Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>.<br />
<br />
So we have <math>a=3, 4, 5</math> or <math>6</math>.<br />
<br />
Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>.<br />
<br />
From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math><br />
<br />
When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.<br />
<br />
When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.<br />
<br />
When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.<br />
<br />
When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.<br />
<br />
Thus, there are ==Solution==<br />
The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.<br />
<br />
Divide both sides by <math>2abc</math>, we get that <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath><br />
<br />
First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.<br />
<br />
Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.<br />
Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>.<br />
<br />
So we have <math>a=3, 4, 5</math> or <math>6</math>.<br />
<br />
Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>.<br />
<br />
From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math><br />
<br />
When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.<br />
<br />
When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.<br />
<br />
When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.<br />
<br />
When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.<br />
<br />
Thus, there are <math>\boxed{\textbf{(B)}\;10}</math> solutions<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&diff=683312015 AMC 12B Problems/Problem 232015-03-04T04:33:24Z<p>Xsrvmy: /* Solution */</p>
<hr />
<div>==Problem==<br />
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?<br />
<br />
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math><br />
<br />
==Solution==<br />
The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.<br />
<br />
Divide both sides by <math>2abc</math>, we get that <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath><br />
<br />
First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.<br />
<br />
Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.<br />
Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>.<br />
<br />
So we have <math>a=3, 4, 5</math> or <math>6</math>.<br />
<br />
Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>.<br />
<br />
From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math><br />
<br />
When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.<br />
<br />
When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.<br />
<br />
When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.<br />
<br />
When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.<br />
<br />
Thus, there are 11 solutions<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_2&diff=682702015 AMC 12B Problems/Problem 22015-03-03T22:20:57Z<p>Xsrvmy: /* Solution (provided by Max) */</p>
<hr />
<div>==Problem==<br />
Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?<br />
<br />
<math> \textbf{(A)}\; \text{3:10 PM} \qquad\textbf{(B)}\; \text{3:30 PM} \qquad\textbf{(C)}\; \text{4:00 PM} \qquad\textbf{(D)}\; \text{4:10 PM} \qquad\textbf{(E)}\; \text{4:30 PM}</math><br />
<br />
==Solution==<br />
The first two tasks took <math>2:40-1:00=100</math> minutes. Thus, each task takes <math>100\div 2=50</math> minutes. So the third task finishes at <math>2:40+50=3:30</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=3|num-b=1}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_25&diff=682692015 AMC 12B Problems/Problem 252015-03-03T22:10:29Z<p>Xsrvmy: /*added the choices*/</p>
<hr />
<div>==Problem==<br />
A bee starts flying from point <math>P_0</math>. She flies <math>1</math> inch due east to point <math>P_1</math>. For <math>j \ge 1</math>, once the bee reaches point <math>P_j</math>, she turns <math>30^{\circ}</math> counterclockwise and then flies <math>j+1</math> inches straight to point <math>P_{j+1}</math>. When the bee reaches <math>P_{2015}</math> she is exactly <math>a \sqrt{b} + c \sqrt{d}</math> inches away from <math>P_0</math>, where <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> are positive integers and <math>b</math> and <math>d</math> are not divisible by the square of any prime. What is <math>a+b+c+d</math> ?<br />
<br />
<math>\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048</math><br />
<br />
==Solution==<br />
<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|after=Last Problem|num-b=24}}<br />
{{MAA Notice}}</div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_2&diff=618492014 USAJMO Problems/Problem 22014-04-30T23:23:14Z<p>Xsrvmy: /* Problem */</p>
<hr />
<div>==Problem==<br />
Let <math>\triangle{ABC}</math> be a non-equilateral, acute triangle with <math>\angle A=60^{\circ}</math>, and let <math>O</math> and <math>H</math> denote the circumcenter and orthocenter of <math>\triangle{ABC}</math>, respectively.<br />
<br />
(a) Prove that line <math>OH</math> intersects both segments <math>AB</math> and <math>AC</math>.<br />
<br />
(b) Line <math>OH</math> intersects segments <math>AB</math> and <math>AC</math> at <math>P</math> and <math>Q</math>, respectively. Denote by <math>s</math> and <math>t</math> the respective areas of triangle <math>APQ</math> and quadrilateral <math>BPQC</math>. Determine the range of possible values for <math>s/t</math>.<br />
<br />
==Solution==<br />
<asy><br />
import olympiad;<br />
unitsize(1inch);<br />
pair A,B,C,O,H,P,Q,i1,i2,i3,i4;<br />
<br />
//define dots<br />
A=3*dir(50);<br />
B=(0,0);<br />
C=right*2.76481496;<br />
<br />
O=circumcenter(A,B,C);<br />
H=orthocenter(A,B,C);<br />
<br />
i1=2*O-H;<br />
i2=2*i1-O;<br />
i3=2*H-O;<br />
i4=2*i3-H;<br />
//These points are for extending PQ. DO NOT DELETE!<br />
<br />
P=intersectionpoint(i2--i4,A--B);<br />
Q=intersectionpoint(i2--i4,A--C);<br />
<br />
//draw<br />
dot(P);<br />
dot(Q);<br />
draw(P--Q);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(O);<br />
dot(H);<br />
draw(A--B--C--cycle);<br />
<br />
//label<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,NW);<br />
label("$Q$",Q,NE);<br />
label("$O$",O,N);<br />
label("$H$",H,N);<br />
//change O and H label positions if interfering with other lines to be added<br />
<br />
//further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistence and are not to be used<br />
</asy></div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_2&diff=618482014 USAJMO Problems/Problem 22014-04-30T23:21:21Z<p>Xsrvmy: /* Diagram */</p>
<hr />
<div>==Problem==<br />
Let <math>\triangle{ABC}</math> be a non-equilateral, acute triangle with <math>\angle A=60\textdegrees</math>, and let <math>O</math> and <math>H</math> denote the circumcenter and orthocenter of <math>\triangle{ABC}</math>, respectively.<br />
<br />
(a) Prove that line <math>OH</math> intersects both segments <math>AB</math> and <math>AC</math>.<br />
<br />
(b) Line <math>OH</math> intersects segments <math>AB</math> and <math>AC</math> at <math>P</math> and <math>Q</math>, respectively. Denote by <math>s</math> and <math>t</math> the respective areas of triangle <math>APQ</math> and quadrilateral <math>BPQC</math>. Determine the range of possible values for <math>s/t</math>.<br />
==Solution==<br />
<asy><br />
import olympiad;<br />
unitsize(1inch);<br />
pair A,B,C,O,H,P,Q,i1,i2,i3,i4;<br />
<br />
//define dots<br />
A=3*dir(50);<br />
B=(0,0);<br />
C=right*2.76481496;<br />
<br />
O=circumcenter(A,B,C);<br />
H=orthocenter(A,B,C);<br />
<br />
i1=2*O-H;<br />
i2=2*i1-O;<br />
i3=2*H-O;<br />
i4=2*i3-H;<br />
//These points are for extending PQ. DO NOT DELETE!<br />
<br />
P=intersectionpoint(i2--i4,A--B);<br />
Q=intersectionpoint(i2--i4,A--C);<br />
<br />
//draw<br />
dot(P);<br />
dot(Q);<br />
draw(P--Q);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(O);<br />
dot(H);<br />
draw(A--B--C--cycle);<br />
<br />
//label<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$P$",P,NW);<br />
label("$Q$",Q,NE);<br />
label("$O$",O,N);<br />
label("$H$",H,N);<br />
//change O and H label positions if interfering with other lines to be added<br />
<br />
//further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistence and are not to be used<br />
</asy></div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2014_USAJMO_Problems/Problem_1&diff=618192014 USAJMO Problems/Problem 12014-04-30T01:45:53Z<p>Xsrvmy: /* More detailed solution */</p>
<hr />
<div>==Problem==<br />
Let <math>a</math>, <math>b</math>, <math>c</math> be real numbers greater than or equal to <math>1</math>. Prove that <cmath>\min{\left (\frac{10a^2-5a+1}{b^2-5b+1},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )}\leq abc </cmath><br />
==Solution==<br />
Since <math>(a-1)^5\geqslant 0</math>,<br />
<cmath>a^5-5a^4+10a^3-10a^2+5a-1\geqslant 0</cmath><br />
or<br />
<cmath>10a^2-5a+1\leqslant a^3(a^2-5a+10)</cmath><br />
Since <math>a^2-5a+10=\left( a-\dfrac{5}{2}\right) +\dfrac{15}{4}\geqslant 0</math>,<br />
<cmath> \frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3 </cmath><br />
Also note that <math>10a^2-5a+1=10\left( a-\dfrac{1}{4}\right)+\dfrac{3}{8}\geqslant 0</math>,<br />
We conclude<br />
<cmath>0\leqslant\frac{10a^2-5a+1}{a^2-5a+10}\leqslant a^3</cmath><br />
Similarly, <br />
<cmath>0\leqslant\frac{10b^2-5b+1}{b^2-5b+10}\leqslant b^3</cmath><br />
<cmath>0\leqslant\frac{10c^2-5c+1}{c^2-5c+10}\leqslant c^3</cmath><br />
So <cmath>\left(\frac{10a^2-5a+1}{a^2-5a+10}\right)\left(\frac{10b^2-5b+1}{b^2-5b+10}\right)\left(\frac{10c^2-5c+1}{c^2-5c+10}\right)\leqslant a^3b^3c^3</cmath><br />
or<br />
<cmath>\left(\frac{10a^2-5a+1}{b^2-5b+10}\right)\left(\frac{10b^2-5b+1}{c^2-5c+10}\right)\left(\frac{10c^2-5c+1}{a^2-5a+10}\right) \leqslant(abc)^3</cmath><br />
Therefore,<br />
<cmath> \min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq abc. </cmath></div>Xsrvmyhttps://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_1&diff=610642014 AIME I Problems/Problem 12014-03-15T18:37:47Z<p>Xsrvmy: Solution added.</p>
<hr />
<div>== Problem 1 ==<br />
<br />
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectrangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.<br />
<br />
DIAGRAM NEEDED<br />
<br />
== Solution ==<br />
<br />
The rectangle is divided into three smaller triangles with a width of 50 mm and a length of <math>\dfrac{80}{3}</math>mm. According the pythagorean theorem, the diagonal of this rectangle is<br />
<math>\sqrt{50^2+\frac{80}{3}^2}=\frac{170}{3}</math>mm. Since that on the lace, there are 6 of these diagonals, a width, and at least 200 mm extension on each side. Therefore, the minimum of the lace in millimeters is<br />
<cmath>6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.</cmath><br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Xsrvmy