https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Geek1ab&feedformat=atom AoPS Wiki - User contributions [en] 2022-08-15T03:36:57Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_27&diff=106160 1992 AHSME Problems/Problem 27 2019-06-07T21:12:46Z <p>Geek1ab: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A circle of radius &lt;math&gt;r&lt;/math&gt; has chords &lt;math&gt;\overline{AB}&lt;/math&gt; of length &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; of length 7. When &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; are extended through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, they intersect at &lt;math&gt;P&lt;/math&gt;, which is outside of the circle. If &lt;math&gt;\angle{APD}=60^\circ&lt;/math&gt; and &lt;math&gt;BP=8&lt;/math&gt;, then &lt;math&gt;r^2=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 70\quad<br /> \text{(B) } 71\quad<br /> \text{(C) } 72\quad<br /> \text{(D) } 73\quad<br /> \text{(E) } 74&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;asy&gt; <br /> import olympiad; <br /> import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); draw(circ); draw(A--P--D); label('$A$', A, N); label('$D$', D, S); label('$C$', C, SE); label('$B$', B, NE); label('$P$', P, E); label('$60^\circ$', P, 2 * (dir(P--A) + dir(P--D))); label('$10$', A--B, S); label('$8$', B--P, NE); label('$7$', C--D, N); <br /> &lt;/asy&gt;<br /> <br /> Applying Power of a Point on &lt;math&gt;P&lt;/math&gt;, we find that &lt;math&gt;PC=9&lt;/math&gt; and thus &lt;math&gt;PD=16&lt;/math&gt;. Observing that &lt;math&gt;PD=2BP&lt;/math&gt; and that &lt;math&gt;\angle BPD=60^{\circ}&lt;/math&gt;, we conclude that &lt;math&gt;BPD&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle with right angle at &lt;math&gt;B&lt;/math&gt;. Thus, &lt;math&gt;BD=8\sqrt{3}&lt;/math&gt; and triangle &lt;math&gt;ABD&lt;/math&gt; is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem &lt;math&gt;AD=2r=2\sqrt{73}&lt;/math&gt;. From here we see that &lt;math&gt;r^2=73&lt;/math&gt;. The answer is thus &lt;math&gt;\fbox{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=26|num-a=28}} <br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_17&diff=97825 2009 AMC 8 Problems/Problem 17 2018-09-15T22:20:42Z <p>Geek1ab: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The positive integers &lt;math&gt; x&lt;/math&gt; and &lt;math&gt; y&lt;/math&gt; are the two smallest positive integers for which the product of &lt;math&gt; 360&lt;/math&gt; and &lt;math&gt; x&lt;/math&gt; is a square and the product of &lt;math&gt; 360&lt;/math&gt; and &lt;math&gt; y&lt;/math&gt; is a cube. What is the sum of &lt;math&gt; x&lt;/math&gt; and &lt;math&gt; y&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 80 \qquad<br /> \textbf{(B)}\ 85 \qquad<br /> \textbf{(C)}\ 115 \qquad<br /> \textbf{(D)}\ 165 \qquad<br /> \textbf{(E)}\ 610&lt;/math&gt;<br /> <br /> ==Solution==<br /> The prime factorization of &lt;math&gt;360=2^3 \cdot 3^2 \cdot 5&lt;/math&gt;. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of x. Similarly, y can be found by making all the exponents divisible by 3, so the minimum possible value of &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;3 \cdot 5^2=75&lt;/math&gt;. Thus, our answer is &lt;math&gt;x+y=10+75=\boxed{\textbf{(B)}\ 85}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=97822 2016 AMC 8 Problems/Problem 22 2018-09-15T14:25:22Z <p>Geek1ab: /* Solution 2 */</p> <hr /> <div>Rectangle &lt;math&gt;DEFA&lt;/math&gt; below is a &lt;math&gt;3 \times 4&lt;/math&gt; rectangle with &lt;math&gt;DC=CB=BA&lt;/math&gt;. What is the area of the &quot;bat wings&quot; (shaded area)?<br /> &lt;asy&gt;<br /> draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br /> draw((3,0)--(1,4)--(0,0));<br /> fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br /> fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br /> label(&quot;$A$&quot;,(3.05,4.2));<br /> label(&quot;$B$&quot;,(2,4.2));<br /> label(&quot;$C$&quot;,(1,4.2));<br /> label(&quot;$D$&quot;,(0,4.2));<br /> label(&quot;$E$&quot;, (0,-0.2));<br /> label(&quot;$F$&quot;, (3,-0.2));<br /> label(&quot;$1$&quot;, (0.5, 4), N);<br /> label(&quot;$1$&quot;, (1.5, 4), N);<br /> label(&quot;$1$&quot;, (2.5, 4), N);<br /> label(&quot;$4$&quot;, (3.2, 2), E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> The area of trapezoid &lt;math&gt;CBFE&lt;/math&gt; is &lt;math&gt;\frac{1+3}2\cdot 4=8&lt;/math&gt;. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a &lt;math&gt;3:1&lt;/math&gt; ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is &lt;math&gt;3,&lt;/math&gt; while the height of the smaller one is &lt;math&gt;1.&lt;/math&gt; Thus, their areas are &lt;math&gt;\frac12&lt;/math&gt; and &lt;math&gt;\frac92&lt;/math&gt;. Subtracting these areas from the trapezoid, we get &lt;math&gt;8-\frac12-\frac92 =\boxed3&lt;/math&gt;. Therefore, the answer to this problem is &lt;math&gt;\boxed{\textbf{(C) }3}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Setting coordinates!<br /> <br /> Let &lt;math&gt;E=(0,0)&lt;/math&gt;, &lt;math&gt;F=(3,0)&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); <br /> draw((3,0)--(1,4)--(0,0)); <br /> fill((0,0)--(1,4)--(1.5,3)--cycle, black); <br /> fill((3,0)--(2,4)--(1.5,3)--cycle, black); <br /> label(scale(0.7)*&quot;$A(3,4)$&quot;,(3.25,4.2)); <br /> label(scale(0.7)*&quot;$B(2,4)$&quot;,(2.1,4.2)); <br /> label(scale(0.7)*&quot;$C(1,4)$&quot;,(0.9,4.2)); <br /> label(scale(0.7)*&quot;$D(0,4)$&quot;,(-0.3,4.2)); <br /> label(scale(0.7)*&quot;$E(0,0)$&quot;, (0,-0.2)); <br /> label(scale(0.7)*&quot;$Z(\frac{3}{2},3)$&quot;, (1.5,1.8)); <br /> label(scale(0.7)*&quot;$F(3,0)$&quot;, (3,-0.2)); <br /> label(scale(0.7)*&quot;$1$&quot;, (0.3, 4), N); <br /> label(scale(0.7)*&quot;$1$&quot;, (1.5, 4), N); <br /> label(scale(0.7)*&quot;$1$&quot;, (2.7, 4), N); <br /> label(scale(0.7)*&quot;$4$&quot;, (3.2, 2), E); <br /> &lt;/asy&gt;<br /> <br /> Now, we easily discover that line &lt;math&gt;CF&lt;/math&gt; has lattice coordinates at &lt;math&gt;(1,4)&lt;/math&gt; and &lt;math&gt;(3,0)&lt;/math&gt;. Hence, the slope of line &lt;math&gt;CF=-2&lt;/math&gt; <br /> <br /> Plugging in the rest of the coordinate points, we find that line &lt;math&gt;CF=-2x+6&lt;/math&gt;<br /> <br /> Doing the same process to line &lt;math&gt;BE&lt;/math&gt;, we find that line &lt;math&gt;BE=2x&lt;/math&gt;.<br /> <br /> Hence, setting them equal to find the intersection point...<br /> <br /> &lt;math&gt;y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3&lt;/math&gt;.<br /> <br /> Hence, we find that the intersection point is &lt;math&gt;(\frac{3}{2},3)&lt;/math&gt;. Call it Z.<br /> <br /> Now, we can see that <br /> <br /> &lt;math&gt;E=(0,0)&lt;/math&gt; <br /> <br /> &lt;math&gt;Z=(\frac{3}{2},3)&lt;/math&gt; <br /> <br /> &lt;math&gt;C=(1,4)&lt;/math&gt;.<br /> <br /> Shoelace!<br /> <br /> Using the well known Shoelace Formula(https://en.m.wikipedia.org/wiki/Shoelace_formula), we find that the area of one of those small shaded triangles is &lt;math&gt;\frac{3}{2}&lt;/math&gt;.<br /> <br /> Now because there are two of them, we multiple that area by &lt;math&gt;2&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }3}&lt;/math&gt;<br /> {{AMC8 box|year=2016|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_6&diff=93660 2014 AMC 8 Problems/Problem 6 2018-03-31T21:46:49Z <p>Geek1ab: /* Solution */</p> <hr /> <div>==Problem==<br /> Six rectangles each with a common base width of &lt;math&gt;2&lt;/math&gt; have lengths of &lt;math&gt;1, 4, 9, 16, 25&lt;/math&gt;, and &lt;math&gt;36&lt;/math&gt;. What is the sum of the areas of the six rectangles?<br /> <br /> &lt;math&gt;\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202&lt;/math&gt;<br /> <br /> ==Solution==<br /> The sum of the areas is equal to &lt;math&gt;2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36&lt;/math&gt;. This is simply equal to &lt;math&gt;2(1+4+9+16+25+36)&lt;/math&gt;, which is equal to &lt;math&gt;2\cdot91&lt;/math&gt;, which is equal to our final answer of &lt;math&gt;\boxed{\textbf{(D)}~182}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2014|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_16&diff=93636 2016 AMC 8 Problems/Problem 16 2018-03-31T14:17:11Z <p>Geek1ab: /* Solution */</p> <hr /> <div>Annie and Bonnie are running laps around a &lt;math&gt;400&lt;/math&gt;-meter oval track. They started together, but Annie has pulled ahead, because she runs &lt;math&gt;25\%&lt;/math&gt; faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br /> <br /> &lt;math&gt;\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> ==Solution==<br /> Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run &lt;math&gt;\boxed{\textbf{(D)}\ 5 }&lt;/math&gt; laps.<br /> {{AMC8 box|year=2016|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_11&diff=93635 2016 AMC 8 Problems/Problem 11 2018-03-31T14:05:46Z <p>Geek1ab: /* Solution */</p> <hr /> <div>Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is &lt;math&gt;132.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> We can write the two digit number in the form of &lt;math&gt;10a+b&lt;/math&gt;; reverse of &lt;math&gt;10a+b&lt;/math&gt; is &lt;math&gt;10b+a&lt;/math&gt;. The sum of those numbers is:<br /> &lt;cmath&gt;(10a+b)+(10b+a)=132&lt;/cmath&gt;&lt;cmath&gt;11a+11b=132&lt;/cmath&gt;&lt;cmath&gt;a+b=12&lt;/cmath&gt;<br /> We can use brute force to find order pairs &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;a+b=12&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both digits, both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have to be integers less than &lt;math&gt;10&lt;/math&gt;. Thus our ordered pairs are &lt;math&gt;(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B)} 7}&lt;/math&gt; ordered pairs.<br /> <br /> {{AMC8 box|year=2016|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=93634 2017 AMC 8 Problems/Problem 17 2018-03-31T12:15:00Z <p>Geek1ab: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=93633 2017 AMC 8 Problems/Problem 17 2018-03-31T12:14:21Z <p>Geek1ab: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=93632 2017 AMC 8 Problems/Problem 17 2018-03-31T12:13:34Z <p>Geek1ab: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C)}45\qquad{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=93631 2017 AMC 8 Problems/Problem 17 2018-03-31T12:12:51Z <p>Geek1ab: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_17&diff=93630 2017 AMC 8 Problems/Problem 17 2018-03-31T12:10:27Z <p>Geek1ab: /* Problem 17 */</p> <hr /> <div>==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> ==Solution==<br /> We can represent the amount of gold with &lt;math&gt;g&lt;/math&gt; and the amount of chests with &lt;math&gt;c&lt;/math&gt;. We can use the problem to make the following equations:<br /> &lt;cmath&gt;9c-18 = g&lt;/cmath&gt;<br /> &lt;cmath&gt;6c+3 = g&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;6c+3 = 9c-18.&lt;/math&gt; This implies that &lt;math&gt;c = 7.&lt;/math&gt; We therefore have &lt;math&gt;g = 45.&lt;/math&gt; So, our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 45}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=80476 2015 AMC 8 Problems/Problem 25 2016-10-01T23:34:31Z <p>Geek1ab: /* Solution 2 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}<br /> Thank you for reading the solutions of the 2015 AMC 8 Problems made by people on AoPS.</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=80475 2015 AMC 8 Problems/Problem 25 2016-10-01T23:33:32Z <p>Geek1ab: /* Solution 2 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> <br /> [asy] <br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> [/asy]<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}<br /> Thank you for reading the solutions of the 2015 AMC 8 Problems made by people on AoPS.</div> Geek1ab https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_5&diff=80465 2015 AMC 8 Problems/Problem 5 2016-10-01T22:14:11Z <p>Geek1ab: /* See I HID THE BODY IN THE VAN */</p> <hr /> <div>Billy's basketball team scored the following points over the course of the first 11 games of the season:<br /> &lt;cmath&gt;42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73&lt;/cmath&gt;<br /> If his team scores 40 in the 12th game, which of the following statistics will show an increase?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \text{range} \qquad<br /> \textbf{(B) } \text{median} \qquad<br /> \textbf{(C) } \text{mean} \qquad<br /> \textbf{(D) } \text{mode} \qquad<br /> \textbf{(E) } \text{mid-range}<br /> &lt;/math&gt;<br /> ===Solution===<br /> <br /> When they score a &lt;math&gt;40&lt;/math&gt; on the next game, the range increases from &lt;math&gt;73-42=31&lt;/math&gt; to &lt;math&gt;73-40=33&lt;/math&gt;. This means the &lt;math&gt;\boxed{\textbf{(A) } \text{range}}&lt;/math&gt; increased.<br /> <br /> ==Solution 2==<br /> <br /> Because &lt;math&gt;40&lt;/math&gt; is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the &lt;math&gt;\boxed{\textbf{(A)}~\text{range}}&lt;/math&gt;, may increase.<br /> <br /> {{AMC8 box|year=2015|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Geek1ab