https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jlee28&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-24T19:04:49Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_11&diff=141853 2016 AMC 10A Problems/Problem 11 2021-01-10T18:38:56Z <p>Jlee28: /* Video Solution */</p> <hr /> <div>== Problem ==<br /> Find the area of the shaded region.<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> <br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> draw((0,5)--(8,0));<br /> <br /> &lt;/asy&gt;<br /> <br /> The bases of these triangles are all &lt;math&gt;1&lt;/math&gt;, and by symmetry, their heights are &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;\frac{5}{2}&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;\frac{5}{2}&lt;/math&gt;. Thus, their areas are &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;\frac{5}{4}&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;\frac{5}{4}&lt;/math&gt;, which add to the area of the shaded region, which is &lt;math&gt;\boxed{6\frac{1}{2}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$4$&quot;,(6,5),dir(90));<br /> label(&quot;$3$&quot;,(5/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$5/2$&quot;,(8,15/4),dir(0));<br /> label(&quot;$3/2$&quot;,(8,7/4),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$4$&quot;,(2,0),dir(270));<br /> label(&quot;$3$&quot;,(11/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$5/2$&quot;,(0,5/4),dir(180));<br /> label(&quot;$3/2$&quot;,(0,13/4),dir(180));<br /> <br /> draw((0,5/2)--(8,5/2));<br /> draw((4,0)--(4,5));<br /> <br /> &lt;/asy&gt;<br /> <br /> Notice that the two added lines bisect each of the &lt;math&gt;4&lt;/math&gt; sides of the large rectangle.<br /> <br /> Subtracting the unshaded area from the total area gives us &lt;math&gt;40-33\frac{1}{2}=\boxed{6\frac{1}{2}}&lt;/math&gt;, so the correct answer is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Notice that we can graph this on the coordinate plane.<br /> <br /> The top-left shaded figure has coordinates of &lt;math&gt;(1,5), (0,5), (0,4), (4,\frac{5}{2})&lt;/math&gt;.<br /> <br /> Notice that we can apply the shoelace method to find the area of this polygon.<br /> <br /> We find that the area of the polygon is &lt;math&gt;\frac{13}{4}&lt;/math&gt;.<br /> <br /> However, notice that the two shaded regions are two congruent polygons.<br /> <br /> Hence, the total area is &lt;math&gt;\frac{13}{2}\implies \boxed{6\frac{1}{2}}&lt;/math&gt; or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> The sum of the heights on each axis is &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. The bases are &lt;math&gt;1&lt;/math&gt;. Hence the shaded area is &lt;math&gt;\frac{(5+8)(1)} 2 = \boxed{6\frac{1}{2}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/dHY8gjoYFXU<br /> <br /> ~IceMatrix<br /> <br /> https://www.youtube.com/watch?v=WojyKGOEk_g<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=10|num-a=12}}<br /> {{AMC12 box|year=2016|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Jlee28 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Answer_Key&diff=115862 2020 AMC 10A Answer Key 2020-02-01T01:02:28Z <p>Jlee28: Created page with &quot;1&quot;</p> <hr /> <div>1</div> Jlee28 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_20&diff=111821 2019 AMC 8 Problems/Problem 20 2019-11-21T00:21:15Z <p>Jlee28: /* Solution 1 */</p> <hr /> <div>==Problem 20==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that to get 16, you can square 4 or -4. Thus, x squared - 5 can have 2 possibilities. x squared is either 9 or 1, leaving x with possiblities 3,-3, 1, and -1.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=19|num-a=21}}<br /> <br /> {{MAA Notice}}</div> Jlee28 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_9&diff=111041 2006 AMC 12B Problems/Problem 9 2019-11-09T23:59:28Z <p>Jlee28: /* Problem */</p> <hr /> <div>== Problem==<br /> How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order? <br /> <br /> &lt;math&gt;<br /> \text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).<br /> <br /> If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.<br /> If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.<br /> If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,<br /> <br /> and so on.<br /> <br /> So, the answer is &lt;math&gt;3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The last digit is 4, 6, or 8.<br /> <br /> If the last digit is &lt;math&gt;x&lt;/math&gt;, the possibilities for the first two digits correspond to 2-element subsets of &lt;math&gt;\{1,2,\dots,x-1\}&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> The answer must be half of a triangular number (evens and decreasing/increasing) so &lt;math&gt;\boxed{34}&lt;/math&gt; or the letter B.<br /> -<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jlee28 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_9&diff=111040 2006 AMC 12B Problems/Problem 9 2019-11-09T23:59:03Z <p>Jlee28: /* Solution 3 */</p> <hr /> <div>== Problem==<br /> How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order? <br /> <br /> &lt;math&gt;<br /> \text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).<br /> <br /> If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.<br /> If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.<br /> If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,<br /> <br /> and so on.<br /> <br /> So, the answer is &lt;math&gt;3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The last digit is 4, 6, or 8.<br /> <br /> If the last digit is &lt;math&gt;x&lt;/math&gt;, the possibilities for the first two digits correspond to 2-element subsets of &lt;math&gt;\{1,2,\dots,x-1\}&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> The answer must be half of a triangular number (evens and decreasing/increasing) so &lt;math&gt;\boxed{34}&lt;/math&gt; or the letter B.<br /> -<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jlee28