https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Qkddud&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-03T13:47:07Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=113146 2019 AMC 8 Problems/Problem 5 2019-12-22T06:25:04Z <p>Qkddud: /* Solution 1 (Using the answer choices) */</p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1 (Using the answer choices)==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the only one left.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=113145 2019 AMC 8 Problems/Problem 5 2019-12-22T06:24:42Z <p>Qkddud: /* Solution 1 (Using the answer choices) */</p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1 (Using the answer choices)==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> \\Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> \\Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> \\Therefore, the answer &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the only one left.<br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_2&diff=113144 2019 AMC 8 Problems/Problem 2 2019-12-22T06:04:21Z <p>Qkddud: /* Solution 2 */</p> <hr /> <div>=Problem 2=<br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 1==<br /> <br /> We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that &lt;math&gt;2\cdot{5}\ = 10&lt;/math&gt; which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is &lt;math&gt;10 + 5 = 15&lt;/math&gt; . Thus, the area is &lt;math&gt;15\cdot{10}\ = 150&lt;/math&gt; for choice &lt;math&gt;\boxed{\textbf{(E)}\ 150}&lt;/math&gt; ~~Saksham27<br /> <br /> ==Solution 2==<br /> Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is &lt;math&gt;5 \cdot 2 = 10&lt;/math&gt;. So the area of the identical rectangles is &lt;math&gt;5 \cdot 10 = 50&lt;/math&gt;. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is &lt;math&gt;50 \cdot 3 = \boxed{\textbf{(E)}\ 150}&lt;/math&gt;. ~~fath2012<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=1|num-a=3}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=112138 User:Dnjhan 2019-11-25T01:49:02Z <p>Qkddud: </p> <hr /> <div>Mateo Daddy</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111462 User:Dnjhan 2019-11-16T00:41:45Z <p>Qkddud: </p> <hr /> <div><br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> <br /> Joshua Han 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Guinea-Pug&diff=111413 User:Guinea-Pug 2019-11-14T04:27:55Z <p>Qkddud: Created page with &quot;Yeet&quot;</p> <hr /> <div>Yeet</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111411 User:Dnjhan 2019-11-14T04:02:50Z <p>Qkddud: </p> <hr /> <div>I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> I AM AN IDIOT AND IM GAY AND LOVE HANNAH AND YENA<br /> <br /> <br /> Joshua Han 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111406 User:Dnjhan 2019-11-14T02:35:57Z <p>Qkddud: </p> <hr /> <div>I love Yena and Tarik.<br /> I'm Gay.<br /> <br /> I'm also a big fluffy unicorn.<br /> <br /> JOSHUA HAN 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111339 User:Dnjhan 2019-11-13T00:36:20Z <p>Qkddud: Replaced content with &quot;2019&quot;</p> <hr /> <div>2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111338 User:Dnjhan 2019-11-13T00:35:52Z <p>Qkddud: </p> <hr /> <div>I'm bisexual I like Tarik.<br /> I love Yena-<br /> <br /> Joshua 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111337 User:Dnjhan 2019-11-13T00:31:38Z <p>Qkddud: </p> <hr /> <div><br /> I love Yena-<br /> <br /> Joshua 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111270 User:Dnjhan 2019-11-12T03:34:48Z <p>Qkddud: </p> <hr /> <div><br /> 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111264 User:Dnjhan 2019-11-12T03:29:27Z <p>Qkddud: Replaced content with &quot; JOSHUA HAN 2019&quot;</p> <hr /> <div><br /> JOSHUA HAN 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111263 User:Dnjhan 2019-11-12T03:27:01Z <p>Qkddud: </p> <hr /> <div>I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND 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KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> <br /> <br /> <br /> JOSHUA HAN 2019</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111261 User:Dnjhan 2019-11-12T03:23:45Z <p>Qkddud: </p> <hr /> <div>I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE 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/> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!<br /> I AM A CUTE FLUFFY UNICORN AND I LOVE HUGS AND KISSES!!!!!</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Qkddud&diff=111216 User:Qkddud 2019-11-11T23:02:50Z <p>Qkddud: </p> <hr /> <div>HI!<br /> <br /> <br /> YOUR A IDIOT!!!<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> GO AWAY NOWWWWWWW</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Qkddud&diff=111215 User:Qkddud 2019-11-11T23:01:45Z <p>Qkddud: Replaced content with &quot;HI! YOUR A IDIOT!!!&quot;</p> <hr /> <div>HI!<br /> <br /> <br /> YOUR A IDIOT!!!</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111214 User:Dnjhan 2019-11-11T22:58:33Z <p>Qkddud: </p> <hr /> <div>Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> <br /> <br /> <br /> <br /> JOSHUA IS AN IDIOT</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=User:Dnjhan&diff=111213 User:Dnjhan 2019-11-11T22:57:50Z <p>Qkddud: </p> <hr /> <div>Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.<br /> Idiot.</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_7&diff=109537 1991 AJHSME Problems/Problem 7 2019-08-29T23:47:20Z <p>Qkddud: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}&lt;/math&gt; is closest to<br /> <br /> &lt;math&gt;\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We can make the approximations<br /> &lt;cmath&gt;\begin{align*}<br /> 487,000 &amp;\approx 500,000 \\ <br /> 12,027,300 &amp;\approx 12,000,000 \\<br /> 9,621,001 &amp;\approx 10,000,000 \\<br /> 19,367 &amp;\approx 20,000.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Using these instead of the original numbers for an estimate, we have<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &amp;= \frac{500,000\times 22,000,000}{1000} \\<br /> &amp;= 500,000 \times 22,000 \\<br /> &amp;= 1.1\times 10^{10} \\<br /> &amp;\approx 10,000,000,000 \rightarrow \boxed{\text{D}}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> ~*If you typed this into a calculator or actually got the answer, it would be AROUND 10 billion.*~<br /> <br /> ~*Calypso*~<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1991|num-b=6|num-a=8}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_7&diff=109535 1991 AJHSME Problems/Problem 7 2019-08-29T23:00:32Z <p>Qkddud: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}&lt;/math&gt; is closest to<br /> <br /> &lt;math&gt;\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We can make the approximations<br /> &lt;cmath&gt;\begin{align*}<br /> 487,000 &amp;\approx 500,000 \\ <br /> 12,027,300 &amp;\approx 12,000,000 \\<br /> 9,621,001 &amp;\approx 10,000,000 \\<br /> 19,367 &amp;\approx 20,000.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Using these instead of the original numbers for an estimate, we have<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &amp;= \frac{500,000\times 22,000,000}{1000} \\<br /> &amp;= 500,000 \times 22,000 \\<br /> &amp;= 1.1\times 10^{10} \\<br /> &amp;\approx 10,000,000,000 \rightarrow \boxed{\text{D}}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> ~*If you typed this into a calculator or actually got the answer, it would be AROUND 10 million.*~<br /> <br /> ~*Calypso*~<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1991|num-b=6|num-a=8}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108373 2000 AMC 12 Problems/Problem 8 2019-08-07T06:09:27Z <p>Qkddud: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108372 2000 AMC 12 Problems/Problem 8 2019-08-07T06:09:18Z <p>Qkddud: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108371 2000 AMC 12 Problems/Problem 8 2019-08-07T06:09:10Z <p>Qkddud: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108370 2000 AMC 12 Problems/Problem 8 2019-08-07T06:09:01Z <p>Qkddud: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108369 2000 AMC 12 Problems/Problem 8 2019-08-07T06:08:49Z <p>Qkddud: /* Problem */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108368 2000 AMC 12 Problems/Problem 8 2019-08-07T06:08:35Z <p>Qkddud: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108367 2000 AMC 12 Problems/Problem 8 2019-08-07T06:08:18Z <p>Qkddud: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ======Solution 4======<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108366 2000 AMC 12 Problems/Problem 8 2019-08-07T06:07:33Z <p>Qkddud: </p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108365 2000 AMC 12 Problems/Problem 8 2019-08-07T06:07:13Z <p>Qkddud: /* Solution */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 5===<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108364 2000 AMC 12 Problems/Problem 8 2019-08-07T06:06:53Z <p>Qkddud: /* Solution */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_17&diff=108361 2006 AMC 10A Problems/Problem 17 2019-08-07T02:14:34Z <p>Qkddud: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> In [[rectangle]] &lt;math&gt;ADEH&lt;/math&gt;, points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; [[trisect]] &lt;math&gt;\overline{AD}&lt;/math&gt;, and points &lt;math&gt;G&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; trisect &lt;math&gt;\overline{HE}&lt;/math&gt;. In addition, &lt;math&gt;AH=AC=2&lt;/math&gt;, and &lt;math&gt;AD=3&lt;/math&gt;. What is the area of [[quadrilateral]] &lt;math&gt;WXYZ&lt;/math&gt; shown in the figure? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad&lt;/math&gt;<br /> <br /> &lt;!-- [[Image:2006_AMC10A-17.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> It is not difficult to see by [[symmetry]] that &lt;math&gt;WXYZ&lt;/math&gt; is a [[square]].<br /> &lt;!-- [[Image:2006_AMC10A-17a.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> MP(&quot;1&quot;,(A+B)/2,2*N); MP(&quot;2&quot;,(A+H)/2,plain.W); D(B--Z); MP(&quot;1&quot;,(B+Z)/2,plain.W); MP(&quot;\frac{\sqrt{2}}{2}&quot;,(W+Z)/2,plain.SE);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> Draw &lt;math&gt;\overline{BZ}&lt;/math&gt;. Clearly &lt;math&gt;BZ = \frac 12AH = 1&lt;/math&gt;. Then &lt;math&gt;\triangle BWZ&lt;/math&gt; is [[isosceles]], and is a &lt;math&gt;45-45-90 \triangle&lt;/math&gt;. Hence &lt;math&gt;WZ = \frac{1}{\sqrt{2}}&lt;/math&gt;, and &lt;math&gt;[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}&lt;/math&gt;.<br /> <br /> There are many different similar ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]].<br /> <br /> === Solution 2 ===<br /> &lt;!-- [[Image:2006_AMC10A-17b.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> D(B--D((A+H)/2)--G);D(C--D((E+D)/2)--F);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> <br /> Draw the lines as shown above, and count the squares. There are 12, so we have &lt;math&gt;\frac{2\cdot 3}{12} = \frac 12&lt;/math&gt;.<br /> ==Solution 3==<br /> We see that if we draw a line to &lt;math&gt;BZ&lt;/math&gt; it is half the width of the rectangle so that length would be &lt;math&gt;1&lt;/math&gt;, and the resulting triangle is a &lt;math&gt;45-45-90&lt;/math&gt; so using the Pythagorean Theorem we can get that each side is &lt;math&gt;\sqrt{\frac{1^2}{2}}&lt;/math&gt; so the area of the middle square would be &lt;math&gt;(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}&lt;/math&gt; which is our answer.<br /> ==Solution 4==<br /> <br /> Since &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are trisection points and &lt;math&gt;AC = 2&lt;/math&gt;, we see that &lt;math&gt;AD = 3&lt;/math&gt;. Also, &lt;math&gt;AC = AH&lt;/math&gt;, so triangle &lt;math&gt;ACH&lt;/math&gt; is a right isosceles triangle, i.e. &lt;math&gt;\angle ACH = \angle AHC = 45^\circ&lt;/math&gt;. By symmetry, triangles &lt;math&gt;AFH&lt;/math&gt;, &lt;math&gt;DEG&lt;/math&gt;, and &lt;math&gt;BED&lt;/math&gt; are also right isosceles triangles. Therefore, &lt;math&gt;\angle WAD = \angle WDA = 45^\circ&lt;/math&gt;, which means triangle &lt;math&gt;AWD&lt;/math&gt; is also a right isosceles triangle. Also, triangle &lt;math&gt;AXC&lt;/math&gt; is a right isosceles triangle.<br /> <br /> Then &lt;math&gt;AW = AD/\sqrt{2} = 3/\sqrt{2}&lt;/math&gt;, and &lt;math&gt;AX = AC/\sqrt{2} = 2/\sqrt{2}&lt;/math&gt;. Hence, &lt;math&gt;XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}&lt;/math&gt;.<br /> <br /> By symmetry, quadrilateral &lt;math&gt;WXYZ&lt;/math&gt; is a square, so its area is<br /> &lt;cmath&gt;XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\frac{1}{2}}.&lt;/cmath&gt;<br /> <br /> ~made by AoPS (somewhere) -put here by qkddud~<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_17&diff=108360 2006 AMC 10A Problems/Problem 17 2019-08-07T02:14:08Z <p>Qkddud: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> In [[rectangle]] &lt;math&gt;ADEH&lt;/math&gt;, points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; [[trisect]] &lt;math&gt;\overline{AD}&lt;/math&gt;, and points &lt;math&gt;G&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; trisect &lt;math&gt;\overline{HE}&lt;/math&gt;. In addition, &lt;math&gt;AH=AC=2&lt;/math&gt;, and &lt;math&gt;AD=3&lt;/math&gt;. What is the area of [[quadrilateral]] &lt;math&gt;WXYZ&lt;/math&gt; shown in the figure? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad&lt;/math&gt;<br /> <br /> &lt;!-- [[Image:2006_AMC10A-17.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> It is not difficult to see by [[symmetry]] that &lt;math&gt;WXYZ&lt;/math&gt; is a [[square]].<br /> &lt;!-- [[Image:2006_AMC10A-17a.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> MP(&quot;1&quot;,(A+B)/2,2*N); MP(&quot;2&quot;,(A+H)/2,plain.W); D(B--Z); MP(&quot;1&quot;,(B+Z)/2,plain.W); MP(&quot;\frac{\sqrt{2}}{2}&quot;,(W+Z)/2,plain.SE);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> Draw &lt;math&gt;\overline{BZ}&lt;/math&gt;. Clearly &lt;math&gt;BZ = \frac 12AH = 1&lt;/math&gt;. Then &lt;math&gt;\triangle BWZ&lt;/math&gt; is [[isosceles]], and is a &lt;math&gt;45-45-90 \triangle&lt;/math&gt;. Hence &lt;math&gt;WZ = \frac{1}{\sqrt{2}}&lt;/math&gt;, and &lt;math&gt;[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}&lt;/math&gt;.<br /> <br /> There are many different similar ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]].<br /> <br /> === Solution 2 ===<br /> &lt;!-- [[Image:2006_AMC10A-17b.png]] --&gt;<br /> &lt;asy&gt;<br /> size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair A,B,C,D,E,F,G,H,W,X,Y,Z;<br /> A=(0,2); B=(1,2); C=(2,2); D=(3,2);<br /> H=(0,0); G=(1,0); F=(2,0); E=(3,0);<br /> D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);<br /> D(A--F); D(B--E); D(D--G); D(C--H);<br /> Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);<br /> D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);<br /> D(B--D((A+H)/2)--G);D(C--D((E+D)/2)--F);<br /> D(A--D--E--H--cycle);<br /> &lt;/asy&gt;<br /> <br /> Draw the lines as shown above, and count the squares. There are 12, so we have &lt;math&gt;\frac{2\cdot 3}{12} = \frac 12&lt;/math&gt;.<br /> ==Solution 3==<br /> We see that if we draw a line to &lt;math&gt;BZ&lt;/math&gt; it is half the width of the rectangle so that length would be &lt;math&gt;1&lt;/math&gt;, and the resulting triangle is a &lt;math&gt;45-45-90&lt;/math&gt; so using the Pythagorean Theorem we can get that each side is &lt;math&gt;\sqrt{\frac{1^2}{2}}&lt;/math&gt; so the area of the middle square would be &lt;math&gt;(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}&lt;/math&gt; which is our answer.<br /> ==Solution 4==<br /> Solution:<br /> Since &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are trisection points and &lt;math&gt;AC = 2&lt;/math&gt;, we see that &lt;math&gt;AD = 3&lt;/math&gt;. Also, &lt;math&gt;AC = AH&lt;/math&gt;, so triangle &lt;math&gt;ACH&lt;/math&gt; is a right isosceles triangle, i.e. &lt;math&gt;\angle ACH = \angle AHC = 45^\circ&lt;/math&gt;. By symmetry, triangles &lt;math&gt;AFH&lt;/math&gt;, &lt;math&gt;DEG&lt;/math&gt;, and &lt;math&gt;BED&lt;/math&gt; are also right isosceles triangles. Therefore, &lt;math&gt;\angle WAD = \angle WDA = 45^\circ&lt;/math&gt;, which means triangle &lt;math&gt;AWD&lt;/math&gt; is also a right isosceles triangle. Also, triangle &lt;math&gt;AXC&lt;/math&gt; is a right isosceles triangle.<br /> <br /> Then &lt;math&gt;AW = AD/\sqrt{2} = 3/\sqrt{2}&lt;/math&gt;, and &lt;math&gt;AX = AC/\sqrt{2} = 2/\sqrt{2}&lt;/math&gt;. Hence, &lt;math&gt;XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}&lt;/math&gt;.<br /> <br /> By symmetry, quadrilateral &lt;math&gt;WXYZ&lt;/math&gt; is a square, so its area is<br /> &lt;cmath&gt;XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\frac{1}{2}}.&lt;/cmath&gt;<br /> <br /> ~made by AoPS HW-put here by qkddud~<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_20&diff=108345 2001 AMC 10 Problems/Problem 20 2019-08-07T01:22:29Z <p>Qkddud: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length &lt;math&gt; 2000 &lt;/math&gt;. What is the length of each side of the octagon?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})}<br /> \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(10,10)--(10,0)--cycle);<br /> draw((0,7)--(3,10));<br /> draw((7,10)--(10,7));<br /> draw((10,3)--(7,0));<br /> draw((3,0)--(0,3));<br /> label(&quot;$x$&quot;,(0,1),W);<br /> label(&quot;$x\sqrt{2}$&quot;,(1.5,1.5),NE);<br /> label(&quot;$2000-2x$&quot;,(5,0),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; 2000 - 2x = x\sqrt2 &lt;/math&gt;<br /> <br /> &lt;math&gt; 2000 = x(2 + \sqrt2) &lt;/math&gt;<br /> <br /> &lt;math&gt; x = \frac {2000}{2 + \sqrt2} =x = \frac {2000(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)}= \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2) &lt;/math&gt; <br /> <br /> &lt;math&gt; x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)} &lt;/math&gt;.<br /> <br /> <br /> ~edited by qkddud~<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_20&diff=108344 2001 AMC 10 Problems/Problem 20 2019-08-07T01:22:05Z <p>Qkddud: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length &lt;math&gt; 2000 &lt;/math&gt;. What is the length of each side of the octagon?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})}<br /> \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(10,10)--(10,0)--cycle);<br /> draw((0,7)--(3,10));<br /> draw((7,10)--(10,7));<br /> draw((10,3)--(7,0));<br /> draw((3,0)--(0,3));<br /> label(&quot;$x$&quot;,(0,1),W);<br /> label(&quot;$x\sqrt{2}$&quot;,(1.5,1.5),NE);<br /> label(&quot;$2000-2x$&quot;,(5,0),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; 2000 - 2x = x\sqrt2 &lt;/math&gt;<br /> <br /> &lt;math&gt; 2000 = x(2 + \sqrt2) &lt;/math&gt;<br /> <br /> &lt;math&gt; x = \frac {2000}{2 + \sqrt2} =x = \frac {2000(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)}= \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2) &lt;/math&gt; <br /> <br /> &lt;math&gt; x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)} &lt;/math&gt;.<br /> <br /> <br /> ~helped by qkddud<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_20&diff=108343 2001 AMC 10 Problems/Problem 20 2019-08-07T01:20:47Z <p>Qkddud: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length &lt;math&gt; 2000 &lt;/math&gt;. What is the length of each side of the octagon?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})}<br /> \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(10,10)--(10,0)--cycle);<br /> draw((0,7)--(3,10));<br /> draw((7,10)--(10,7));<br /> draw((10,3)--(7,0));<br /> draw((3,0)--(0,3));<br /> label(&quot;$x$&quot;,(0,1),W);<br /> label(&quot;$x\sqrt{2}$&quot;,(1.5,1.5),NE);<br /> label(&quot;$2000-2x$&quot;,(5,0),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; 2000 - 2x = x\sqrt2 &lt;/math&gt;<br /> <br /> &lt;math&gt; 2000 = x(2 + \sqrt2) &lt;/math&gt;<br /> <br /> &lt;math&gt; x = \frac {2000}{2 + \sqrt2} = \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2) &lt;/math&gt; <br /> <br /> &lt;math&gt; x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)} &lt;/math&gt;.<br /> <br /> <br /> ~helped by qkddud<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_24&diff=107354 2016 AMC 8 Problems/Problem 24 2019-07-04T22:48:17Z <p>Qkddud: /* Solution */</p> <hr /> <div>The digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are each used once to write a five-digit number &lt;math&gt;PQRST&lt;/math&gt;. The three-digit number &lt;math&gt;PQR&lt;/math&gt; is divisible by &lt;math&gt;4&lt;/math&gt;, the three-digit number &lt;math&gt;QRS&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt;, and the three-digit number &lt;math&gt;RST&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;. What is &lt;math&gt;P&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We see that since &lt;math&gt;QRS&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; must equal either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;, but it cannot equal &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;S=5&lt;/math&gt;. We notice that since &lt;math&gt;PQR&lt;/math&gt; must be even, &lt;math&gt;R&lt;/math&gt; must be either &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;. However, when &lt;math&gt;R=2&lt;/math&gt;, we see that &lt;math&gt;T \equiv 2 \pmod{3}&lt;/math&gt;, which cannot happen because &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; are already used up; so &lt;math&gt;R=4&lt;/math&gt;. This gives &lt;math&gt;T \equiv 3 \pmod{4}&lt;/math&gt;, meaning &lt;math&gt;T=3&lt;/math&gt;. Now, we see that &lt;math&gt;Q&lt;/math&gt; could be either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;, but &lt;math&gt;14&lt;/math&gt; is not divisible by &lt;math&gt;4&lt;/math&gt;, but &lt;math&gt;24&lt;/math&gt; is. This means that &lt;math&gt;R=4&lt;/math&gt; and &lt;math&gt;P=\boxed{\textbf{(A)}\ 1}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> We know that out of &lt;math&gt;PQRST&lt;/math&gt; &lt;math&gt;QRS&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt;. Therefore &lt;math&gt;S&lt;/math&gt; is obviously 5 because &lt;math&gt;QRS&lt;/math&gt; is divisible by 5. So we now have &lt;math&gt;PQR5T&lt;/math&gt; as our number. Next, lets move on to the second piece of information that was given to us. RST is divisible by 3. So, according to the divisibility of 3 rule the sum of &lt;math&gt;RST&lt;/math&gt; has to be a multiple of 3. The only 2 big enough is 9 and 12 and since 5 is already given. The possible sums of &lt;math&gt;RT&lt;/math&gt; is 4 and 7. So, the possible values for &lt;math&gt;R&lt;/math&gt; are 1,3,4,3 and the possible values of &lt;math&gt;T&lt;/math&gt; is 3,1,3,4. So, using this we can move on to the fact that &lt;math&gt;PQR&lt;/math&gt; is divisible by 4. So, using that we know that &lt;math&gt;R&lt;/math&gt; has to be even so 4 is the only possible value for &lt;math&gt;R&lt;/math&gt;. Using that we also know that 3 is the only possible value for 3. So, we know have &lt;math&gt;PQRST&lt;/math&gt; = &lt;math&gt;PQ453&lt;/math&gt; so the possible values are 1 and 2 for &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Using the divisibility rule of 4 we know that &lt;math&gt;QR&lt;/math&gt; has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for &lt;math&gt;P&lt;/math&gt; is 1. &lt;math&gt;P=\boxed{\textbf{(A)}\ 1}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=107353 2016 AMC 8 Problems/Problem 23 2019-07-04T22:44:41Z <p>Qkddud: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram:<br /> <br /> &lt;asy&gt;<br /> <br /> label(&quot;$D$&quot;, D, SE);<br /> label(&quot;$E$&quot;, E, N);<br /> &lt;/asy&gt;<br /> <br /> we see that &lt;math&gt;\triangle EAB&lt;/math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;math&gt;\overarc{EB}=m\angle EAB=60^\circ&lt;/math&gt;. Therefore, since it is an inscribed angle<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=107352 2016 AMC 8 Problems/Problem 23 2019-07-04T22:44:22Z <p>Qkddud: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram:<br /> <br /> &lt;asy<br /> <br /> label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, D, SE);<br /> label(&quot;&lt;math&gt;E&lt;/math&gt;&quot;, E, N);<br /> &lt;/asy&gt;<br /> <br /> we see that &lt;math&gt;\triangle EAB&lt;/math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;math&gt;\overarc{EB}=m\angle EAB=60^\circ&lt;/math&gt;. Therefore, since it is an inscribed angle<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_2&diff=107152 2016 AMC 8 Problems/Problem 2 2019-06-29T23:55:35Z <p>Qkddud: /* Solution */</p> <hr /> <div>In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;AD=8&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (6,8), NE);<br /> label(&quot;$D$&quot;, (0, 8), NW);<br /> label(&quot;$M$&quot;, (0, 4), W);<br /> label(&quot;$4$&quot;, (0, 2), W);<br /> label(&quot;$6$&quot;, (3, 0), S);&lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> <br /> Use the triangle area formula for triangles: &lt;math&gt;A = \frac{bh}{2},&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt; is the area, &lt;math&gt;b&lt;/math&gt; is the base, and &lt;math&gt;h&lt;/math&gt; is the height. This equation gives us &lt;math&gt;A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{\textbf{(A) } 12}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> A triangle with the same height and base as a rectangle is half of the rectangle's area. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Therefore, we get &lt;math&gt;\frac{48}{4} =\boxed{\textbf{(A) } 12}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_17&diff=107074 2018 AMC 8 Problems/Problem 17 2019-06-28T23:32:45Z <p>Qkddud: /* Solution */</p> <hr /> <div>==Problem 17==<br /> Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is &lt;math&gt;2&lt;/math&gt; miles, which is &lt;math&gt;10,560&lt;/math&gt; feet, and Bella covers &lt;math&gt;2 \tfrac{1}{2}&lt;/math&gt; feet with each step. How many steps will Bella take by the time she meets Ella?<br /> <br /> &lt;math&gt;\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since Ella rides 5 times as fast as Bella, Ella rides at a rate of &lt;math&gt;\frac{25}{2}&lt;/math&gt; or &lt;math&gt;12 \tfrac{1}{2}&lt;/math&gt;. Together, they move &lt;math&gt;15&lt;/math&gt; feet towards each other every unit. Dividing &lt;math&gt;10560&lt;/math&gt; by &lt;math&gt;15&lt;/math&gt; to find the number of steps Ella takes results in the answer of &lt;math&gt;\boxed{\textbf{(A) }704}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since Ella rides 5 times faster than Bella, the ratio of their speeds is 5:1. This means that Bella travels 1/6 of the way, and 1/6 of 10560 feet is 1760 feet. Bella also walks 2.5 feet in a step, and 1760 divided by 2.5 is 704.<br /> <br /> -UnstoppableGoddess<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_6&diff=107046 2018 AMC 8 Problems/Problem 6 2019-06-28T01:52:01Z <p>Qkddud: /* Solution 2 */</p> <hr /> <div>==Problem 6==<br /> On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?<br /> <br /> &lt;math&gt;\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is &lt;math&gt;r=\frac dt=\frac{10}{0.5}=20&lt;/math&gt;mph. His speed on the highway then is &lt;math&gt;60&lt;/math&gt;mph. He drives &lt;math&gt;50&lt;/math&gt; miles, so he also drives &lt;math&gt;50&lt;/math&gt; minutes. The total amount of minutes spent on his trip is &lt;math&gt;30+50\implies \boxed{\textbf{(C) }80}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since Anh drives 3 times as fast on the highway, it takes him 1/3 of the time to drive 10 miles on the highway than on the coastal road. 1/3 of 30 is 10, and since he drives 50 miles on the highway, we multiply 10 by 5 to get 50. This means it took him 50 minutes to drive on the highway, and if we add the 30 minutes it took for him to drive on the coastal road, we would get 80.<br /> <br /> -UnstoppableGoddess<br /> (helped by qkddud)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=5|num-a=7}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_6&diff=107045 2018 AMC 8 Problems/Problem 6 2019-06-28T01:51:41Z <p>Qkddud: /* Solution */</p> <hr /> <div>==Problem 6==<br /> On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?<br /> <br /> &lt;math&gt;\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is &lt;math&gt;r=\frac dt=\frac{10}{0.5}=20&lt;/math&gt;mph. His speed on the highway then is &lt;math&gt;60&lt;/math&gt;mph. He drives &lt;math&gt;50&lt;/math&gt; miles, so he also drives &lt;math&gt;50&lt;/math&gt; minutes. The total amount of minutes spent on his trip is &lt;math&gt;30+50\implies \boxed{\textbf{(C) }80}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since Anh drives 3 times as fast on the highway, it takes him 1/3 of the time to drive 10 miles on the highway than on the coastal road. 1/3 of 30 is 10, and since he drives 50 miles on the highway, we multiply 10 by 5 to get 50. This means it took him 50 minutes to drive on the highway, and if we add the 30 minutes it took for him to drive on the coastal road, we would get 80.<br /> <br /> -UnstoppableGoddess<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=5|num-a=7}}<br /> <br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=99505 2013 AMC 12B Problems/Problem 4 2018-12-17T08:13:38Z <p>Qkddud: /* Problem */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> ==Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;==<br /> <br /> ==Solution==<br /> ==Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;==<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=99504 2013 AMC 12B Problems/Problem 4 2018-12-17T08:13:23Z <p>Qkddud: /* Problem */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> ===Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;===<br /> <br /> ==Solution==<br /> ==Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;==<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=99503 2013 AMC 12B Problems/Problem 4 2018-12-17T08:13:00Z <p>Qkddud: /* Solution */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution==<br /> ==Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;==<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_8&diff=98872 2002 AMC 12A Problems/Problem 8 2018-11-21T23:23:13Z <p>Qkddud: /* Problem */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #8]] and [[2002 AMC 10A Problems|2002 AMC 10A #8]]}}<br /> <br /> <br /> ==Problem==<br /> Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let &lt;math&gt;B&lt;/math&gt; be the total area of the blue triangles, &lt;math&gt;W&lt;/math&gt; the total area of the white squares, and &lt;math&gt;P&lt;/math&gt; the area of the pink square. Which of the following is correct?<br /> <br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);<br /> fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,pink);<br /> path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;<br /> path divider=(-2,2)--(-3,3)--cycle;<br /> fill(onewhite,white);<br /> fill(rotate(90)*onewhite,white);<br /> fill(rotate(180)*onewhite,white);<br /> fill(rotate(270)*onewhite,white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ B = W \qquad \text{(B)}\ W = P \qquad \text{(C)}\ B = P \qquad \text{(D)}\ 3B = 2P \qquad \text{(E)}\ 2P = W&lt;/math&gt;<br /> <br /> ==Solution==<br /> The blue that's touching the center pink square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have &lt;math&gt;\boxed{B=W\Rightarrow \text{(A)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_8&diff=98871 2002 AMC 12A Problems/Problem 8 2018-11-21T23:22:32Z <p>Qkddud: /* Solution */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #8]] and [[2002 AMC 10A Problems|2002 AMC 10A #8]]}}<br /> <br /> <br /> ==Problem==<br /> Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let &lt;math&gt;B&lt;/math&gt; be the total area of the blue triangles, &lt;math&gt;W&lt;/math&gt; the total area of the white squares, and &lt;math&gt;P&lt;/math&gt; the area of the pink square. Which of the following is correct?<br /> <br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);<br /> fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,pink);<br /> path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;<br /> path divider=(-2,2)--(-3,3)--cycle;<br /> fill(onewhite,white);<br /> fill(rotate(90)*onewhite,white);<br /> fill(rotate(180)*onewhite,white);<br /> fill(rotate(270)*onewhite,white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = W&lt;/math&gt;<br /> <br /> ==Solution==<br /> The blue that's touching the center pink square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have &lt;math&gt;\boxed{B=W\Rightarrow \text{(A)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_8&diff=98869 2002 AMC 12A Problems/Problem 8 2018-11-21T23:21:33Z <p>Qkddud: /* Problem */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #8]] and [[2002 AMC 10A Problems|2002 AMC 10A #8]]}}<br /> <br /> <br /> ==Problem==<br /> Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let &lt;math&gt;B&lt;/math&gt; be the total area of the blue triangles, &lt;math&gt;W&lt;/math&gt; the total area of the white squares, and &lt;math&gt;P&lt;/math&gt; the area of the pink square. Which of the following is correct?<br /> <br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);<br /> fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,pink);<br /> path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;<br /> path divider=(-2,2)--(-3,3)--cycle;<br /> fill(onewhite,white);<br /> fill(rotate(90)*onewhite,white);<br /> fill(rotate(180)*onewhite,white);<br /> fill(rotate(270)*onewhite,white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = W&lt;/math&gt;<br /> <br /> ==Solution==<br /> The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have &lt;math&gt;\boxed{B=W\Rightarrow \text{(A)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2002|ab=A|num-b=7|num-a=9}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Qkddud https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_2&diff=98834 2002 AMC 10A Problems/Problem 2 2018-11-21T22:07:51Z <p>Qkddud: /* Solution */</p> <hr /> <div>==Problem==<br /> Given that a, b, and c are non-zero real numbers, define &lt;math&gt;(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}&lt;/math&gt;, find &lt;math&gt;(2, 12, 9)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt;(2, 12, 9)=\frac{2}{12}+\frac{12}{9}+\frac{9}{2}=\frac{1}{6}+\frac{4}{3}+\frac{9}{2}=\frac{1}{6}+\frac{8}{6}+\frac{27}{6}=\frac{36}{6}=6&lt;/math&gt;. Our answer is then &lt;math&gt;\boxed{\text{(C)}\ 6}&lt;/math&gt;.<br /> <br /> Alternate solution for the lazy: Without computing the answer exactly, we see that &lt;math&gt;2/12=\text{a little}&lt;/math&gt;, &lt;math&gt;12/9=\text{more than }1&lt;/math&gt;, and &lt;math&gt;9/2=4.5&lt;/math&gt;.<br /> The sum is &lt;math&gt;4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})&lt;/math&gt;, and as all the options are integers, the correct one is &lt;math&gt;6&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=A|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Qkddud