https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Lcz&feedformat=atom AoPS Wiki - User contributions [en] 2020-01-19T02:22:38Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_25&diff=114436 2013 AMC 10A Problems/Problem 25 2020-01-07T16:14:55Z <p>Lcz: /* Solution 2 (Working Backwards) */</p> <hr /> <div>==Problem==<br /> <br /> All &lt;math&gt;20&lt;/math&gt; diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1 (Drawing)==<br /> <br /> If you draw a clear diagram like the one below, it is easy to see that there are &lt;math&gt;\boxed{\textbf{(A) }49}&lt;/math&gt; points.<br /> <br /> &lt;asy&gt;<br /> size(14cm);<br /> pathpen = brown + 1.337;<br /> // Initialize octagon<br /> pair[] A;<br /> for (int i=0; i&lt;8; ++i) {<br /> A[i] = dir(45*i);<br /> }<br /> D(CR( (0,0), 1));<br /> // Draw diagonals<br /> // choose pen colors<br /> pen[] colors;<br /> colors = orange + 1.337;<br /> colors = purple;<br /> colors = green;<br /> colors = black;<br /> for (int d=1; d&lt;=4; ++d) {<br /> pathpen = colors[d];<br /> for (int j=0; j&lt;8; ++j) {<br /> D(A[j]--A[(j+d) % 8]);<br /> }<br /> }<br /> pathpen = blue + 2;<br /> // Draw all the intersections<br /> pointpen = red + 7;<br /> for (int x1=0; x1&lt;8; ++x1) {<br /> for (int x2=x1+1; x2&lt;8; ++x2) {<br /> for (int x3=x2+1; x3&lt;8; ++x3) {<br /> for (int x4=x3+1; x4&lt;8; ++x4) {<br /> D(IP(A[x1]--A[x2], A[x3]--A[x4]));<br /> D(IP(A[x1]--A[x3], A[x4]--A[x2]));<br /> D(IP(A[x1]--A[x4], A[x2]--A[x3]));<br /> }<br /> }<br /> }<br /> }&lt;/asy&gt;<br /> <br /> ==Solution 2 (Working Backwards)==<br /> <br /> Let the number of intersections be &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;x\le \dbinom{8}{4} = 70&lt;/math&gt;, as every &lt;math&gt;4&lt;/math&gt; vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract &lt;math&gt;\dbinom{4}{2} -1 = 5&lt;/math&gt; from this count, &lt;math&gt;70-5 = 65&lt;/math&gt;. Note that diagonals like &lt;math&gt;\overline{AD}&lt;/math&gt;, &lt;math&gt;\overline{CG}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; all intersect at the same point. There are &lt;math&gt;8&lt;/math&gt; of this type with three diagonals intersecting at the same point, so we need to subtract &lt;math&gt;2&lt;/math&gt; of the &lt;math&gt;\dbinom{3}{2}&lt;/math&gt; (one is kept as the actual intersection). In the end, we obtain &lt;math&gt;65 - 16 = \boxed{\textbf{(A) }49}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Using the answer choices)==<br /> We know that the amount of intersection points is at most &lt;math&gt;\dbinom{8}{4} = 70&lt;/math&gt;, as in solution &lt;math&gt;2&lt;/math&gt;. There's probably going to be more than &lt;math&gt;5&lt;/math&gt; (to get &lt;math&gt;(B) 65&lt;/math&gt;), leading us to the only reasonable answer, &lt;math&gt;\boxed{\textbf{(A) }49}&lt;/math&gt;.<br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&diff=112062 2019 AMC 8 Problems/Problem 9 2019-11-22T22:42:16Z <p>Lcz: /* Solution 2 */</p> <hr /> <div>Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Using the formula for the volume of a cylinder, we get that the volume of Alex's can is &lt;math&gt;3^2\cdot12\cdot\pi&lt;/math&gt;, and that the volume of Felicia's can is &lt;math&gt;6^2\cdot6\cdot\pi&lt;/math&gt;. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; ~~SmileKat32<br /> <br /> ==Solution 2==<br /> <br /> The ratio of the numbers is &lt;math&gt;1/2&lt;/math&gt;. Looking closely at the formula &lt;math&gt;r^2 * h * \pi&lt;/math&gt;, we see that the &lt;math&gt;r * h * \pi&lt;/math&gt; will cancel, meaning that the ratio of them will be &lt;math&gt;1(2)/2(2) = \boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&diff=111973 2019 AMC 8 Problems/Problem 9 2019-11-21T20:40:12Z <p>Lcz: /* Solution 1 */</p> <hr /> <div>Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Using the formula for the volume of a cylinder, we get that the volume of Alex's can is &lt;math&gt;3^2\cdot12\cdot\pi&lt;/math&gt;, and that the volume of Felicia's can is &lt;math&gt;6^2\cdot6\cdot\pi&lt;/math&gt;. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; ~~SmileKat32<br /> <br /> ==Solution 2==<br /> <br /> The ratio of them is &lt;math&gt;1/2&lt;/math&gt;, meaning that the ratio of them will be &lt;math&gt;1(2)/2(2)&lt;cmath&gt;=&lt;/cmath&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=111972 2019 AMC 8 Problems/Problem 5 2019-11-21T20:34:07Z <p>Lcz: /* 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 is the only one left.<br /> <br /> -Lcz<br /> <br /> ==Solution 2==<br /> The tortoise walks at a constant rate, when the hare is resting, the distance will stay the same, and if the hare finds the tortoise already there, the constant line will reach the end of the distance first, meaning that the answer is &lt;math&gt;(B)&lt;/math&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=111971 2019 AMC 8 Problems/Problem 5 2019-11-21T20:31:10Z <p>Lcz: /* 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, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer is the only one left, &lt;math&gt;(B)&lt;/math&gt;<br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=111970 2019 AMC 8 Problems/Problem 5 2019-11-21T20:30:50Z <p>Lcz: /* Solution 1 */</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, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer is the only one left, &lt;math&gt;(B)&lt;/math&gt; QED<br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=111969 2019 AMC 8 Problems/Problem 5 2019-11-21T20:26:37Z <p>Lcz: /* Problem 5 */</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==<br /> <br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=111106 2011 AIME I Problems/Problem 4 2019-11-11T00:27:42Z <p>Lcz: /* Solution 1 */</p> <hr /> <div>== Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=125&lt;/math&gt;, &lt;math&gt;AC=117&lt;/math&gt; and &lt;math&gt;BC=120&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at point &lt;math&gt;L&lt;/math&gt;, and the angle bisector of angle &lt;math&gt;B&lt;/math&gt; intersects &lt;math&gt; \overline{AC} &lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt; \overline{BK}&lt;/math&gt; and &lt;math&gt; \overline{AL}&lt;/math&gt;, respectively. Find &lt;math&gt;MN&lt;/math&gt;.<br /> <br /> <br /> == Solution 1 == <br /> Extend &lt;math&gt;{CM}&lt;/math&gt; and &lt;math&gt;{CN}&lt;/math&gt; such that they intersect line &lt;math&gt;{AB}&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. <br /> Since &lt;math&gt;{BM}&lt;/math&gt; is the angle bisector of angle &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;{CM}&lt;/math&gt; is perpendicular to &lt;math&gt;{BM}&lt;/math&gt;, so &lt;math&gt;BP=BC=120&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. But &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{056}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> [There seem to be some mislabeled points going on here but the idea is sound.]<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;ABC&lt;/math&gt;. Now, since &lt;math&gt;AM \perp LC&lt;/math&gt; and &lt;math&gt;AN \perp KB&lt;/math&gt;, we have &lt;math&gt;AMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = AI&lt;/math&gt;. Since &lt;math&gt;\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK&lt;/math&gt;, we have that &lt;math&gt;MN = AI \cdot \cos \angle IAK&lt;/math&gt;. Letting &lt;math&gt;X&lt;/math&gt; be the point of contact of the incircle of &lt;math&gt;ABC&lt;/math&gt; with side &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;AX=MN&lt;/math&gt; thus &lt;math&gt;MN=\frac{117+120-125}{2}=\boxed{056}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) == <br /> Project &lt;math&gt;I&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; as &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;ID&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; are both in-radii of &lt;math&gt;\triangle ABC&lt;/math&gt; so we get right triangles with legs &lt;math&gt;r&lt;/math&gt; (the in-radius length) and &lt;math&gt;s - c = 56&lt;/math&gt;. Since &lt;math&gt;IC&lt;/math&gt; is the hypotenuse for the 4 triangles (&lt;math&gt;\triangle INC, \triangle IMC, \triangle IDC,&lt;/math&gt; and &lt;math&gt;\triangle IEC&lt;/math&gt;), &lt;math&gt;C, D, M, I, N, E&lt;/math&gt; are con-cyclic on a circle we shall denote as &lt;math&gt;\omega&lt;/math&gt; which is also the circumcircle of &lt;math&gt;\triangle CMN&lt;/math&gt; and &lt;math&gt;\triangle CDE&lt;/math&gt;. To find &lt;math&gt;MN&lt;/math&gt;, we can use the Law of Cosines on &lt;math&gt;\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;. Now, the circumradius &lt;math&gt;R&lt;/math&gt; can be found with Pythagorean Theorem with &lt;math&gt;\triangle CDI&lt;/math&gt; or &lt;math&gt;\triangle CEI&lt;/math&gt;: &lt;math&gt;r^2 + 56^2 = (2R)^2&lt;/math&gt;. To find &lt;math&gt;r&lt;/math&gt;, we can use the formula &lt;math&gt;rs = [ABC]&lt;/math&gt; and by Heron's, &lt;math&gt;[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}&lt;/math&gt;. To find &lt;math&gt;\angle MCN&lt;/math&gt;, we can find &lt;math&gt;\angle MIN&lt;/math&gt; since &lt;math&gt;\angle MCN = 180 - \angle MIN&lt;/math&gt;. &lt;math&gt;\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}&lt;/math&gt;. Thus, &lt;math&gt;\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}&lt;/math&gt; and since &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt;, we have &lt;math&gt;\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}&lt;/math&gt;. Plugging this into our Law of Cosines formula gives &lt;math&gt;MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})&lt;/math&gt;. To find &lt;math&gt;\cos{\angle C}&lt;/math&gt;, we use LoC on &lt;math&gt;\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}&lt;/math&gt;. Our formula now becomes &lt;math&gt;MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}&lt;/math&gt;. After simplifying, we get &lt;math&gt;MN^2 = 3136 \implies MN = \boxed{056}&lt;/math&gt;.<br /> <br /> --lucasxia01<br /> <br /> == Solution 4==<br /> <br /> Because &lt;math&gt;\angle CMI = \angle CNI = 90&lt;/math&gt;, &lt;math&gt;CMIN&lt;/math&gt; is cyclic. <br /> <br /> Ptolemy on CMIN:<br /> <br /> &lt;math&gt;CN*MI+CM*IN=CI*MN&lt;/math&gt;<br /> <br /> &lt;math&gt;CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN&lt;/math&gt;<br /> <br /> &lt;math&gt;MN = CI \sin \angle MCN&lt;/math&gt; by angle addition formula.<br /> <br /> &lt;math&gt;\angle MCN = 180 - \angle MIN = 90 - \angle BCI&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be where the incircle touches &lt;math&gt;BC&lt;/math&gt;, then &lt;math&gt;CI \cos \angle BCI = CH = \frac{a+b-c}{2}&lt;/math&gt;.<br /> &lt;math&gt;a=120, b=117, c=125&lt;/math&gt;, for a final answer of &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_8&diff=111084 2011 AIME II Problems/Problem 8 2019-11-10T16:47:20Z <p>Lcz: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;z_1,z_2,z_3,\dots,z_{12}&lt;/math&gt; be the 12 zeroes of the polynomial &lt;math&gt;z^{12}-2^{36}&lt;/math&gt;. For each &lt;math&gt;j&lt;/math&gt;, let &lt;math&gt;w_j&lt;/math&gt; be one of &lt;math&gt;z_j&lt;/math&gt; or &lt;math&gt;i z_j&lt;/math&gt;. Then the maximum possible value of the real part of &lt;math&gt;\displaystyle\sum_{j=1}^{12} w_j&lt;/math&gt; can be written as &lt;math&gt;m+\sqrt{n}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;center&gt;[[File:2011_AIME_II_-8.png‎]]&lt;/center&gt;<br /> <br /> The twelve dots above represent the &lt;math&gt;12&lt;/math&gt; roots of the equation &lt;math&gt;z^{12}-2^{36}=0&lt;/math&gt;. If we write &lt;math&gt;z=a+bi&lt;/math&gt;, then the real part of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;a&lt;/math&gt; and the real part of &lt;math&gt;iz&lt;/math&gt; is &lt;math&gt;-b&lt;/math&gt;. The blue dots represent those roots &lt;math&gt;z&lt;/math&gt; for which the real part of &lt;math&gt;z&lt;/math&gt; is greater than the real part of &lt;math&gt;iz&lt;/math&gt;, and the red dots represent those roots &lt;math&gt;z&lt;/math&gt; for which the real part of &lt;math&gt;iz&lt;/math&gt; is greater than the real part of &lt;math&gt;z&lt;/math&gt;. Now, the sum of the real parts of the blue dots is easily seen to be &lt;math&gt;8+16\cos\frac{\pi}{6}=8+8\sqrt{3}&lt;/math&gt; and the negative of the sum of the imaginary parts of the red dots is easily seen to also be &lt;math&gt;8+8\sqrt{3}&lt;/math&gt;. Hence our desired sum is &lt;math&gt;16+16\sqrt{3}=16+\sqrt{768}&lt;/math&gt;, giving the answer &lt;math&gt;\boxed{784}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Complex Number Problems]]<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102434 2019 AMC 10B Problems/Problem 11 2019-02-14T20:08:59Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the amount of marbles in each jar &lt;math&gt;x&lt;/math&gt;, because they are equivalent. Thus, &lt;math&gt;x/10&lt;/math&gt; is the amount of green marbles in &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;x/9&lt;/math&gt; is the amount of green marbles in &lt;math&gt;2&lt;/math&gt;. &lt;math&gt;x/9+x/10=19x/90&lt;/math&gt;, &lt;math&gt;19x/90=95&lt;/math&gt;, and &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. Because the &lt;math&gt;9/10&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;8/9&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;9x/10-8x/9=x/90&lt;/math&gt;, so there must be &lt;math&gt;5&lt;/math&gt; more marbles in jar &lt;math&gt;1&lt;/math&gt; than jar &lt;math&gt;2&lt;/math&gt;. The answer is &lt;math&gt;A&lt;/math&gt;<br /> <br /> (Edited by Lcz)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_20&diff=101987 2015 AMC 10B Problems/Problem 20 2019-02-13T13:45:08Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?<br /> &lt;math&gt;\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); <br /> label(&quot;2&quot;,(0,0,0),S);<br /> label(&quot;A&quot;,(1,0,0),W);<br /> label(&quot;B&quot;,(0,0,1),N);<br /> label(&quot;1&quot;,(1,0,1),NW);<br /> label(&quot;3&quot;,(1,1,0),S);<br /> label(&quot;C&quot;,(0,1,0),E);<br /> label(&quot;D&quot;,(1,1,1),SE);<br /> label(&quot;4&quot;,(0,1,1),NE);<br /> &lt;/asy&gt;<br /> <br /> We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled &lt;math&gt;A&lt;/math&gt;.<br /> <br /> If we define a &quot;move&quot; as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from &lt;math&gt;A&lt;/math&gt; (since Erin cannot crawl back to &lt;math&gt;A&lt;/math&gt;), this vertex must be &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Therefore, we now just need to count the number of paths from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt;. To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to &lt;math&gt;4&lt;/math&gt;, and 2 choices for which vertex Erin was at 2 moves before &lt;math&gt;4&lt;/math&gt;. All of Erin's previous moves were forced, so the total number of legal paths from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3 \cdot 2 = \boxed{\textbf{(A)}\; 6}&lt;/math&gt;.<br /> <br /> ==Solution 2 (3D Geo)==<br /> Lets say that this cube is an unit cube and the given corner is &lt;math&gt;(0,0,0)&lt;/math&gt;. Because Erin can not return back to its starting point, he can not be on &lt;math&gt;(0,0,1)&lt;/math&gt;, &lt;math&gt;(0,1,0)&lt;/math&gt;, or &lt;math&gt;(1,0,0)&lt;/math&gt;. He can not be on &lt;math&gt;(1,1,0)&lt;/math&gt; , &lt;math&gt;(1,0,1)&lt;/math&gt;, or &lt;math&gt;(0,1,1,)&lt;/math&gt; because after &lt;math&gt;7&lt;/math&gt; moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at &lt;math&gt;(1,1,1)&lt;/math&gt;. Now, we draw a net and see that there are &lt;math&gt;3&lt;/math&gt; choices for the first move, &lt;math&gt;2&lt;/math&gt; for the second, and the rest are forced. Thus the answer is &lt;math&gt;3*2 = \boxed{\textbf{(A)}\; 6}&lt;/math&gt;.<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=101916 2015 AMC 10B Problems/Problem 22 2019-02-12T16:48:02Z <p>Lcz: /* Solution 5 */</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. &lt;math&gt;FJ = FG&lt;/math&gt; since &lt;math&gt;\triangle FGJ&lt;/math&gt; is also isosceles. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle JHG \cong \triangle AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{JH}{AF+FJ}=\frac{FG}{FA}&lt;/cmath&gt;.<br /> &lt;cmath&gt;\frac{1}{1+FG} = \frac{FG}1&lt;/cmath&gt;<br /> &lt;cmath&gt;1 = FG^2 + FG&lt;/cmath&gt;<br /> &lt;cmath&gt;FG^2+FG-1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;FG = \frac{-1 \pm \sqrt{5} }{2}&lt;/cmath&gt;<br /> <br /> So, &lt;math&gt;FG=\frac{-1 + \sqrt{5}}{2}&lt;/math&gt; since &lt;math&gt;FG&lt;/math&gt; must be greater than 0.<br /> <br /> Notice that &lt;math&gt;CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 2 (Trigonometry)==<br /> Note that since &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon, all of its interior angles are &lt;math&gt;108^\circ&lt;/math&gt;. We can say that pentagon &lt;math&gt;FGHIJ&lt;/math&gt; is also regular by symmetry. So, all of the interior angles of &lt;math&gt;FGHIJ&lt;/math&gt; are &lt;math&gt;108^\circ&lt;/math&gt;. Now, we can angle chase and use trigonometry to get that &lt;math&gt;FG=2\sin18^\circ&lt;/math&gt;, &lt;math&gt;JH=2\sin18^\circ*(2\sin18^\circ+1)&lt;/math&gt;, and &lt;math&gt;DC=2\sin18^\circ*(2\sin18^\circ+2)&lt;/math&gt;. Adding these together, we get that &lt;math&gt;FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to &lt;math&gt;8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;, but we can find that this is closest to &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> When you first see this problem you can't help but see similar triangles. But this shape is filled with &lt;math&gt;36 - 72 - 72&lt;/math&gt; triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of &lt;math&gt;FG&lt;/math&gt; so we can apply similar triangles easily. To simplify the process lets write &lt;math&gt;FG&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;.<br /> <br /> First what is &lt;math&gt;JH&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;, also remember &lt;math&gt;AJ = 1+x&lt;/math&gt;: &lt;cmath&gt;\frac{JH}{1+x}=\frac{x}{1}&lt;/cmath&gt;&lt;math&gt;JH&lt;/math&gt; = &lt;math&gt;{x}^2+x&lt;/math&gt;<br /> <br /> Next, find &lt;math&gt;DC&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;, also remember &lt;math&gt;AD = 2+x&lt;/math&gt;: &lt;cmath&gt;\frac{DC}{2+x}=\frac{x}{1}&lt;/cmath&gt;&lt;math&gt;DC&lt;/math&gt; = &lt;math&gt;{x}^2+2x&lt;/math&gt;<br /> <br /> So adding all the &lt;math&gt;FG + JH + CD&lt;/math&gt; we get &lt;math&gt;2{x}^2+4x&lt;/math&gt;. Now we have to find out what x is. For this, we break out a bit of trig. Let's look at &lt;math&gt;\triangle AFG&lt;/math&gt; By the law of sines: <br /> &lt;cmath&gt;\frac{x}{sin(36)}=\frac{1}{sin(72)}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\frac{sin(36)}{sin(72)}&lt;/cmath&gt;<br /> <br /> Now by the double angle identities in trig. &lt;math&gt;sin(72) = 2sin(36)cos(36)&lt;/math&gt; <br /> substituting in &lt;cmath&gt;x=\frac{1}{2cos(36)}&lt;/cmath&gt; <br /> <br /> A good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that:<br /> &lt;math&gt;cos(36)&lt;/math&gt;= &lt;cmath&gt;\frac{1 + \sqrt{5}}{4}&lt;/cmath&gt; <br /> <br /> so now we know: <br /> &lt;cmath&gt;x = \frac{2}{1+\sqrt{5}} = \frac{-1+\sqrt{5}}{2}&lt;/cmath&gt;<br /> <br /> Substituting back into &lt;math&gt;2{x}^2+4x&lt;/math&gt; we get &lt;math&gt;FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 4 (Answer choices)==<br /> <br /> Notice that &lt;math&gt;A&lt;/math&gt; is trisected, meaning that &lt;cmath&gt;AG=BH=EJ=JH=1&lt;/cmath&gt;. <br /> Since &lt;math&gt;JH=1&lt;/math&gt;, and the other lines we are supposed to solve for do not look like they can contribute to the whole number value, it is likely that &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt; is our answer.<br /> <br /> Note: Only do this if low on time since there could potentially be a weird figure affecting the &lt;math&gt;1&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Because basically everything in pentagon stuff is in ratio of big Phi or small Phi, we use logic to find that the answer is big Phi &lt;math&gt;+ 1 +&lt;/math&gt; small Phi = &lt;math&gt;(\sqrt5+1)/2 + 1 + (\sqrt5-1)/2&lt;/math&gt; = &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=101915 2015 AMC 10B Problems/Problem 22 2019-02-12T16:47:52Z <p>Lcz: /* Solution 4 (Answer choices) */</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. &lt;math&gt;FJ = FG&lt;/math&gt; since &lt;math&gt;\triangle FGJ&lt;/math&gt; is also isosceles. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle JHG \cong \triangle AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{JH}{AF+FJ}=\frac{FG}{FA}&lt;/cmath&gt;.<br /> &lt;cmath&gt;\frac{1}{1+FG} = \frac{FG}1&lt;/cmath&gt;<br /> &lt;cmath&gt;1 = FG^2 + FG&lt;/cmath&gt;<br /> &lt;cmath&gt;FG^2+FG-1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;FG = \frac{-1 \pm \sqrt{5} }{2}&lt;/cmath&gt;<br /> <br /> So, &lt;math&gt;FG=\frac{-1 + \sqrt{5}}{2}&lt;/math&gt; since &lt;math&gt;FG&lt;/math&gt; must be greater than 0.<br /> <br /> Notice that &lt;math&gt;CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 2 (Trigonometry)==<br /> Note that since &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon, all of its interior angles are &lt;math&gt;108^\circ&lt;/math&gt;. We can say that pentagon &lt;math&gt;FGHIJ&lt;/math&gt; is also regular by symmetry. So, all of the interior angles of &lt;math&gt;FGHIJ&lt;/math&gt; are &lt;math&gt;108^\circ&lt;/math&gt;. Now, we can angle chase and use trigonometry to get that &lt;math&gt;FG=2\sin18^\circ&lt;/math&gt;, &lt;math&gt;JH=2\sin18^\circ*(2\sin18^\circ+1)&lt;/math&gt;, and &lt;math&gt;DC=2\sin18^\circ*(2\sin18^\circ+2)&lt;/math&gt;. Adding these together, we get that &lt;math&gt;FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to &lt;math&gt;8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;, but we can find that this is closest to &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> When you first see this problem you can't help but see similar triangles. But this shape is filled with &lt;math&gt;36 - 72 - 72&lt;/math&gt; triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of &lt;math&gt;FG&lt;/math&gt; so we can apply similar triangles easily. To simplify the process lets write &lt;math&gt;FG&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;.<br /> <br /> First what is &lt;math&gt;JH&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;, also remember &lt;math&gt;AJ = 1+x&lt;/math&gt;: &lt;cmath&gt;\frac{JH}{1+x}=\frac{x}{1}&lt;/cmath&gt;&lt;math&gt;JH&lt;/math&gt; = &lt;math&gt;{x}^2+x&lt;/math&gt;<br /> <br /> Next, find &lt;math&gt;DC&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;, also remember &lt;math&gt;AD = 2+x&lt;/math&gt;: &lt;cmath&gt;\frac{DC}{2+x}=\frac{x}{1}&lt;/cmath&gt;&lt;math&gt;DC&lt;/math&gt; = &lt;math&gt;{x}^2+2x&lt;/math&gt;<br /> <br /> So adding all the &lt;math&gt;FG + JH + CD&lt;/math&gt; we get &lt;math&gt;2{x}^2+4x&lt;/math&gt;. Now we have to find out what x is. For this, we break out a bit of trig. Let's look at &lt;math&gt;\triangle AFG&lt;/math&gt; By the law of sines: <br /> &lt;cmath&gt;\frac{x}{sin(36)}=\frac{1}{sin(72)}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\frac{sin(36)}{sin(72)}&lt;/cmath&gt;<br /> <br /> Now by the double angle identities in trig. &lt;math&gt;sin(72) = 2sin(36)cos(36)&lt;/math&gt; <br /> substituting in &lt;cmath&gt;x=\frac{1}{2cos(36)}&lt;/cmath&gt; <br /> <br /> A good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that:<br /> &lt;math&gt;cos(36)&lt;/math&gt;= &lt;cmath&gt;\frac{1 + \sqrt{5}}{4}&lt;/cmath&gt; <br /> <br /> so now we know: <br /> &lt;cmath&gt;x = \frac{2}{1+\sqrt{5}} = \frac{-1+\sqrt{5}}{2}&lt;/cmath&gt;<br /> <br /> Substituting back into &lt;math&gt;2{x}^2+4x&lt;/math&gt; we get &lt;math&gt;FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 4 (Answer choices)==<br /> <br /> Notice that &lt;math&gt;A&lt;/math&gt; is trisected, meaning that &lt;cmath&gt;AG=BH=EJ=JH=1&lt;/cmath&gt;. <br /> Since &lt;math&gt;JH=1&lt;/math&gt;, and the other lines we are supposed to solve for do not look like they can contribute to the whole number value, it is likely that &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt; is our answer.<br /> <br /> Note: Only do this if low on time since there could potentially be a weird figure affecting the &lt;math&gt;1&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Because basically everything in pentagon stuff is in ratio of big Phi or small Phi, we use logic to find that the answer is big Phi &lt;math&gt;+ 1 +&lt;/math&gt; small Phi = &lt;math&gt;(\sqrt5+1)/2 + 1 + (\sqrt5-1)/2&lt;/math&gt; = &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=101914 2017 AMC 10A Problems/Problem 11 2019-02-12T16:39:26Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all points in three-dimensional space within 3 units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume 216&lt;math&gt;\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within &lt;math&gt;3&lt;/math&gt; units of a point would be a sphere with radius &lt;math&gt;3&lt;/math&gt;. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal &lt;math&gt;216 \pi&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is equal to the length of our line segment.<br /> <br /> Solving, we find that &lt;math&gt;x = \boxed{\textbf{(D)}\ 20}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Because this is just a cylinder and &lt;math&gt;2&lt;/math&gt; &quot;half spheres&quot;, and the radius is &lt;math&gt;3&lt;/math&gt;, the volume of the &lt;math&gt;2&lt;/math&gt; half spheres is &lt;math&gt;4(3^3)/3 \pi = 36 \pi&lt;/math&gt;. Since we also know that the volume of this whole thing is &lt;math&gt;216 \pi&lt;/math&gt;, we do &lt;math&gt;216-36&lt;/math&gt; to get &lt;math&gt;180 \pi&lt;/math&gt; as the area of the cylinder. Thus the height is &lt;math&gt;180 \pi&lt;/math&gt; over the base, or &lt;math&gt;180 \pi/9\pi=20&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;<br /> <br /> ==Diagram for Solution==<br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_12&diff=101913 2017 AMC 10B Problems/Problem 12 2019-02-12T16:28:27Z <p>Lcz: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Elmer's new car gives &lt;math&gt;50\%&lt;/math&gt; percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is &lt;math&gt;20\%&lt;/math&gt; more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?<br /> <br /> &lt;math&gt;\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose that his old car runs at &lt;math&gt;x&lt;/math&gt; km per liter. Then his new car runs at &lt;math&gt;\frac{3}{2}x&lt;/math&gt; km per liter, or &lt;math&gt;x&lt;/math&gt; km per &lt;math&gt;\frac{2}{3}&lt;/math&gt; of a liter. Let the cost of the old car's fuel be &lt;math&gt;c&lt;/math&gt;, so the trip in the old car takes &lt;math&gt;xc&lt;/math&gt; dollars, while the trip in the new car takes &lt;math&gt;\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc&lt;/math&gt;. He saves &lt;math&gt;\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Because they do not give you a given amount of distance, we'll just make that distance &lt;math&gt;3x&lt;/math&gt; miles. Then, we find that the new car will use &lt;math&gt;2*1.2=2.4x&lt;/math&gt;. The old car will use &lt;math&gt;3x&lt;/math&gt;. Thus the answer is &lt;math&gt;(3-2.4)/3=.6/3=20/100= \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> -Lcz<br /> <br /> <br /> {{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_12&diff=101912 2017 AMC 10B Problems/Problem 12 2019-02-12T16:27:02Z <p>Lcz: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Elmer's new car gives &lt;math&gt;50\%&lt;/math&gt; percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is &lt;math&gt;20\%&lt;/math&gt; more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?<br /> <br /> &lt;math&gt;\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose that his old car runs at &lt;math&gt;x&lt;/math&gt; km per liter. Then his new car runs at &lt;math&gt;\frac{3}{2}x&lt;/math&gt; km per liter, or &lt;math&gt;x&lt;/math&gt; km per &lt;math&gt;\frac{2}{3}&lt;/math&gt; of a liter. Let the cost of the old car's fuel be &lt;math&gt;c&lt;/math&gt;, so the trip in the old car takes &lt;math&gt;xc&lt;/math&gt; dollars, while the trip in the new car takes &lt;math&gt;\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc&lt;/math&gt;. He saves &lt;math&gt;\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Because they do not give you a given amount of distance, we'll just make that distance &lt;math&gt;3x&lt;/math&gt; miles. Then, we find that the new car will use &lt;math&gt;2*1.2=2.4x&lt;/math&gt;. The old car will use &lt;math&gt;3x&lt;/math&gt;. Thus the answer is &lt;math&gt;(3-2.4)/3=.6/3=20/100= \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> <br /> {{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_12&diff=101911 2017 AMC 10B Problems/Problem 12 2019-02-12T16:26:46Z <p>Lcz: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Elmer's new car gives &lt;math&gt;50\%&lt;/math&gt; percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is &lt;math&gt;20\%&lt;/math&gt; more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?<br /> <br /> &lt;math&gt;\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose that his old car runs at &lt;math&gt;x&lt;/math&gt; km per liter. Then his new car runs at &lt;math&gt;\frac{3}{2}x&lt;/math&gt; km per liter, or &lt;math&gt;x&lt;/math&gt; km per &lt;math&gt;\frac{2}{3}&lt;/math&gt; of a liter. Let the cost of the old car's fuel be &lt;math&gt;c&lt;/math&gt;, so the trip in the old car takes &lt;math&gt;xc&lt;/math&gt; dollars, while the trip in the new car takes &lt;math&gt;\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc&lt;/math&gt;. He saves &lt;math&gt;\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> <br /> {{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_12&diff=101910 2017 AMC 10B Problems/Problem 12 2019-02-12T16:26:15Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> Elmer's new car gives &lt;math&gt;50\%&lt;/math&gt; percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is &lt;math&gt;20\%&lt;/math&gt; more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?<br /> <br /> &lt;math&gt;\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose that his old car runs at &lt;math&gt;x&lt;/math&gt; km per liter. Then his new car runs at &lt;math&gt;\frac{3}{2}x&lt;/math&gt; km per liter, or &lt;math&gt;x&lt;/math&gt; km per &lt;math&gt;\frac{2}{3}&lt;/math&gt; of a liter. Let the cost of the old car's fuel be &lt;math&gt;c&lt;/math&gt;, so the trip in the old car takes &lt;math&gt;xc&lt;/math&gt; dollars, while the trip in the new car takes &lt;math&gt;\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc&lt;/math&gt;. He saves &lt;math&gt;\frac{\frac{1}{5}xc}{xc} = \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.<br /> <br /> <br /> {{AMC10 box|year=2017|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}<br /> ==Solution 2==<br /> Because they do not give you a given amount of distance, we'll just make that distance &lt;math&gt;3x&lt;/math&gt; miles. Then, we find that the new car will use &lt;math&gt;2*1.2=2.4x&lt;/math&gt;. The old car will use &lt;math&gt;3x&lt;/math&gt;. Thus the answer is &lt;math&gt;(3-2.4)/3=.6/3=20/100= \boxed{\textbf{(A)}\ 20\%}&lt;/math&gt;.</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_14&diff=101519 2019 AMC 10A Problems/Problem 14 2019-02-09T22:35:49Z <p>Lcz: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #14]] and [[2019 AMC 12A Problems|2019 AMC 12A #8]]}}<br /> <br /> For a set of four distinct lines in a plane, there are exactly &lt;math&gt;N&lt;/math&gt; distinct points that lie on two or more of the lines. What is the sum of all possible values of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21&lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing &lt;math&gt;4&lt;/math&gt; lines, you see that the maximum points of intersection is &lt;math&gt;6&lt;/math&gt;. Then, we see that we can not make &lt;math&gt;5&lt;/math&gt; intersection points and can just make a quadrilateral, triangle, two T's, and one pair of coordinate axis to make &lt;math&gt;4,3,2,&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; points of intersection. Thus the answer is &lt;math&gt;1+2+3+4+6&lt;/math&gt; to get &lt;math&gt;16, B&lt;/math&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=13|num-a=15}}<br /> {{AMC12 box|year=2019|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_10&diff=101518 2019 AMC 10A Problems/Problem 10 2019-02-09T22:29:39Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A rectangular floor that is &lt;math&gt;10&lt;/math&gt; feet wide and &lt;math&gt;17&lt;/math&gt; feet long is tiled with &lt;math&gt;170&lt;/math&gt; one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?<br /> <br /> &lt;math&gt;\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28&lt;/math&gt;<br /> <br /> ==Solution==<br /> Because this is a &lt;math&gt;10&lt;/math&gt; by &lt;math&gt;17&lt;/math&gt; grid, the number of tiles that the bug visits is &lt;math&gt;10+17-1=26&lt;/math&gt;. The answer is &lt;math&gt;B&lt;/math&gt;<br /> <br /> Note: You will find that the general formula for this is &lt;math&gt;a+b-lcm(a,b)&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_9&diff=101517 2019 AMC 10A Problems/Problem 9 2019-02-09T22:26:57Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> What is the greatest three-digit positive integer &lt;math&gt;n&lt;/math&gt; for which the sum of the first &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;\underline{not}&lt;/math&gt; a divisor of the product of the first &lt;math&gt;n&lt;/math&gt; positive integers?<br /> <br /> &lt;math&gt;\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> Because the sum of &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;(n)(n+1)/2&lt;/math&gt;, and we want this to not be a divisor of the &lt;math&gt;n!&lt;/math&gt;, &lt;math&gt;n+1&lt;/math&gt; must be prime. The greatest three-digit integer that is prime is &lt;math&gt;997&lt;/math&gt;. Subtract &lt;math&gt;1&lt;/math&gt; to get &lt;math&gt;996 =&gt; B&lt;/math&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_9&diff=101516 2019 AMC 10A Problems/Problem 9 2019-02-09T22:26:44Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> What is the greatest three-digit positive integer &lt;math&gt;n&lt;/math&gt; for which the sum of the first &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;\underline{not}&lt;/math&gt; a divisor of the product of the first &lt;math&gt;n&lt;/math&gt; positive integers?<br /> <br /> &lt;math&gt;\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> Because the sum of &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;(n)(n+1)/2&lt;/math&gt;, and we want this to not be a divisor of the &lt;math&gt;n!&lt;/math&gt;, &lt;math&gt;n+1&lt;/math&gt; must be prime. The greatest three-digit integer that is prime is &lt;math&gt;997&lt;/math&gt;. Subtract &lt;math&gt;1&lt;/math&gt; to get &lt;math&gt;996 =&gt; B&lt;/math&gt;<br /> <br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_9&diff=101515 2019 AMC 10A Problems/Problem 9 2019-02-09T22:26:20Z <p>Lcz: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> What is the greatest three-digit positive integer &lt;math&gt;n&lt;/math&gt; for which the sum of the first &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;\underline{not}&lt;/math&gt; a divisor of the product of the first &lt;math&gt;n&lt;/math&gt; positive integers?<br /> <br /> &lt;math&gt;\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> Because the sum of &lt;math&gt;n&lt;/math&gt; positive integers is &lt;math&gt;(n)(n+1)/2&lt;/math&gt;, and we want this to not be a divisor of the &lt;math&gt;n!&lt;/math&gt;, &lt;math&gt;n+1&lt;/math&gt; must be prime. The greatest three-digit integer that is prime is &lt;math&gt;997&lt;/math&gt;. Subtract &lt;math&gt;1&lt;/math&gt; to get &lt;math&gt;996 =&gt; B&lt;/math&gt;<br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_12&diff=101502 2019 AMC 10A Problems/Problem 12 2019-02-09T22:16:08Z <p>Lcz: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #12]] and [[2019 AMC 12A Problems|2019 AMC 12A #7]]}}<br /> <br /> ==Problem==<br /> <br /> Melanie computes the mean &lt;math&gt;\mu&lt;/math&gt;, the median &lt;math&gt;M&lt;/math&gt;, and the modes of the &lt;math&gt;365&lt;/math&gt; values that are the dates in the months of &lt;math&gt;2019&lt;/math&gt;. Thus her data consist of &lt;math&gt;12&lt;/math&gt; &lt;math&gt;1\text{s}&lt;/math&gt;, &lt;math&gt;12&lt;/math&gt; &lt;math&gt;2\text{s}&lt;/math&gt;, . . . , &lt;math&gt;12&lt;/math&gt; &lt;math&gt;28\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;29\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;30\text{s}&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt; &lt;math&gt;31\text{s}&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the median of the modes. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A) } \mu &lt; d &lt; M \qquad\textbf{(B) } M &lt; d &lt; \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d &lt; M &lt; \mu \qquad\textbf{(E) } d &lt; \mu &lt; M&lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;365&lt;/math&gt; values which means that the median is the &lt;math&gt;183&lt;/math&gt;rd biggest value. Because there are &lt;math&gt;12&lt;/math&gt; of each of the first &lt;math&gt;28&lt;/math&gt; numbers, the median is &lt;math&gt;183/12=15.25&lt;/math&gt;.The mean of these numbers is a little under &lt;math&gt;16&lt;/math&gt; because there are only &lt;math&gt;7&lt;/math&gt; &lt;math&gt;31&lt;/math&gt;'s. The median of the modes is the median of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;28&lt;/math&gt;, yielding 14.5. Because &lt;math&gt;14.5&lt;15.25&lt;16&lt;/math&gt;, the answer is &lt;math&gt;B&lt;/math&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2019|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_12&diff=101501 2019 AMC 10A Problems/Problem 12 2019-02-09T22:15:44Z <p>Lcz: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #12]] and [[2019 AMC 12A Problems|2019 AMC 12A #7]]}}<br /> <br /> ==Problem==<br /> <br /> Melanie computes the mean &lt;math&gt;\mu&lt;/math&gt;, the median &lt;math&gt;M&lt;/math&gt;, and the modes of the &lt;math&gt;365&lt;/math&gt; values that are the dates in the months of &lt;math&gt;2019&lt;/math&gt;. Thus her data consist of &lt;math&gt;12&lt;/math&gt; &lt;math&gt;1\text{s}&lt;/math&gt;, &lt;math&gt;12&lt;/math&gt; &lt;math&gt;2\text{s}&lt;/math&gt;, . . . , &lt;math&gt;12&lt;/math&gt; &lt;math&gt;28\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;29\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;30\text{s}&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt; &lt;math&gt;31\text{s}&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the median of the modes. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A) } \mu &lt; d &lt; M \qquad\textbf{(B) } M &lt; d &lt; \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d &lt; M &lt; \mu \qquad\textbf{(E) } d &lt; \mu &lt; M&lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;365&lt;/math&gt; values which means that the median is the &lt;math&gt;183&lt;/math&gt;rd biggest value. Because there are &lt;math&gt;12&lt;/math&gt; of each of the first &lt;math&gt;28&lt;/math&gt; numbers, the median is &lt;math&gt;183/12=15.25&lt;/math&gt;.The mean of these numbers is a little under &lt;math&gt;16&lt;/math&gt; because there are only &lt;math&gt;7&lt;/math&gt; &lt;math&gt;31&lt;/math&gt;'s. The median of the modes is the median of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;28&lt;/math&gt;, yielding 14.5. Because &lt;math&gt;14.5&lt;15.25&lt;16&lt;/math&gt;, the answer is &lt;cmath&gt;B&lt;/cmath&gt;<br /> <br /> -Lcz<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2019|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Lcz https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_19&diff=100846 2012 AMC 10A Problems/Problem 19 2019-01-26T00:05:25Z <p>Lcz: /* Solution */</p> <hr /> <div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #13]] and [[2012 AMC 10A Problems|2012 AMC 10A #19]]}}<br /> <br /> ==Problem 19==<br /> <br /> Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let Paula work at a rate of &lt;math&gt;p&lt;/math&gt;, the two helpers work at a combined rate of &lt;math&gt;h&lt;/math&gt;, and the time it takes to eat lunch be &lt;math&gt;L&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:<br /> <br /> &lt;cmath&gt;(8-L)(p+h)=50&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(6.2-L)h=24&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(11.2-L)p=26&lt;/cmath&gt;<br /> <br /> With three equations and three variables, we need to find the value of &lt;math&gt;L&lt;/math&gt;.<br /> Adding the second and third equations together gives us &lt;math&gt;6.2h+11.2p-L(p+h)=50&lt;/math&gt;. Subtracting the first equation from this new one gives us &lt;math&gt;-1.8h+3.2p=0&lt;/math&gt;, so we get &lt;math&gt;h=\frac{16}{9}p&lt;/math&gt;. <br /> Plugging into the second equation:<br /> <br /> &lt;cmath&gt;(6.2-L)\frac{16}{9}p=24&lt;/cmath&gt;<br /> &lt;cmath&gt;(6.2-L)p=\frac{27}{2}&lt;/cmath&gt;<br /> <br /> We can then subtract this from the third equation:<br /> <br /> &lt;cmath&gt;5p=26-\frac{27}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;p=\frac{5}{2}&lt;/cmath&gt;<br /> Plugging &lt;math&gt;p&lt;/math&gt; into our third equation gives: &lt;cmath&gt;L=\frac{4}{5}&lt;/cmath&gt;<br /> <br /> Converting &lt;math&gt;L&lt;/math&gt; from hours to minutes gives us &lt;math&gt;L=48&lt;/math&gt; minutes, which is &lt;math&gt;\boxed{\textbf{(D)}\ 48}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Easy process of elimination)<br /> Because Paula worked from &lt;cmath&gt;8:00 A.M.&lt;/cmath&gt; to &lt;cmath&gt;7:12 P.M.&lt;/cmath&gt;, she worked for 11 hours and 12 minutes = 672 minutes. Since there is %100-50-24=%26 left, we get the equation 26a=672. Because 672 is 22 mod 26, the only answer that works is &lt;math&gt;\boxed{\textbf{(D)}\ 48}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2012|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Lcz