https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Virjoy2001&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-23T15:37:41Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_21&diff=141162 2011 AMC 10A Problems/Problem 21 2020-12-31T18:01:48Z <p>Virjoy2001: /* Problem */</p> <hr /> <div>== Problem ==<br /> Two counterfeit coins of equal weight are mixed with &lt;math&gt;8&lt;/math&gt; identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the &lt;math&gt;10&lt;/math&gt; coins. A second pair is selected at random without replacement from the remaining &lt;math&gt;8&lt;/math&gt; coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all &lt;math&gt;4&lt;/math&gt; selected coins are genuine?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Note that we are trying to find the conditional probability &lt;math&gt;P(A \vert B) = \frac{P(A \cap B)}{P(B)}&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt; is the &lt;math&gt;4&lt;/math&gt; coins being genuine and &lt;math&gt;B&lt;/math&gt; is the sum of the weight of the coins being equal. The only possibilities for &lt;math&gt;B&lt;/math&gt; are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) &lt;math&gt;(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)&lt;/math&gt;. We see that &lt;math&gt;A \cap B&lt;/math&gt; happens with probability &lt;math&gt;\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; happens with probability &lt;math&gt;\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}&lt;/math&gt;, hence &lt;math&gt;P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> If we pick &lt;math&gt;4&lt;/math&gt; indistinguishable real coins from the set of &lt;math&gt;8&lt;/math&gt; real coins, there are &lt;math&gt;\binom{8}{4}&lt;/math&gt; ways to pick the coins. If we then place the coins in four distinguishable slots on the scale, there are &lt;math&gt;4!&lt;/math&gt; ways to arrange them, giving &lt;math&gt;4!\cdot \binom{8}{4}&lt;/math&gt; ways to choose and place &lt;math&gt;8&lt;/math&gt; real coins. This gives &lt;math&gt;1680&lt;/math&gt; desirable combinations.<br /> <br /> If we pick &lt;math&gt;2&lt;/math&gt; real coins and &lt;math&gt;2&lt;/math&gt; fake coins, there are &lt;math&gt;\binom{8}{2}\binom{2}{2}&lt;/math&gt; ways to choose the coins. There are &lt;math&gt;4&lt;/math&gt; choices for the first slot on the left side of the scale. Whichever type of coin is placed in that first slot, there are &lt;math&gt;2&lt;/math&gt; choices for the second slot on the left side of the scale, since it must be of the opposite type of coin. There are &lt;math&gt;2&lt;/math&gt; choices for the first slot on the right side of the scale, and only &lt;math&gt;1&lt;/math&gt; choice for the last slot on the right side.<br /> <br /> Thus, there are &lt;math&gt;4\cdot 2\cdot 2\cdot 1 = 16&lt;/math&gt; ways to arrange the coins, and &lt;math&gt;\binom{8}{2}\binom{2}{2} = 28&lt;/math&gt; sets of possible coins, for a total of &lt;math&gt;16\cdot 28 = 448&lt;/math&gt; combinations that are legal, yet undesirable.<br /> <br /> The overall probability is thus &lt;math&gt;\frac{1680}{1680 + 448} = \boxed{\frac{15}{19} \ \mathbf{(D)}}&lt;/math&gt;.<br /> <br /> Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essence, this &quot;overcounts&quot; all numbers by a factor of &lt;math&gt;8&lt;/math&gt;, since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability.<br /> <br /> == Solution 3==<br /> <br /> Place the two coins from the first pair in a line followed by the two coins from the second pair followed by the six left-over coins. We can do that in &lt;math&gt;\binom{10}{2} = 45&lt;/math&gt; different ways.<br /> <br /> We need to exclude those cases where the weight shows a difference. There are two cases where a pair has both counterfeit coins and there are &lt;math&gt;4\cdot6&lt;/math&gt; cases where one counterfeit coin is chosen and one is in the left-over. That leaves &lt;math&gt;45-2-24=19&lt;/math&gt; cases.<br /> <br /> Of these, &lt;math&gt;\binom{6}{2}=15&lt;/math&gt; cases has both counterfeit coins in the left-over. The probability of having chosen four genuine coins therefore is &lt;math&gt;\boxed{\frac{15}{19} \ \mathbf{(D)}}&lt;/math&gt;.<br /> <br /> == Solution 4==<br /> WLOG, allow for all the coins to be distinguishable.<br /> We split this up into cases. Case &lt;math&gt;1&lt;/math&gt; being the weight of &lt;math&gt;2&lt;/math&gt; genuine coins together and Case &lt;math&gt;2&lt;/math&gt; being the weight of &lt;math&gt;1&lt;/math&gt; genuine coin and &lt;math&gt;1&lt;/math&gt; counterfeit coin.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: All Genuine coins chosen.<br /> This happens in &lt;math&gt;\frac{\binom{8}{2}\cdot\binom{6}{2}}{2}=210&lt;/math&gt; ways<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Genuine coin and Counterfeit coin both chosen.<br /> This happens in &lt;math&gt;8\cdot7=56&lt;/math&gt; ways.<br /> <br /> Hence, the answer is &lt;math&gt;\boxed{\frac{15}{19} \ \mathbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Video Solution by the Beauty of Math==<br /> https://www.youtube.com/watch?v=W5NlV2B_83U&amp;feature=push-u-sub&amp;attr_tag=XAAsHxmGZAHEdZgT%3A6<br /> <br /> ~icematrix<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_17&diff=139451 2002 AMC 12B Problems/Problem 17 2020-12-11T15:59:35Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #17]] and [[2002 AMC 10B Problems|2002 AMC 10B #21]]}}<br /> == Problem ==<br /> Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \text{Andy}<br /> \qquad\mathrm{(B)}\ \text{Beth}<br /> \qquad\mathrm{(C)}\ \text{Carlos}<br /> \qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.}<br /> \qquad\mathrm{(E)}\ \text{All\ three\ tie.}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We say Andy's lawn has an area of &lt;math&gt;x&lt;/math&gt;. Beth's lawn thus has an area of &lt;math&gt;\frac{x}{2}&lt;/math&gt;, and Carlos's lawn has an area of &lt;math&gt;\frac{x}{3}&lt;/math&gt;. <br /> <br /> We say Andy's lawn mower cuts at a speed of &lt;math&gt;y&lt;/math&gt;. Carlos's cuts at a speed of &lt;math&gt;\frac{y}{3}&lt;/math&gt;, and Beth's cuts at a speed &lt;math&gt;\frac{2y}{3}&lt;/math&gt;.<br /> <br /> Each person's lawn is cut at a time of &lt;math&gt;\frac{\text{area}}{\text{rate}}&lt;/math&gt;, so Andy's is cut in &lt;math&gt;\frac{x}{y}&lt;/math&gt; time, as is Carlos's. Beth's is cut in &lt;math&gt;\frac{3}{4}\times\frac{x}{y}&lt;/math&gt;, so the first one to finish is &lt;math&gt;\boxed{\mathrm{(B)}\ \text{Beth}}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> WLOG, we can set values of their lawns' areas and their owners' speeds. Let the area of Andy's lawn be &lt;math&gt;6&lt;/math&gt; units, Beth's lawn be &lt;math&gt;3&lt;/math&gt; units, and Carlos's lawn be &lt;math&gt;2&lt;/math&gt; units. Let Carlos's mowing area per hour (honestly the time you set won't matter) be &lt;math&gt;1&lt;/math&gt;, let Beth's mowing area per hour be &lt;math&gt;2&lt;/math&gt;, and let Andy's mowing area per hour be &lt;math&gt;3&lt;/math&gt;. Now, we can easily calculate their time by dividing their lawns' areas by their respective owners' speeds. Andy's time is &lt;math&gt;\frac{6}{3} = 2&lt;/math&gt; hours, Beth's time is &lt;math&gt;\frac{3}{2} = 1.5&lt;/math&gt; hours, and Carlos's time is &lt;math&gt;\frac{6}{3} = 2&lt;/math&gt; hours. Our answer is clearly &lt;math&gt;\boxed{\mathrm{(B)}\ \text{Beth}}&lt;/math&gt;.<br /> <br /> <br /> ~Solution by virjoy2001<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=B|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_17&diff=139432 2002 AMC 12B Problems/Problem 17 2020-12-11T02:23:56Z <p>Virjoy2001: /* Solution */</p> <hr /> <div>{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #17]] and [[2002 AMC 10B Problems|2002 AMC 10B #21]]}}<br /> == Problem ==<br /> Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \text{Andy}<br /> \qquad\mathrm{(B)}\ \text{Beth}<br /> \qquad\mathrm{(C)}\ \text{Carlos}<br /> \qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.}<br /> \qquad\mathrm{(E)}\ \text{All\ three\ tie.}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We say Andy's lawn has an area of &lt;math&gt;x&lt;/math&gt;. Beth's lawn thus has an area of &lt;math&gt;\frac{x}{2}&lt;/math&gt;, and Carlos's lawn has an area of &lt;math&gt;\frac{x}{3}&lt;/math&gt;. <br /> <br /> We say Andy's lawn mower cuts at a speed of &lt;math&gt;y&lt;/math&gt;. Carlos's cuts at a speed of &lt;math&gt;\frac{y}{3}&lt;/math&gt;, and Beth's cuts at a speed &lt;math&gt;\frac{2y}{3}&lt;/math&gt;.<br /> <br /> Each person's lawn is cut at a time of &lt;math&gt;\frac{\text{area}}{\text{rate}}&lt;/math&gt;, so Andy's is cut in &lt;math&gt;\frac{x}{y}&lt;/math&gt; time, as is Carlos's. Beth's is cut in &lt;math&gt;\frac{3}{4}\times\frac{x}{y}&lt;/math&gt;, so the first one to finish is &lt;math&gt;\boxed{\mathrm{(B)}\ \text{Beth}}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> WLOG, we can set values of their lawns' areas and their owners' speeds. Let the area of Andy's lawn be &lt;math&gt;6&lt;/math&gt; units, Beth's lawn be &lt;math&gt;3&lt;/math&gt; units, and Carlos's lawn be &lt;math&gt;2&lt;/math&gt; units. Let Carlos's mowing area per hour (honestly the time you set won't matter) be &lt;math&gt;1&lt;/math&gt;, let Beth's mowing area per hour be &lt;math&gt;2&lt;/math&gt;, and let Andy's mowing area per hour be &lt;math&gt;3&lt;/math&gt;. Now, we can easily calculate their time by dividing their lawns' areas by their respective owners' speeds. Andy's time is &lt;math&gt;\frac{6}{3} = 2&lt;/math&gt; hours, Beth's time is &lt;math&gt;\frac{3}{2} 1.5&lt;/math&gt; hours, and Carlos's time is &lt;math&gt;\frac{6}{3} = 2&lt;/math&gt; hours. Our answer is clearly &lt;math&gt;\boxed{\mathrm{(B)}\ \text{Beth}}&lt;/math&gt;.<br /> <br /> <br /> ~Solution by virjoy2001<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=B|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=139161 2001 AMC 10 Problems/Problem 19 2020-12-06T23:18:04Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> Let's use [[stars and bars]].<br /> Let the donuts be represented by &lt;math&gt; O &lt;/math&gt;s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us &lt;math&gt; 4 &lt;/math&gt; in all. The four donuts we want can be represented as &lt;math&gt; OOOO &lt;/math&gt;. Notice that we can add two &quot;dividers&quot; to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, &lt;math&gt; O|OO|O &lt;/math&gt; represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in &lt;math&gt; \binom{6}{2}=15 &lt;/math&gt; ways. Our answer is hence &lt;math&gt; \boxed{\textbf{(D)}\ 15} &lt;/math&gt;. Notice that this can be generalized to get the balls and urn (stars and bars) identity.<br /> <br /> ==Solution 2==<br /> <br /> Simple casework works here as well:<br /> Set up the following ratios:<br /> &lt;cmath&gt;4:0:0&lt;/cmath&gt;<br /> &lt;cmath&gt;3:1:0&lt;/cmath&gt;<br /> &lt;cmath&gt;2:2:0&lt;/cmath&gt;<br /> &lt;cmath&gt;2:1:1&lt;/cmath&gt;<br /> <br /> In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get &lt;math&gt;\frac{4!}{3!2!}&lt;/math&gt; for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is &lt;math&gt;3 \cdot 3+6 = \boxed{15}&lt;/math&gt;.<br /> <br /> Solution by IronicNinja<br /> <br /> Edit by virjoy2001 (Reason LaTeX mistake)<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_18&diff=138896 2017 AMC 10A Problems/Problem 18 2020-12-02T01:00:27Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Amelia has a coin that lands heads with probability &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and Blaine has a coin that lands on heads with probability &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;q-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 /> Let &lt;math&gt;P&lt;/math&gt; be the probability Amelia wins. Note that &lt;math&gt;P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P&lt;/math&gt;, as if she gets to her turn again, she is back where she started with probability of winning &lt;math&gt;P&lt;/math&gt;. The chance she wins on her first turn is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The chance she makes it to her turn again is a combination of her failing to win the first turn - &lt;math&gt;\frac{2}{3}&lt;/math&gt; and Blaine failing to win - &lt;math&gt;\frac{3}{5}&lt;/math&gt;. Multiplying gives us &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Thus,<br /> &lt;cmath&gt;P = \frac{1}{3} + \frac{2}{5}P&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;P = \frac{5}{9}&lt;/math&gt;, so the answer is &lt;math&gt;9-5=\boxed{\textbf{(D)}\ 4}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P&lt;/math&gt; be the probability Amelia wins. Note that &lt;math&gt;P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...&lt;/math&gt;This can be represented by an infinite geometric series: &lt;cmath&gt;P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.&lt;/cmath&gt; <br /> Therefore, &lt;math&gt;P = \frac{5}{9}&lt;/math&gt;, so the answer is &lt;math&gt;9-5 = \boxed{\textbf{(D)}\ 4}.&lt;/math&gt;<br /> <br /> Solution by ktong<br /> <br /> ~minor LaTeX edit by virjoy2001<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Probability Problems]]</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_15&diff=138861 2017 AMC 10A Problems/Problem 15 2020-12-01T02:31:46Z <p>Virjoy2001: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Chloe chooses a real number uniformly at random from the interval &lt;math&gt;[0, 2017]&lt;/math&gt;. Independently, Laurent chooses a real number uniformly at random from the interval &lt;math&gt;[0, 4034]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloe's number? <br /> &lt;math&gt; \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Denote &quot;winning&quot; to mean &quot;picking a greater number&quot;.<br /> There is a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance that Laurent chooses a number in the interval &lt;math&gt;[2017, 4034]&lt;/math&gt;. In this case, Chloé cannot possibly win, since the maximum number she can pick is &lt;math&gt;2017&lt;/math&gt;. Otherwise, if Laurent picks a number in the interval &lt;math&gt;[0, 2017]&lt;/math&gt;, with probability &lt;math&gt;\frac{1}{2}&lt;/math&gt;, then the two people are symmetric, and each has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of winning. Then, the total probability is &lt;math&gt;\frac{1}{2}\times1 + \frac{1}{2}\times\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}.&lt;/math&gt;<br /> <br /> ~Small grammar mistake corrected by virjoy2001 (missing period)<br /> <br /> ==Solution 2==<br /> We can use geometric probability to solve this.<br /> Suppose a point &lt;math&gt;(x,y)&lt;/math&gt; lies in the &lt;math&gt;xy&lt;/math&gt;-plane. Let &lt;math&gt;x&lt;/math&gt; be Chloe's number and &lt;math&gt;y&lt;/math&gt; be Laurent's number. Then obviously we want &lt;math&gt;y&gt;x&lt;/math&gt;, which basically gives us a region above a line. We know that Chloe's number is in the interval &lt;math&gt;[0,2017]&lt;/math&gt; and Laurent's number is in the interval &lt;math&gt;[0,4034]&lt;/math&gt;, so we can create a rectangle in the plane, whose length is &lt;math&gt;2017&lt;/math&gt; and whose width is &lt;math&gt;4034&lt;/math&gt;. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from &lt;math&gt;1&lt;/math&gt;. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line &lt;math&gt;y&gt;x&lt;/math&gt;, which is &lt;math&gt;\frac{2017 \cdot 2017}{2}&lt;/math&gt;. Instead of bashing this out we know that the rectangle has area &lt;math&gt;2017 \cdot 4034&lt;/math&gt;. So the probability that Laurent has a smaller number is &lt;math&gt;\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}&lt;/math&gt;. Simplifying the expression yields &lt;math&gt;\frac{1}{4}&lt;/math&gt; and so &lt;math&gt;1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Scale down by &lt;math&gt;2017&lt;/math&gt; to get that Chloe picks from &lt;math&gt;[0,1]&lt;/math&gt; and Laurent picks from &lt;math&gt;[0,2]&lt;/math&gt;. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of &lt;math&gt;0.5&lt;/math&gt;. Therefore, Laurent has a range of &lt;math&gt;0.5&lt;/math&gt; to &lt;math&gt;2&lt;/math&gt; to pick from, on average, which is a length of &lt;math&gt;2-0.5=1.5&lt;/math&gt; out of a total length of &lt;math&gt;2-0=2&lt;/math&gt;. Therefore, the probability is &lt;math&gt;1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/s4vnGlwwHHw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Probability Problems]]</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_15&diff=138860 2017 AMC 10A Problems/Problem 15 2020-12-01T02:31:14Z <p>Virjoy2001: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Chloe chooses a real number uniformly at random from the interval &lt;math&gt;[0, 2017]&lt;/math&gt;. Independently, Laurent chooses a real number uniformly at random from the interval &lt;math&gt;[0, 4034]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloe's number? <br /> &lt;math&gt; \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Denote &quot;winning&quot; to mean &quot;picking a greater number&quot;.<br /> There is a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance that Laurent chooses a number in the interval &lt;math&gt;[2017, 4034]&lt;/math&gt;. In this case, Chloé cannot possibly win, since the maximum number she can pick is &lt;math&gt;2017&lt;/math&gt;. Otherwise, if Laurent picks a number in the interval &lt;math&gt;[0, 2017]&lt;/math&gt;, with probability &lt;math&gt;\frac{1}{2}&lt;/math&gt;, then the two people are symmetric, and each has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of winning. Then, the total probability is &lt;math&gt;\frac{1}{2}\times1 + \frac{1}{2}\times\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}.&lt;/math&gt;<br /> <br /> ~Small grammar mistake corrected by virjoy2001 (missing period)<br /> <br /> ==Solution 2==<br /> We can use geometric probability to solve this.<br /> Suppose a point &lt;math&gt;(x,y)&lt;/math&gt; lies in the &lt;math&gt;xy&lt;/math&gt;-plane. Let &lt;math&gt;x&lt;/math&gt; be Chloe's number and &lt;math&gt;y&lt;/math&gt; be Laurent's number. Then obviously we want &lt;math&gt;y&gt;x&lt;/math&gt;, which basically gives us a region above a line. We know that Chloe's number is in the interval &lt;math&gt;[0,2017]&lt;/math&gt; and Laurent's number is in the interval &lt;math&gt;[0,4034]&lt;/math&gt;, so we can create a rectangle in the plane, whose length is &lt;math&gt;2017&lt;/math&gt; and whose width is &lt;math&gt;4034&lt;/math&gt;. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from &lt;math&gt;1&lt;/math&gt;. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line &lt;math&gt;y&gt;x&lt;/math&gt;, which is &lt;math&gt;\frac{2017 \cdot 2017}{2}&lt;/math&gt;. Instead of bashing this out we know that the rectangle has area &lt;math&gt;2017 \cdot 4034&lt;/math&gt;. So the probability that Laurent has a smaller number is &lt;math&gt;\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}&lt;/math&gt;. Simplifying the expression yields &lt;math&gt;\frac{1}{4}&lt;/math&gt; and so &lt;math&gt;1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Scale down by &lt;math&gt;2017&lt;/math&gt; to get that Chloe picks from &lt;math&gt;[0,1]&lt;/math&gt; and Laurent picks from &lt;math&gt;[0,2]&lt;/math&gt;. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of &lt;math&gt;0.5&lt;/math&gt;. Therefore, Laurent has a range of &lt;math&gt;0.5&lt;/math&gt; to &lt;math&gt;2&lt;/math&gt; to pick from, on average, which is a length of &lt;math&gt;2-0.5=1.5&lt;/math&gt; out of a total length of &lt;math&gt;2-0=2&lt;/math&gt;. Therefore, the probability is &lt;math&gt;1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}&lt;/math&gt;<br /> ==Video Solution==<br /> A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/s4vnGlwwHHw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Probability Problems]]</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=138859 2017 AMC 10A Problems/Problem 11 2020-12-01T02:28:59Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216\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 1==<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 &lt;math&gt;3&lt;/math&gt; 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;\frac{4(3^3)\pi}{3} = 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;, so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 20}.&lt;/math&gt;<br /> <br /> ~Minor edit by virjoy2001<br /> <br /> ==Diagram for Solution==<br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==Video Solution==<br /> https://youtu.be/s4vnGlwwHHw<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> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=138858 2017 AMC 10A Problems/Problem 11 2020-12-01T02:28:44Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216\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 1==<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 &lt;math&gt;3&lt;/math&gt; 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;\frac{4(3^3)\pi}{3} = 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;,so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 20}.&lt;/math&gt;<br /> <br /> ~Minor edit by virjoy2001<br /> <br /> ==Diagram for Solution==<br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==Video Solution==<br /> https://youtu.be/s4vnGlwwHHw<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> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=138857 2017 AMC 10A Problems/Problem 11 2020-12-01T02:28:11Z <p>Virjoy2001: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216\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 1==<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 &lt;math&gt;3&lt;/math&gt; 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;\frac{4(3^3)\pi}{3} = 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;,so our answer is \boxed{\textbf{(D)}\ 20}.$<br /> ~Minor edit by virjoy2001<br /> <br /> ==Diagram for Solution==<br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==Video Solution==<br /> https://youtu.be/s4vnGlwwHHw<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> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=138702 2018 AMC 10A Problems/Problem 19 2020-11-29T02:39:51Z <p>Virjoy2001: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the units digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}.&lt;/math&gt;<br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}.&lt;/math&gt; We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> <br /> ~very minor edits by virjoy2001<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 as we would be double counting if we did). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> ~vsamc<br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=138701 2018 AMC 10A Problems/Problem 19 2020-11-29T02:39:38Z <p>Virjoy2001: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the units digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}.&lt;/math&gt;<br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}.&lt;/math&gt; We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> ~very minor edits by virjoy2001<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 as we would be double counting if we did). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> ~vsamc<br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_4&diff=135988 2021 AMC 10A Problems/Problem 4 2020-10-28T17:58:28Z <p>Virjoy2001: Problem 4</p> <hr /> <div>==Problem==</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=1987_AJHSME_Problems/Problem_7&diff=135987 1987 AJHSME Problems/Problem 7 2020-10-28T17:51:58Z <p>Virjoy2001: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The large cube shown is made up of &lt;math&gt;27&lt;/math&gt; identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is <br /> <br /> &lt;math&gt;\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(36);<br /> draw((0,0)--(3,0)--(3,3)--(0,3)--cycle);<br /> draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3));<br /> draw((0,3)--(2.2,4.4)--(5.2,4.4));<br /> fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black);<br /> fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black);<br /> fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black);<br /> fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black);<br /> fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black);<br /> draw((1,3)--(3.2,4.4));<br /> draw((2,3)--(4.2,4.4));<br /> draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666));<br /> draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333));<br /> fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black);<br /> fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black);<br /> fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black);<br /> fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black);<br /> fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; painted unit squares on the half of the cube shown, so there are &lt;math&gt;10\cdot 2=20&lt;/math&gt; unit cubes with at least one face painted, thus our answer is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1987|num-b=6|num-a=8}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=1987_AJHSME_Problems/Problem_7&diff=135986 1987 AJHSME Problems/Problem 7 2020-10-28T17:51:09Z <p>Virjoy2001: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The large cube shown is made up of &lt;math&gt;27&lt;/math&gt; identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is <br /> <br /> &lt;math&gt;\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(36);<br /> draw((0,0)--(3,0)--(3,3)--(0,3)--cycle);<br /> draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3));<br /> draw((0,3)--(2.2,4.4)--(5.2,4.4));<br /> fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black);<br /> fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black);<br /> fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black);<br /> fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black);<br /> fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black);<br /> draw((1,3)--(3.2,4.4));<br /> draw((2,3)--(4.2,4.4));<br /> draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666));<br /> draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333));<br /> fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black);<br /> fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black);<br /> fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black);<br /> fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black);<br /> fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> Clearly no cube has more than one face painted. Therefore, the number of cubes with at least one face painted is equal to the number of painted unit squares.<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; painted unit squares on the half of the cube shown, so there are &lt;math&gt;10\cdot 2=20&lt;/math&gt; unit cubes with at least one face painted, thus our answer is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1987|num-b=6|num-a=8}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=109066 2018 AMC 8 Problems/Problem 22 2019-08-20T19:27:21Z <p>Virjoy2001: /* =Solution 1 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> =Solution 1=<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=109065 2018 AMC 8 Problems/Problem 22 2019-08-20T19:27:07Z <p>Virjoy2001: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1=<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=109064 2018 AMC 8 Problems/Problem 22 2019-08-20T19:26:53Z <p>Virjoy2001: /* See Also */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_19&diff=109053 2018 AMC 8 Problems/Problem 19 2019-08-20T17:35:48Z <p>Virjoy2001: /* Problem 19 */</p> <hr /> <div>=Problem 19=<br /> In a sign pyramid a cell gets a &quot;+&quot; if the two cells below it have the same sign, and it gets a &quot;-&quot; if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a &quot;+&quot; at the top of the pyramid?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle;<br /> draw(box); label(&quot;$+$&quot;,(0,0));<br /> draw(shift(1,0)*box); label(&quot;$-$&quot;,(1,0));<br /> draw(shift(2,0)*box); label(&quot;$+$&quot;,(2,0));<br /> draw(shift(3,0)*box); label(&quot;$-$&quot;,(3,0));<br /> draw(shift(0.5,0.4)*box); label(&quot;$-$&quot;,(0.5,0.4));<br /> draw(shift(1.5,0.4)*box); label(&quot;$-$&quot;,(1.5,0.4));<br /> draw(shift(2.5,0.4)*box); label(&quot;$-$&quot;,(2.5,0.4));<br /> draw(shift(1,0.8)*box); label(&quot;$+$&quot;,(1,0.8));<br /> draw(shift(2,0.8)*box); label(&quot;$+$&quot;,(2,0.8));<br /> draw(shift(1.5,1.2)*box); label(&quot;$+\$&quot;,(1.5,1.2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Instead of + and -, let us use 1 and 0, respectively. If we let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; be the values of the four cells on the bottom row, then the three cells on the next row are equal to &lt;math&gt;a+b&lt;/math&gt;, &lt;math&gt;b+c&lt;/math&gt;, and &lt;math&gt;c+d&lt;/math&gt; taken modulo 2 (this is exactly the same as finding &lt;math&gt;a \text{ XOR } b&lt;/math&gt;, and so on). The two cells on the next row are &lt;math&gt;a+2b+c&lt;/math&gt; and &lt;math&gt;b+2c+d&lt;/math&gt; taken modulo 2, and lastly, the cell on the top row gets &lt;math&gt;a+3b+3c+d \pmod{2}&lt;/math&gt;.<br /> <br /> Thus, we are looking for the number of assignments of 0's and 1's for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a+3b+3c+d \equiv 1 \pmod{2}&lt;/math&gt;, or in other words, is odd. As &lt;math&gt;3 \equiv 1 \pmod{2}&lt;/math&gt;, this is the same as finding the number of assignments such that &lt;math&gt;a+b+c+d \equiv 1 \pmod{2}&lt;/math&gt;. Notice that, no matter what &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are, this uniquely determines &lt;math&gt;d&lt;/math&gt;. There are &lt;math&gt;2^3 = 8&lt;/math&gt; ways to assign 0's and 1's arbitrarily to &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 8}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Each row is fully determined by its leftmost cell (the other cells in the row are restricted from the cells above). Since we are given the first row already, we still need to decide the other &lt;math&gt;3&lt;/math&gt; rows. The first cell in each row has only &lt;math&gt;2&lt;/math&gt; possibilities (+ and -), so we have &lt;math&gt;2^3=\boxed{\textbf{(C) }8}&lt;/math&gt; ways.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2018|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_15&diff=109052 2018 AMC 8 Problems/Problem 15 2019-08-20T17:34:18Z <p>Virjoy2001: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of &lt;math&gt;1&lt;/math&gt; square unit, then what is the area of the shaded region, in square units?<br /> <br /> &lt;asy&gt;<br /> size(4cm);<br /> filldraw(scale(2)*unitcircle,gray,black);<br /> filldraw(shift(-1,0)*unitcircle,white,black);<br /> filldraw(shift(1,0)*unitcircle,white,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let the radius of the large circle be &lt;math&gt;R&lt;/math&gt;. Then the radii of the smaller circles are &lt;math&gt;\frac R2&lt;/math&gt;. The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is &lt;math&gt;\frac 14&lt;/math&gt;. This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is &lt;math&gt;\boxed{\textbf{(D) } 1}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let the radius of the two smaller circles be &lt;math&gt;r&lt;/math&gt;. It follows that the area of one of the smaller circles is &lt;math&gt;{\pi}r^2&lt;/math&gt;. Thus, the area of the two inner circles combined would evaluate to &lt;math&gt;2{\pi}r^2&lt;/math&gt; which is &lt;math&gt;1&lt;/math&gt;. Since the radius of the bigger circle is two times that of the smaller circles(the diameter), the radius of the larger circle in terms of &lt;math&gt;r&lt;/math&gt; would be &lt;math&gt;2r&lt;/math&gt;. The area of the larger circle would come to &lt;math&gt;(2r)^2{\pi} = 4{\pi}r^2&lt;/math&gt;. Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have &lt;cmath&gt;4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.&lt;/cmath&gt;<br /> Therefore, the area of the shaded region is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;\boxed{D}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_15&diff=109051 2018 AMC 8 Problems/Problem 15 2019-08-20T17:33:33Z <p>Virjoy2001: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of &lt;math&gt;1&lt;/math&gt; square unit, then what is the area of the shaded region, in square units?<br /> <br /> &lt;asy&gt;<br /> size(4cm);<br /> filldraw(scale(2)*unitcircle,gray,black);<br /> filldraw(shift(-1,0)*unitcircle,white,black);<br /> filldraw(shift(1,0)*unitcircle,white,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}&lt;/math&gt;<br /> <br /> Solution 1<br /> <br /> Let the radius of the large circle be &lt;math&gt;R&lt;/math&gt;. Then the radii of the smaller circles are &lt;math&gt;\frac R2&lt;/math&gt;. The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is &lt;math&gt;\frac 14&lt;/math&gt;. This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is &lt;math&gt;\boxed{\textbf{(D) } 1}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let the radius of the two smaller circles be &lt;math&gt;r&lt;/math&gt;. It follows that the area of one of the smaller circles is &lt;math&gt;{\pi}r^2&lt;/math&gt;. Thus, the area of the two inner circles combined would evaluate to &lt;math&gt;2{\pi}r^2&lt;/math&gt; which is &lt;math&gt;1&lt;/math&gt;. Since the radius of the bigger circle is two times that of the smaller circles(the diameter), the radius of the larger circle in terms of &lt;math&gt;r&lt;/math&gt; would be &lt;math&gt;2r&lt;/math&gt;. The area of the larger circle would come to &lt;math&gt;(2r)^2{\pi} = 4{\pi}r^2&lt;/math&gt;. Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have &lt;cmath&gt;4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.&lt;/cmath&gt;<br /> Therefore, the area of the shaded region is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;\boxed{D}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> Virjoy2001 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_15&diff=109050 2018 AMC 8 Problems/Problem 15 2019-08-20T17:32:45Z <p>Virjoy2001: /* Solution */</p> <hr /> <div>==Problem 15==<br /> In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of &lt;math&gt;1&lt;/math&gt; square unit, then what is the area of the shaded region, in square units?<br /> <br /> &lt;asy&gt;<br /> size(4cm);<br /> filldraw(scale(2)*unitcircle,gray,black);<br /> filldraw(shift(-1,0)*unitcircle,white,black);<br /> filldraw(shift(1,0)*unitcircle,white,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let the radius of the large circle be &lt;math&gt;R&lt;/math&gt;. Then the radii of the smaller circles are &lt;math&gt;\frac R2&lt;/math&gt;. The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is &lt;math&gt;\frac 14&lt;/math&gt;. This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is &lt;math&gt;\boxed{\textbf{(D) } 1}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let the radius of the two smaller circles be &lt;math&gt;r&lt;/math&gt;. It follows that the area of one of the smaller circles is &lt;math&gt;{\pi}r^2&lt;/math&gt;. Thus, the area of the two inner circles combined would evaluate to &lt;math&gt;2{\pi}r^2&lt;/math&gt; which is &lt;math&gt;1&lt;/math&gt;. Since the radius of the bigger circle is two times that of the smaller circles(the diameter), the radius of the larger circle in terms of &lt;math&gt;r&lt;/math&gt; would be &lt;math&gt;2r&lt;/math&gt;. The area of the larger circle would come to &lt;math&gt;(2r)^2{\pi} = 4{\pi}r^2&lt;/math&gt;. Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have &lt;cmath&gt;4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.&lt;/cmath&gt;<br /> Therefore, the area of the shaded region is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;\boxed{D}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> Virjoy2001