https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Starshooter11&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-25T14:23:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_21&diff=132765 2014 AMC 10A Problems/Problem 21 2020-08-30T22:31:59Z <p>Starshooter11: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that when &lt;math&gt;y=0&lt;/math&gt;, the &lt;math&gt;x&lt;/math&gt; values of the equations should be equal by the problem statement. We have that<br /> &lt;cmath&gt;0 = ax + 5 \implies x = -\dfrac{5}{a}&lt;/cmath&gt; &lt;cmath&gt;0 = 3x+b \implies x= -\dfrac{b}{3}&lt;/cmath&gt;<br /> Which means that &lt;cmath&gt;-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15&lt;/cmath&gt;<br /> The only possible pairs &lt;math&gt;(a,b)&lt;/math&gt; then are &lt;math&gt;(a,b) = (1,15), (3,5), (5,3), (15, 1)&lt;/math&gt;. These pairs give respective &lt;math&gt;x&lt;/math&gt;-values of &lt;math&gt;-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}&lt;/math&gt; which have a sum of &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Going off of Solution 1, for the first equation, notice that the value of &lt;math&gt;x&lt;/math&gt; cannot be less than &lt;math&gt;-5&lt;/math&gt;. We also know for the first equation that the values of &lt;math&gt;x&lt;/math&gt; have to be &lt;math&gt;5&lt;/math&gt; divided by something. Also, for the second equation, the values of &lt;math&gt;x&lt;/math&gt; can only be &lt;math&gt;-\frac13,-\frac23,-\frac33, \dots&lt;/math&gt;. Therefore, we see that, the only values common between the two sequences are &lt;math&gt;-1, -5, -\frac13,-\frac53&lt;/math&gt;, and adding them up, we get for our answer, &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Similar to the above solutions except we're using the equations<br /> &lt;math&gt;a = -\frac{-5}{x}&lt;/math&gt; <br /> and<br /> &lt;math&gt;b = -3x&lt;/math&gt;<br /> With this, we know that c has to be negative. Doing some math, we find &lt;math&gt;x&lt;/math&gt; to be <br /> &lt;math&gt;-1, -5, -\frac{1}{3}, &lt;/math&gt;and&lt;math&gt; -\frac{5}{3}&lt;/math&gt;<br /> <br /> Adding them up gives you our answer:&lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ~Starshooter11<br /> <br /> <br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2014amc10a/375<br /> <br /> ~ dolphin7<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_21&diff=132764 2014 AMC 10A Problems/Problem 21 2020-08-30T22:31:44Z <p>Starshooter11: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that when &lt;math&gt;y=0&lt;/math&gt;, the &lt;math&gt;x&lt;/math&gt; values of the equations should be equal by the problem statement. We have that<br /> &lt;cmath&gt;0 = ax + 5 \implies x = -\dfrac{5}{a}&lt;/cmath&gt; &lt;cmath&gt;0 = 3x+b \implies x= -\dfrac{b}{3}&lt;/cmath&gt;<br /> Which means that &lt;cmath&gt;-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15&lt;/cmath&gt;<br /> The only possible pairs &lt;math&gt;(a,b)&lt;/math&gt; then are &lt;math&gt;(a,b) = (1,15), (3,5), (5,3), (15, 1)&lt;/math&gt;. These pairs give respective &lt;math&gt;x&lt;/math&gt;-values of &lt;math&gt;-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}&lt;/math&gt; which have a sum of &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Going off of Solution 1, for the first equation, notice that the value of &lt;math&gt;x&lt;/math&gt; cannot be less than &lt;math&gt;-5&lt;/math&gt;. We also know for the first equation that the values of &lt;math&gt;x&lt;/math&gt; have to be &lt;math&gt;5&lt;/math&gt; divided by something. Also, for the second equation, the values of &lt;math&gt;x&lt;/math&gt; can only be &lt;math&gt;-\frac13,-\frac23,-\frac33, \dots&lt;/math&gt;. Therefore, we see that, the only values common between the two sequences are &lt;math&gt;-1, -5, -\frac13,-\frac53&lt;/math&gt;, and adding them up, we get for our answer, &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2014amc10a/375<br /> <br /> ~ dolphin7<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_21&diff=132763 2014 AMC 10A Problems/Problem 21 2020-08-30T22:31:18Z <p>Starshooter11: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that when &lt;math&gt;y=0&lt;/math&gt;, the &lt;math&gt;x&lt;/math&gt; values of the equations should be equal by the problem statement. We have that<br /> &lt;cmath&gt;0 = ax + 5 \implies x = -\dfrac{5}{a}&lt;/cmath&gt; &lt;cmath&gt;0 = 3x+b \implies x= -\dfrac{b}{3}&lt;/cmath&gt;<br /> Which means that &lt;cmath&gt;-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15&lt;/cmath&gt;<br /> The only possible pairs &lt;math&gt;(a,b)&lt;/math&gt; then are &lt;math&gt;(a,b) = (1,15), (3,5), (5,3), (15, 1)&lt;/math&gt;. These pairs give respective &lt;math&gt;x&lt;/math&gt;-values of &lt;math&gt;-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}&lt;/math&gt; which have a sum of &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Going off of Solution 1, for the first equation, notice that the value of &lt;math&gt;x&lt;/math&gt; cannot be less than &lt;math&gt;-5&lt;/math&gt;. We also know for the first equation that the values of &lt;math&gt;x&lt;/math&gt; have to be &lt;math&gt;5&lt;/math&gt; divided by something. Also, for the second equation, the values of &lt;math&gt;x&lt;/math&gt; can only be &lt;math&gt;-\frac13,-\frac23,-\frac33, \dots&lt;/math&gt;. Therefore, we see that, the only values common between the two sequences are &lt;math&gt;-1, -5, -\frac13,-\frac53&lt;/math&gt;, and adding them up, we get for our answer, &lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2014amc10a/375<br /> <br /> ~ dolphin7<br /> <br /> ==Solution 3==<br /> Similar to the above solutions except we're using the equations<br /> &lt;math&gt;a = -\frac{-5}{x}&lt;/math&gt; <br /> and<br /> &lt;math&gt;b = -3x&lt;/math&gt;<br /> With this, we know that c has to be negative. Doing some math, we find &lt;math&gt;x&lt;/math&gt; to be <br /> &lt;math&gt;-1, -5, -\frac{1}{3}, &lt;/math&gt;and&lt;math&gt; -\frac{5}{3}&lt;/math&gt;<br /> <br /> Adding them up gives you our answer:&lt;math&gt;\boxed{\textbf{(E)} \: -8}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_23&diff=132438 2018 AMC 10A Problems/Problem 23 2020-08-24T02:10:32Z <p>Starshooter11: /* Video Solution by Richard Rusczyk */</p> <hr /> <div>== Problem ==<br /> <br /> Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square &lt;math&gt;S&lt;/math&gt; so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from &lt;math&gt;S&lt;/math&gt; to the hypotenuse is 2 units. What fraction of the field is planted?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), N);<br /> label(&quot;$3$&quot;, (0,1.5), E);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the square have side length &lt;math&gt;x&lt;/math&gt;. Connect the upper-right vertex of square &lt;math&gt;S&lt;/math&gt; with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is &lt;math&gt;6&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{x}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{x}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Square &lt;math&gt;S&lt;/math&gt; has area &lt;math&gt;x^2&lt;/math&gt;, and the two thin triangle regions have area &lt;math&gt;\dfrac{x(3-x)}{2}&lt;/math&gt; and &lt;math&gt;\dfrac{x(4-x)}{2}&lt;/math&gt;. The final triangular region with the hypotenuse as its base and height &lt;math&gt;2&lt;/math&gt; has area &lt;math&gt;5&lt;/math&gt;. Thus, we have &lt;cmath&gt;x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6&lt;/cmath&gt;<br /> <br /> Solving gives &lt;math&gt;x=\dfrac{2}{7}&lt;/math&gt;. The area of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\dfrac{4}{49}&lt;/math&gt; and the desired ratio is &lt;math&gt;\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> Alternatively, once you get &lt;math&gt;x=\frac{2}{7}&lt;/math&gt;, you can avoid computation by noticing that there is a denominator of &lt;math&gt;7&lt;/math&gt;, so the answer must have a factor of &lt;math&gt;7&lt;/math&gt; in the denominator, which only &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt; does.<br /> <br /> ==Solution 2==<br /> Let the square have side length &lt;math&gt;s&lt;/math&gt;. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is &lt;math&gt;\frac{5}{3}(2)=\frac{10}{3}&lt;/math&gt;. Now, let's extend this larger similar right triangle to the left until it hits the side of length 3. Now, the length is &lt;math&gt;\frac{10}{3}+s&lt;/math&gt;, and using the ratios of the side lengths, the height is &lt;math&gt;\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}&lt;/math&gt;. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so &lt;cmath&gt;\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}&lt;/cmath&gt;<br /> <br /> Now comes the easy part: finding the ratio of the areas: &lt;math&gt;\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We use coordinate geometry. Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt; and the hypotenuse be the line &lt;math&gt;3x+4y = 12&lt;/math&gt; for &lt;math&gt;0\le x\le 3&lt;/math&gt;. Denote the position of &lt;math&gt;S&lt;/math&gt; as &lt;math&gt;(s,s)&lt;/math&gt;, and by the point to line distance formula, we know that &lt;cmath&gt;\frac{|3s+4s-12|}{5} = 2&lt;/cmath&gt; &lt;cmath&gt;\Rightarrow |7s-12| = 10&lt;/cmath&gt; Obviously &lt;math&gt;s&lt;\frac{22}{7}&lt;/math&gt;, so &lt;math&gt;s = \frac{2}{7}&lt;/math&gt;, and from here the rest of the solution follows to get &lt;math&gt;\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let the side length of the square be &lt;math&gt;x&lt;/math&gt;. First off, let us make a similar triangle with the segment of length &lt;math&gt;2&lt;/math&gt; and the top-right corner of &lt;math&gt;S&lt;/math&gt;. Therefore, the longest side of the smaller triangle must be &lt;math&gt;2 \cdot \frac54 = \frac52&lt;/math&gt;. We then do operations with that side in terms of &lt;math&gt;x&lt;/math&gt;. We subtract &lt;math&gt;x&lt;/math&gt; from the bottom, and &lt;math&gt;\frac{3x}{4}&lt;/math&gt; from the top. That gives us the equation of &lt;math&gt;3-\frac{7x}{4} = \frac{5}{2}&lt;/math&gt;. Solving, &lt;cmath&gt;12-7x = 10 \implies x = \frac{2}{7}.&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;x^2 = \frac{4}{49}&lt;/math&gt;, so the fraction of the triangle (area &lt;math&gt;6&lt;/math&gt;) covered by the square is &lt;math&gt;\frac{2}{147}&lt;/math&gt;. The answer is then &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> draw((0.3,0.3)--(3.6,0.3), dashed);<br /> draw((0.3,2.7)--(0.3,0.3), dashed);<br /> label(&quot;$S$&quot;, (-0.05,-0.05), NE);<br /> draw((0.3,0.3)--(1.41,1.91));<br /> draw((1.63,1.78)--(1.48,1.56));<br /> draw((1.28,1.70)--(1.48,1.56));<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$\frac{10}{3}$&quot;, (2,0.3), N);<br /> label(&quot;$\frac{5}{2}$&quot;, (0.3,1.5), E);<br /> label(&quot;$2$&quot;, (1,1.2), E);<br /> draw((3.6,0)--(3.6,0.3), dashed);<br /> draw((0,2.7)--(0.3,2.7), dashed);<br /> label(&quot;$\small{l}$&quot;, (3.6,0.15), W);<br /> label(&quot;$\small{l}$&quot;, (0.15,2.7), S);<br /> label(&quot;$\small{l}$&quot;, (0.3,0.15), E);<br /> label(&quot;$\small{l}$&quot;, (0.15,0.3), N);<br /> &lt;/asy&gt;<br /> <br /> On the diagram above, find two smaller triangles similar to the large one with side lengths &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;; consequently, the segments with length &lt;math&gt;\frac{5}{2}&lt;/math&gt; and &lt;math&gt;\frac{10}{3}&lt;/math&gt;.<br /> <br /> With &lt;math&gt;l&lt;/math&gt; being the side length of the square, we need to find an expression for &lt;math&gt;l&lt;/math&gt;. Using the hypotenuse, we can see that &lt;math&gt;\frac{3}{2}+\frac{8}{3}+\frac{5}{4}l+\frac{5}{3}l=5&lt;/math&gt;. Simplifying, &lt;math&gt;\frac{35}{12}l=\frac{5}{6}&lt;/math&gt;, or &lt;math&gt;l=2/7&lt;/math&gt;.<br /> <br /> A different calculation would yield &lt;math&gt;l+\frac{3}{4}l+\frac{5}{2}=3&lt;/math&gt;, so &lt;math&gt;\frac{7}{4}l=\frac{1}{2}&lt;/math&gt;. In other words, &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;, while to check, &lt;math&gt;l+\frac{4}{3}l+\frac{10}{3}=4&lt;/math&gt;. As such, &lt;math&gt;\frac{7}{3}l=\frac{2}{3}&lt;/math&gt;, and &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;.<br /> <br /> Finally, we get &lt;math&gt;A(\Square S)=l^2=\frac{4}{49}&lt;/math&gt;, to finish. As a proportion of the triangle with area &lt;math&gt;6&lt;/math&gt;, the answer would be &lt;math&gt;1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}&lt;/math&gt;, so &lt;math&gt;\boxed{\textit D}&lt;/math&gt; is correct.<br /> <br /> == Solution 6: Also Coordinate Geo ==<br /> Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt;, the point &lt;math&gt;(x,x)&lt;/math&gt; be the far edge of the unplanted square and the hypotenuse be the line &lt;math&gt;y=-\frac{3}{4}x+3&lt;/math&gt;. Since the line from &lt;math&gt;(x,x)&lt;/math&gt; to the hypotenuse is the shortest possible distance, we know this line, call it line &lt;math&gt;\l&lt;/math&gt;, is perpendicular to the hypotenuse and therefore has a slope of &lt;math&gt;\frac{4}{3}&lt;/math&gt;. <br /> <br /> Since we know &lt;math&gt;m=\frac{4}{3}&lt;/math&gt; , we can see that the line rises by &lt;math&gt;\frac{8}{5}&lt;/math&gt; and moves to the right by &lt;math&gt;\frac{6}{5}&lt;/math&gt; to meet the hypotenuse. (Let &lt;math&gt;2 = 5x&lt;/math&gt; and the rise be &lt;math&gt;4x&lt;/math&gt; and the run be &lt;math&gt;3x&lt;/math&gt; and then solve.) Therefore, line &lt;math&gt;\l&lt;/math&gt; intersects the hypotenuse at the point &lt;math&gt;(x+\frac{6}{5}, x+\frac{8}{5})&lt;/math&gt;. Plugging into the equation for the hypotenuse we have &lt;math&gt;x=\frac{2}{7}&lt;/math&gt; , and after a bit of computation we get &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://www.youtube.com/watch?v=p9npzq4FY_Y<br /> <br /> ~ dolphin7<br /> <br /> <br /> == Solution 7(slightly different from first solution):<br /> Same drawing as before:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{a}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{a}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Let's assign &lt;math&gt;a&lt;/math&gt; as the side length of box S. We then get each of the smaller triangle areas.<br /> The sum of all the triangular areas(not including the box) is equal to<br /> &lt;math&gt;\frac{(3-a)*a}{2}&lt;/math&gt; + &lt;math&gt;\frac{(4-a)*a}{2}&lt;/math&gt; + &lt;math&gt;\frac{(5*2)}{2}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{(3*4)}{2}&lt;/math&gt; - &lt;math&gt;a^2&lt;/math&gt;<br /> <br /> You can solve for &lt;math&gt;a=\frac{2}{7}&lt;/math&gt;<br /> <br /> Then, the ratio would be &lt;math&gt;1-\frac{\frac{2}{7}^2}{6}&lt;/math&gt; which is equal to &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> ~Starshooter11<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_23&diff=132437 2018 AMC 10A Problems/Problem 23 2020-08-24T02:07:42Z <p>Starshooter11: /* Solution 6: Also Coordinate Geo */</p> <hr /> <div>== Problem ==<br /> <br /> Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square &lt;math&gt;S&lt;/math&gt; so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from &lt;math&gt;S&lt;/math&gt; to the hypotenuse is 2 units. What fraction of the field is planted?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), N);<br /> label(&quot;$3$&quot;, (0,1.5), E);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the square have side length &lt;math&gt;x&lt;/math&gt;. Connect the upper-right vertex of square &lt;math&gt;S&lt;/math&gt; with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is &lt;math&gt;6&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{x}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{x}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Square &lt;math&gt;S&lt;/math&gt; has area &lt;math&gt;x^2&lt;/math&gt;, and the two thin triangle regions have area &lt;math&gt;\dfrac{x(3-x)}{2}&lt;/math&gt; and &lt;math&gt;\dfrac{x(4-x)}{2}&lt;/math&gt;. The final triangular region with the hypotenuse as its base and height &lt;math&gt;2&lt;/math&gt; has area &lt;math&gt;5&lt;/math&gt;. Thus, we have &lt;cmath&gt;x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6&lt;/cmath&gt;<br /> <br /> Solving gives &lt;math&gt;x=\dfrac{2}{7}&lt;/math&gt;. The area of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\dfrac{4}{49}&lt;/math&gt; and the desired ratio is &lt;math&gt;\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> Alternatively, once you get &lt;math&gt;x=\frac{2}{7}&lt;/math&gt;, you can avoid computation by noticing that there is a denominator of &lt;math&gt;7&lt;/math&gt;, so the answer must have a factor of &lt;math&gt;7&lt;/math&gt; in the denominator, which only &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt; does.<br /> <br /> ==Solution 2==<br /> Let the square have side length &lt;math&gt;s&lt;/math&gt;. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is &lt;math&gt;\frac{5}{3}(2)=\frac{10}{3}&lt;/math&gt;. Now, let's extend this larger similar right triangle to the left until it hits the side of length 3. Now, the length is &lt;math&gt;\frac{10}{3}+s&lt;/math&gt;, and using the ratios of the side lengths, the height is &lt;math&gt;\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}&lt;/math&gt;. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so &lt;cmath&gt;\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}&lt;/cmath&gt;<br /> <br /> Now comes the easy part: finding the ratio of the areas: &lt;math&gt;\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We use coordinate geometry. Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt; and the hypotenuse be the line &lt;math&gt;3x+4y = 12&lt;/math&gt; for &lt;math&gt;0\le x\le 3&lt;/math&gt;. Denote the position of &lt;math&gt;S&lt;/math&gt; as &lt;math&gt;(s,s)&lt;/math&gt;, and by the point to line distance formula, we know that &lt;cmath&gt;\frac{|3s+4s-12|}{5} = 2&lt;/cmath&gt; &lt;cmath&gt;\Rightarrow |7s-12| = 10&lt;/cmath&gt; Obviously &lt;math&gt;s&lt;\frac{22}{7}&lt;/math&gt;, so &lt;math&gt;s = \frac{2}{7}&lt;/math&gt;, and from here the rest of the solution follows to get &lt;math&gt;\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let the side length of the square be &lt;math&gt;x&lt;/math&gt;. First off, let us make a similar triangle with the segment of length &lt;math&gt;2&lt;/math&gt; and the top-right corner of &lt;math&gt;S&lt;/math&gt;. Therefore, the longest side of the smaller triangle must be &lt;math&gt;2 \cdot \frac54 = \frac52&lt;/math&gt;. We then do operations with that side in terms of &lt;math&gt;x&lt;/math&gt;. We subtract &lt;math&gt;x&lt;/math&gt; from the bottom, and &lt;math&gt;\frac{3x}{4}&lt;/math&gt; from the top. That gives us the equation of &lt;math&gt;3-\frac{7x}{4} = \frac{5}{2}&lt;/math&gt;. Solving, &lt;cmath&gt;12-7x = 10 \implies x = \frac{2}{7}.&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;x^2 = \frac{4}{49}&lt;/math&gt;, so the fraction of the triangle (area &lt;math&gt;6&lt;/math&gt;) covered by the square is &lt;math&gt;\frac{2}{147}&lt;/math&gt;. The answer is then &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> draw((0.3,0.3)--(3.6,0.3), dashed);<br /> draw((0.3,2.7)--(0.3,0.3), dashed);<br /> label(&quot;$S$&quot;, (-0.05,-0.05), NE);<br /> draw((0.3,0.3)--(1.41,1.91));<br /> draw((1.63,1.78)--(1.48,1.56));<br /> draw((1.28,1.70)--(1.48,1.56));<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$\frac{10}{3}$&quot;, (2,0.3), N);<br /> label(&quot;$\frac{5}{2}$&quot;, (0.3,1.5), E);<br /> label(&quot;$2$&quot;, (1,1.2), E);<br /> draw((3.6,0)--(3.6,0.3), dashed);<br /> draw((0,2.7)--(0.3,2.7), dashed);<br /> label(&quot;$\small{l}$&quot;, (3.6,0.15), W);<br /> label(&quot;$\small{l}$&quot;, (0.15,2.7), S);<br /> label(&quot;$\small{l}$&quot;, (0.3,0.15), E);<br /> label(&quot;$\small{l}$&quot;, (0.15,0.3), N);<br /> &lt;/asy&gt;<br /> <br /> On the diagram above, find two smaller triangles similar to the large one with side lengths &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;; consequently, the segments with length &lt;math&gt;\frac{5}{2}&lt;/math&gt; and &lt;math&gt;\frac{10}{3}&lt;/math&gt;.<br /> <br /> With &lt;math&gt;l&lt;/math&gt; being the side length of the square, we need to find an expression for &lt;math&gt;l&lt;/math&gt;. Using the hypotenuse, we can see that &lt;math&gt;\frac{3}{2}+\frac{8}{3}+\frac{5}{4}l+\frac{5}{3}l=5&lt;/math&gt;. Simplifying, &lt;math&gt;\frac{35}{12}l=\frac{5}{6}&lt;/math&gt;, or &lt;math&gt;l=2/7&lt;/math&gt;.<br /> <br /> A different calculation would yield &lt;math&gt;l+\frac{3}{4}l+\frac{5}{2}=3&lt;/math&gt;, so &lt;math&gt;\frac{7}{4}l=\frac{1}{2}&lt;/math&gt;. In other words, &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;, while to check, &lt;math&gt;l+\frac{4}{3}l+\frac{10}{3}=4&lt;/math&gt;. As such, &lt;math&gt;\frac{7}{3}l=\frac{2}{3}&lt;/math&gt;, and &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;.<br /> <br /> Finally, we get &lt;math&gt;A(\Square S)=l^2=\frac{4}{49}&lt;/math&gt;, to finish. As a proportion of the triangle with area &lt;math&gt;6&lt;/math&gt;, the answer would be &lt;math&gt;1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}&lt;/math&gt;, so &lt;math&gt;\boxed{\textit D}&lt;/math&gt; is correct.<br /> <br /> == Solution 6: Also Coordinate Geo ==<br /> Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt;, the point &lt;math&gt;(x,x)&lt;/math&gt; be the far edge of the unplanted square and the hypotenuse be the line &lt;math&gt;y=-\frac{3}{4}x+3&lt;/math&gt;. Since the line from &lt;math&gt;(x,x)&lt;/math&gt; to the hypotenuse is the shortest possible distance, we know this line, call it line &lt;math&gt;\l&lt;/math&gt;, is perpendicular to the hypotenuse and therefore has a slope of &lt;math&gt;\frac{4}{3}&lt;/math&gt;. <br /> <br /> Since we know &lt;math&gt;m=\frac{4}{3}&lt;/math&gt; , we can see that the line rises by &lt;math&gt;\frac{8}{5}&lt;/math&gt; and moves to the right by &lt;math&gt;\frac{6}{5}&lt;/math&gt; to meet the hypotenuse. (Let &lt;math&gt;2 = 5x&lt;/math&gt; and the rise be &lt;math&gt;4x&lt;/math&gt; and the run be &lt;math&gt;3x&lt;/math&gt; and then solve.) Therefore, line &lt;math&gt;\l&lt;/math&gt; intersects the hypotenuse at the point &lt;math&gt;(x+\frac{6}{5}, x+\frac{8}{5})&lt;/math&gt;. Plugging into the equation for the hypotenuse we have &lt;math&gt;x=\frac{2}{7}&lt;/math&gt; , and after a bit of computation we get &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://www.youtube.com/watch?v=p9npzq4FY_Y<br /> <br /> ~ dolphin7<br /> <br /> <br /> == Solution 7(slightly different from first solution):<br /> Same drawing as before:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{a}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{a}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Let's assign &lt;math&gt;a&lt;/math&gt; as the side length of box S. We then get each of the smaller triangle areas.<br /> The sum of all the triangular areas(not including the box) is equal to<br /> &lt;math&gt;\frac{(3-a)*a}{2}&lt;/math&gt; + &lt;math&gt;\frac{(4-a)*a}{2}&lt;/math&gt; + &lt;math&gt;\frac{(5*2)}{2}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{(3*4)}{2}&lt;/math&gt; - &lt;math&gt;a^2&lt;/math&gt;<br /> <br /> You can solve for &lt;math&gt;a=\frac{2}{7}&lt;/math&gt;<br /> <br /> Then, the ratio would be &lt;math&gt;1-\frac{\frac{2}{7}^2}{6}&lt;/math&gt; which is equal to &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_4&diff=124486 2019 AMC 10B Problems/Problem 4 2020-06-08T19:49:50Z <p>Starshooter11: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> All lines with equation &lt;math&gt;ax+by=c&lt;/math&gt; such that &lt;math&gt;a,b,c&lt;/math&gt; form an arithmetic progression pass through a common point. What are the coordinates of that point?<br /> <br /> &lt;math&gt;\textbf{(A) } (-1,2)<br /> \qquad\textbf{(B) } (0,1)<br /> \qquad\textbf{(C) } (1,-2)<br /> \qquad\textbf{(D) } (1,0)<br /> \qquad\textbf{(E) } (1,2)&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> If all lines satisfy the condition, then we can just plug in values for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; that form an arithmetic progression. Let's use &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt;, &lt;math&gt;c=3&lt;/math&gt;, and &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt;. Then the two lines we get are: &lt;cmath&gt;x+2y=3&lt;/cmath&gt; &lt;cmath&gt;x+3y=5&lt;/cmath&gt;<br /> Use elimination to deduce &lt;cmath&gt;y = 2&lt;/cmath&gt; and plug this into one of the previous line equations. We get &lt;cmath&gt;x+4 = 3 \Rightarrow x=-1&lt;/cmath&gt; Thus the common point is &lt;math&gt;\boxed{\textbf{(A) } (-1,2)}&lt;/math&gt;.<br /> <br /> ~IronicNinja<br /> <br /> ===Solution 2===<br /> We know that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; form an arithmetic progression, so if the common difference is &lt;math&gt;d&lt;/math&gt;, we can say &lt;math&gt;a,b,c = a, a+d, a+2d.&lt;/math&gt; Now we have &lt;math&gt;ax+ (a+d)y = a+2d&lt;/math&gt;, and expanding gives &lt;math&gt;ax + ay + dy = a + 2d.&lt;/math&gt; Factoring gives &lt;math&gt;a(x+y-1)+d(y-2) = 0&lt;/math&gt;. Since this must always be true (regardless of the values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;), we must have &lt;math&gt;x+y-1 = 0&lt;/math&gt; and &lt;math&gt;y-2 = 0&lt;/math&gt;, so &lt;math&gt;x,y = -1, 2,&lt;/math&gt; and the common point is &lt;math&gt;\boxed{\textbf{(A) } (-1,2)}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> We use process of elimination. &lt;math&gt;\textbf{B}&lt;/math&gt; doesn't necessarily work because &lt;math&gt;b = c&lt;/math&gt; isn't always true. &lt;math&gt;\textbf{C, D, E}&lt;/math&gt; also doesn't necessarily work because the x-value is &lt;math&gt;1&lt;/math&gt;, but the y-value is an integer. So by process of elimination, &lt;math&gt;\boxed{\textbf{(A) } (-1, 2)}&lt;/math&gt; is our answer. ~Baolan<br /> <br /> ==Solution 4==<br /> We know that in &lt;math&gt;ax + by = c&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are in an arithmetic progression. We can simplify any arithmetic progression to be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;.<br /> <br /> For example, the progression &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt; can be rewritten as &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt; by going back by one value. We can then divide all 3 numbers by 2 which gives us &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;.<br /> <br /> Now, we substitute &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; with &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; respectively. This gives us<br /> <br /> &lt;math&gt;y = 2&lt;/math&gt; and &lt;math&gt;-x = 1&lt;/math&gt; which can be written as &lt;math&gt;x = -1&lt;/math&gt;. The only point of intersection is &lt;math&gt;(-1,2)&lt;/math&gt;. So, our answer is<br /> <br /> &lt;math&gt;\boxed{\textbf{(A) } (-1, 2)}&lt;/math&gt;.<br /> ~Starshooter11<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=124482 2019 AMC 10B Problems/Problem 11 2020-06-08T19:40:51Z <p>Starshooter11: /* Solution 2(Completely Solve) */</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 &lt;math&gt;1&lt;/math&gt; the ratio of blue to green marbles is &lt;math&gt;9:1&lt;/math&gt;, and the ratio of blue to green marbles in Jar &lt;math&gt;2&lt;/math&gt; is &lt;math&gt;8:1&lt;/math&gt;. There are &lt;math&gt;95&lt;/math&gt; green marbles in all. How many more blue marbles are in Jar &lt;math&gt;1&lt;/math&gt; than in Jar &lt;math&gt;2&lt;/math&gt;?<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 number of marbles in each jar &lt;math&gt;x&lt;/math&gt; (because the problem specifies that they each contain the same number). Thus, &lt;math&gt;\frac{x}{10}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{x}{9}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;2&lt;/math&gt;. Since &lt;math&gt;\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}&lt;/math&gt;, we have &lt;math&gt;\frac{19x}{90}=95&lt;/math&gt;, so there are &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. <br /> <br /> Because &lt;math&gt;\frac{9x}{10}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{8x}{9}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5&lt;/math&gt; more marbles in Jar &lt;math&gt;1&lt;/math&gt; than Jar &lt;math&gt;2&lt;/math&gt;. This means the answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> <br /> ==Solution 2(Completely Solve)==<br /> Let &lt;math&gt;b_1&lt;/math&gt;, &lt;math&gt;g_1&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;g_2&lt;/math&gt;, represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the <br /> the amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations, <br /> &lt;math&gt;\frac{b_1}{g_1} = \frac{9}{1}&lt;/math&gt;, <br /> &lt;math&gt;\frac{b_2}{g_2} = \frac{8}{1}&lt;/math&gt;, <br /> &lt;math&gt;g_1 + g_2 =95&lt;/math&gt;, and <br /> &lt;math&gt;b_1 + g_1 = b_2 + g_2&lt;/math&gt;.<br /> Since &lt;math&gt;b_1 = 9g_1&lt;/math&gt; and &lt;math&gt;b_2 = 8g_2&lt;/math&gt;, we substitue that in to obtain &lt;math&gt;10g_1 = 9g_2&lt;/math&gt;. <br /> Coupled with our third equation, we find that &lt;math&gt;g_1 = 45&lt;/math&gt;, and that &lt;math&gt;g_2 = 50&lt;/math&gt;. We now use this information to find &lt;math&gt;b_1 = 405&lt;/math&gt;<br /> and &lt;math&gt;b_2 = 400&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;b_1 - b_2 = 5&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> ~Binderclips1<br /> <br /> ~LaTeX fixed by Starshooter11<br /> <br /> ==Video Solution==<br /> https://youtu.be/mXvetCMMzpU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=124481 2019 AMC 10B Problems/Problem 11 2020-06-08T19:38:24Z <p>Starshooter11: /* Solution 2(Completely Solve) */</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 &lt;math&gt;1&lt;/math&gt; the ratio of blue to green marbles is &lt;math&gt;9:1&lt;/math&gt;, and the ratio of blue to green marbles in Jar &lt;math&gt;2&lt;/math&gt; is &lt;math&gt;8:1&lt;/math&gt;. There are &lt;math&gt;95&lt;/math&gt; green marbles in all. How many more blue marbles are in Jar &lt;math&gt;1&lt;/math&gt; than in Jar &lt;math&gt;2&lt;/math&gt;?<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 number of marbles in each jar &lt;math&gt;x&lt;/math&gt; (because the problem specifies that they each contain the same number). Thus, &lt;math&gt;\frac{x}{10}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{x}{9}&lt;/math&gt; is the number of green marbles in Jar &lt;math&gt;2&lt;/math&gt;. Since &lt;math&gt;\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}&lt;/math&gt;, we have &lt;math&gt;\frac{19x}{90}=95&lt;/math&gt;, so there are &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. <br /> <br /> Because &lt;math&gt;\frac{9x}{10}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{8x}{9}&lt;/math&gt; is the number of blue marbles in Jar &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5&lt;/math&gt; more marbles in Jar &lt;math&gt;1&lt;/math&gt; than Jar &lt;math&gt;2&lt;/math&gt;. This means the answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> <br /> ==Solution 2(Completely Solve)==<br /> Let &lt;math&gt;b_1&lt;/math&gt;, &lt;math&gt;g_1&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;g_2&lt;/math&gt;, represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the <br /> the amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations, <br /> &lt;math&gt;\frac{b_1}{g_1} = \frac{9}{1}&lt;/math&gt;, <br /> &lt;math&gt;\frac{b_2}{g_2} = \frac{8}{1}&lt;/math&gt;, <br /> &lt;math&gt;g_1 + g_2 =95&lt;/math&gt;, and <br /> &lt;math&gt;b_1 + g_1 = b_2 + g_2&lt;/math&gt;.<br /> Since &lt;math&gt;b_1 = 9g_1&lt;/math&gt; and &lt;math&gt;b_2 = 8g_2&lt;/math&gt;, we substitue that in to obtain &lt;math&gt;10g_1 = 9g_2&lt;/math&gt;. <br /> Coupled with our third equation, we find that &lt;math&gt;g_1 = 45&lt;/math&gt;, and that &lt;math&gt;g_2 = 50&lt;/math&gt;. We now use this information to find &lt;math&gt;b_1 = 405&lt;/math&gt;<br /> and &lt;math&gt;b_2 = 400&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;b_1 - b_2 = 5&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(A) } 5}&lt;/math&gt;.<br /> ~Binderclips1<br /> ~edited by Starshooter11<br /> <br /> ==Video Solution==<br /> https://youtu.be/mXvetCMMzpU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_8&diff=124478 2019 AMC 10B Problems/Problem 8 2020-06-08T19:32:33Z <p>Starshooter11: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length &lt;math&gt;2&lt;/math&gt; and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> pen white = gray(1);<br /> pen gray = gray(0.5);<br /> draw((0,0)--(2sqrt(3),0)--(2sqrt(3),2sqrt(3))--(0,2sqrt(3))--cycle);<br /> fill((0,0)--(2sqrt(3),0)--(2sqrt(3),2sqrt(3))--(0,2sqrt(3))--cycle, gray);<br /> draw((sqrt(3)-1,0)--(sqrt(3),sqrt(3))--(sqrt(3)+1,0)--cycle);<br /> fill((sqrt(3)-1,0)--(sqrt(3),sqrt(3))--(sqrt(3)+1,0)--cycle, white);<br /> draw((sqrt(3)-1,2sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3)+1,2sqrt(3))--cycle);<br /> fill((sqrt(3)-1,2sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3)+1,2sqrt(3))--cycle, white);<br /> draw((0,sqrt(3)-1)--(sqrt(3),sqrt(3))--(0,sqrt(3)+1)--cycle);<br /> fill((0,sqrt(3)-1)--(sqrt(3),sqrt(3))--(0,sqrt(3)+1)--cycle, white);<br /> draw((2sqrt(3),sqrt(3)-1)--(sqrt(3),sqrt(3))--(2sqrt(3),sqrt(3)+1)--cycle);<br /> fill((2sqrt(3),sqrt(3)-1)--(sqrt(3),sqrt(3))--(2sqrt(3),sqrt(3)+1)--cycle, white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 12 - 4\sqrt{3} \qquad \textbf{(C) } 3\sqrt{3}\qquad \textbf{(D) } 4\sqrt{3} \qquad \textbf{(E) } 16 - 4\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We notice that the square can be split into &lt;math&gt;4&lt;/math&gt; congruent smaller squares, with the altitude of the equilateral triangle being the side of this smaller square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (which has already been split in half). #storm<br /> <br /> When we split an equilateral triangle in half, we get two &lt;math&gt;30^{\circ}-60^{\circ}-90^{\circ}&lt;/math&gt; triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is &lt;math&gt;\sqrt{3}&lt;/math&gt;. We can then compute the area of the two triangles as &lt;math&gt;2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}&lt;/math&gt;.<br /> <br /> The area of the each small squares is the square of the side length, i.e. &lt;math&gt;\left(\sqrt{3}\right)^2 = 3&lt;/math&gt;. Therefore, the area of the shaded region in each of the four squares is &lt;math&gt;3 - \sqrt{3}&lt;/math&gt;.<br /> <br /> Since there are &lt;math&gt;4&lt;/math&gt; of these squares, we multiply this by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;4\left(3 - \sqrt{3}\right) = \boxed{\textbf{(B) } 12 - 4\sqrt{3}}&lt;/math&gt; as our answer.<br /> <br /> ==Solution 2==<br /> We can see that the side length of the square is &lt;math&gt;2\sqrt{3}&lt;/math&gt; by considering the altitude of the equilateral triangle as in Solution 1. Using the Pythagorean Theorem, the diagonal of the square is thus &lt;math&gt;\sqrt{12+12}=\sqrt{24}=2\sqrt{6}&lt;/math&gt;. Because of this, the height of one of the four shaded kites is &lt;math&gt;\sqrt{6}&lt;/math&gt;. Now, we just need to find the length of that kite. By the Pythagorean Theorem again, this length is &lt;math&gt;\frac{2\sqrt{3} - 2}{2} \times \sqrt{2} = \sqrt{3} - 1 = \sqrt{6} - \sqrt{2}&lt;/math&gt;. Now using &lt;math&gt;\text{area} = \frac{1}{2} \cdot \text{length} \cdot \text{width}&lt;/math&gt;, the area of one of the four kites is &lt;math&gt;2 \sqrt{6} \times \left(\sqrt{6}-\sqrt{2}\right) = 12 - 2\sqrt{12} = \boxed{\textbf{(B) } 12 - 4\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Based on the previous solutions, we know that the side length of the square is &lt;math&gt;2\sqrt{3}&lt;/math&gt;. That means that the area of the square is &lt;math&gt;2\sqrt{3} \times 2\sqrt{3} = 12&lt;/math&gt;<br /> <br /> We also know that the area of one of the triangles is &lt;math&gt;\sqrt{3} \times 2 \div 2 = \sqrt{3}&lt;/math&gt;. That means that all four triangles have a total area of &lt;math&gt;\sqrt{3} \times 4 = 4\sqrt{3}&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;12 - 4\sqrt{3} = \boxed{\textbf{(B) } 12 - 4\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/7xf_g3YQk00<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=124143 2020 AMC 10A Problems/Problem 13 2020-06-06T20:43:57Z <p>Starshooter11: /* Solution 5(Last Resort) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} * 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} * \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} * \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Solution 3==<br /> If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since it starts on &lt;math&gt;(1,2)&lt;/math&gt;, there is a &lt;math&gt;\frac{3}{4}&lt;/math&gt; chance (up, down, or right) it will reach a diagonal on the first jump and &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance (left) it will reach the vertical side. The probablity of landing on a vertical is &lt;math&gt;\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}.}&lt;/math&gt;<br /> - Lingjun.<br /> <br /> ==Solution 4 (Complete States)==<br /> Let &lt;math&gt;P_{(x,y)}&lt;/math&gt; denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at &lt;math&gt;(x,y)&lt;/math&gt;. Note that &lt;math&gt;P_{(1,2)}=P_{(3,2)}&lt;/math&gt; by reflective symmetry over the line &lt;math&gt;x=2&lt;/math&gt;. Similarly, &lt;math&gt;P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}&lt;/math&gt;, and &lt;math&gt;P_{(2,1)}=P_{(2,3)}&lt;/math&gt;. <br /> Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> We have a system of &lt;math&gt;4&lt;/math&gt; equations in &lt;math&gt;4&lt;/math&gt; variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> Plugging in the third equation into this gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}&lt;/cmath&gt;<br /> Next, plugging in the second and third equation into the first equation yields <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> Now plugging in (*) into this, we get <br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -mathisawesome2169<br /> <br /> ==Solution 5(Last Resort)==<br /> Please note that this is only a simple solution that is not guaranteed to work for similar problems. This is also my first time writing a solution.<br /> <br /> In a 4 x 4 square, there are a total of 5 + 3 + 5 + 3 = 16 points on the edges. <br /> Of these points, 10 of them are on the vertical sides.<br /> So the possibility of hitting a vertical sides given a random series of jumps is &lt;math&gt;\frac{10}{16}&lt;/math&gt; which simplifies to &lt;math&gt;\frac{5}{8}&lt;/math&gt;.<br /> <br /> Therefore the answer is<br /> &lt;cmath&gt;\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -Starshooter11<br /> <br /> ==Video Solution 1==<br /> [https://www.youtube.com/watch?v=ZGwAasE32Y4&amp;t=280s IceMatrix's Solution (Starts at 4:40)]<br /> <br /> ==Video Solution 2==<br /> <br /> [https://youtu.be/xGs7BjQbGYU On The Spot STEM's Solution]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=124142 2020 AMC 10A Problems/Problem 13 2020-06-06T20:43:43Z <p>Starshooter11: /* Solution 5(Last Resort) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} * 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} * \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} * \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Solution 3==<br /> If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since it starts on &lt;math&gt;(1,2)&lt;/math&gt;, there is a &lt;math&gt;\frac{3}{4}&lt;/math&gt; chance (up, down, or right) it will reach a diagonal on the first jump and &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance (left) it will reach the vertical side. The probablity of landing on a vertical is &lt;math&gt;\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}.}&lt;/math&gt;<br /> - Lingjun.<br /> <br /> ==Solution 4 (Complete States)==<br /> Let &lt;math&gt;P_{(x,y)}&lt;/math&gt; denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at &lt;math&gt;(x,y)&lt;/math&gt;. Note that &lt;math&gt;P_{(1,2)}=P_{(3,2)}&lt;/math&gt; by reflective symmetry over the line &lt;math&gt;x=2&lt;/math&gt;. Similarly, &lt;math&gt;P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}&lt;/math&gt;, and &lt;math&gt;P_{(2,1)}=P_{(2,3)}&lt;/math&gt;. <br /> Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> We have a system of &lt;math&gt;4&lt;/math&gt; equations in &lt;math&gt;4&lt;/math&gt; variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> Plugging in the third equation into this gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}&lt;/cmath&gt;<br /> Next, plugging in the second and third equation into the first equation yields <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> Now plugging in (*) into this, we get <br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -mathisawesome2169<br /> <br /> ==Solution 5(Last Resort)==<br /> Please note that this is only a simple solution that is not guaranteed to work for similar problems. This is also my first time writing a solution.<br /> In a 4 x 4 square, there are a total of 5 + 3 + 5 + 3 = 16 points on the edges. <br /> Of these points, 10 of them are on the vertical sides.<br /> So the possibility of hitting a vertical sides given a random series of jumps is &lt;math&gt;\frac{10}{16}&lt;/math&gt; which simplifies to &lt;math&gt;\frac{5}{8}&lt;/math&gt;.<br /> <br /> Therefore the answer is<br /> &lt;cmath&gt;\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -Starshooter11<br /> <br /> ==Video Solution 1==<br /> [https://www.youtube.com/watch?v=ZGwAasE32Y4&amp;t=280s IceMatrix's Solution (Starts at 4:40)]<br /> <br /> ==Video Solution 2==<br /> <br /> [https://youtu.be/xGs7BjQbGYU On The Spot STEM's Solution]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=124141 2020 AMC 10A Problems/Problem 13 2020-06-06T20:42:53Z <p>Starshooter11: /* Solution 5(Last Resort) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} * 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} * \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} * \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Solution 3==<br /> If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since it starts on &lt;math&gt;(1,2)&lt;/math&gt;, there is a &lt;math&gt;\frac{3}{4}&lt;/math&gt; chance (up, down, or right) it will reach a diagonal on the first jump and &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance (left) it will reach the vertical side. The probablity of landing on a vertical is &lt;math&gt;\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}.}&lt;/math&gt;<br /> - Lingjun.<br /> <br /> ==Solution 4 (Complete States)==<br /> Let &lt;math&gt;P_{(x,y)}&lt;/math&gt; denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at &lt;math&gt;(x,y)&lt;/math&gt;. Note that &lt;math&gt;P_{(1,2)}=P_{(3,2)}&lt;/math&gt; by reflective symmetry over the line &lt;math&gt;x=2&lt;/math&gt;. Similarly, &lt;math&gt;P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}&lt;/math&gt;, and &lt;math&gt;P_{(2,1)}=P_{(2,3)}&lt;/math&gt;. <br /> Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> We have a system of &lt;math&gt;4&lt;/math&gt; equations in &lt;math&gt;4&lt;/math&gt; variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> Plugging in the third equation into this gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}&lt;/cmath&gt;<br /> Next, plugging in the second and third equation into the first equation yields <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> Now plugging in (*) into this, we get <br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -mathisawesome2169<br /> <br /> ==Solution 5(Last Resort)==<br /> Please note that this is only a simple solution that is not guaranteed to work for similar problems. This is also my first time writing a solution.<br /> In a 4 x 4 square, there are a total of 5 + 3 + 5 + 3 = 16 points on the edges. <br /> Of these points, 10 of them are on the vertical sides.<br /> So the possibility of hitting a vertical sides given a random series of jumps is &lt;math&gt;\frac{10}{16}&lt;/math&gt; which simplifies to &lt;math&gt;\frac{5}{8}&lt;/math&gt;<br /> Therefore the answer is<br /> &lt;cmath&gt;\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -Starshooter11<br /> <br /> ==Video Solution 1==<br /> [https://www.youtube.com/watch?v=ZGwAasE32Y4&amp;t=280s IceMatrix's Solution (Starts at 4:40)]<br /> <br /> ==Video Solution 2==<br /> <br /> [https://youtu.be/xGs7BjQbGYU On The Spot STEM's Solution]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=123850 2020 AMC 10A Problems/Problem 13 2020-06-05T03:09:56Z <p>Starshooter11: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} * 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} * \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} * \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Solution 3==<br /> If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since it starts on &lt;math&gt;(1,2)&lt;/math&gt;, there is a &lt;math&gt;\frac{3}{4}&lt;/math&gt; chance (up, down, or right) it will reach a diagonal on the first jump and &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance (left) it will reach the vertical side. The probablity of landing on a vertical is &lt;math&gt;\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}.}&lt;/math&gt;<br /> - Lingjun.<br /> <br /> ==Solution 4 (Complete States)==<br /> Let &lt;math&gt;P_{(x,y)}&lt;/math&gt; denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at &lt;math&gt;(x,y)&lt;/math&gt;. Note that &lt;math&gt;P_{(1,2)}=P_{(3,2)}&lt;/math&gt; by reflective symmetry over the line &lt;math&gt;x=2&lt;/math&gt;. Similarly, &lt;math&gt;P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}&lt;/math&gt;, and &lt;math&gt;P_{(2,1)}=P_{(2,3)}&lt;/math&gt;. <br /> Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> We have a system of &lt;math&gt;4&lt;/math&gt; equations in &lt;math&gt;4&lt;/math&gt; variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> Plugging in the third equation into this gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}&lt;/cmath&gt;<br /> Next, plugging in the second and third equation into the first equation yields <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> Now plugging in (*) into this, we get <br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -mathisawesome2169<br /> <br /> ==Solution 5(Last Resort)==<br /> Please note that this is only a simple solution that is not guaranteed to work for similar problems. This is also my first time writing a solution.<br /> In a 4 x 4 square, there are a total of 5 + 3 + 5 + 3 = 16 points on the edges. <br /> Of these points, 10 of them are on the vertical sides.<br /> So the possibility of hitting a vertical sides given a random series of jumps is \frac{10}{16}\ which simplifies to \frac{5}{8}\<br /> Therefore the answer is\boxed{\textbf{(B) }\frac{5}{8}}&lt;math&gt;&lt;/math&gt;<br /> -Starshooter11<br /> <br /> ==Video Solution 1==<br /> [https://www.youtube.com/watch?v=ZGwAasE32Y4&amp;t=280s IceMatrix's Solution (Starts at 4:40)]<br /> <br /> ==Video Solution 2==<br /> <br /> [https://youtu.be/xGs7BjQbGYU On The Spot STEM's Solution]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_5&diff=123376 2020 AMC 10B Problems/Problem 5 2020-06-01T00:07:29Z <p>Starshooter11: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to order &lt;math&gt;7&lt;/math&gt; objects. However, since there's &lt;math&gt;3!=6&lt;/math&gt; ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and &lt;math&gt;2!=2&lt;/math&gt; ways to order the green tiles, we have to divide out these possibilities.<br /> <br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We can repeat chooses extensively to find the answer.<br /> There are &lt;math&gt;7&lt;/math&gt; choose &lt;math&gt;3&lt;/math&gt; ways to arrange the yellow tiles which is &lt;math&gt;35&lt;/math&gt;.<br /> Then from the remaining tiles there are &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways to arrange the green tiles.<br /> And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles,<br /> giving us an answer of &lt;math&gt;35*6*2=420&lt;/math&gt;<br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; <br /> - noahdavid<br /> - (edited)starshooter11<br /> <br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Starshooter11 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_5&diff=123375 2020 AMC 10B Problems/Problem 5 2020-06-01T00:05:23Z <p>Starshooter11: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to order &lt;math&gt;7&lt;/math&gt; objects. However, since there's &lt;math&gt;3!=6&lt;/math&gt; ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and &lt;math&gt;2!=2&lt;/math&gt; ways to order the green tiles, we have to divide out these possibilities.<br /> <br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> <br /> We can repeat chooses extensively to find the answer.<br /> There are &lt;math&gt;7&lt;/math&gt; choose &lt;math&gt;3&lt;/math&gt; ways to arrange the yellow tiles which is &lt;math&gt;35&lt;/math&gt;.<br /> Then from the remaining tiles there are &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways to arrange the green tiles.<br /> And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles,<br /> giving us an answer of &lt;math&gt;35*6*2=420&lt;/math&gt;<br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; <br /> - noahdavid<br /> <br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Starshooter11