https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hydroquantum&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-01T22:38:32Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=91455 2018 AMC 10B Problems/Problem 15 2018-02-16T20:25:49Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad&lt;/math&gt;.<br /> <br /> Solution by HydroQuantum</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=91454 2018 AMC 10B Problems/Problem 15 2018-02-16T20:25:33Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \boxed{\textbf{(A) } 2(w+h)^2 \qquad}&lt;/math&gt;.<br /> <br /> Solution by HydroQuantum</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=91447 2018 AMC 10B Problems/Problem 15 2018-02-16T20:21:08Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \textbf{(A) } 2(w+h)^2 \qquad&lt;/math&gt;.<br /> <br /> Solution by HydroQuantum</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=91446 2018 AMC 10B Problems/Problem 15 2018-02-16T20:20:55Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \textbf{(A) } 2(w+h)^2 \qquad&lt;/math&gt;.<br /> Solution by HydroQuantum</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=91445 2018 AMC 10B Problems/Problem 15 2018-02-16T20:20:00Z <p>Hydroquantum: </p> <hr /> <div>A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \textbf{(A) } 2(w+h)^2 \qquad&lt;/math&gt;.</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=84071 2017 AMC 12B Problems 2017-02-18T03:28:27Z <p>Hydroquantum: /* See also */</p> <hr /> <div>WORK IN PROGRESS<br /> <br /> {{AMC12 Problems|year=2017|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> Kymbrea's comic book collection currently has &lt;math&gt;30&lt;/math&gt; comic books in it, and she is adding to her collection at the rate of &lt;math&gt;2&lt;/math&gt; comic books per month. LaShawn's collection currently has &lt;math&gt;10&lt;/math&gt; comic books in it, and he is adding to his collection at the rate of &lt;math&gt;6&lt;/math&gt; comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; satify the inequalities<br /> &lt;math&gt;0&lt;x&lt;1&lt;/math&gt;, &lt;math&gt;-1&lt;y&lt;0&lt;/math&gt;, and &lt;math&gt;1&lt;z&lt;2&lt;/math&gt;.<br /> Which of the following numbers is necessarily positive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Supposed that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are nonzero real numbers such that &lt;math&gt;\frac{3x+y}{x-3y}=-2&lt;/math&gt;. What is the value of &lt;math&gt;\frac{x+3y}{3x-y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 4==<br /> Samia set off on her bicycle to visit her friend, traveling at an average speed of &lt;math&gt;17&lt;/math&gt; kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at &lt;math&gt;5&lt;/math&gt; kilometers per hour. In all it took her &lt;math&gt;44&lt;/math&gt; minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> The data set &lt;math&gt;[6,19,33,33,39,41,41,43,51,57]&lt;/math&gt; has median &lt;math&gt;Q_2 = 40&lt;/math&gt;, first quartile &lt;math&gt;Q_1 = 33&lt;/math&gt;, and third quartile &lt;math&gt;Q_3=43&lt;/math&gt;. An outlier in a data set is a value that is more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range below the first quartile &lt;math&gt;(Q_1)&lt;/math&gt; or more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range above the third quartile &lt;math&gt;(Q_3)&lt;/math&gt;, where the interquartile range is defined as &lt;math&gt;Q_3 - Q_1&lt;/math&gt;. How many outliers does this data set have?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The circle having &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(8,6)&lt;/math&gt; as the endpoints of a diameter intersects the &lt;math&gt;x&lt;/math&gt;-axis at a second point. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of this point? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The functions &lt;math&gt;\sin(x)&lt;/math&gt; and &lt;math&gt;\cos(x)&lt;/math&gt; are periodic with least period &lt;math&gt;2\pi&lt;/math&gt;. What is the least period of the function &lt;math&gt;\cos(\sin(x))&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}&lt;/math&gt; It's not periodic.<br /> <br /> [[2017 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A circle has center &lt;math&gt;(-10,-4)&lt;/math&gt; and radius &lt;math&gt;13&lt;/math&gt;. Another circle has center &lt;math&gt;(3,9)&lt;/math&gt; and radius &lt;math&gt;\sqrt{65}&lt;/math&gt;. The line passing through the two points of intersection of the two circles has equation &lt;math&gt;x + y = c&lt;/math&gt;. What is &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> At Typico High School, &lt;math&gt;60\%&lt;/math&gt; of the students like dancing, and the rest dislike it. Of those who like dancing, &lt;math&gt;80\%&lt;/math&gt; say that they like it, and the rest say that they dislike it. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt; say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Call a positive integer &lt;math&gt;monotonous&lt;/math&gt; if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;23578&lt;/math&gt;, and &lt;math&gt;987620&lt;/math&gt; are monotonous, but &lt;math&gt;88&lt;/math&gt;, &lt;math&gt;7434&lt;/math&gt;, and &lt;math&gt;23557&lt;/math&gt; are not. How many monotonous positive integers are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 12==<br /> What is the sum of the roots of &lt;math&gt;z^{12}=64&lt;/math&gt; that have a positive real part? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> In the figure below, &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; disks are to be painted blue, &lt;math&gt;2&lt;/math&gt; are to be painted red, and &lt;math&gt;1&lt;/math&gt; is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br /> <br /> &lt;asy&gt;<br /> size(100);<br /> pair A, B, C, D, E, F;<br /> A = (0,0);<br /> B = (1,0);<br /> C = (2,0);<br /> D = rotate(60, A)*B;<br /> E = B + D;<br /> F = rotate(60, A)*C;<br /> draw(Circle(A, 0.5));<br /> draw(Circle(B, 0.5));<br /> draw(Circle(C, 0.5));<br /> draw(Circle(D, 0.5));<br /> draw(Circle(E, 0.5));<br /> draw(Circle(F, 0.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The number &lt;math&gt;21!=51,090,942,171,709,440,000&lt;/math&gt; has over &lt;math&gt;60,000&lt;/math&gt; positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> A coin is biased in such a way that on each toss the probability of heads is &lt;math&gt;\frac{2}{3}&lt;/math&gt; and the probability of tails is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; The probabilities are the same.<br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> [[2017 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle <br /> ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0,1)&lt;/math&gt;. What is the probability that &lt;math&gt;\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor&lt;/math&gt;, where &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Last year Isabella took &lt;math&gt;7&lt;/math&gt; math tests and received &lt;math&gt;7&lt;/math&gt; different scores, each an integer between &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was &lt;math&gt;95&lt;/math&gt;. What was her score on the sixth test?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}&lt;/math&gt;<br /> <br /> ==Problem 23==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;\triangle ABC \sim \triangle BCD&lt;/math&gt;, and &lt;math&gt;AB &gt; BC&lt;/math&gt;. There is a point &lt;math&gt;E&lt;/math&gt; in the interior of &lt;math&gt;ABCD&lt;/math&gt; such that &lt;math&gt;\triangle ABC \sim \triangle CEB&lt;/math&gt; and the area of &lt;math&gt;\triangle AED&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt; times the area of &lt;math&gt;\triangle CEB&lt;/math&gt;. What is &lt;math&gt;\frac{AB}{BC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2} \qquad \textbf{(B)}\ 2 + \sqrt{2} \qquad \textbf{(C)}\ \sqrt{17} \qquad \textbf{(D)}\ 2 + \sqrt{5} \qquad\textbf{(E)}\ 1 + 2\sqrt{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A set of &lt;math&gt;n&lt;/math&gt; people participate in an online video basketball tournament. Each person may be a member of any number of &lt;math&gt;5&lt;/math&gt;-player teams, but no teams may have exactly the same &lt;math&gt;5&lt;/math&gt; members. The site statistics show a curious fact: The average, over all subsets of size &lt;math&gt;9&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size &lt;math&gt;8&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those &lt;math&gt;8&lt;/math&gt; people. How many values &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;9 \leq n \leq 2017&lt;/math&gt;, can be the number of participants?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 477 \qquad \textbf{(B)}\ 482 \qquad \textbf{(C)}\ 487 \qquad \textbf{(D)}\ 557 \qquad\textbf{(E)}\ 562&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=84070 2017 AMC 12B Problems 2017-02-18T03:27:34Z <p>Hydroquantum: /* See also */</p> <hr /> <div>WORK IN PROGRESS<br /> <br /> {{AMC12 Problems|year=2017|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> Kymbrea's comic book collection currently has &lt;math&gt;30&lt;/math&gt; comic books in it, and she is adding to her collection at the rate of &lt;math&gt;2&lt;/math&gt; comic books per month. LaShawn's collection currently has &lt;math&gt;10&lt;/math&gt; comic books in it, and he is adding to his collection at the rate of &lt;math&gt;6&lt;/math&gt; comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; satify the inequalities<br /> &lt;math&gt;0&lt;x&lt;1&lt;/math&gt;, &lt;math&gt;-1&lt;y&lt;0&lt;/math&gt;, and &lt;math&gt;1&lt;z&lt;2&lt;/math&gt;.<br /> Which of the following numbers is necessarily positive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Supposed that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are nonzero real numbers such that &lt;math&gt;\frac{3x+y}{x-3y}=-2&lt;/math&gt;. What is the value of &lt;math&gt;\frac{x+3y}{3x-y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 4==<br /> Samia set off on her bicycle to visit her friend, traveling at an average speed of &lt;math&gt;17&lt;/math&gt; kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at &lt;math&gt;5&lt;/math&gt; kilometers per hour. In all it took her &lt;math&gt;44&lt;/math&gt; minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> The data set &lt;math&gt;[6,19,33,33,39,41,41,43,51,57]&lt;/math&gt; has median &lt;math&gt;Q_2 = 40&lt;/math&gt;, first quartile &lt;math&gt;Q_1 = 33&lt;/math&gt;, and third quartile &lt;math&gt;Q_3=43&lt;/math&gt;. An outlier in a data set is a value that is more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range below the first quartile &lt;math&gt;(Q_1)&lt;/math&gt; or more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range above the third quartile &lt;math&gt;(Q_3)&lt;/math&gt;, where the interquartile range is defined as &lt;math&gt;Q_3 - Q_1&lt;/math&gt;. How many outliers does this data set have?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The circle having &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(8,6)&lt;/math&gt; as the endpoints of a diameter intersects the &lt;math&gt;x&lt;/math&gt;-axis at a second point. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of this point? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The functions &lt;math&gt;\sin(x)&lt;/math&gt; and &lt;math&gt;\cos(x)&lt;/math&gt; are periodic with least period &lt;math&gt;2\pi&lt;/math&gt;. What is the least period of the function &lt;math&gt;\cos(\sin(x))&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}&lt;/math&gt; It's not periodic.<br /> <br /> [[2017 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A circle has center &lt;math&gt;(-10,-4)&lt;/math&gt; and radius &lt;math&gt;13&lt;/math&gt;. Another circle has center &lt;math&gt;(3,9)&lt;/math&gt; and radius &lt;math&gt;\sqrt{65}&lt;/math&gt;. The line passing through the two points of intersection of the two circles has equation &lt;math&gt;x + y = c&lt;/math&gt;. What is &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> At Typico High School, &lt;math&gt;60\%&lt;/math&gt; of the students like dancing, and the rest dislike it. Of those who like dancing, &lt;math&gt;80\%&lt;/math&gt; say that they like it, and the rest say that they dislike it. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt; say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Call a positive integer &lt;math&gt;monotonous&lt;/math&gt; if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;23578&lt;/math&gt;, and &lt;math&gt;987620&lt;/math&gt; are monotonous, but &lt;math&gt;88&lt;/math&gt;, &lt;math&gt;7434&lt;/math&gt;, and &lt;math&gt;23557&lt;/math&gt; are not. How many monotonous positive integers are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 12==<br /> What is the sum of the roots of &lt;math&gt;z^{12}=64&lt;/math&gt; that have a positive real part? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> In the figure below, &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; disks are to be painted blue, &lt;math&gt;2&lt;/math&gt; are to be painted red, and &lt;math&gt;1&lt;/math&gt; is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br /> <br /> &lt;asy&gt;<br /> size(100);<br /> pair A, B, C, D, E, F;<br /> A = (0,0);<br /> B = (1,0);<br /> C = (2,0);<br /> D = rotate(60, A)*B;<br /> E = B + D;<br /> F = rotate(60, A)*C;<br /> draw(Circle(A, 0.5));<br /> draw(Circle(B, 0.5));<br /> draw(Circle(C, 0.5));<br /> draw(Circle(D, 0.5));<br /> draw(Circle(E, 0.5));<br /> draw(Circle(F, 0.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The number &lt;math&gt;21!=51,090,942,171,709,440,000&lt;/math&gt; has over &lt;math&gt;60,000&lt;/math&gt; positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> A coin is biased in such a way that on each toss the probability of heads is &lt;math&gt;\frac{2}{3}&lt;/math&gt; and the probability of tails is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; The probabilities are the same.<br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> [[2017 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle <br /> ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0,1)&lt;/math&gt;. What is the probability that &lt;math&gt;\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor&lt;/math&gt;, where &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Last year Isabella took &lt;math&gt;7&lt;/math&gt; math tests and received &lt;math&gt;7&lt;/math&gt; different scores, each an integer between &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was &lt;math&gt;95&lt;/math&gt;. What was her score on the sixth test?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}&lt;/math&gt;<br /> <br /> ==Problem 23==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;\triangle ABC \sim \triangle BCD&lt;/math&gt;, and &lt;math&gt;AB &gt; BC&lt;/math&gt;. There is a point &lt;math&gt;E&lt;/math&gt; in the interior of &lt;math&gt;ABCD&lt;/math&gt; such that &lt;math&gt;\triangle ABC \sim \triangle CEB&lt;/math&gt; and the area of &lt;math&gt;\triangle AED&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt; times the area of &lt;math&gt;\triangle CEB&lt;/math&gt;. What is &lt;math&gt;\frac{AB}{BC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2} \qquad \textbf{(B)}\ 2 + \sqrt{2} \qquad \textbf{(C)}\ \sqrt{17} \qquad \textbf{(D)}\ 2 + \sqrt{5} \qquad\textbf{(E)}\ 1 + 2\sqrt{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A set of &lt;math&gt;n&lt;/math&gt; people participate in an online video basketball tournament. Each person may be a member of any number of &lt;math&gt;5&lt;/math&gt;-player teams, but no teams may have exactly the same &lt;math&gt;5&lt;/math&gt; members. The site statistics show a curious fact: The average, over all subsets of size &lt;math&gt;9&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size &lt;math&gt;8&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those &lt;math&gt;8&lt;/math&gt; people. How many values &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;9 \leq n \leq 2017&lt;/math&gt;, can be the number of participants?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 477 \qquad \textbf{(B)}\ 482 \qquad \textbf{(C)}\ 487 \qquad \textbf{(D)}\ 557 \qquad\textbf{(E)}\ 562&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2017|ab=B|before=[[2016 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_19&diff=84049 2017 AMC 10B Problems/Problem 19 2017-02-17T19:46:31Z <p>Hydroquantum: /* See Also */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> ==Solution==<br /> <br /> ===Solution 1===<br /> Note that by symmetry, &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is also equilateral. Therefore, we only need to find one of the sides of &lt;math&gt;A'B'C'&lt;/math&gt; to determine the area ratio. WLOG, let &lt;math&gt;AB = BC = CA = 1&lt;/math&gt;. Therefore, &lt;math&gt;BB' = 3&lt;/math&gt; and &lt;math&gt;BC' = 4&lt;/math&gt;. Also, &lt;math&gt;\angle B'BC' = 120^{\circ}&lt;/math&gt;, so by the Law of Cosines, &lt;math&gt;B'C' = \sqrt{37}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;<br /> <br /> ===Solution 2 ===<br /> As mentioned in the first solution, &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is equilateral. WLOG, let &lt;math&gt;AB=2&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be on the line passing through &lt;math&gt;AB&lt;/math&gt; such that &lt;math&gt;A'D&lt;/math&gt; is perpendicular to &lt;math&gt;AB&lt;/math&gt;. Note that &lt;math&gt;\triangle A'DA&lt;/math&gt; is a 30-60-90 with right angle at &lt;math&gt;D&lt;/math&gt;. Since &lt;math&gt;AA'=6&lt;/math&gt;, &lt;math&gt;AD=3&lt;/math&gt; and &lt;math&gt;A'D=3\sqrt{3}&lt;/math&gt;. So we know that &lt;math&gt;DB'=11&lt;/math&gt;. Note that &lt;math&gt;\triangle A'DB'&lt;/math&gt; is a right triangle with right angle at &lt;math&gt;D&lt;/math&gt;. So by the Pythagorean theorem, we find &lt;math&gt;A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.&lt;/math&gt; Therefore, the answer is &lt;math&gt;(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_8&diff=84048 2017 AMC 12B Problems/Problem 8 2017-02-17T19:45:04Z <p>Hydroquantum: /* Solution 1: Cross Multiplying */</p> <hr /> <div>==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1: Cross-Multiplication==<br /> Let &lt;math&gt;a&lt;/math&gt; be the short side of the rectangle, and &lt;math&gt;b&lt;/math&gt; be the long side of the rectangle. The diagonal, therefore, is &lt;math&gt;\sqrt{a^2 + b^2}&lt;/math&gt;. We can get the equation &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;a\sqrt{a^2 + b^2} = b^2&lt;/math&gt;. Squaring both sides of the equation, we get &lt;math&gt;a^2 (a^2 + b^2) = b^4&lt;/math&gt;, which simplifies to &lt;math&gt;a^4 + a^2b^2 - b^4 = 0&lt;/math&gt;. Solving for a quadratic in &lt;math&gt;a^2&lt;/math&gt;, using the quadratic formula we get &lt;math&gt;a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}&lt;/math&gt; which gives us &lt;math&gt;\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}&lt;/math&gt;. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is &lt;math&gt;\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> Solution by: vedadehhc<br /> <br /> ==Solution 2: Substitution==<br /> <br /> Solution by HydroQuantum<br /> <br /> <br /> Let the short side of the rectangle be &lt;math&gt;a&lt;/math&gt; and let the long side of the rectangle be &lt;math&gt;b&lt;/math&gt;. Then, the diagonal, according to the Pythagorean Theorem, is &lt;math&gt;\sqrt{a+b}&lt;/math&gt;. Therefore, we can write the equation:<br /> <br /> &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;.<br /> <br /> We are trying to find the square of the ratio of &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;. Let's let our answer, &lt;math&gt;\frac{a^2}{b^2}&lt;/math&gt;, be &lt;math&gt;k&lt;/math&gt;. Then, squaring the above equation,<br /> <br /> &lt;math&gt;\frac{a^2}{b^2}=k=\frac{b^2}{a^2 + b^2}=\frac{b^2}{a^2}-\frac{b^2}{b^2}=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;k=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Multiplying each side of the equation by &lt;math&gt;k&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2=1-k&lt;/math&gt;.<br /> <br /> Adding each side by &lt;math&gt;k-1&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2+k-1=0&lt;/math&gt;.<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; using the Quadratic Formula,<br /> <br /> &lt;math&gt;k=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Since the ratio of lengths and diagonals of a rectangle cannot be negative, and &lt;math&gt;\sqrt{5}&gt;1&lt;/math&gt;, the &lt;math&gt;\pm&lt;/math&gt; symbol can only take on the &lt;math&gt;+&lt;/math&gt;. Therefore,<br /> <br /> &lt;math&gt;k=\frac{-1+\sqrt{5}}{2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_8&diff=84047 2017 AMC 12B Problems/Problem 8 2017-02-17T19:42:33Z <p>Hydroquantum: /* Solution 2: Substitution */</p> <hr /> <div>==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1: Cross Multiplying==<br /> Let &lt;math&gt;a&lt;/math&gt; be the short side of the rectangle, and &lt;math&gt;b&lt;/math&gt; be the long side of the rectangle. The diagonal, therefore, is &lt;math&gt;\sqrt{a^2 + b^2}&lt;/math&gt;. We can get the equation &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;a\sqrt{a^2 + b^2} = b^2&lt;/math&gt;. Squaring both sides of the equation, we get &lt;math&gt;a^2 (a^2 + b^2) = b^4&lt;/math&gt;, which simplifies to &lt;math&gt;a^4 + a^2b^2 - b^4 = 0&lt;/math&gt;. Solving for a quadratic in &lt;math&gt;a^2&lt;/math&gt;, using the quadratic formula we get &lt;math&gt;a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}&lt;/math&gt; which gives us &lt;math&gt;\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}&lt;/math&gt;. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is &lt;math&gt;\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> Solution by: vedadehhc<br /> <br /> ==Solution 2: Substitution==<br /> <br /> Solution by HydroQuantum<br /> <br /> <br /> Let the short side of the rectangle be &lt;math&gt;a&lt;/math&gt; and let the long side of the rectangle be &lt;math&gt;b&lt;/math&gt;. Then, the diagonal, according to the Pythagorean Theorem, is &lt;math&gt;\sqrt{a+b}&lt;/math&gt;. Therefore, we can write the equation:<br /> <br /> &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;.<br /> <br /> We are trying to find the square of the ratio of &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;. Let's let our answer, &lt;math&gt;\frac{a^2}{b^2}&lt;/math&gt;, be &lt;math&gt;k&lt;/math&gt;. Then, squaring the above equation,<br /> <br /> &lt;math&gt;\frac{a^2}{b^2}=k=\frac{b^2}{a^2 + b^2}=\frac{b^2}{a^2}-\frac{b^2}{b^2}=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;k=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Multiplying each side of the equation by &lt;math&gt;k&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2=1-k&lt;/math&gt;.<br /> <br /> Adding each side by &lt;math&gt;k-1&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2+k-1=0&lt;/math&gt;.<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; using the Quadratic Formula,<br /> <br /> &lt;math&gt;k=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Since the ratio of lengths and diagonals of a rectangle cannot be negative, and &lt;math&gt;\sqrt{5}&gt;1&lt;/math&gt;, the &lt;math&gt;\pm&lt;/math&gt; symbol can only take on the &lt;math&gt;+&lt;/math&gt;. Therefore,<br /> <br /> &lt;math&gt;k=\frac{-1+\sqrt{5}}{2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_8&diff=84046 2017 AMC 12B Problems/Problem 8 2017-02-17T19:41:11Z <p>Hydroquantum: /* Solution 2: Substitution */</p> <hr /> <div>==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1: Cross Multiplying==<br /> Let &lt;math&gt;a&lt;/math&gt; be the short side of the rectangle, and &lt;math&gt;b&lt;/math&gt; be the long side of the rectangle. The diagonal, therefore, is &lt;math&gt;\sqrt{a^2 + b^2}&lt;/math&gt;. We can get the equation &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;a\sqrt{a^2 + b^2} = b^2&lt;/math&gt;. Squaring both sides of the equation, we get &lt;math&gt;a^2 (a^2 + b^2) = b^4&lt;/math&gt;, which simplifies to &lt;math&gt;a^4 + a^2b^2 - b^4 = 0&lt;/math&gt;. Solving for a quadratic in &lt;math&gt;a^2&lt;/math&gt;, using the quadratic formula we get &lt;math&gt;a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}&lt;/math&gt; which gives us &lt;math&gt;\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}&lt;/math&gt;. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is &lt;math&gt;\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> Solution by: vedadehhc<br /> <br /> ==Solution 2: Substitution==<br /> <br /> Solution by HydroQuantum<br /> <br /> <br /> Let the short side of the rectangle be &lt;math&gt;a&lt;/math&gt; and let the long side of the rectangle be &lt;math&gt;b&lt;/math&gt;. Then, the diagonal, according to the Pythagorean Theorem, is &lt;math&gt;sqrt{a+b)}&lt;/math&gt;. Therefore, we can write the equation:<br /> <br /> &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;.<br /> <br /> We are trying to find the square of the ratio of &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;. Let's let our answer, &lt;math&gt;\frac{a^2}{b^2}&lt;/math&gt;, be &lt;math&gt;k&lt;/math&gt;. Then, squaring the above equation,<br /> <br /> &lt;math&gt;\frac{a^2}{b^2}=k=\frac{b^2}{a^2 + b^2}=\frac{b^2}{a^2}-\frac{b^2}{b^2}=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;k=\frac{1}{k}-1&lt;/math&gt;.<br /> <br /> Multiplying each side of the equation by &lt;math&gt;k&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2=1-k&lt;/math&gt;.<br /> <br /> Adding each side by &lt;math&gt;k-1&lt;/math&gt;,<br /> <br /> &lt;math&gt;k^2+k-1=0&lt;/math&gt;.<br /> <br /> Solving for &lt;math&gt;k&lt;/math&gt; using the Quadratic Formula,<br /> <br /> &lt;math&gt;k=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Since the ratio of lengths and diagonals of a rectangle cannot be negative, and &lt;math&gt;sqrt{5}&gt;1&lt;/math&gt;, the &lt;math&gt;\pm&lt;/math&gt; symbol can only take on the &lt;math&gt;+&lt;/math&gt;. Therefore,<br /> <br /> &lt;math&gt;k=\frac{-1+\sqrt{5}}{2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_8&diff=84045 2017 AMC 12B Problems/Problem 8 2017-02-17T19:26:05Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1: Cross Multiplying==<br /> Let &lt;math&gt;a&lt;/math&gt; be the short side of the rectangle, and &lt;math&gt;b&lt;/math&gt; be the long side of the rectangle. The diagonal, therefore, is &lt;math&gt;\sqrt{a^2 + b^2}&lt;/math&gt;. We can get the equation &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;a\sqrt{a^2 + b^2} = b^2&lt;/math&gt;. Squaring both sides of the equation, we get &lt;math&gt;a^2 (a^2 + b^2) = b^4&lt;/math&gt;, which simplifies to &lt;math&gt;a^4 + a^2b^2 - b^4 = 0&lt;/math&gt;. Solving for a quadratic in &lt;math&gt;a^2&lt;/math&gt;, using the quadratic formula we get &lt;math&gt;a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}&lt;/math&gt; which gives us &lt;math&gt;\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}&lt;/math&gt;. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is &lt;math&gt;\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> Solution by: vedadehhc<br /> <br /> ==Solution 2: Substitution==<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_8&diff=84044 2017 AMC 12B Problems/Problem 8 2017-02-17T19:16:12Z <p>Hydroquantum: /* Problem 8 */</p> <hr /> <div>==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a&lt;/math&gt; be the short side of the rectangle, and &lt;math&gt;b&lt;/math&gt; be the long side of the rectangle. The diagonal, therefore, is &lt;math&gt;\sqrt{a^2 + b^2}&lt;/math&gt;. We can get the equation &lt;math&gt;\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}&lt;/math&gt;. Cross-multiplying, we get &lt;math&gt;a\sqrt{a^2 + b^2} = b^2&lt;/math&gt;. Squaring both sides of the equation, we get &lt;math&gt;a^2 (a^2 + b^2) = b^4&lt;/math&gt;, which simplifies to &lt;math&gt;a^4 + a^2b^2 - b^4 = 0&lt;/math&gt;. Solving for a quadratic in &lt;math&gt;a^2&lt;/math&gt;, using the quadratic formula we get &lt;math&gt;a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}&lt;/math&gt; which gives us &lt;math&gt;\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}&lt;/math&gt;. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is &lt;math&gt;\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}&lt;/math&gt;.<br /> <br /> Solution by: vedadehhc<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=84043 2017 AMC 12B Problems/Problem 15 2017-02-17T19:03:53Z <p>Hydroquantum: /* Solution 2: Inspection */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution 1: Law of Cosines==<br /> Solution by HydroQuantum<br /> <br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;cmath&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = &lt;/cmath&gt; &lt;cmath&gt;(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.&lt;/cmath&gt; Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> ==Solution 2: Inspection==<br /> Note that the height and base of &lt;math&gt;\triangle A'CC'&lt;/math&gt; are respectively 4 times and 3 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore the area of &lt;math&gt;\triangle A'CC'&lt;/math&gt; is 12 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> By symmetry, &lt;math&gt;\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'&lt;/math&gt;. Adding the areas of these three triangles and &lt;math&gt;\triangle ABC&lt;/math&gt; for the total area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; gives a ratio of &lt;math&gt;(12 + 12 + 12 + 1) : 1&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> <br /> {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=83995 2016 AMC 8 Problems/Problem 25 2017-02-17T03:08:17Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2 (Quicker Solution, basically solution one)==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> Notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AC}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> Solution by HydroQuantum<br /> <br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB = BC = CD = DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh = (16)(15) = 240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34r = 240&lt;/math&gt;. Solving this, &lt;math&gt;r = \frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83986 2017 AMC 12B Problems/Problem 15 2017-02-17T02:57:08Z <p>Hydroquantum: /* Solution 1: Law of Cosines */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution 1: Law of Cosines==<br /> Solution by HydroQuantum<br /> <br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;cmath&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = &lt;/cmath&gt; &lt;cmath&gt;(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.&lt;/cmath&gt; Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> ==Solution 2: Inspection==<br /> Note that the height and base of &lt;math&gt;\triangle A'CC'&lt;/math&gt; are respectively 4 times and 3 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore the area of &lt;math&gt;\triangle A'CC'&lt;/math&gt; is 12 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> By symmetry, &lt;math&gt;\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'&lt;/math&gt;. Adding the areas of these three triangles and &lt;math&gt;\triangle ABC&lt;/math&gt; for the total area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; gives a ratio of &lt;math&gt;(12 + 12 + 12 + 1) : 1&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83984 2017 AMC 12B Problems/Problem 15 2017-02-17T02:56:41Z <p>Hydroquantum: /* Solution 2: Inspection */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution 1: Law of Cosines==<br /> Solution by HydroQuantum<br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;cmath&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = &lt;/cmath&gt; &lt;cmath&gt;(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.&lt;/cmath&gt; Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> ==Solution 2: Inspection==<br /> Note that the height and base of &lt;math&gt;\triangle A'CC'&lt;/math&gt; are respectively 4 times and 3 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore the area of &lt;math&gt;\triangle A'CC'&lt;/math&gt; is 12 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> By symmetry, &lt;math&gt;\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'&lt;/math&gt;. Adding the areas of these three triangles and &lt;math&gt;\triangle ABC&lt;/math&gt; for the total area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; gives a ratio of &lt;math&gt;(12 + 12 + 12 + 1) : 1&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83981 2017 AMC 12B Problems/Problem 15 2017-02-17T02:56:01Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution 1: Law of Cosines==<br /> Solution by HydroQuantum<br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;cmath&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = &lt;/cmath&gt; &lt;cmath&gt;(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.&lt;/cmath&gt; Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> ==Solution 2 (inspection)==<br /> Note that the height and base of &lt;math&gt;\triangle A'CC'&lt;/math&gt; are respectively 4 times and 3 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore the area of &lt;math&gt;\triangle A'CC'&lt;/math&gt; is 12 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> By symmetry, &lt;math&gt;\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'&lt;/math&gt;. Adding the areas of these three triangles and &lt;math&gt;\triangle ABC&lt;/math&gt; for the total area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; gives a ratio of &lt;math&gt;(12 + 12 + 12 + 1) : 1&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=User_talk:Hydroquantum&diff=83871 User talk:Hydroquantum 2017-02-16T23:37:14Z <p>Hydroquantum: </p> <hr /> <div>Hey there! I'm HydroQuantum.<br /> <br /> I'm an ambitious math student and learner. Right now, my goal is reaching a near-perfect or perfect score on the AMC 10 and AMC 12. Feel free to contact me at quantumdoge@gmail.com!</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=User_talk:Hydroquantum&diff=83868 User talk:Hydroquantum 2017-02-16T23:24:52Z <p>Hydroquantum: </p> <hr /> <div>Hey there! I'm HydroQuantum.<br /> <br /> I'm an ambitious math student and learner. Right now, my goal is reaching a near-perfect or perfect score on the AMC 10 and AMC 12. Feel free to contact me at hydroquantum@gmail.com!</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=User_talk:Hydroquantum&diff=83867 User talk:Hydroquantum 2017-02-16T23:24:20Z <p>Hydroquantum: Created page with &quot;Hey there! I'm HydroQuantum. I'm an ambitious math student and learner. Right now, my goal is reaching a near-perfect or perfect score on the AMC 10 and AMC 12.&quot;</p> <hr /> <div>Hey there! I'm HydroQuantum.<br /> <br /> I'm an ambitious math student and learner. Right now, my goal is reaching a near-perfect or perfect score on the AMC 10 and AMC 12.</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83839 2017 AMC 12B Problems/Problem 15 2017-02-16T22:15:59Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Solution by HydroQuantum<br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;math&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)&lt;/math&gt;. This simplifies to &lt;math&gt;y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2&lt;/math&gt;. Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83838 2017 AMC 12B Problems/Problem 15 2017-02-16T22:15:13Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Solution by HydroQuantum<br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;. Then, the area of the small (inside) equilateral triangle is &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;. Therefore the denominator of the ratio must be &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;math&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)&lt;/math&gt;. This simplifies to &lt;math&gt;y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2&lt;/math&gt;. Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83837 2017 AMC 12B Problems/Problem 15 2017-02-16T22:14:21Z <p>Hydroquantum: /* Solution by HydroQuantum */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Solution by HydroQuantum<br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;. Then, the area of the small (inside) equilateral triangle is &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;. Therefore the denominator of the ratio must be &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;math&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)&lt;/math&gt;. This simplifies to &lt;math&gt;y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2&lt;/math&gt;. Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83836 2017 AMC 12B Problems/Problem 15 2017-02-16T22:13:43Z <p>Hydroquantum: /* Solution */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution by HydroQuantum==<br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;. Then, the area of the small (inside) equilateral triangle is &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;. Therefore the denominator of the ratio must be &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;math&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)&lt;/math&gt;. This simplifies to &lt;math&gt;y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2&lt;/math&gt;. Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83835 2017 AMC 12B Problems/Problem 15 2017-02-16T22:13:12Z <p>Hydroquantum: </p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;. Then, the area of the small (inside) equilateral triangle is &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;. Therefore the denominator of the ratio must be &lt;math&gt;\frac{x^2\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;math&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)&lt;/math&gt;. This simplifies to &lt;math&gt;y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2&lt;/math&gt;. Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83833 2017 AMC 12B Problems/Problem 15 2017-02-16T21:53:54Z <p>Hydroquantum: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83832 2017 AMC 12B Problems/Problem 15 2017-02-16T21:52:29Z <p>Hydroquantum: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=83831 2017 AMC 12B Problems/Problem 15 2017-02-16T21:51:31Z <p>Hydroquantum: Created page with &quot;==Problem 15== Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/...&quot;</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 15|Solution]]<br /> [[2017 AMC 10B Problems/Problem 19|Solution]]</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=Blaise_Pascal&diff=82565 Blaise Pascal 2017-01-26T01:14:52Z <p>Hydroquantum: Created page with &quot;https://en.wikipedia.org/wiki/Blaise_Pascal&quot;</p> <hr /> <div>https://en.wikipedia.org/wiki/Blaise_Pascal</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81613 2016 AMC 8 Problems/Problem 25 2016-11-25T17:21:35Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB = BC = CD = DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh = (16)(15) = 240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34r = 240&lt;/math&gt;. Solving this, &lt;math&gt;r = \frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81612 2016 AMC 8 Problems/Problem 25 2016-11-25T14:23:04Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81610 2016 AMC 8 Problems/Problem 25 2016-11-25T14:10:33Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81609 2016 AMC 8 Problems/Problem 25 2016-11-25T14:07:55Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81608 2016 AMC 8 Problems/Problem 25 2016-11-25T14:07:25Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81607 2016 AMC 8 Problems/Problem 25 2016-11-25T14:04:22Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81606 2016 AMC 8 Problems/Problem 25 2016-11-25T14:03:49Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81605 2016 AMC 8 Problems/Problem 25 2016-11-25T14:03:12Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB /cong DA&lt;/math&gt;, &lt;math&gt;BC /cong CD&lt;/math&gt;. &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;CD&lt;/math&gt; = &lt;math&gt;DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81604 2016 AMC 8 Problems/Problem 25 2016-11-25T14:00:17Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81603 2016 AMC 8 Problems/Problem 25 2016-11-25T14:00:06Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{(bh}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;bh&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81602 2016 AMC 8 Problems/Problem 25 2016-11-25T13:59:39Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{(b)(h)}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;(b)(h)&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81601 2016 AMC 8 Problems/Problem 25 2016-11-25T13:59:19Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt; which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{(b)(h)}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;(b)(h)&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81600 2016 AMC 8 Problems/Problem 25 2016-11-25T13:58:04Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{(b)(h)}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;(b)(h)&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81599 2016 AMC 8 Problems/Problem 25 2016-11-25T13:44:23Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt; and let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{(b)(h)}{2}&lt;/math&gt;, the area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is twice that, which is &lt;math&gt;(b)(h)&lt;/math&gt; = &lt;math&gt;(16)(15)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81592 2016 AMC 8 Problems/Problem 25 2016-11-24T20:40:00Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt;. Half the base &lt;math&gt;BD&lt;/math&gt; of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81591 2016 AMC 8 Problems/Problem 25 2016-11-24T20:37:21Z <p>Hydroquantum: /* Solution 4: Inscribed Circle */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt;. Half the base &lt;math&gt;BD&lt;/math&gt; of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}&lt;/math&gt; = &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81590 2016 AMC 8 Problems/Problem 25 2016-11-24T20:34:04Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Call this triangle &lt;math&gt;ABD&lt;/math&gt;. Half the base &lt;math&gt;BD&lt;/math&gt; of triangle &lt;math&gt;ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81589 2016 AMC 8 Problems/Problem 25 2016-11-24T20:29:01Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Half the base of the triangle is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each leg is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus &lt;math&gt;[ABCD]&lt;/math&gt; is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81581 2016 AMC 8 Problems/Problem 25 2016-11-24T20:09:03Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> <br /> ==Solution 4: Inscribed Circle==<br /> Half the base of the triangle is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each leg is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle inside a rhombus.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81580 2016 AMC 8 Problems/Problem 25 2016-11-24T20:08:36Z <p>Hydroquantum: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> ==Solution 4: Inscribed Circle==<br /> Half the base of the triangle is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each leg is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle inside a rhombus.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{(17)(4)}{2} = 34&lt;/math&gt;. The area of the rhombus is &lt;math&gt;(15)(16)&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;.<br /> <br /> The formula for the radius of an inscribed circle tangent to all sides of a quadrilateral &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;240&lt;/math&gt;. Solving this, &lt;math&gt;r&lt;/math&gt; = &lt;math&gt;\frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hydroquantum