https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=A.shyam.25&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-02T09:32:55Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_15&diff=164301 2000 AMC 12 Problems/Problem 15 2021-10-30T17:24:33Z <p>A.shyam.25: added note</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #15]] and [[2000 AMC 10 Problems|2000 AMC 10 #24]]}}<br /> <br /> == Problem ==<br /> Let &lt;math&gt;f&lt;/math&gt; be a [[function]] for which &lt;math&gt;f\left(\dfrac{x}{3}\right) = x^2 + x + 1&lt;/math&gt;. Find the sum of all values of &lt;math&gt;z&lt;/math&gt; for which &lt;math&gt;f(3z) = 7&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;y = \frac{x}{3}&lt;/math&gt;; then &lt;math&gt;f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1&lt;/math&gt;. Thus &lt;math&gt;f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0&lt;/math&gt;, and &lt;math&gt;z = -\frac{1}{3}, \frac{2}{9}&lt;/math&gt;. These sum up to &lt;math&gt;\boxed{\textbf{(B) }-\frac19}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Similar to Solution 1, we have &lt;math&gt;=81z^2+9z-6=0.&lt;/math&gt; The answer is the sum of the roots, which by [[Vieta's Formulas]] is &lt;math&gt;-\frac{b}{a}=-\frac{9}{81}=\boxed{\textbf{(B) }-\frac19}&lt;/math&gt;.<br /> <br /> ~dolphin7<br /> <br /> ==Solution 3==<br /> Set &lt;math&gt;f\left(\frac{x}{3} \right) = x^2+x+1=7&lt;/math&gt; to get &lt;math&gt;x^2+x-6=0.&lt;/math&gt; From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be &lt;math&gt;-1.&lt;/math&gt; Each root of this equation is &lt;math&gt;9&lt;/math&gt; times greater than a corresponding root of &lt;math&gt;f(3z) = 7&lt;/math&gt; (because &lt;math&gt;\frac{x}{3} = 3z&lt;/math&gt; gives &lt;math&gt;x = 9z&lt;/math&gt;), thus the sum of the roots in the equation &lt;math&gt;f(3z)=7&lt;/math&gt; is &lt;math&gt;-\frac{1}{9}&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B) }-\frac19}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Since we have &lt;math&gt;f(x/3)&lt;/math&gt;, &lt;math&gt;f(3z)&lt;/math&gt; occurs at &lt;math&gt;x=9z.&lt;/math&gt; Thus, &lt;math&gt;f(9z/3) = f(3z) = (9z)^2 + 9z + 1&lt;/math&gt;. We set this equal to 7:<br /> <br /> &lt;math&gt;81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0&lt;/math&gt;. For any quadratic &lt;math&gt;ax^2 + bx +c = 0&lt;/math&gt;, the sum of the roots is &lt;math&gt;-\frac{b}{a}&lt;/math&gt;. Thus, the sum of the roots of this equation is &lt;math&gt;-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}&lt;/math&gt;.<br /> <br /> ==Note==<br /> All solutions that apply Vieta must check if the discriminant is zero, which in this case it isn't.<br /> <br /> == Video Solutions ==<br /> <br /> https://youtu.be/qR85EBnpWV8<br /> <br /> <br /> https://m.youtube.com/watch?v=NyoLydoc3j8&amp;feature=youtu.be<br /> <br /> == Video Solution 2==<br /> https://youtu.be/3dfbWzOfJAI?t=1300<br /> <br /> ~ pi_is_3.14<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2000|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=163598 2018 AMC 10A Problems/Problem 16 2021-10-16T19:09:31Z <p>A.shyam.25: </p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, NW);<br /> &lt;/asy&gt;<br /> <br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; distinct line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> <br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths (since these circles intersect the hypotenuse twice) from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=163597 2018 AMC 10A Problems/Problem 16 2021-10-16T19:02:33Z <p>A.shyam.25: </p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, NW);<br /> &lt;/asy&gt;<br /> <br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; distinct line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> <br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths (since these circles intersect the hypotenuse twice) from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Solution 3 - Triangle Inequality==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> dot(&quot;$R$&quot;, (-6,14), N);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=163596 2018 AMC 10A Problems/Problem 16 2021-10-16T19:02:01Z <p>A.shyam.25: </p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, NW);<br /> &lt;/asy&gt;<br /> <br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; distinct line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> <br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths (since these circles intersect the hypotenuse twice) from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Solution 3 - Triangle Inequality==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> dot(&quot;$R$&quot;, (-6,12), N);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=163595 2018 AMC 10A Problems/Problem 16 2021-10-16T19:01:09Z <p>A.shyam.25: </p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, NW);<br /> &lt;/asy&gt;<br /> <br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; distinct line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> <br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths (since these circles intersect the hypotenuse twice) from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Solution 3 - Triangle Inequality==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> dot(&quot;$R$&quot;, (-3,10), N);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=163594 2018 AMC 10A Problems/Problem 16 2021-10-16T19:00:44Z <p>A.shyam.25: </p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, NW);<br /> &lt;/asy&gt;<br /> <br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; distinct line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> <br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths (since these circles intersect the hypotenuse twice) from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Solution 3 - Triangle Inequality==<br /> <br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> dot(&quot;$R$&quot;, (0,2), N);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_18&diff=163358 2009 AMC 10B Problems/Problem 18 2021-10-09T22:44:11Z <p>A.shyam.25: /* Solution 5 (Pythagorean Theorem) */</p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=8&lt;/math&gt; and &lt;math&gt;BC=6&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;\overline{ME}\perp\overline{AC}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AME&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{65}{8}<br /> \qquad<br /> \text{(B) } \frac{25}{3}<br /> \qquad<br /> \text{(C) } 9<br /> \qquad<br /> \text{(D) } \frac{75}{8}<br /> \qquad<br /> \text{(E) } \frac{85}{8}<br /> &lt;/math&gt;<br /> <br /> == Solution 1 (Coordinate Geo)==<br /> <br /> Set &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;(0,0)&lt;/math&gt;. Since &lt;math&gt;M&lt;/math&gt; is the midpoint of the diagonal, it would be &lt;math&gt;(4,-3)&lt;/math&gt;. The diagonal &lt;math&gt;AC&lt;/math&gt; would be the line &lt;math&gt;y = -\frac{3x}{4}&lt;/math&gt;. Since &lt;math&gt;ME&lt;/math&gt; is perpendicular to &lt;math&gt;AC&lt;/math&gt;, its line would be in the form &lt;math&gt;y = \frac{4x}{3} + b&lt;/math&gt;. Plugging in &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;-3&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; would give &lt;math&gt;b = \frac{25}{3}&lt;/math&gt;. To find the x-intercept of &lt;math&gt;y = \frac{4x}{3} + \frac{25}{3}&lt;/math&gt; we plug in &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt; and get &lt;math&gt;x = \frac{25}{4}&lt;/math&gt;. Then, using the Shoelace Formula for &lt;math&gt;(0,0)&lt;/math&gt; , &lt;math&gt;(4,-3)&lt;/math&gt;, and &lt;math&gt;(\frac{25}{4}, 0)&lt;/math&gt;, we find the area is &lt;math&gt;\frac{75}{8}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean theorem]] we have &lt;math&gt;AC=10&lt;/math&gt;, hence &lt;math&gt;AM=5&lt;/math&gt;. <br /> <br /> The triangles &lt;math&gt;AME&lt;/math&gt; and &lt;math&gt;ABC&lt;/math&gt; have the same angle at &lt;math&gt;A&lt;/math&gt; and a right angle, thus all their angles are equal, and therefore these two triangles are similar.<br /> <br /> The ratio of their sides is &lt;math&gt;\frac{AM}{AB} = \frac 58&lt;/math&gt;, hence the ratio of their areas is &lt;math&gt;\left( \frac 58 \right)^2 = \frac{25}{64}&lt;/math&gt;.<br /> <br /> And as the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{6\cdot 8}2 = 24&lt;/math&gt;, the area of triangle &lt;math&gt;AME&lt;/math&gt; is &lt;math&gt;24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }&lt;/math&gt;.<br /> <br /> == Solution 3 (Only Pythagorean Theorem) == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> Draw &lt;math&gt;EC&lt;/math&gt; as shown from the diagram. Since &lt;math&gt;AC&lt;/math&gt; is of length &lt;math&gt;10&lt;/math&gt;, we have that &lt;math&gt;AM&lt;/math&gt; is of length &lt;math&gt;5&lt;/math&gt;, because of the midpoint &lt;math&gt;M&lt;/math&gt;. Through the Pythagorean theorem, we know that &lt;math&gt;AE^2 = AM^2 + ME^2 \implies 25 + ME^2&lt;/math&gt;, which means &lt;math&gt;AE = \sqrt{25 + ME^2}&lt;/math&gt;. Define &lt;math&gt;ME&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt; for the sake of clarity. We know that &lt;math&gt;EB = 8 - \sqrt{25 + x^2}&lt;/math&gt;. From here, we know that &lt;math&gt;CE^2 = CB^2 + BE^2 = ME^2 + MC^2&lt;/math&gt;. From here, we can write the expression &lt;math&gt;6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}&lt;/math&gt;. Now, remember &lt;math&gt;CE \neq \frac{15}{4}&lt;/math&gt;. &lt;math&gt;x = \frac{15}{4} = ME&lt;/math&gt;, since we set &lt;math&gt;x = ME&lt;/math&gt; in the start of the solution. Now to find the area &lt;math&gt;\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We know &lt;math&gt;AM = \frac{10}{2} = 5&lt;/math&gt; by the Pythagorean theorem, and furthermore, &lt;math&gt;\triangle AME&lt;/math&gt; is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore, &lt;math&gt;ME = 5 \cdot \frac{6}{8} = \frac{15}{4}&lt;/math&gt;, and the area of the triangle is &lt;math&gt;5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Pythagorean Theorem)==<br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean Theorem]], we claim that &lt;math&gt;AC = 10&lt;/math&gt;. It then follows that &lt;math&gt;AM \cong MC = 5.&lt;/math&gt; <br /> <br /> Because we have &lt;math&gt;AM \cong MC, \angle AME \cong \angle CME,&lt;/math&gt; and reflexive side &lt;math&gt;EM&lt;/math&gt;, it follows that &lt;math&gt;\triangle AMC \cong \triangle CME.&lt;/math&gt; By CPCTC, we have &lt;math&gt;AE \cong EC.&lt;/math&gt; For the sake of simplicity, we'll call those side lengths &lt;math&gt;x&lt;/math&gt;. Also, since &lt;math&gt;AE = x,&lt;/math&gt; we get &lt;math&gt;BE = 8 - x.&lt;/math&gt; We can now set up the Pythagorean theorem on &lt;math&gt;\triangle EBC&lt;/math&gt;: &lt;cmath&gt;(8 - x)^2 + 6^2 = x^2.&lt;/cmath&gt; Combining like terms and simplifying gives &lt;math&gt;-16x + 100 = 0&lt;/math&gt; so &lt;math&gt;x = \frac{25}{4}.&lt;/math&gt; <br /> <br /> It helps to think that in order to find &lt;math&gt;[AME],&lt;/math&gt; we must have &lt;math&gt;\overline{MC}&lt;/math&gt; and &lt;math&gt;\overline{EM}.&lt;/math&gt; Let &lt;math&gt;EM = y.&lt;/math&gt; Applying the Pythagorean Theorem to &lt;math&gt;\triangle CME&lt;/math&gt; gives &lt;cmath&gt;5^2 + y^2 = \left(\frac{25}{4} \right)^2.&lt;/cmath&gt; Solving for &lt;math&gt;y&lt;/math&gt; (this is not that difficult) gives &lt;math&gt;y = \frac{15}{4}.&lt;/math&gt; So, the area of &lt;math&gt;\triangle AME&lt;/math&gt; is &lt;math&gt;\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_18&diff=163357 2009 AMC 10B Problems/Problem 18 2021-10-09T22:43:40Z <p>A.shyam.25: /* Solution 5 (Pythagorean Theorem) */</p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=8&lt;/math&gt; and &lt;math&gt;BC=6&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;\overline{ME}\perp\overline{AC}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AME&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{65}{8}<br /> \qquad<br /> \text{(B) } \frac{25}{3}<br /> \qquad<br /> \text{(C) } 9<br /> \qquad<br /> \text{(D) } \frac{75}{8}<br /> \qquad<br /> \text{(E) } \frac{85}{8}<br /> &lt;/math&gt;<br /> <br /> == Solution 1 (Coordinate Geo)==<br /> <br /> Set &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;(0,0)&lt;/math&gt;. Since &lt;math&gt;M&lt;/math&gt; is the midpoint of the diagonal, it would be &lt;math&gt;(4,-3)&lt;/math&gt;. The diagonal &lt;math&gt;AC&lt;/math&gt; would be the line &lt;math&gt;y = -\frac{3x}{4}&lt;/math&gt;. Since &lt;math&gt;ME&lt;/math&gt; is perpendicular to &lt;math&gt;AC&lt;/math&gt;, its line would be in the form &lt;math&gt;y = \frac{4x}{3} + b&lt;/math&gt;. Plugging in &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;-3&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; would give &lt;math&gt;b = \frac{25}{3}&lt;/math&gt;. To find the x-intercept of &lt;math&gt;y = \frac{4x}{3} + \frac{25}{3}&lt;/math&gt; we plug in &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt; and get &lt;math&gt;x = \frac{25}{4}&lt;/math&gt;. Then, using the Shoelace Formula for &lt;math&gt;(0,0)&lt;/math&gt; , &lt;math&gt;(4,-3)&lt;/math&gt;, and &lt;math&gt;(\frac{25}{4}, 0)&lt;/math&gt;, we find the area is &lt;math&gt;\frac{75}{8}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean theorem]] we have &lt;math&gt;AC=10&lt;/math&gt;, hence &lt;math&gt;AM=5&lt;/math&gt;. <br /> <br /> The triangles &lt;math&gt;AME&lt;/math&gt; and &lt;math&gt;ABC&lt;/math&gt; have the same angle at &lt;math&gt;A&lt;/math&gt; and a right angle, thus all their angles are equal, and therefore these two triangles are similar.<br /> <br /> The ratio of their sides is &lt;math&gt;\frac{AM}{AB} = \frac 58&lt;/math&gt;, hence the ratio of their areas is &lt;math&gt;\left( \frac 58 \right)^2 = \frac{25}{64}&lt;/math&gt;.<br /> <br /> And as the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{6\cdot 8}2 = 24&lt;/math&gt;, the area of triangle &lt;math&gt;AME&lt;/math&gt; is &lt;math&gt;24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }&lt;/math&gt;.<br /> <br /> == Solution 3 (Only Pythagorean Theorem) == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> Draw &lt;math&gt;EC&lt;/math&gt; as shown from the diagram. Since &lt;math&gt;AC&lt;/math&gt; is of length &lt;math&gt;10&lt;/math&gt;, we have that &lt;math&gt;AM&lt;/math&gt; is of length &lt;math&gt;5&lt;/math&gt;, because of the midpoint &lt;math&gt;M&lt;/math&gt;. Through the Pythagorean theorem, we know that &lt;math&gt;AE^2 = AM^2 + ME^2 \implies 25 + ME^2&lt;/math&gt;, which means &lt;math&gt;AE = \sqrt{25 + ME^2}&lt;/math&gt;. Define &lt;math&gt;ME&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt; for the sake of clarity. We know that &lt;math&gt;EB = 8 - \sqrt{25 + x^2}&lt;/math&gt;. From here, we know that &lt;math&gt;CE^2 = CB^2 + BE^2 = ME^2 + MC^2&lt;/math&gt;. From here, we can write the expression &lt;math&gt;6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}&lt;/math&gt;. Now, remember &lt;math&gt;CE \neq \frac{15}{4}&lt;/math&gt;. &lt;math&gt;x = \frac{15}{4} = ME&lt;/math&gt;, since we set &lt;math&gt;x = ME&lt;/math&gt; in the start of the solution. Now to find the area &lt;math&gt;\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We know &lt;math&gt;AM = \frac{10}{2} = 5&lt;/math&gt; by the Pythagorean theorem, and furthermore, &lt;math&gt;\triangle AME&lt;/math&gt; is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore, &lt;math&gt;ME = 5 \cdot \frac{6}{8} = \frac{15}{4}&lt;/math&gt;, and the area of the triangle is &lt;math&gt;5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Pythagorean Theorem)==<br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean Theorem]], we claim that &lt;math&gt;AC = 10&lt;/math&gt;. It then follows that &lt;math&gt;AM \cong MC = 5.&lt;/math&gt; <br /> <br /> Because we have &lt;math&gt;AM \cong MC, \angle AME \cong \angle CME,&lt;/math&gt; and reflexive side &lt;math&gt;EM&lt;/math&gt;, it follows that &lt;math&gt;\triangle AMC \cong \triangle CME.&lt;/math&gt; By CPCTC, we have &lt;math&gt;AE \cong EC.&lt;/math&gt; For the sake of simplicity, we'll call those side lengths &lt;math&gt;x&lt;/math&gt;. Also, since &lt;math&gt;AE = x,&lt;/math&gt; we get &lt;math&gt;BE = 8 - x.&lt;/math&gt; We can now set up the Pythagorean theorem on &lt;math&gt;\triangle EBC&lt;/math&gt;: &lt;math&gt;(8 - x)^2 + 6^2 = x^2.&lt;/math&gt; Combining like terms and simplifying gives &lt;math&gt;-16x + 100 = 0&lt;/math&gt; so &lt;math&gt;x = \frac{25}{4}.&lt;/math&gt; <br /> <br /> It helps to think that in order to find &lt;math&gt;[AME],&lt;/math&gt; we must have &lt;math&gt;\overline{MC}&lt;/math&gt; and &lt;math&gt;\overline{EM}.&lt;/math&gt; Let &lt;math&gt;EM = y.&lt;/math&gt; Applying the Pythagorean Theorem to &lt;math&gt;\triangle CME&lt;/math&gt; gives &lt;math&gt;5^2 + y^2 = \left(\frac{25}{4} \right)^2.&lt;/math&gt; Solving for &lt;math&gt;y&lt;/math&gt; (this is not that difficult) gives &lt;math&gt;y = \frac{15}{4}.&lt;/math&gt; So, the area of &lt;math&gt;\triangle AME&lt;/math&gt; is &lt;math&gt;\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_18&diff=163356 2009 AMC 10B Problems/Problem 18 2021-10-09T22:43:05Z <p>A.shyam.25: </p> <hr /> <div>== Problem ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=8&lt;/math&gt; and &lt;math&gt;BC=6&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;\overline{ME}\perp\overline{AC}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AME&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{65}{8}<br /> \qquad<br /> \text{(B) } \frac{25}{3}<br /> \qquad<br /> \text{(C) } 9<br /> \qquad<br /> \text{(D) } \frac{75}{8}<br /> \qquad<br /> \text{(E) } \frac{85}{8}<br /> &lt;/math&gt;<br /> <br /> == Solution 1 (Coordinate Geo)==<br /> <br /> Set &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;(0,0)&lt;/math&gt;. Since &lt;math&gt;M&lt;/math&gt; is the midpoint of the diagonal, it would be &lt;math&gt;(4,-3)&lt;/math&gt;. The diagonal &lt;math&gt;AC&lt;/math&gt; would be the line &lt;math&gt;y = -\frac{3x}{4}&lt;/math&gt;. Since &lt;math&gt;ME&lt;/math&gt; is perpendicular to &lt;math&gt;AC&lt;/math&gt;, its line would be in the form &lt;math&gt;y = \frac{4x}{3} + b&lt;/math&gt;. Plugging in &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;-3&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; would give &lt;math&gt;b = \frac{25}{3}&lt;/math&gt;. To find the x-intercept of &lt;math&gt;y = \frac{4x}{3} + \frac{25}{3}&lt;/math&gt; we plug in &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt; and get &lt;math&gt;x = \frac{25}{4}&lt;/math&gt;. Then, using the Shoelace Formula for &lt;math&gt;(0,0)&lt;/math&gt; , &lt;math&gt;(4,-3)&lt;/math&gt;, and &lt;math&gt;(\frac{25}{4}, 0)&lt;/math&gt;, we find the area is &lt;math&gt;\frac{75}{8}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean theorem]] we have &lt;math&gt;AC=10&lt;/math&gt;, hence &lt;math&gt;AM=5&lt;/math&gt;. <br /> <br /> The triangles &lt;math&gt;AME&lt;/math&gt; and &lt;math&gt;ABC&lt;/math&gt; have the same angle at &lt;math&gt;A&lt;/math&gt; and a right angle, thus all their angles are equal, and therefore these two triangles are similar.<br /> <br /> The ratio of their sides is &lt;math&gt;\frac{AM}{AB} = \frac 58&lt;/math&gt;, hence the ratio of their areas is &lt;math&gt;\left( \frac 58 \right)^2 = \frac{25}{64}&lt;/math&gt;.<br /> <br /> And as the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{6\cdot 8}2 = 24&lt;/math&gt;, the area of triangle &lt;math&gt;AME&lt;/math&gt; is &lt;math&gt;24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }&lt;/math&gt;.<br /> <br /> == Solution 3 (Only Pythagorean Theorem) == <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> Draw &lt;math&gt;EC&lt;/math&gt; as shown from the diagram. Since &lt;math&gt;AC&lt;/math&gt; is of length &lt;math&gt;10&lt;/math&gt;, we have that &lt;math&gt;AM&lt;/math&gt; is of length &lt;math&gt;5&lt;/math&gt;, because of the midpoint &lt;math&gt;M&lt;/math&gt;. Through the Pythagorean theorem, we know that &lt;math&gt;AE^2 = AM^2 + ME^2 \implies 25 + ME^2&lt;/math&gt;, which means &lt;math&gt;AE = \sqrt{25 + ME^2}&lt;/math&gt;. Define &lt;math&gt;ME&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt; for the sake of clarity. We know that &lt;math&gt;EB = 8 - \sqrt{25 + x^2}&lt;/math&gt;. From here, we know that &lt;math&gt;CE^2 = CB^2 + BE^2 = ME^2 + MC^2&lt;/math&gt;. From here, we can write the expression &lt;math&gt;6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}&lt;/math&gt;. Now, remember &lt;math&gt;CE \neq \frac{15}{4}&lt;/math&gt;. &lt;math&gt;x = \frac{15}{4} = ME&lt;/math&gt;, since we set &lt;math&gt;x = ME&lt;/math&gt; in the start of the solution. Now to find the area &lt;math&gt;\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We know &lt;math&gt;AM = \frac{10}{2} = 5&lt;/math&gt; by the Pythagorean theorem, and furthermore, &lt;math&gt;\triangle AME&lt;/math&gt; is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore, &lt;math&gt;ME = 5 \cdot \frac{6}{8} = \frac{15}{4}&lt;/math&gt;, and the area of the triangle is &lt;math&gt;5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Pythagorean Theorem)==<br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;<br /> path ortho = shift(M)*rotate(-90)*(A--C);<br /> pair Ep = intersectionpoint(ortho, A--B);<br /> draw( A--B--C--D--cycle );<br /> draw( A--C );<br /> draw( M--Ep );<br /> filldraw( A--M--Ep--cycle, lightgray, black );<br /> draw( rightanglemark(A,M,Ep) );<br /> draw( C--Ep );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,S);<br /> label(&quot;$M$&quot;,M,NW);<br /> &lt;/asy&gt;<br /> <br /> By the [[Pythagorean Theorem]], we claim that &lt;math&gt;AC = 10&lt;/math&gt;. It then follows that &lt;math&gt;AM \cong MC = 5.&lt;/math&gt; <br /> <br /> Because we have &lt;math&gt;AM \cong MC, \angle AME \cong \angle CME,&lt;/math&gt; and reflexive side &lt;math&gt;EM&lt;/math&gt;, it follows that &lt;math&gt;\triangle AMC \cong \triangle CME.&lt;/math&gt; By CPCTC, we have &lt;math&gt;AE \cong EC.&lt;/math&gt; For the sake of simplicity, we'll call those side lengths &lt;math&gt;x&lt;/math&gt;. Also, since &lt;math&gt;AE = x,&lt;/math&gt; we get &lt;math&gt;BE = 8 - x.&lt;/math&gt; We can now set up the Pythagorean theorem on &lt;math&gt;\triangle EBC&lt;/math&gt;: &lt;math&gt;(8 - x)^2 + 6^2 = x^2.&lt;/math&gt; Combining like terms and simplifying gives &lt;math&gt;-16x + 100 = 0&lt;/math&gt; so &lt;math&gt;x = \frac{25}{4}.&lt;/math&gt; <br /> <br /> It helps to think that in order to find &lt;math&gt;[AME],&lt;/math&gt; we must have &lt;math&gt;\overline{MC}&lt;/math&gt; and &lt;math&gt;\overline{EM}.&lt;/math&gt; Let &lt;math&gt;EM = y.&lt;/math&gt; Applying the Pythagorean Theorem to &lt;math&gt;\triangle CME&lt;/math&gt; gives &lt;math&gt;5^2 + y^2 = \left(\frac{25}{4} \right)^2.&lt;/math&gt; Solving for &lt;math&gt;y&lt;/math&gt; (this is not that difficult) gives &lt;math&gt;y = \frac{15}{4}.&lt;/math&gt; So, the area of &lt;math&gt;\triangle AME&lt;/math&gt; is &lt;math&gt;\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}}.&lt;/math&gt;<br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=Mathematics_websites&diff=151340 Mathematics websites 2021-04-11T16:41:35Z <p>A.shyam.25: </p> <hr /> <div>There are many great '''Mathematics websites''' around the [[internet]]. Here we organize a list of those sites we feel are best for students with high interest in [[mathematics]]. Note that [[mathematics forums]] are listed and discussed seperately.<br /> <br /> <br /> <br /> == Internet Resource Websites ==<br /> <br /> * [http://mathforum.org/library/ Internet Mathematics Library]<br /> * [http://archives.math.utk.edu/index.html Math Archives]<br /> * [http://www.math-atlas.org/ Math Atlas]<br /> <br /> == Websites For High School Students ==<br /> <br /> * [https://www.pims.math.ca/resources/publications/pi-sky Pi in the Sky] is a mathematics magazine for high school students.<br /> * [https://mathcsr.org Math and CS Research] is a math and computer science publication with a wide range of in-depth articles for high school students.<br /> <br /> === Websites for Olympiad students ===<br /> <br /> * [[Komal]] is a storied Hungarian [[math]] and [[physics]] journal. [http://www.komal.hu/info/bemutatkozas.e.shtml website].<br /> * [http://www.math.ust.hk/excalibur/ Mathematical Excalibur].<br /> * [http://reflections.awesomemath.org/ Mathematical Reflections] is a new online journal for Olympiad and collegiate mathematics.<br /> * [http://www.geometer.org/mathcircles/ Tom Davis's] site for [[math circles]] topics.<br /> * [http://mathworld.wolfram.com/ MathWorld] is a vast and well-maintained resource for math, science, and computer science professionals and students studying at a high level.<br /> <br /> == Websites For Math Enthusiasts ==<br /> <br /> * [[AoPS]] -- That's where you are now! [http://www.artofproblemsolving.com Home].<br /> * [https://brilliant.org/ Brilliant] contains a lot of problems and has a vast community of math lovers.<br /> * [https://www.expii.com/ Expii] is a relatively new addition but now it contains thousands of topics and members.<br /> * [https://www.omegalearn.org/middle-competition-math/ Middle School Competition Math Resources]<br /> * [https://www.omegalearn.org/high-competition-math/ High School Competition Math Resources]<br /> * [[Cut-the-knot]], a.k.a. Interactive Mathematics Miscellany and Puzzles, is a large and amazing site put together by [[Alexander Bogomolny]]. It includes an enormous number of [[mathematics articles]] and [[math games]] that are well-designed for teaching mathematical concepts. [http://www.cut-the-knot.org/index.shtml website].<br /> * [http://www.geometer.org/mathcircles/ Tom Davis's] site for [[math circles]] topics.<br /> * [https://mathvault.ca Math Vault] is a resource hub for people interested in learning more about higher mathematics.<br /> * [https://www.mathschase.com/ Math Chase] has times tables games, along with many other mathematics skill-based games.<br /> <br /> == Websites for Math Teachers ==<br /> * [http://www.ct4me.net/ CT4ME] is dedicated to promoting the use of [[technology]] in [[mathematics education]].<br /> * [https://nrich.maths.org/ NRICH] provides resources for mathematics teachers, parents and children.<br /> <br /> <br /> <br /> == Websites For Math History ==<br /> <br /> * [http://turnbull.dcs.st-and.ac.uk/history/index.html MacTudor History of Mathematics]<br /> * [[Wikipedia]] includes an enormous amount of information on the [http://en.wikipedia.org/wiki/History_of_math history of mathematics].<br /> <br /> == Websites For Mathematicians ==<br /> <br /> * [http://mathworld.wolfram.com/ MathWorld] is a vast and well-maintained resource for math, science, and computer science professionals and students studying at a high level.<br /> * [[Wikipedia]] includes articles about noncontroversial, published [http://en.wikipedia.org/wiki/Mathematics mathematics].<br /> <br /> == See also ==<br /> <br /> * [[Mathematics competitions]]<br /> * [[Mathematics forums]]<br /> * [[Mathematics news]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=Mathematics_websites&diff=151339 Mathematics websites 2021-04-11T16:39:39Z <p>A.shyam.25: Khan Academy</p> <hr /> <div>There are many great '''Mathematics websites''' around the [[internet]]. Here we organize a list of those sites we feel are best for students with high interest in [[mathematics]]. Note that [[mathematics forums]] are listed and discussed seperately.<br /> <br /> <br /> <br /> == Internet Resource Websites ==<br /> <br /> * [http://mathforum.org/library/ Internet Mathematics Library]<br /> * [http://archives.math.utk.edu/index.html Math Archives]<br /> * [http://www.math-atlas.org/ Math Atlas]<br /> <br /> == Websites For High School Students ==<br /> <br /> * [https://www.pims.math.ca/resources/publications/pi-sky Pi in the Sky] is a mathematics magazine for high school students.<br /> * [https://mathcsr.org Math and CS Research] is a math and computer science publication with a wide range of in-depth articles for high school students.<br /> <br /> === Websites for Olympiad students ===<br /> <br /> * [[Komal]] is a storied Hungarian [[math]] and [[physics]] journal. [http://www.komal.hu/info/bemutatkozas.e.shtml website].<br /> * [http://www.math.ust.hk/excalibur/ Mathematical Excalibur].<br /> * [http://reflections.awesomemath.org/ Mathematical Reflections] is a new online journal for Olympiad and collegiate mathematics.<br /> * [http://www.geometer.org/mathcircles/ Tom Davis's] site for [[math circles]] topics.<br /> * [http://mathworld.wolfram.com/ MathWorld] is a vast and well-maintained resource for math, science, and computer science professionals and students studying at a high level.<br /> <br /> == Websites For Math Enthusiasts ==<br /> <br /> * [[AoPS]] -- That's where you are now! [http://www.artofproblemsolving.com Home].<br /> * [https://brilliant.org/ Brilliant] contains a lot of problems and has a vast community of math lovers.<br /> * [https://www.expii.com/ Expii] is a relatively new addition but now it contains thousands of topics and members.<br /> * [https://www.omegalearn.org/middle-competition-math/ Middle School Competition Math Resources]<br /> * [https://www.omegalearn.org/high-competition-math/ High School Competition Math Resources]<br /> * [[Cut-the-knot]], a.k.a. Interactive Mathematics Miscellany and Puzzles, is a large and amazing site put together by [[Alexander Bogomolny]]. It includes an enormous number of [[mathematics articles]] and [[math games]] that are well-designed for teaching mathematical concepts. [http://www.cut-the-knot.org/index.shtml website].<br /> * [http://www.geometer.org/mathcircles/ Tom Davis's] site for [[math circles]] topics.<br /> * [https://mathvault.ca Math Vault] is a resource hub for people interested in learning more about higher mathematics.<br /> * [https://www.mathschase.com/ Math Chase] has times tables games, along with many other mathematics skill-based games.<br /> * [https://www.khanacademy.org/] is a great resource for learning the basic concepts of a topic, and for review. <br /> <br /> == Websites for Math Teachers ==<br /> * [http://www.ct4me.net/ CT4ME] is dedicated to promoting the use of [[technology]] in [[mathematics education]].<br /> * [https://nrich.maths.org/ NRICH] provides resources for mathematics teachers, parents and children.<br /> <br /> <br /> <br /> == Websites For Math History ==<br /> <br /> * [http://turnbull.dcs.st-and.ac.uk/history/index.html MacTudor History of Mathematics]<br /> * [[Wikipedia]] includes an enormous amount of information on the [http://en.wikipedia.org/wiki/History_of_math history of mathematics].<br /> <br /> == Websites For Mathematicians ==<br /> <br /> * [http://mathworld.wolfram.com/ MathWorld] is a vast and well-maintained resource for math, science, and computer science professionals and students studying at a high level.<br /> * [[Wikipedia]] includes articles about noncontroversial, published [http://en.wikipedia.org/wiki/Mathematics mathematics].<br /> <br /> == See also ==<br /> <br /> * [[Mathematics competitions]]<br /> * [[Mathematics forums]]<br /> * [[Mathematics news]]</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=147239 User:Piphi 2021-02-17T03:49:34Z <p>A.shyam.25: </p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;460&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;&lt;font color=&quot;black&quot;&gt;<br /> Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> <br /> Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#333333;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;#f0f2f3&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;#f0f2f3&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|91.6|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|97|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|50.5|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|21.65|width=100%}}&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=User:Onafets&diff=144295 User:Onafets 2021-02-01T02:48:37Z <p>A.shyam.25: </p> <hr /> <div>&lt;center&gt;<br /> Current status: Not Postbanned :D<br /> <br /> Activity status: Active<br /> <br /> PMs: Open<br /> -----<br /> Pronouns: He/Him/His<br /> <br /> Nicknames: SK<br /> -----<br /> <br /> Forums:<br /> <br /> Everything Makes No Cents (Peak 10th) [ADMIN]<br /> <br /> The Brawl Stars Forum (Peak 20th) [Active/Trusted User]<br /> -----<br /> <br /> Facts about Onafets:<br /> <br /> 1. Onafets loves centslordm &amp; Gmaas. But he cant decide who is more pr0.<br /> <br /> 2. Onafets has been postbanned twice now. (And Messaged by Sheriff at least 10 times)<br /> <br /> 3. Onafets plays brawl stars, now at 11,500 trophies. (but centslordm is at 20,000 so he sad)<br /> <br /> 4. Onafets rickrolls. <br /> <br /> 5. Onafets writes facts about himself because he is a lonely boi.<br /> <br /> 6. Onafets has never taken AMC, MATHCOUNTS, Never played FTW! (if anyone claims I have taken this, they are wrong!)<br /> <br /> 7. Onafets has a derp blog that died a month ago. Onafets removed the css because he didn't think it was good enough, and now he will never get it back.<br /> <br /> 8. Onafets is an active participant on cents' blog (won't say name because it keeps changing lol)</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=User:Centslordm&diff=143084 User:Centslordm 2021-01-24T18:17:41Z <p>A.shyam.25: </p> <hr /> <div>Facts about The Lord of the ¢:<br /> <br /> 0. centslordm has an epic blog with cewl CSS [https://artofproblemsolving.com/community/c1229127 here]<br /> <br /> 1. centslordm owns everything in the world, second only to the great GMAAS. <br /> <br /> 2. centslordmwants you to call him ¢, it makes sense.<br /> <br /> 3. ¢ is the embodiment of stupidity.<br /> <br /> 4. ¢ says the word &quot;intriguing&quot; a lot. A LOT.<br /> <br /> 5. ¢ has said the word &quot;intriguing&quot; ∞ * TREE(3) times in 3 seconds.<br /> <br /> 6. ¢ has over 4,000 upvotes received. He is one of only a few users to have done so, even with about only ~2,200 posts. <br /> <br /> 7. ¢ is bad at Brawl Stars. He can't even get a rank 30.<br /> <br /> 8. ¢ has some really cool CSS located [https://artofproblemsolving.com/community/c1621357h2356796_test_post here]<br /> <br /> His nicknames include:<br /> <br /> - cents<br /> - ¢<br /> - Burning Water<br /> - Embodiment of Stupidity<br /> - DollarGod<br /> - MyNameIsNot<br /> - EatYeetDeleteRepeat<br /> - The Dum One<br /> - simp lord</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=139497 User:Piphi 2020-12-12T21:40:13Z <p>A.shyam.25: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;409&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#B1B2B3;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> <br /> Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#727373;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;#f0f2f3&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;#f0f2f3&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|80.2|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|64.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|47.5|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|20.36|width=100%}}&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=Distributive_property&diff=128947 Distributive property 2020-07-23T14:13:19Z <p>A.shyam.25: </p> <hr /> <div>Given two [[binary operation]]s &lt;math&gt;\times&lt;/math&gt; and &lt;math&gt;+&lt;/math&gt; acting on a set &lt;math&gt;S&lt;/math&gt;, we say that &lt;math&gt;\times&lt;/math&gt; has the '''distributive property''' over &lt;math&gt;+&lt;/math&gt; (or &lt;math&gt;\times&lt;/math&gt; ''distributes over'' &lt;math&gt;+&lt;/math&gt;) if, for all &lt;math&gt;a, b, c \in S&lt;/math&gt; we have &lt;math&gt;a\times(b + c) = (a\times b) + (a \times c)&lt;/math&gt; and &lt;math&gt;(a + b) \times c = (a \times c) + (b \times c)&lt;/math&gt;. <br /> <br /> Note that if the [[operation]] &lt;math&gt;\times&lt;/math&gt; is [[commutative property | commutative]], these two conditions are the same, but if &lt;math&gt;\times&lt;/math&gt; does not commute then we could have operations which ''left-distribute'' but do not ''right-distribute'', or vice-versa.<br /> <br /> Key Note - This isn't an example of the Distributive Property!<br /> &lt;cmath&gt;a(b \times c) = ab \times ac.&lt;/cmath&gt; This is actually using the Associative Property, not the Distributive Property.<br /> <br /> Also note that there is no particular reason that distributivity should be one-way, as it is with conventional multiplication and addition. For example, the [[set]] operations [[union]] (&lt;math&gt;\cup&lt;/math&gt;) and [[intersection]] (&lt;math&gt;\cap&lt;/math&gt;) distribute over each other: for any sets &lt;math&gt;A, B, C&lt;/math&gt; we have &lt;math&gt;A \cup (B \cap C) = (A \cup B) \cap (A \cup C)&lt;/math&gt; and &lt;math&gt;A \cap(B \cup C) = (A \cap B) \cup (A \cap C)&lt;/math&gt;. <br /> <br /> (In fact, this is a special case of a more general setting: in a [[distributive lattice]], each of the operations [[meet]] and [[join]] distributes over the other. Meet and join correspond to union and intersection when the lattice is a [[boolean lattice]].)<br /> <br /> <br /> <br /> {{stub}}</div> A.shyam.25 https://artofproblemsolving.com/wiki/index.php?title=Distributive_property&diff=128946 Distributive property 2020-07-23T14:11:36Z <p>A.shyam.25: </p> <hr /> <div>Given two [[binary operation]]s &lt;math&gt;\times&lt;/math&gt; and &lt;math&gt;+&lt;/math&gt; acting on a set &lt;math&gt;S&lt;/math&gt;, we say that &lt;math&gt;\times&lt;/math&gt; has the '''distributive property''' over &lt;math&gt;+&lt;/math&gt; (or &lt;math&gt;\times&lt;/math&gt; ''distributes over'' &lt;math&gt;+&lt;/math&gt;) if, for all &lt;math&gt;a, b, c \in S&lt;/math&gt; we have &lt;math&gt;a\times(b + c) = (a\times b) + (a \times c)&lt;/math&gt; and &lt;math&gt;(a + b) \times c = (a \times c) + (b \times c)&lt;/math&gt;. <br /> <br /> Note that if the [[operation]] &lt;math&gt;\times&lt;/math&gt; is [[commutative property | commutative]], these two conditions are the same, but if &lt;math&gt;\times&lt;/math&gt; does not commute then we could have operations which ''left-distribute'' but do not ''right-distribute'', or vice-versa.<br /> <br /> Key Note - This isn't an example of the Distributive Property!<br /> &lt;cmath&gt;a(b \times c) = ab \times ac.&lt;/cmath&gt; This is actually using the Associative Property, NOT THE DISTRIBUTIVE PROPERTY!<br /> <br /> Also note that there is no particular reason that distributivity should be one-way, as it is with conventional multiplication and addition. For example, the [[set]] operations [[union]] (&lt;math&gt;\cup&lt;/math&gt;) and [[intersection]] (&lt;math&gt;\cap&lt;/math&gt;) distribute over each other: for any sets &lt;math&gt;A, B, C&lt;/math&gt; we have &lt;math&gt;A \cup (B \cap C) = (A \cup B) \cap (A \cup C)&lt;/math&gt; and &lt;math&gt;A \cap(B \cup C) = (A \cap B) \cup (A \cap C)&lt;/math&gt;. <br /> <br /> (In fact, this is a special case of a more general setting: in a [[distributive lattice]], each of the operations [[meet]] and [[join]] distributes over the other. Meet and join correspond to union and intersection when the lattice is a [[boolean lattice]].)<br /> <br /> <br /> <br /> {{stub}}</div> A.shyam.25