https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Superls&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-01T08:53:47Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem&diff=141764 Shoelace Theorem 2021-01-09T17:24:08Z <p>Superls: /* Proof 2 */</p> <hr /> <div>The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the [[Cartesian coordinate system | coordinates]] of its [[vertex|vertices]].<br /> <br /> ==Theorem==<br /> Suppose the polygon &lt;math&gt;P&lt;/math&gt; has vertices &lt;math&gt;(a_1, b_1)&lt;/math&gt;, &lt;math&gt;(a_2, b_2)&lt;/math&gt;, ... , &lt;math&gt;(a_n, b_n)&lt;/math&gt;, listed in clockwise order. Then the area (&lt;math&gt;A&lt;/math&gt;) of &lt;math&gt;P&lt;/math&gt; is<br /> <br /> &lt;cmath&gt;A = \dfrac{1}{2} \left|(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1) \right|&lt;/cmath&gt;<br /> <br /> You can also go counterclockwise order, as long as you find the absolute value of the answer.<br /> <br /> The Shoelace Theorem gets its name because if one lists the coordinates in a column,<br /> &lt;cmath&gt;\begin{align*}<br /> (a_1 &amp;, b_1) \\<br /> (a_2 &amp;, b_2) \\<br /> &amp; \vdots \\<br /> (a_n &amp;, b_n) \\<br /> (a_1 &amp;, b_1) \\<br /> \end{align*}&lt;/cmath&gt;<br /> and marks the pairs of coordinates to be multiplied, <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> string[] subscripts={&quot;$1$&quot;,&quot;$2$&quot;,&quot; &quot;,&quot;$n$&quot;,&quot;$1$&quot;};<br /> for(int i=1; i &lt; 6; ++i)<br /> {<br /> label(i==3 ? &quot;$\vdots$&quot; : &quot;$a$&quot;,(0,-i*.7));<br /> label(i==3 ? &quot;$\vdots$&quot; : &quot;$b$&quot;,(1.2,-i*.7));<br /> label(subscripts[i-1],(0,-i*.7),SE,fontsize(9pt)); <br /> label(subscripts[i-1],(1.2,-i*.7),SE,fontsize(9pt)); <br /> }<br /> for(int i=1; i&lt;5; ++i)<br /> draw((0.3,-i*.7)--(1,-(i+1)*.7));<br /> <br /> pair c=(1.2,0);<br /> label(&quot;$-$&quot;,shift(c)*(1.2,-2.1));<br /> label(&quot;$A=\frac12$&quot;,shift(-c)*(0,-2.1));<br /> draw(shift(-1/3*c)*((0,-.5)--(0,-3.9)));<br /> draw(shift(13/3*c)*((0,-.5)--(0,-3.9)));<br /> <br /> for(int i=1; i &lt; 6; ++i)<br /> {<br /> label(i==3 ? &quot;$\vdots$&quot; : &quot;$a$&quot;,shift(3*c)*(0,-i*.7));<br /> label(i==3 ? &quot;$\vdots$&quot; : &quot;$b$&quot;,shift(3*c)*(1.2,-i*.7));<br /> label(subscripts[i-1],shift(3*c)*(0,-i*.7),SE,fontsize(9pt)); <br /> label(subscripts[i-1],shift(3*c)*(1.2,-i*.7),SE,fontsize(9pt)); <br /> }<br /> for(int i=1; i&lt;5; ++i)<br /> draw(shift(3*c)*(0.3,-(i+1)*.7)--shift(3*c)*(1,-i*.7));<br /> &lt;/asy&gt;<br /> the resulting image looks like laced-up shoes.<br /> <br /> ==Other Forms==<br /> This can also be written in form of a summation &lt;cmath&gt;A = \dfrac{1}{2} \left|\sum_{i=1}^n{(x_{i+1}+x_i)(y_{i+1}-y_i)}\right|&lt;/cmath&gt;<br /> or in terms of determinants as &lt;cmath&gt;A = \dfrac{1}{2} \left|\sum_{i=1}^n{\det\begin{pmatrix}x_i&amp;x_{i+1}\\y_i&amp;y_{i+1}\end{pmatrix}}\right|&lt;/cmath&gt;<br /> which is useful in the &lt;math&gt;3D&lt;/math&gt; variant of the Shoelace theorem,<br /> <br /> or as the special case of Green's Theorem <br /> &lt;center&gt;<br /> &lt;math&gt;\tilde{A}=\iint\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=&lt;/math&gt;&lt;font size=&quot;6&quot;&gt;∳&lt;/font&gt;&lt;math&gt;(Ldx+Mdy)&lt;/math&gt;<br /> &lt;/center&gt;<br /> where &lt;math&gt;L=-y&lt;/math&gt; and &lt;math&gt;M=0&lt;/math&gt; so &lt;math&gt;\tilde{A}=A&lt;/math&gt;.<br /> <br /> ==Proof 1==<br /> Claim 1: The area of a triangle with coordinates &lt;math&gt;A(x_1, y_1)&lt;/math&gt;, &lt;math&gt;B(x_2, y_2)&lt;/math&gt;, and &lt;math&gt;C(x_3, y_3)&lt;/math&gt; is &lt;math&gt;\frac{|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|}{2}&lt;/math&gt;.<br /> <br /> ===Proof of claim 1:===<br /> <br /> Writing the coordinates in 3D and translating &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(0, 0, 0)&lt;/math&gt; we get the new coordinates &lt;math&gt;A'(0, 0, 0)&lt;/math&gt;, &lt;math&gt;B(x_2-x_1, y_2-y_1, 0)&lt;/math&gt;, and &lt;math&gt;C(x_3-x_1, y_3-y_1, 0)&lt;/math&gt;. Now if we let &lt;math&gt;\vec{b}=(x_2-x_1 \quad y_2-y_1 \quad 0)&lt;/math&gt; and &lt;math&gt;\vec{c}=(x_3-x_1 \quad y_3-y_1 \quad 0)&lt;/math&gt; then by definition of the cross product &lt;math&gt;[ABC]=\frac{||\vec{b} \times \vec{c}||}{2}=\frac{1}{2}||(0 \quad 0 \quad x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)||=\frac{x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2}{2}&lt;/math&gt;.<br /> <br /> ===Proof:===<br /> <br /> We will proceed with induction.<br /> <br /> By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon &lt;math&gt;A_1A_2A_3...A_n&lt;/math&gt; then it is also true for &lt;math&gt;A_1A_2A_3...A_nA_{n+1}&lt;/math&gt;.<br /> <br /> We cut &lt;math&gt;A_1A_2A_3...A_nA_{n+1}&lt;/math&gt; into two polygons, &lt;math&gt;A_1A_2A_3...A_n&lt;/math&gt; and &lt;math&gt;A_1A_nA_{n+1}&lt;/math&gt;. Let the coordinates of point &lt;math&gt;A_i&lt;/math&gt; be &lt;math&gt;(x_i, y_i)&lt;/math&gt;. Then, applying the shoelace theorem on &lt;math&gt;A_1A_2A_3...A_n&lt;/math&gt; and &lt;math&gt;A_1A_nA_{n+1}&lt;/math&gt; we get<br /> <br /> &lt;cmath&gt;[A_1A_2A_3...A_n]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)&lt;/cmath&gt;<br /> &lt;cmath&gt;[A_1A_nA_{n+1}]=\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)&lt;/cmath&gt;<br /> <br /> Hence<br /> <br /> &lt;cmath&gt;[A_1A_2A_3...A_nA_{n+1}]=[A_1A_2A_3...A_n]+[A_1A_nA_{n+1}]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)+\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)&lt;/cmath&gt;<br /> &lt;cmath&gt;=\frac{1}{2}((x_2y_1+x_3y_2+...+x_{n+1}y_n+x_1y_{n+1})-(x_1y_2+x_2y_3+...+x_ny_{n+1}+x_{n+1}y_1))=\boxed{\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i)}&lt;/cmath&gt;<br /> <br /> As claimed.<br /> <br /> ~ShreyJ<br /> <br /> ==Proof 2==<br /> Let &lt;math&gt;\Omega&lt;/math&gt; be the set of points belonging to the polygon.<br /> We have that<br /> &lt;cmath&gt;<br /> A=\int_{\Omega}\alpha,<br /> &lt;/cmath&gt;<br /> where &lt;math&gt;\alpha=dx\wedge dy&lt;/math&gt;.<br /> The volume form &lt;math&gt;\alpha&lt;/math&gt; is an exact form since &lt;math&gt;d\omega=\alpha&lt;/math&gt;, where<br /> &lt;cmath&gt;<br /> \omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}<br /> &lt;/cmath&gt;<br /> Using this substitution, we have<br /> &lt;cmath&gt;<br /> \int_{\Omega}\alpha=\int_{\Omega}d\omega.<br /> &lt;/cmath&gt;<br /> Next, we use the Theorem of Stokes to obtain<br /> &lt;cmath&gt;<br /> \int_{\Omega}d\omega=\int_{\partial\Omega}\omega.<br /> &lt;/cmath&gt;<br /> We can write &lt;math&gt;\partial \Omega=\bigcup A(i)&lt;/math&gt;, where &lt;math&gt;A(i)&lt;/math&gt; is the line<br /> segment from &lt;math&gt;(x_i,y_i)&lt;/math&gt; to &lt;math&gt;(x_{i+1},y_{i+1})&lt;/math&gt;. With this notation,<br /> we may write<br /> &lt;cmath&gt;<br /> \int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega.<br /> &lt;/cmath&gt;<br /> If we substitute for &lt;math&gt;\omega&lt;/math&gt;, we obtain<br /> &lt;cmath&gt;<br /> \sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.<br /> &lt;/cmath&gt;<br /> If we parameterize, we get <br /> &lt;cmath&gt;<br /> \frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.<br /> &lt;/cmath&gt;<br /> Performing the integration, we get <br /> &lt;cmath&gt;<br /> \frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)-<br /> (y_{i}+y_{i+1})(x_{i+1}-x_i)].<br /> &lt;/cmath&gt;<br /> More algebra yields the result<br /> &lt;cmath&gt;<br /> \frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).<br /> &lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.<br /> <br /> See page 281 in this book (in the Polygon Area section.)<br /> https://cses.fi/book/book.pdf<br /> <br /> (The only thing that needs to be modified in this proof is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)<br /> <br /> == Problems ==<br /> === Introductory ===<br /> In right triangle &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\angle ACB=90^{\circ}&lt;/math&gt;, &lt;math&gt;AC=2&lt;/math&gt;, and &lt;math&gt;BC=3&lt;/math&gt;. [[Median]]s &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; are drawn to sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at point &lt;math&gt;F&lt;/math&gt;. Find the area of &lt;math&gt;\triangle ABF&lt;/math&gt;.<br /> === Exploratory ===<br /> Observe that &lt;cmath&gt;\frac12\left|\det\begin{pmatrix}<br /> x_1 &amp; y_1\\<br /> x_2 &amp; y_2<br /> \end{pmatrix}\right|&lt;/cmath&gt; is the area of a triangle with vertices &lt;math&gt;(x_1,y_1),(x_2,y_2),(0,0)&lt;/math&gt; and &lt;cmath&gt;\frac16\left|\det\begin{pmatrix}<br /> x_1 &amp; y_1 &amp; z_1\\<br /> x_2 &amp; y_2 &amp; z_2\\<br /> x_3 &amp; y_3 &amp; z_3<br /> \end{pmatrix}\right|&lt;/cmath&gt; is the volume of a tetrahedron with vertices &lt;math&gt;(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)&lt;/math&gt;. Does a similar formula hold for &lt;math&gt;n&lt;/math&gt;Dimensional triangles for any &lt;math&gt;n&lt;/math&gt;? If so how can we use this to derive the &lt;math&gt;n&lt;/math&gt;D Shoelace Formula?<br /> <br /> == External Links==<br /> A good explanation and exploration into why the theorem works by James Tanton:<br /> [http://www.jamestanton.com/wp-content/uploads/2012/03/Cool-Math-Essay_June-2014_SHOELACE-FORMULA.pdf]<br /> <br /> Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: [http://media.icys2018.com/2018/04/IndonesiaPatrickNicholas105190.pdf]<br /> <br /> Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: [https://mathematica.stackexchange.com/a/26015]<br /> <br /> <br /> <br /> <br /> <br /> [[Category:Geometry]]<br /> [[Category:Theorems]]<br /> AOPS</div> Superls https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=135761 2015 AMC 8 Problems/Problem 17 2020-10-24T21:20:32Z <p>Superls: /* Solution 4 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> Somehow we get &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school.<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;, giving<br /> &lt;cmath&gt;(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> Hence, &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have &lt;math&gt;2x/3 = 18&lt;/math&gt; miles per hour. Solving for x gives us 27 miles per hour. Because &lt;math&gt;20&lt;/math&gt; minutes is a third of an hour, the distance would then be &lt;math&gt;9&lt;/math&gt; miles (&lt;math&gt;(D)9&lt;/math&gt;).<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Superls https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=135760 2015 AMC 8 Problems/Problem 17 2020-10-24T21:19:34Z <p>Superls: /* Solution 2 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> Somehow we get &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school.<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;, giving<br /> &lt;cmath&gt;(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> Hence, &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have &lt;math&gt;2x/3 = 18&lt;/math&gt; miles per hour. Solving for x gives us 27 miles per hour. Because &lt;math&gt;20&lt;/math&gt; minutes is a third of an hour, the distance would then be &lt;math&gt;9&lt;/math&gt; miles. &lt;math&gt;(D)9&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Superls https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=135759 2015 AMC 8 Problems/Problem 17 2020-10-24T21:19:03Z <p>Superls: /* Solution 1 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> Somehow we get &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;, giving<br /> &lt;cmath&gt;(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> Hence, &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have &lt;math&gt;2x/3 = 18&lt;/math&gt; miles per hour. Solving for x gives us 27 miles per hour. Because &lt;math&gt;20&lt;/math&gt; minutes is a third of an hour, the distance would then be &lt;math&gt;9&lt;/math&gt; miles. &lt;math&gt;(D)9&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Superls https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=135758 2015 AMC 8 Problems/Problem 25 2020-10-24T21:14:06Z <p>Superls: /* Solution 1 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown:<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the &lt;math&gt;4&lt;/math&gt; big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by &lt;math&gt;\mathrm{AA}&lt;/math&gt; similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be &lt;math&gt;x&lt;/math&gt;; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;. Note that the other solution we got, namely, &lt;math&gt;x=\dfrac{5+\sqrt 5} 2&lt;/math&gt;, is the length of the segment &lt;math&gt;5-x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;5-x&lt;/math&gt; together sum to &lt;math&gt;5&lt;/math&gt;, the side of the length of the large square, and similarly, the sum of the solutions is &lt;math&gt;5&lt;/math&gt;. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be &lt;math&gt;x&lt;/math&gt;, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> Thus the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ===Solution 2(Contest Solution)=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of &lt;math&gt;25-(4\cdot 4)=9&lt;/math&gt; so it has length &lt;math&gt;\sqrt{9}=3&lt;/math&gt; and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely &lt;math&gt;\frac{3 \cdot 1}{2}&lt;/math&gt;. Thus, the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. Furthermore, the area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ~Clarification on the process of solution by mathboy282<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> So the area of the square we need is &lt;math&gt;25- (4+6) = 15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/WNiJWmKCfj0 - Happytwin<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Superls