https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Joeya&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-13T10:06:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=147229 2021 AMC 12B Problems/Problem 8 2021-02-17T01:50:44Z <p>Joeya: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);<br /> draw(O--A--B);<br /> draw(O--R);<br /> draw(O--L);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$B$&quot;, B, N);<br /> label(&quot;$R$&quot;, R, NE);<br /> label(&quot;$L$&quot;, L, N);<br /> label(&quot;$O$&quot;, O, S);<br /> label(&quot;$d$&quot;, O--A, W);<br /> label(&quot;$2d$&quot;, A--B, W*2+0.5*N);<br /> label(&quot;$r$&quot;, O--R, S);<br /> label(&quot;$r$&quot;, O--L, S*0.5 + 1.5 * E);<br /> dot(O);<br /> dot(A);<br /> dot(B);<br /> dot(R);<br /> dot(L);<br /> <br /> draw(circle((0, 4), 4));<br /> draw((-3.873, 3) -- (3.873, 3));<br /> draw((-3.873, 5) -- (3.873, 5));<br /> draw((-2.645, 7) -- (2.645, 7));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> Since two parallel chords have the same length (&lt;math&gt;38&lt;/math&gt;), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle RAO&lt;/math&gt; on the diagram)<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle LBO&lt;/math&gt; on the diagram)<br /> <br /> By the Pythagorean theorem, we can create the following system of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{\textbf{(B)}\ 6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya and diagram by Jamess2022(burntTacos).<br /> (Someone fix the diagram if possible. -&lt;i&gt; Done. &lt;/i&gt;)<br /> <br /> ==Solution 2 (Coordinates)==<br /> <br /> Because we know that the equation of a circle is &lt;math&gt;(x-a)^2 + (y-b)^2 = r^2&lt;/math&gt; where the center of the circle is &lt;math&gt;(a, b)&lt;/math&gt; and the radius is &lt;math&gt;r&lt;/math&gt;, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is &lt;math&gt;x^2 + y^2 = r^2&lt;/math&gt;. Now, we can set the distance between the chords as &lt;math&gt;2d&lt;/math&gt; so the distance from the chord with length 38 to the diameter is &lt;math&gt;d&lt;/math&gt;. <br /> <br /> Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:<br /> <br /> &lt;math&gt;(19, d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(19, -d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(17, -3d)&lt;/math&gt;<br /> <br /> <br /> Now, we can plug one of the first two value in as well as the last one to get the following equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Subtracting these two equations, we get &lt;math&gt;19^2 - 17^2 = 8d^2&lt;/math&gt; - therefore, we get &lt;math&gt;72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3&lt;/math&gt;. We want to find &lt;math&gt;2d = 6&lt;/math&gt; because that's the distance between two chords. So, our answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ~Tony_Li2007<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=VzwxbsuSQ80<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtu.be/yxt8-rUUosI<br /> <br /> == Video Solution by OmegaLearn (Circular Geometry) ==<br /> https://youtu.be/XNYq4ZMBtBU<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2021|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_9&diff=146536 2011 AMC 10B Problems/Problem 9 2021-02-14T01:36:37Z <p>Joeya: /* Solution 3 (Shortcut) */</p> <hr /> <div>== Problem==<br /> <br /> The area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt;. Segment &lt;math&gt;DE&lt;/math&gt; is perpendicular to segment &lt;math&gt;AB&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;? &lt;p&gt;<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3));<br /> pair[] ps={A,B,C,D,E};<br /> <br /> draw(A--B--C--cycle);<br /> draw(E--D);<br /> draw(rightanglemark(E,D,B));<br /> <br /> dot(ps);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,NE);<br /> label(&quot;$3$&quot;,midpoint(A--C),NW);<br /> label(&quot;$4$&quot;,midpoint(C--B),NE);<br /> label(&quot;$5$&quot;,midpoint(A--B),SW);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Therefore &lt;math&gt;DE = \frac{3}{4} BD&lt;/math&gt;. Find the areas of the triangles.<br /> &lt;cmath&gt;\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6&lt;/cmath&gt;<br /> &lt;cmath&gt;\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2&lt;/cmath&gt;<br /> The area of &lt;math&gt;\triangle EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{3}{8} BD^2 &amp;= 6 \times \frac{1}{3}\\<br /> 9BD^2 &amp;= 48\\<br /> BD^2 &amp;= \frac{16}{3}\\<br /> BD &amp;= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Since the area of &lt;math&gt;\triangle EBD&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus,<br /> &lt;cmath&gt;\frac{1}{3}[\triangle ABC] = [\triangle EBD]&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB&lt;/cmath&gt;<br /> In order to scale the sides of ED and DB to make &lt;math&gt;\frac{1}{3}&lt;/math&gt; (since the ratios of sides are the same), we take the square root of &lt;math&gt;\frac{1}{3} = \frac{\sqrt(3)}{3}&lt;/math&gt; to scale each side by the same amount. <br /> <br /> Thus &lt;math&gt;BD = 4 \times \frac{\sqrt(3)}{3}&lt;/math&gt; and the answer is &lt;math&gt;BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3 (Shortcut)==<br /> <br /> The ratio of the areas of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; and &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1 : 3&lt;/math&gt;, meaning the ratio of the sides is &lt;math&gt;1 : \sqrt{3}&lt;/math&gt;. The only answer choice involving &lt;math&gt;\sqrt{3}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=146535 2021 AMC 12B Problems/Problem 8 2021-02-14T01:35:56Z <p>Joeya: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);<br /> draw(O--A--B);<br /> draw(O--R);<br /> draw(O--L);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$B$&quot;, B, N);<br /> label(&quot;$R$&quot;, R, NE);<br /> label(&quot;$L$&quot;, L, N);<br /> label(&quot;$O$&quot;, O, S);<br /> label(&quot;$d$&quot;, O--A, W);<br /> label(&quot;$2d$&quot;, A--B, W*2+0.5*N);<br /> label(&quot;$r$&quot;, O--R, S);<br /> label(&quot;$r$&quot;, O--L, S*0.5 + 1.5 * E);<br /> dot(O);<br /> dot(A);<br /> dot(B);<br /> dot(R);<br /> dot(L);<br /> <br /> draw(circle((0, 4), 4));<br /> draw((-3.873, 3) -- (3.873, 3));<br /> draw((-3.873, 5) -- (3.873, 5));<br /> draw((-2.645, 7) -- (2.645, 7));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle RAO&lt;/math&gt; on the diagram)<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle LBO&lt;/math&gt; on the diagram)<br /> <br /> By the Pythagorean theorem, we can create the following system of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{\textbf{(B)}\ 6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya and diagram by Jamess2022(burntTacos).<br /> (Someone fix the diagram if possible. -&lt;i&gt; Done. &lt;/i&gt;)<br /> <br /> ==Solution 2 (Coordinates)==<br /> <br /> Because we know that the equation of a circle is &lt;math&gt;(x-a)^2 + (y-b)^2 = r^2&lt;/math&gt; where the center of the circle is &lt;math&gt;(a, b)&lt;/math&gt; and the radius is &lt;math&gt;r&lt;/math&gt;, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is &lt;math&gt;x^2 + y^2 = r^2&lt;/math&gt;. Now, we can set the distance between the chords as &lt;math&gt;2d&lt;/math&gt; so the distance from the chord with length 38 to the diameter is &lt;math&gt;d&lt;/math&gt;. <br /> <br /> Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:<br /> <br /> &lt;math&gt;(19, d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(19, -d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(17, -3d)&lt;/math&gt;<br /> <br /> <br /> Now, we can plug one of the first two value in as well as the last one to get the following equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Subtracting these two equations, we get &lt;math&gt;19^2 - 17^2 = 8d^2&lt;/math&gt; - therefore, we get &lt;math&gt;72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3&lt;/math&gt;. We want to find &lt;math&gt;2d = 6&lt;/math&gt; because that's the distance between two chords. So, our answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ~Tony_Li2007<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=VzwxbsuSQ80<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtu.be/yxt8-rUUosI<br /> <br /> == Video Solution by OmegaLearn (Circular Geometry) ==<br /> https://youtu.be/XNYq4ZMBtBU<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2021|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=146534 2021 AMC 12B Problems/Problem 8 2021-02-14T01:33:58Z <p>Joeya: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);<br /> draw(O--A--B);<br /> draw(O--R);<br /> draw(O--L);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$B$&quot;, B, N);<br /> label(&quot;$R$&quot;, R, NE);<br /> label(&quot;$L$&quot;, L, N);<br /> label(&quot;$O$&quot;, O, S);<br /> label(&quot;$d$&quot;, O--A, W);<br /> label(&quot;$2d$&quot;, A--B, W*2+0.5*N);<br /> label(&quot;$r$&quot;, O--R, S);<br /> label(&quot;$r$&quot;, O--L, S*0.5 + 1.5 * E);<br /> dot(O);<br /> dot(A);<br /> dot(B);<br /> dot(R);<br /> dot(L);<br /> <br /> draw(circle((0, 4), 4));<br /> draw((-3.873, 3) -- (3.873, 3));<br /> draw((-3.873, 5) -- (3.873, 5));<br /> draw((-2.645, 7) -- (2.645, 7));<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle RAO&lt;/math&gt; on the diagram)<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; (&lt;math&gt;\triangle LBO&lt;/math&gt; on the diagram)<br /> <br /> By the Pythagorean theorem, we can create the following system of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;<br /> <br /> -Solution by Joeya and diagram by Jamess2022(burntTacos).<br /> (Someone fix the diagram if possible. -&lt;i&gt; Done. &lt;/i&gt;)<br /> <br /> ==Solution 2 (Coordinates)==<br /> <br /> Because we know that the equation of a circle is &lt;math&gt;(x-a)^2 + (y-b)^2 = r^2&lt;/math&gt; where the center of the circle is &lt;math&gt;(a, b)&lt;/math&gt; and the radius is &lt;math&gt;r&lt;/math&gt;, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is &lt;math&gt;x^2 + y^2 = r^2&lt;/math&gt;. Now, we can set the distance between the chords as &lt;math&gt;2d&lt;/math&gt; so the distance from the chord with length 38 to the diameter is &lt;math&gt;d&lt;/math&gt;. <br /> <br /> Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:<br /> <br /> &lt;math&gt;(19, d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(19, -d)&lt;/math&gt;<br /> <br /> &lt;math&gt;(17, -3d)&lt;/math&gt;<br /> <br /> <br /> Now, we can plug one of the first two value in as well as the last one to get the following equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Subtracting these two equations, we get &lt;math&gt;19^2 - 17^2 = 8d^2&lt;/math&gt; - therefore, we get &lt;math&gt;72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3&lt;/math&gt;. We want to find &lt;math&gt;2d = 6&lt;/math&gt; because that's the distance between two chords. So, our answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ~Tony_Li2007<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=VzwxbsuSQ80<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtu.be/yxt8-rUUosI<br /> <br /> == Video Solution by OmegaLearn (Circular Geometry) ==<br /> https://youtu.be/XNYq4ZMBtBU<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2021|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=146532 2017 AMC 12B Problems/Problem 18 2021-02-14T01:27:22Z <p>Joeya: /* Solution 4: Coordinate geometry */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Coordinate Geometry)==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, therefore &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;\left(\frac{7}{5}y-2\right)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and line &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;\left(\frac{49}{25}y^2-\frac{28}{5}y+4\right)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == Video Solution (Similar Triangles) ==<br /> https://youtu.be/NsQbhYfGh1Q?t=512<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_17&diff=146531 2015 AMC 12B Problems/Problem 17 2021-02-14T01:24:59Z <p>Joeya: /* Solution 2.5 */</p> <hr /> <div>==Problem==<br /> An unfair coin lands on heads with a probability of &lt;math&gt;\tfrac{1}{4}&lt;/math&gt;. When tossed &lt;math&gt;n&gt;1&lt;/math&gt; times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> When tossed &lt;math&gt;n&lt;/math&gt; times, the probability of getting exactly 2 heads and the rest tails is<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.&lt;/cmath&gt;<br /> <br /> Similarly, the probability of getting exactly 3 heads is<br /> <br /> &lt;cmath&gt;\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.&lt;/cmath&gt;<br /> <br /> Now set the two probabilities equal to each other and solve for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;3 = \frac{n-2}{3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n-2 = 9&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \fbox{\textbf{(D)}\; 11}&lt;/cmath&gt;<br /> <br /> <br /> Note: the original problem did not specify &lt;math&gt;n&gt;1&lt;/math&gt;, so &lt;math&gt;n=1&lt;/math&gt; was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)<br /> <br /> ==Solution 2==<br /> <br /> Bash it out with the answer choices! (not really a rigorous solution)<br /> <br /> ==Solution 2.5==<br /> <br /> In order to test the answer choices efficiently, realize that the probability &lt;math&gt;n&lt;/math&gt; flips yielding two heads is of the form:<br /> <br /> &lt;math&gt;\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}&lt;/math&gt;<br /> <br /> Similarly, the form for the probability of three heads is:<br /> <br /> &lt;math&gt;\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}&lt;/math&gt;<br /> <br /> The probability of getting three heads (comapred to the probability of getting two) from &lt;math&gt;n&lt;/math&gt; flips is missing a factor of &lt;math&gt;3&lt;/math&gt; in the numerator. Thus, we need &lt;math&gt;\dbinom{n}{3}&lt;/math&gt; to add a factor of &lt;math&gt;3&lt;/math&gt; to the numerator of the probability of getting three heads.<br /> Our testing equation becomes<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} \times 3 = \dbinom{n}{3}&lt;/cmath&gt;<br /> <br /> since after factoring out the &lt;math&gt;3&lt;/math&gt; from &lt;math&gt;\dbinom{n}{3}&lt;/math&gt;, the remaining factorizations should be equal. <br /> <br /> The only answer choice satisfying this condition is &lt;math&gt;\fbox{\textbf{(D)}\;11}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_14&diff=146298 2021 AMC 12A Problems/Problem 14 2021-02-13T02:23:06Z <p>Joeya: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> What is the value of&lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000&lt;/math&gt;<br /> ==Solution==<br /> This equals<br /> &lt;cmath&gt;\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> ~JHawk0224<br /> <br /> ==Solution 2==<br /> First, we can get rid of the &lt;math&gt;k&lt;/math&gt; exponents using properties of logarithms:<br /> <br /> &lt;cmath&gt;\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k&lt;/cmath&gt; (Leaving the single &lt;math&gt;k&lt;/math&gt; in the exponent will come in handy later). Similarly,<br /> <br /> &lt;cmath&gt;\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2&lt;/cmath&gt;<br /> <br /> Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{(1 + 2 + \dots + 20)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}&lt;/cmath&gt;<br /> <br /> To evaluate the exponent of the &lt;math&gt;3&lt;/math&gt; in the first logarithm, we use the triangular numbers equation:<br /> <br /> &lt;cmath&gt;1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210&lt;/cmath&gt;<br /> <br /> Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:<br /> <br /> &lt;cmath&gt;\log_{a} b\log_{x} y = \log_{a} y\log_{x} b&lt;/cmath&gt;<br /> <br /> Thus,<br /> <br /> &lt;cmath&gt;\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> <br /> -Solution by Joeya<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=FD9BE7hpRvg&amp;t=322s<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==<br /> https://youtu.be/vgFPZ-hyd-I<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_14&diff=146296 2021 AMC 12A Problems/Problem 14 2021-02-13T02:19:53Z <p>Joeya: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> What is the value of&lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000&lt;/math&gt;<br /> ==Solution==<br /> This equals<br /> &lt;cmath&gt;\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> ~JHawk0224<br /> <br /> ==Solution 2==<br /> First, we can get rid of the &lt;math&gt;k&lt;/math&gt; exponents using properties of logarithms:<br /> <br /> &lt;cmath&gt;\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k&lt;/cmath&gt; (Leaving the single &lt;math&gt;k&lt;/math&gt; in the exponent will come in handy later). Similarly,<br /> <br /> &lt;cmath&gt;\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2&lt;/cmath&gt;<br /> <br /> Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{(1 + 2 + \dots + 20)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}&lt;/cmath&gt;<br /> <br /> To evaluate the exponent of the &lt;math&gt;3&lt;/math&gt; in the first logarithm, we use the triangular numbers equation:<br /> <br /> &lt;cmath&gt;\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210&lt;/cmath&gt;<br /> <br /> Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:<br /> <br /> &lt;cmath&gt;\log_{a} b\log_{x} y = \log_{a} y\log_{x} b&lt;/cmath&gt;<br /> <br /> Thus,<br /> <br /> &lt;cmath&gt;\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> <br /> -Solution by Joeya<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=FD9BE7hpRvg&amp;t=322s<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==<br /> https://youtu.be/vgFPZ-hyd-I<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_14&diff=146294 2021 AMC 12A Problems/Problem 14 2021-02-13T02:18:46Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> What is the value of&lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000&lt;/math&gt;<br /> ==Solution==<br /> This equals<br /> &lt;cmath&gt;\left(\sum_{k=1}^{20}k\log_5(3)\right)\left(\sum_{k=1}^{100}\log_9(25)\right)=\frac{20\cdot21}{2}\cdot\log_5(3)\cdot100\log_3(5)=\boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> ~JHawk0224<br /> <br /> ==Solution 2==<br /> First, we can get rid of the &lt;math&gt;k&lt;/math&gt; exponents using properties of logarithms:<br /> <br /> &lt;cmath&gt;\left(\log_{5^k} 3^{k^2}\right) = k^2 * \frac{1}{k} * \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k&lt;/cmath&gt; (Leaving the single &lt;math&gt;k&lt;/math&gt; in the exponent will come in handy later). Similarly,<br /> <br /> &lt;cmath&gt;\left(\log_{9^k} 25^{k}\right) = k * \frac{1}{k} * \log_{9} 25 = \log_{9} 5^2&lt;/cmath&gt;<br /> <br /> Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5} 3^k\right) = \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} = \log_{5} 3^{1 + 2 + \dots + 20}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\left(\sum_{k=1}^{100} \log_{9} 5^2\right) = \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2= \log_{9} 5^{2(100)} = \log_{9} 5^{200}&lt;/cmath&gt;<br /> <br /> To evaluate the exponent of the &lt;math&gt;3&lt;/math&gt; in the first logarithm, we use the triangular numbers equation:<br /> <br /> &lt;cmath&gt;\frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210&lt;/cmath&gt;<br /> <br /> Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:<br /> <br /> &lt;cmath&gt;\log_{a} b\log_{x} y = \log_{a} y\log_{x} b&lt;/cmath&gt;<br /> <br /> Thus,<br /> <br /> &lt;cmath&gt;\left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) = \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) = (200)(105) = \boxed{\textbf{(E)} 21000}&lt;/cmath&gt;<br /> <br /> -Solution by Joeya<br /> <br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=FD9BE7hpRvg&amp;t=322s<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Using Logarithmic Manipulations) ==<br /> https://youtu.be/vgFPZ-hyd-I<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=145544 2021 AMC 12B Problems/Problem 8 2021-02-12T01:14:08Z <p>Joeya: /* Solution */</p> <hr /> <div>==Problem==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt;<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; <br /> <br /> By the Pythagorean theorem, we can create the following systems of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;<br /> <br /> -Solution by Joeya (someone draw a diagram n fix my latex please)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems&diff=145167 2021 AMC 12B Problems 2021-02-11T20:48:52Z <p>Joeya: /* Problem 16 */</p> <hr /> <div>{{AMC12 Problems|year=2021|ab=B}}<br /> ==Problem 1==<br /> How many integer values of &lt;math&gt;x&lt;/math&gt; satisfy &lt;math&gt;|x|&lt;3\pi?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> At a math contest, &lt;math&gt;57&lt;/math&gt; students are wearing blue shirts, and another &lt;math&gt;75&lt;/math&gt; students are wearing yellow shirts. The &lt;math&gt;132&lt;/math&gt; students are assigned into &lt;math&gt;66&lt;/math&gt; points. In exactly &lt;math&gt;23&lt;/math&gt; of these pairs, both students are wearing blue shirts. In how many pairs are both studets wearing yellow shirts?<br /> <br /> &lt;math&gt;\textbf{(A) }23 \qquad \textbf{(B) }32 \qquad \textbf{(C) }37 \qquad \textbf{(D) }41 \qquad \textbf{(E) }64&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Suppose&lt;cmath&gt;2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.&lt;/cmath&gt;What is the value of &lt;math&gt;x?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is &lt;math&gt;84&lt;/math&gt;, and the afternoon class's mean score is &lt;math&gt;70&lt;/math&gt;. The ratio of the number of students in the morning clas to the number of students in the afternoon class is &lt;math&gt;\frac34&lt;/math&gt;. What is the mean of the score of all the students?<br /> <br /> &lt;math&gt;\textbf{(A) }74 \qquad \textbf{(B) }75 \qquad \textbf{(C) }76 \qquad \textbf{(D) }77 \qquad \textbf{(E) }78&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90^\circ&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y=-x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b-a?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }3 \qquad \textbf{(C) }5 \qquad \textbf{(D) }7 \qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> An inverted cone with base radius &lt;math&gt;12 \text{cm}&lt;/math&gt; and height &lt;math&gt;18\text{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of &lt;math&gt;24\text{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Let &lt;math&gt;N=34\cdot34\cdot63\cdot270.&lt;/math&gt; What is the ratio of the sum of the odd divisors of &lt;math&gt;N&lt;/math&gt; to the sum of the even divisors of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1:16 \qquad \textbf{(B) }1:15 \qquad \textbf{(C) }1:14 \qquad \textbf{(D) }1:8 \qquad \textbf{(E) }1:3&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> What is the value of&lt;cmath&gt;\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Two distinct numbers are selected from the set &lt;math&gt;\{1,2,3,4,\dots,36,37\}&lt;/math&gt; so that the sum of the remaining &lt;math&gt;35&lt;/math&gt; numbers is the product of these two numbers. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=13,BC=14&lt;/math&gt; and &lt;math&gt;AC=15&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point on &lt;math&gt;\overline{AC}&lt;/math&gt; such that &lt;math&gt;PC=10&lt;/math&gt;. There are exactly two points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; on line &lt;math&gt;BP&lt;/math&gt; such that quadrilaterals &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;ABCE&lt;/math&gt; are trapezoids. What is the distance &lt;math&gt;DE?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Suppose that &lt;math&gt;S&lt;/math&gt; is a finite set of positive integers. If the greatest integer in &lt;math&gt;S&lt;/math&gt; is removed from &lt;math&gt;S&lt;/math&gt;, then the average value (arithmetic mean) of the integers remaining is &lt;math&gt;32&lt;/math&gt;. If the least integer in &lt;math&gt;S&lt;/math&gt; is also removed, then the average value of the integers remaining is &lt;math&gt;35&lt;/math&gt;. If the great integer is then returned to the set, the average value of the integers rises to &lt;math&gt;40.&lt;/math&gt; The greatest integer in the original set &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;72&lt;/math&gt; greater than the least integer in &lt;math&gt;S&lt;/math&gt;. What is the average value of all the integers in the set &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> How many values of &lt;math&gt;\theta&lt;/math&gt; in the interval &lt;math&gt;0&lt;\theta\le 2\pi&lt;/math&gt; satisfy&lt;cmath&gt;1-3\sin\theta+5\cos3\theta?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a rectangle and let &lt;math&gt;\overline{DM}&lt;/math&gt; be a segment perpendicular to the plane of &lt;math&gt;ABCD&lt;/math&gt;. Suppose that &lt;math&gt;\overline{DM}&lt;/math&gt; has integer length, and the lengths of &lt;math&gt;\overline{MA},\overline{MC},&lt;/math&gt; and &lt;math&gt;\overline{MB}&lt;/math&gt; are consecutive odd positive integers (in this order). What is the volume of pyramid &lt;math&gt;MACD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Let &lt;math&gt;g(x)&lt;/math&gt; be a polynomial with leading coefficient &lt;math&gt;1,&lt;/math&gt; whose three roots are the reciprocals of the three roots of &lt;math&gt;f(x)=x^3+ax^2+bx+c,&lt;/math&gt; where &lt;math&gt;1&lt;a&lt;b&lt;c.&lt;/math&gt; What is &lt;math&gt;g(1)&lt;/math&gt; in terms of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isoceles trapezoid having parallel bases &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; with &lt;math&gt;AB&gt;CD.&lt;/math&gt; Line segments from a point inside &lt;math&gt;ABCD&lt;/math&gt; to the vertices divide the trapezoid into four triangles whose areas are &lt;math&gt;2, 3, 4,&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; starting with the triangle with base &lt;math&gt;\overline{CD}&lt;/math&gt; and moving clockwise as shown in the diagram below. What is the ratio &lt;math&gt;\frac{AB}{CD}?&lt;/math&gt;<br /> &lt;asy&gt;<br /> unitsize(100);<br /> pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); <br /> draw(A--B--C--D--cycle, black); <br /> draw(A--P, black);<br /> draw(B--P, black);<br /> draw(C--P, black);<br /> draw(D--P, black);<br /> label(&quot;$A$&quot;,A,(-1,0));<br /> label(&quot;$B$&quot;,B,(1,0));<br /> label(&quot;$C$&quot;,C,(1,-0));<br /> label(&quot;$D$&quot;,D,(-1,0));<br /> label(&quot;$2$&quot;,E,(0,0));<br /> label(&quot;$3$&quot;,F,(0,0));<br /> label(&quot;$4$&quot;,G,(0,0));<br /> label(&quot;$5$&quot;,H,(0,0));<br /> dot(A^^B^^C^^D^^P);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let &lt;math&gt;z&lt;/math&gt; be a complex number satisfying &lt;math&gt;12|z|^2=2|z+2|^2+|z^2+1|^2+31.&lt;/math&gt; What is the value if &lt;math&gt;z+\frac 6z?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Two fair dice, each with at least &lt;math&gt;6&lt;/math&gt; faces are rolled. On each face of each dice is printed a distinct integer from &lt;math&gt;1&lt;/math&gt; to the number of faces on that die, inclusive. The probability of rolling a sum if &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;\frac34&lt;/math&gt; of the probability of rolling a sum of &lt;math&gt;10,&lt;/math&gt; and the probability of rolling a sum of &lt;math&gt;12&lt;/math&gt; is &lt;math&gt;\frac{1}{12}&lt;/math&gt;. What is the least possible number of faces on the two dice combined?<br /> <br /> &lt;math&gt;\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Let &lt;math&gt;Q(z)&lt;/math&gt; and &lt;math&gt;R(z)&lt;/math&gt; be the unique polynomials such that&lt;cmath&gt;z^{2021}+1=(z^2+z+1)Q(z)+R(z)&lt;/cmath&gt;and the degree of &lt;math&gt;R&lt;/math&gt; is less than &lt;math&gt;2.&lt;/math&gt; What is &lt;math&gt;R(z)?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Let &lt;math&gt;S&lt;/math&gt; be the sum of all positive real numbers &lt;math&gt;x&lt;/math&gt; for which&lt;cmath&gt;x^{2^{\sqrt2}}=\sqrt2^{2^x}.&lt;/cmath&gt;Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A) }S&lt;\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2&lt;S&lt;2\qquad \textbf{(D) }2\le S&lt;6 \qquad \textbf{(E) }S\ge 6&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;i&lt;/math&gt; is &lt;math&gt;2^{-i}&lt;/math&gt; for &lt;math&gt;i=1,2,3,....&lt;/math&gt; More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is &lt;math&gt;\frac pq,&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins &lt;math&gt;3,17,&lt;/math&gt; and &lt;math&gt;10.&lt;/math&gt;) What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> These problems will not be posted until the 2021 AMC 12B is released on Wednesday, February 10, 2021.<br /> <br /> [[2021 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of lattice points in the coordinate plane, both of whose coordinates are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;30,&lt;/math&gt; inclusive. Exactly &lt;math&gt;300&lt;/math&gt; points in &lt;math&gt;S&lt;/math&gt; lie on or below a line with equation &lt;math&gt;y=mx.&lt;/math&gt; The possible values of &lt;math&gt;m&lt;/math&gt; lie in an interval of length &lt;math&gt;\frac ab,&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a+b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }31 \qquad \textbf{(B) }47 \qquad \textbf{(C) }62\qquad \textbf{(D) }72 \qquad \textbf{(E) }85&lt;/math&gt;<br /> <br /> [[2021 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=B|before=[[2021 AMC 12A Problems]]|after=[[2022 AMC 12A Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=145149 2021 AMC 12B Problems/Problem 8 2021-02-11T20:42:02Z <p>Joeya: /* Solution */</p> <hr /> <div>==Problem 8==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the chords touch the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt;<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; <br /> <br /> By the Pythagorean theorem, we can create the following systems of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;<br /> <br /> -Solution by Joeya (someone draw a diagram n fix my latex please)</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=145144 2021 AMC 12B Problems/Problem 8 2021-02-11T20:39:47Z <p>Joeya: /* Solution */</p> <hr /> <div>==Problem 8==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center of the circle to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the chords touch the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt;<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; <br /> <br /> By the Pythagorean theorem, we can create the following systems of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;<br /> <br /> -Solution by Joeya (someone draw a diagram please)</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=145134 2021 AMC 12B Problems/Problem 8 2021-02-11T20:36:04Z <p>Joeya: </p> <hr /> <div>==Problem 8==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the chords touch the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt;<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; <br /> <br /> By the Pythagorean theorem, we can create the following systems of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;<br /> <br /> -Solution by Joeya (someone draw a diagram please)</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_8&diff=145128 2021 AMC 12B Problems/Problem 8 2021-02-11T20:33:28Z <p>Joeya: Created page with &quot;==Problem 8== Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between tw...&quot;</p> <hr /> <div>==Problem 8==<br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38,38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Since the two chords of length &lt;math&gt;38&lt;/math&gt; have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be &lt;math&gt;d&lt;/math&gt;. Thus, the distance from the center to the chord of length &lt;math&gt;34&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;2d + d = 3d&lt;/cmath&gt;<br /> <br /> and the distance between each of the chords is just &lt;math&gt;2d&lt;/math&gt;. Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;. Drawing radii to the points where the chords touch the circle, we create two different right triangles: <br /> <br /> - One with base &lt;math&gt;\frac{38}{2}= 19&lt;/math&gt;, height &lt;math&gt;d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt;<br /> <br /> - Another with base &lt;math&gt;\frac{34}{2} = 17&lt;/math&gt;, height &lt;math&gt;3d&lt;/math&gt;, and hypotenuse &lt;math&gt;r&lt;/math&gt; <br /> <br /> By the Pythagorean theorem, we can create the following systems of equations:<br /> <br /> &lt;cmath&gt;19^2 + d^2 = r^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;17^2 + (3d)^2 = r^2&lt;/cmath&gt;<br /> <br /> Solving, we find &lt;math&gt;d = 3&lt;/math&gt;, so &lt;math&gt;2d = \boxed{(B) 6}&lt;/math&gt;</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=User:Joeya&diff=144133 User:Joeya 2021-01-31T07:26:01Z <p>Joeya: /* Yooooooooooooooooooooo */</p> <hr /> <div>==Better Call Saul==</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_9&diff=143326 2011 AMC 10B Problems/Problem 9 2021-01-26T18:38:47Z <p>Joeya: Undo revision 143321 by Sugar rush (talk)</p> <hr /> <div>== Problem==<br /> <br /> The area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt;. Segment &lt;math&gt;DE&lt;/math&gt; is perpendicular to segment &lt;math&gt;AB&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;? &lt;p&gt;<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3));<br /> pair[] ps={A,B,C,D,E};<br /> <br /> draw(A--B--C--cycle);<br /> draw(E--D);<br /> draw(rightanglemark(E,D,B));<br /> <br /> dot(ps);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,NE);<br /> label(&quot;$3$&quot;,midpoint(A--C),NW);<br /> label(&quot;$4$&quot;,midpoint(C--B),NE);<br /> label(&quot;$5$&quot;,midpoint(A--B),SW);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Therefore &lt;math&gt;DE = \frac{3}{4} BD&lt;/math&gt;. Find the areas of the triangles.<br /> &lt;cmath&gt;\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6&lt;/cmath&gt;<br /> &lt;cmath&gt;\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2&lt;/cmath&gt;<br /> The area of &lt;math&gt;\triangle EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{3}{8} BD^2 &amp;= 6 \times \frac{1}{3}\\<br /> 9BD^2 &amp;= 48\\<br /> BD^2 &amp;= \frac{16}{3}\\<br /> BD &amp;= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Since the area of &lt;math&gt;\triangle EBD&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus,<br /> &lt;cmath&gt;\frac{1}{3}[\triangle ABC] = [\triangle EBD]&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB&lt;/cmath&gt;<br /> In order to scale the sides of ED and DB to make &lt;math&gt;\frac{1}{3}&lt;/math&gt; (since the ratios of sides are the same), we take the square root of &lt;math&gt;\frac{1}{3} = \frac{\sqrt(3)}{3}&lt;/math&gt; to scale each side by the same amount. <br /> <br /> Thus &lt;math&gt;BD = 4 \times \frac{\sqrt(3)}{3}&lt;/math&gt; and the answer is &lt;math&gt;BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3 (Shortcut)==<br /> <br /> The ratio of the areas of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; and &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1 : 3&lt;/math&gt;, meaning the ratio of the sides is &lt;math&gt;1 : \sqrt{3}&lt;/math&gt;. The only answer choice involving &lt;math&gt;\sqrt{3}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> -Solution by Joeya<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_9&diff=143320 2011 AMC 10B Problems/Problem 9 2021-01-26T18:29:13Z <p>Joeya: </p> <hr /> <div>== Problem==<br /> <br /> The area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt;. Segment &lt;math&gt;DE&lt;/math&gt; is perpendicular to segment &lt;math&gt;AB&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;? &lt;p&gt;<br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3));<br /> pair[] ps={A,B,C,D,E};<br /> <br /> draw(A--B--C--cycle);<br /> draw(E--D);<br /> draw(rightanglemark(E,D,B));<br /> <br /> dot(ps);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,S);<br /> label(&quot;$E$&quot;,E,NE);<br /> label(&quot;$3$&quot;,midpoint(A--C),NW);<br /> label(&quot;$4$&quot;,midpoint(C--B),NE);<br /> label(&quot;$5$&quot;,midpoint(A--B),SW);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Therefore &lt;math&gt;DE = \frac{3}{4} BD&lt;/math&gt;. Find the areas of the triangles.<br /> &lt;cmath&gt;\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6&lt;/cmath&gt;<br /> &lt;cmath&gt;\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2&lt;/cmath&gt;<br /> The area of &lt;math&gt;\triangle EBD&lt;/math&gt; is one third of the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{3}{8} BD^2 &amp;= 6 \times \frac{1}{3}\\<br /> 9BD^2 &amp;= 48\\<br /> BD^2 &amp;= \frac{16}{3}\\<br /> BD &amp;= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;\triangle ABC \sim \triangle EBD&lt;/math&gt; by AA Similarity. Since the area of &lt;math&gt;\triangle EBD&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus,<br /> &lt;cmath&gt;\frac{1}{3}[\triangle ABC] = [\triangle EBD]&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB&lt;/cmath&gt;<br /> In order to scale the sides of ED and DB to make &lt;math&gt;\frac{1}{3}&lt;/math&gt; (since the ratios of sides are the same), we take the square root of &lt;math&gt;\frac{1}{3} = \frac{\sqrt(3)}{3}&lt;/math&gt; to scale each side by the same amount. <br /> <br /> Thus &lt;math&gt;BD = 4 \times \frac{\sqrt(3)}{3}&lt;/math&gt; and the answer is &lt;math&gt;BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 3 (Shortcut)==<br /> <br /> The ratio of the areas of &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EBD&lt;/math&gt; and &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;1 : 3&lt;/math&gt;, meaning the ratio of the sides is &lt;math&gt;1 : \sqrt{3}&lt;/math&gt;. The only answer choice involving &lt;math&gt;\sqrt{3}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> -Solution by Joeya<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=User:Joeya&diff=143315 User:Joeya 2021-01-26T17:53:03Z <p>Joeya: Created page with &quot;==Yooooooooooooooooooooo==&quot;</p> <hr /> <div>==Yooooooooooooooooooooo==</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143314 2011 AMC 12B Problems/Problem 6 2021-01-26T17:35:31Z <p>Joeya: Undo revision 143312 by Sugar rush (talk)</p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. <br /> <br /> Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt; since quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143301 2011 AMC 12B Problems/Problem 6 2021-01-26T06:39:23Z <p>Joeya: Undo revision 143278 by Sugar rush (talk)</p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. <br /> <br /> Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt; since quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143275 2011 AMC 12B Problems/Problem 6 2021-01-26T01:26:28Z <p>Joeya: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. <br /> <br /> Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt; since quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143274 2011 AMC 12B Problems/Problem 6 2021-01-26T01:24:42Z <p>Joeya: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. <br /> <br /> Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143273 2011 AMC 12B Problems/Problem 6 2021-01-26T01:24:06Z <p>Joeya: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_6&diff=143272 2011 AMC 12B Problems/Problem 6 2021-01-26T01:22:57Z <p>Joeya: </p> <hr /> <div>== Problem ==<br /> Two tangents to a circle are drawn from a point &lt;math&gt;A&lt;/math&gt;. The points of contact &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; divide the circle into arcs with lengths in the ratio &lt;math&gt;2 : 3&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle{BAC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).<br /> <br /> In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. <br /> <br /> Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.<br /> <br /> Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:<br /> <br /> 1/2 (216°-144°) = 1/2 (72°) &lt;math&gt; =\boxed{36 \textbf{(C)}}. &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is &lt;math&gt;\frac{3x-2x}2 = \frac {x} 2&lt;/math&gt; by the arc length formula. Also note that &lt;math&gt;3x + 2x = 360&lt;/math&gt;, which simplifies to &lt;math&gt;x= 72.&lt;/math&gt; Hence the angle formed by the tangents is equal to &lt;math&gt;\boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. The radii extending from &lt;math&gt;O&lt;/math&gt; to the points of tangency &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are both perpendicular to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. Thus &lt;math&gt;\angle ABO = \angle ACO = 90^{\circ}&lt;/math&gt;, and quadrilateral &lt;math&gt;ABOC&lt;/math&gt; is cyclic, since the opposite angles add to &lt;math&gt;180^{\circ}&lt;/math&gt;. Since the ratio of the arc lengths is &lt;math&gt;2 : 3&lt;/math&gt;, the shorter arc (the arc cut off by &lt;math&gt;\angle BOC&lt;/math&gt;) is &lt;math&gt;144^{\circ}&lt;/math&gt;. &lt;math&gt;\angle BOC&lt;/math&gt; and &lt;math&gt;\angle BAC&lt;/math&gt; must add up to &lt;math&gt;180^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle BAC = 180 - 144 = \boxed{36 \textbf{(C)}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See also==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_16&diff=142584 2013 AMC 12B Problems/Problem 16 2021-01-18T04:14:16Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be an equiangular convex pentagon of perimeter &lt;math&gt;1&lt;/math&gt;. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let &lt;math&gt;s&lt;/math&gt; be the perimeter of this star. What is the difference between the maximum and the minimum possible values of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure &lt;math&gt;\frac{180^\circ (5-2)}{5}=108^\circ&lt;/math&gt;, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure &lt;math&gt;180^\circ - 108^\circ = 72^\circ&lt;/math&gt;. The base angles are equal, so the triangles must be isosceles.<br /> <br /> Let one of the sides of the pentagon have length &lt;math&gt;x_1&lt;/math&gt; (and the others &lt;math&gt;x_2, x_3, x_4, x_5&lt;/math&gt;). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length &lt;math&gt;\frac{x_1}{2} \sec 72^\circ&lt;/math&gt;, and so the two sides together have length &lt;math&gt;x_1 \sec 72^\circ&lt;/math&gt;. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get &lt;math&gt;(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ&lt;/math&gt; (because the perimeter of the pentagon is &lt;math&gt;1&lt;/math&gt;). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is &lt;math&gt;\boxed{\textbf{(A)} \ 0}&lt;/math&gt;.<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_16&diff=142583 2013 AMC 12B Problems/Problem 16 2021-01-18T04:12:58Z <p>Joeya: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be an equiangular convex pentagon of perimeter &lt;math&gt;1&lt;/math&gt;. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let &lt;math&gt;s&lt;/math&gt; be the perimeter of this star. What is the difference between the maximum and the minimum possible values of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure &lt;math&gt;\frac{180^\circ (5-2)}{5}=108^\circ&lt;/math&gt;, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure &lt;math&gt;180^\circ - 108^\circ = 72^\circ&lt;/math&gt;. The base angles are equal, so the triangles must be isosceles.<br /> <br /> Let one of the sides of the pentagon have length &lt;math&gt;x_1&lt;/math&gt; (and the others &lt;math&gt;x_2, x_3, x_4, x_5&lt;/math&gt;). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length &lt;math&gt;\frac{x_1}{2} \sec 72^\circ&lt;/math&gt;, and so the two sides together have length &lt;math&gt;x_1 \sec 72^\circ&lt;/math&gt;. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get &lt;math&gt;(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ&lt;/math&gt; (because the perimeter of the pentagon is &lt;math&gt;1&lt;/math&gt;). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is &lt;math&gt;\boxed{\textbf{(A)} \ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> With a bit of cognitive effort, it is easy to visualize that readjusting the side lengths of any odd-number sided polygon will change the angle measures no matter what (unless you are upscaling all the side lengths equally, but we are working with a fixed perimiter). Since the star is equiangular, this obviously means there is only one possible perimeter for the star, so the difference between the maximum and minimum perimeters is &lt;math&gt;\boxed{\textbf{(A)} \ 0}&lt;/math&gt;.<br /> <br /> -Joeya<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_16&diff=142582 2013 AMC 12B Problems/Problem 16 2021-01-18T04:12:47Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be an equiangular convex pentagon of perimeter &lt;math&gt;1&lt;/math&gt;. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let &lt;math&gt;s&lt;/math&gt; be the perimeter of this star. What is the difference between the maximum and the minimum possible values of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure &lt;math&gt;\frac{180^\circ (5-2)}{5}=108^\circ&lt;/math&gt;, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure &lt;math&gt;180^\circ - 108^\circ = 72^\circ&lt;/math&gt;. The base angles are equal, so the triangles must be isosceles.<br /> <br /> Let one of the sides of the pentagon have length &lt;math&gt;x_1&lt;/math&gt; (and the others &lt;math&gt;x_2, x_3, x_4, x_5&lt;/math&gt;). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length &lt;math&gt;\frac{x_1}{2} \sec 72^\circ&lt;/math&gt;, and so the two sides together have length &lt;math&gt;x_1 \sec 72^\circ&lt;/math&gt;. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get &lt;math&gt;(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ&lt;/math&gt; (because the perimeter of the pentagon is &lt;math&gt;1&lt;/math&gt;). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is &lt;math&gt;\boxed{\textbf{(A)} \ 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> With a bit of cognitive effort, it is easy to visualize that readjusting the side lengths of any odd-number sided polygon will change the angle measures no matter what (unless you are upscaling all the side lengths equally, but we are working with a fixed perimiter). Since the star is equiangular, this obviously means there is only one possible perimeter for the star, so the difference between the maximum and minimum perimeters is &lt;math&gt;\boxed{\textbf{(A)} \ 0}&lt;/math&gt;. <br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=142581 2013 AMC 12B Problems/Problem 4 2021-01-18T04:03:24Z <p>Joeya: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Taking the harmonic mean of the two rates, we get &lt;cmath&gt;\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}.&lt;/cmath&gt;<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=142580 2013 AMC 12B Problems/Problem 4 2021-01-18T04:03:02Z <p>Joeya: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Taking the harmonic mean of the two rates, we get &lt;cmath&gt;\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}&lt;/cmath&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142497 2014 AMC 12B Problems/Problem 10 2021-01-17T23:46:38Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142496 2014 AMC 12B Problems/Problem 10 2021-01-17T23:46:23Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=142494 2013 AMC 12B Problems/Problem 4 2021-01-17T23:41:45Z <p>Joeya: </p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Taking the harmonic mean of the two rates, we get &lt;cmath&gt;\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{\textbf{(B) }16}&lt;/cmath&gt;<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142165 2014 AMC 12B Problems/Problem 10 2021-01-16T07:06:44Z <p>Joeya: /* Solution 3 (Testing Values) */</p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3 (Testing Values)==<br /> We know that since adding a multiple of &lt;math&gt;55&lt;/math&gt; will increase the number, we know that &lt;math&gt;a \leq{c}&lt;/math&gt;, since these values will flip. After testing values for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of &lt;math&gt;a^2 + b^2 + c^2&lt;/math&gt; as 27, specifically the combination &lt;math&gt;(a, b, c)&lt;/math&gt; of &lt;math&gt;(1, 1, 5)&lt;/math&gt;. However, we see that no multiple of &lt;math&gt;55&lt;/math&gt; can be added to &lt;math&gt;115&lt;/math&gt; to achieve &lt;math&gt;511&lt;/math&gt;. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only &lt;math&gt;b&lt;/math&gt; can be zero, we first try to see if &lt;math&gt;(1, 0, 5)&lt;/math&gt; (which creates 26 from the expression &lt;math&gt;1^2 + 0^2 + 5^2&lt;/math&gt;, answer choice A) can be added to a multiple of &lt;math&gt;55&lt;/math&gt; to create &lt;math&gt;501&lt;/math&gt;, which we find it cannot. However, we quickly realize that the next consecutive trio, &lt;math&gt;(1, 0, 6)&lt;/math&gt;, does satisfy this condition, and &lt;math&gt;1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}&lt;/math&gt;. We come to this quickly after realizing that if the multiple of &lt;math&gt;55&lt;/math&gt; was an even multiple, it would have to end in &lt;math&gt;0&lt;/math&gt; and thus the &lt;math&gt;c&lt;/math&gt; digit would remain unchanged, so it must be an odd multiple, which will carry over the &lt;math&gt;c&lt;/math&gt; digit, so &lt;math&gt;c \geq{6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142164 2014 AMC 12B Problems/Problem 10 2021-01-16T07:05:43Z <p>Joeya: /* Solution 3 (Testing Values) */</p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3 (Testing Values)==<br /> We know that since adding a multiple of &lt;math&gt;55&lt;/math&gt; will increase the number, we know that &lt;math&gt;a \leq{c}&lt;/math&gt;, since these values will flip. After testing values for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of &lt;math&gt;a^2 + b^2 + c^2&lt;/math&gt; as 27, specifically the combination &lt;math&gt;(a, b, c)&lt;/math&gt; of &lt;math&gt;(1, 1, 5)&lt;/math&gt;. However, we see that no multiple of &lt;math&gt;55&lt;/math&gt; can be added to &lt;math&gt;115&lt;/math&gt; to achieve &lt;math&gt;511&lt;/math&gt;. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only &lt;math&gt;b&lt;/math&gt; can be zero, we first try to see if &lt;math&gt;(1, 0, 5)&lt;/math&gt; (which creates 26 from the expression &lt;math&gt;1^2 + 0^2 + 5^2&lt;/math&gt;, answer choice A) can be added to a multiple of &lt;math&gt;55&lt;/math&gt; to create &lt;math&gt;501&lt;/math&gt;, which we find it cannot. However, we quickly realize that the next consecutive pair &lt;math&gt;(1, 0, 6)&lt;/math&gt; does satisfy this condition, and &lt;math&gt;1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}&lt;/math&gt;. We come to this quickly after realizing that if the multiple of &lt;math&gt;55&lt;/math&gt; was an even multiple, it would have to end in &lt;math&gt;0&lt;/math&gt; and thus the &lt;math&gt;c&lt;/math&gt; digit would remain unchanged, so it must be an odd multiple, which will carry over the &lt;math&gt;c&lt;/math&gt; digit, so &lt;math&gt;c \geq{6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142163 2014 AMC 12B Problems/Problem 10 2021-01-16T07:05:09Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3 (Testing Values)==<br /> We know that since adding a multiple of &lt;math&gt;55&lt;/math&gt; will increase the number, we know that &lt;math&gt;a \leq{c}&lt;/math&gt;. After testing values for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of &lt;math&gt;a^2 + b^2 + c^2&lt;/math&gt; as 27, specifically the combination &lt;math&gt;(a, b, c)&lt;/math&gt; of &lt;math&gt;(1, 1, 5)&lt;/math&gt;. However, we see that no multiple of &lt;math&gt;55&lt;/math&gt; can be added to &lt;math&gt;115&lt;/math&gt; to achieve &lt;math&gt;511&lt;/math&gt;. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only &lt;math&gt;b&lt;/math&gt; can be zero, we first try to see if &lt;math&gt;(1, 0, 5)&lt;/math&gt; (which creates 26 from the expression &lt;math&gt;1^2 + 0^2 + 5^2&lt;/math&gt;, answer choice A) can be added to a multiple of &lt;math&gt;55&lt;/math&gt; to create &lt;math&gt;501&lt;/math&gt;, which we find it cannot. However, we quickly realize that the next consecutive pair &lt;math&gt;(1, 0, 6)&lt;/math&gt; does satisfy this condition, and &lt;math&gt;1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}&lt;/math&gt;. We come to this quickly after realizing that if the multiple of &lt;math&gt;55&lt;/math&gt; was an even multiple, it would have to end in &lt;math&gt;0&lt;/math&gt; and thus the &lt;math&gt;c&lt;/math&gt; digit would remain unchanged, so it must be an odd multiple, which will carry over the &lt;math&gt;c&lt;/math&gt; digit, so &lt;math&gt;c \geq{6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_10&diff=142162 2014 AMC 12B Problems/Problem 10 2021-01-16T07:04:44Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> <br /> Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, &lt;math&gt;abc&lt;/math&gt; miles was displayed on the odometer, where &lt;math&gt;abc&lt;/math&gt; is a 3-digit number with &lt;math&gt;a \geq{1}&lt;/math&gt; and &lt;math&gt;a+b+c \leq{7}&lt;/math&gt;. At the end of the trip, the odometer showed &lt;math&gt;cba&lt;/math&gt; miles. What is &lt;math&gt;a^2+b^2+c^2?&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the number of miles she drove is divisible by &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must either be the equal or differ by &lt;math&gt;5&lt;/math&gt;. We can quickly conclude that the former is impossible, so &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must be &lt;math&gt;5&lt;/math&gt; apart. Because we know that &lt;math&gt;c &gt; a&lt;/math&gt; and &lt;math&gt;a + c \le 7&lt;/math&gt; and &lt;math&gt;a \ge 1&lt;/math&gt;, we find that the only possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, respectively. Because &lt;math&gt;a + b + c \le 7&lt;/math&gt;, &lt;math&gt;b = 0&lt;/math&gt;. Therefore, we have<br /> &lt;cmath&gt;a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Let the number of hours Danica drove be &lt;math&gt;k&lt;/math&gt;. Then we know that &lt;math&gt;100a + 10b + c + 55k&lt;/math&gt; = &lt;math&gt;100c + 10b + a&lt;/math&gt;. Simplifying, we have &lt;math&gt;99c - 99a = 55k&lt;/math&gt;, or &lt;math&gt;9c - 9a = 5k&lt;/math&gt;. Thus, k is divisible by &lt;math&gt;9&lt;/math&gt;. Because &lt;math&gt;55 * 18 = 990&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;9&lt;/math&gt;, and therefore &lt;math&gt;c - a = 5&lt;/math&gt;. Because &lt;math&gt;a + b + c \leq{7}&lt;/math&gt; and &lt;math&gt;a \geq{1}&lt;/math&gt;, &lt;math&gt;a = 1&lt;/math&gt;, &lt;math&gt;c = 6&lt;/math&gt; and &lt;math&gt;b = 0&lt;/math&gt;, and our answer is &lt;math&gt;a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37&lt;/math&gt;, or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Testing Values)==<br /> We know that since adding a multiple of &lt;math&gt;55&lt;/math&gt; will increase the number, we know that &lt;math&gt;a \leq{c}&lt;/math&gt;. After testing values for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of &lt;math&gt;a^2 + b^2 + c^2&lt;/math&gt; as 27, specifically the combination &lt;math&gt;(a, b, c)&lt;/math&gt; of &lt;math&gt;(1, 1, 5)&lt;/math&gt;. However, we see that no multiple of &lt;math&gt;55&lt;/math&gt; can be added to &lt;math&gt;115&lt;/math&gt; to achieve &lt;math&gt;511&lt;/math&gt;. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only &lt;math&gt;b&lt;/math&gt; can be zero, we first try to see if &lt;math&gt;(1, 0, 5)&lt;/math&gt; (which creates 26 from the expression &lt;math&gt;1^2 + 0^2 + 5^2&lt;/math&gt;, answer choice A) can be added to a multiple of &lt;math&gt;55&lt;/math&gt; to create &lt;math&gt;501&lt;/math&gt;, which we find it cannot. However, we quickly realize that the next consecutive pair &lt;math&gt;(1, 0, 6)&lt;/math&gt; does satisfy this condition, and &lt;math&gt;1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}&lt;/math&gt;. We come to this quickly after realizing that if the multiple of &lt;math&gt;55&lt;/math&gt; was an even multiple, it would have to end in &lt;math&gt;0&lt;/math&gt; and thus the &lt;math&gt;c&lt;/math&gt; digit would remain unchanged, so it must be an odd multiple, which will carry over the &lt;math&gt;c&lt;/math&gt; digit, so &lt;math&gt;c \geq{6}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_17&diff=141899 2015 AMC 12B Problems/Problem 17 2021-01-11T19:32:11Z <p>Joeya: /* Solution 2.5 */</p> <hr /> <div>==Problem==<br /> An unfair coin lands on heads with a probability of &lt;math&gt;\tfrac{1}{4}&lt;/math&gt;. When tossed &lt;math&gt;n&gt;1&lt;/math&gt; times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> When tossed &lt;math&gt;n&lt;/math&gt; times, the probability of getting exactly 2 heads and the rest tails is<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.&lt;/cmath&gt;<br /> <br /> Similarly, the probability of getting exactly 3 heads is<br /> <br /> &lt;cmath&gt;\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.&lt;/cmath&gt;<br /> <br /> Now set the two probabilities equal to each other and solve for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;3 = \frac{n-2}{3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n-2 = 9&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \fbox{\textbf{(D)}\; 11}&lt;/cmath&gt;<br /> <br /> <br /> Note: the original problem did not specify &lt;math&gt;n&gt;1&lt;/math&gt;, so &lt;math&gt;n=1&lt;/math&gt; was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)<br /> <br /> ==Solution 2==<br /> <br /> Bash it out with the answer choices! (not really a rigorous solution)<br /> <br /> ==Solution 2.5==<br /> <br /> In order to test the answer choices efficiently, realize that the probability &lt;math&gt;n&lt;/math&gt; flips yielding two heads is of the form:<br /> <br /> &lt;math&gt;\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}&lt;/math&gt;<br /> <br /> Similarly, the form for the probability of three heads is:<br /> <br /> &lt;math&gt;\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}&lt;/math&gt;.<br /> <br /> The probability of getting three heads (comapred to the probability of getting two) from &lt;math&gt;n&lt;/math&gt; flips is missing a factor of &lt;math&gt;3&lt;/math&gt; in the numerator. Thus, we need &lt;math&gt;\dbinom{n}{3}&lt;/math&gt; to add a factor of &lt;math&gt;3&lt;/math&gt; to the numerator of the probability of getting three heads.<br /> Our testing equation becomes<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} \times 3 = \dbinom{n}{3}&lt;/cmath&gt;<br /> <br /> since after factoring out the &lt;math&gt;3&lt;/math&gt; from &lt;math&gt;\dbinom{n}{3}&lt;/math&gt;, the remaining factorizations should be equal. <br /> <br /> The only answer choice satisfying this condition is &lt;math&gt;\fbox{\textbf{(D)}\;11}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_17&diff=141898 2015 AMC 12B Problems/Problem 17 2021-01-11T19:20:27Z <p>Joeya: /* Solution 2.5 */</p> <hr /> <div>==Problem==<br /> An unfair coin lands on heads with a probability of &lt;math&gt;\tfrac{1}{4}&lt;/math&gt;. When tossed &lt;math&gt;n&gt;1&lt;/math&gt; times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> When tossed &lt;math&gt;n&lt;/math&gt; times, the probability of getting exactly 2 heads and the rest tails is<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.&lt;/cmath&gt;<br /> <br /> Similarly, the probability of getting exactly 3 heads is<br /> <br /> &lt;cmath&gt;\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.&lt;/cmath&gt;<br /> <br /> Now set the two probabilities equal to each other and solve for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;3 = \frac{n-2}{3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n-2 = 9&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \fbox{\textbf{(D)}\; 11}&lt;/cmath&gt;<br /> <br /> <br /> Note: the original problem did not specify &lt;math&gt;n&gt;1&lt;/math&gt;, so &lt;math&gt;n=1&lt;/math&gt; was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)<br /> <br /> ==Solution 2==<br /> <br /> Bash it out with the answer choices! (not really a rigorous solution)<br /> <br /> ==Solution 2.5==<br /> <br /> In order to test the answer choices efficiently, realize that the probability &lt;math&gt;n&lt;/math&gt; flips yielding two heads is of the form:<br /> <br /> &lt;math&gt;\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}&lt;/math&gt;<br /> <br /> Similarly, the form for the probability of three heads is:<br /> <br /> &lt;math&gt;\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}&lt;/math&gt;.<br /> <br /> The probability of getting three heads (comapred to the probability of getting two) from &lt;math&gt;n&lt;/math&gt; flips is missing a factor of &lt;math&gt;3&lt;/math&gt; in the numerator. Thus, we need &lt;math&gt;\dbinom{n}{3}&lt;/math&gt; to add a factor of &lt;math&gt;3&lt;/math&gt; to the numerator of the probability of getting three heads.<br /> Our testing equation becomes<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} \times 3 = \dbinom{n}{3}&lt;/cmath&gt;.<br /> <br /> since after factoring out the &lt;math&gt;3&lt;/math&gt; from &lt;math&gt;\dbinom{n}{3}&lt;/math&gt;, the remaining factorizations should be equal. <br /> <br /> The only answer choice satisfying this condition is &lt;math&gt;\fbox{\textbf{(D)}\;11}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141897 2017 AMC 12B Problems/Problem 18 2021-01-11T19:16:54Z <p>Joeya: /* Solution 4: Coordinate geometry */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate geometry==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, therefore &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;\left(\frac{7}{5}y-2\right)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and line &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;\left(\frac{49}{25}y^2-\frac{28}{5}y+4\right)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_17&diff=141893 2015 AMC 12B Problems/Problem 17 2021-01-11T17:11:54Z <p>Joeya: /* Solution 2.5 */</p> <hr /> <div>==Problem==<br /> An unfair coin lands on heads with a probability of &lt;math&gt;\tfrac{1}{4}&lt;/math&gt;. When tossed &lt;math&gt;n&gt;1&lt;/math&gt; times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> When tossed &lt;math&gt;n&lt;/math&gt; times, the probability of getting exactly 2 heads and the rest tails is<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.&lt;/cmath&gt;<br /> <br /> Similarly, the probability of getting exactly 3 heads is<br /> <br /> &lt;cmath&gt;\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.&lt;/cmath&gt;<br /> <br /> Now set the two probabilities equal to each other and solve for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;3 = \frac{n-2}{3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n-2 = 9&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \fbox{\textbf{(D)}\; 11}&lt;/cmath&gt;<br /> <br /> <br /> Note: the original problem did not specify &lt;math&gt;n&gt;1&lt;/math&gt;, so &lt;math&gt;n=1&lt;/math&gt; was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)<br /> <br /> ==Solution 2==<br /> <br /> Bash it out with the answer choices! (not really a rigorous solution)<br /> <br /> ==Solution 2.5==<br /> <br /> In order to test the answer choices efficiently, realize that the probability &lt;math&gt;n&lt;/math&gt; flips yielding two heads is of the form:<br /> <br /> &lt;math&gt;\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}&lt;/math&gt;<br /> <br /> Similarly, the form for the probability of three heads is:<br /> <br /> &lt;math&gt;\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}&lt;/math&gt;.<br /> <br /> The probability of getting three heads (comapred to the probability of getting two) from &lt;math&gt;n&lt;/math&gt; flips is missing a factor of three in the numerator. Thus, we need &lt;math&gt;\dbinom{n}{3}&lt;/math&gt; to add a factor of 3 to the numerator of the probability of getting three heads.<br /> Our testing equation becomes<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} \times 3 = \dbinom{n}{3}&lt;/cmath&gt;.<br /> <br /> since after factoring out the three from &lt;math&gt;\dbinom{n}{3}&lt;/math&gt;, the remaining factorizations should be equal. <br /> <br /> The only answer choice satisfying this condition is &lt;math&gt;\fbox{\textbf{(D)}\;11}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_17&diff=141892 2015 AMC 12B Problems/Problem 17 2021-01-11T17:10:47Z <p>Joeya: </p> <hr /> <div>==Problem==<br /> An unfair coin lands on heads with a probability of &lt;math&gt;\tfrac{1}{4}&lt;/math&gt;. When tossed &lt;math&gt;n&gt;1&lt;/math&gt; times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> When tossed &lt;math&gt;n&lt;/math&gt; times, the probability of getting exactly 2 heads and the rest tails is<br /> <br /> &lt;cmath&gt;\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.&lt;/cmath&gt;<br /> <br /> Similarly, the probability of getting exactly 3 heads is<br /> <br /> &lt;cmath&gt;\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.&lt;/cmath&gt;<br /> <br /> Now set the two probabilities equal to each other and solve for &lt;math&gt;n&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;3 = \frac{n-2}{3}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n-2 = 9&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \fbox{\textbf{(D)}\; 11}&lt;/cmath&gt;<br /> <br /> <br /> Note: the original problem did not specify &lt;math&gt;n&gt;1&lt;/math&gt;, so &lt;math&gt;n=1&lt;/math&gt; was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:23, 19 February 2016 (EST)<br /> <br /> ==Solution 2==<br /> <br /> Bash it out with the answer choices! (not really a rigorous solution)<br /> <br /> ==Solution 2.5==<br /> <br /> In order to test the answer choices efficiently, realize that the probability &lt;math&gt;n&lt;/math&gt; flips yielding two heads is of the form:<br /> <br /> &lt;math&gt;\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}&lt;/math&gt;<br /> <br /> Similarly, the form for the probability of three heads is:<br /> <br /> &lt;math&gt;\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}&lt;/math&gt;.<br /> <br /> The probability of getting three heads (comapred to the probability of getting two) from &lt;math&gt;n&lt;/math&gt; flips is missing a factor of three in the numerator. Thus, we need &lt;math&gt;\dbinom{n}{3}&lt;/math&gt; to add a factor of 3 to the numerator of the probability of getting three heads.<br /> Our testing equation becomes:<br /> <br /> &lt;math&gt;\dbinom{n}{2} \times 3 = \dbinom{n}{3}&lt;/math&gt;.<br /> <br /> Since after factoring out the three from &lt;math&gt;\dbinom{n}{3}&lt;/math&gt;, the remaining factorizations should be equal. <br /> <br /> The only answer choice satisfying this condition is &lt;math&gt;\fbox{\textbf{(D)}\;11}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=141442 2020 AMC 10A Problems/Problem 16 2021-01-03T20:10:24Z <p>Joeya: /* Solution 5 (Estimating but differently again) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br /> <br /> == Problem ==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> === Diagram ===<br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray);<br /> draw(arc((1,0), 0.3989, 90, 180));<br /> filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray);<br /> draw(arc((1,1), 0.3989, 180, 270));<br /> filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray);<br /> draw(arc((0,1), 0.3989, 270, 360));<br /> filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> Diagram by [[User:Mathandski|MathandSki]] Using Asymptote<br /> <br /> Note: The diagram represents each unit square of the given &lt;math&gt;2020 * 2020&lt;/math&gt; square.<br /> <br /> ===Solution===<br /> <br /> We consider an individual one-by-one block.<br /> <br /> If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius &lt;math&gt;d&lt;/math&gt;, the area covered by the circles should be &lt;math&gt;0.5&lt;/math&gt;. Because of this, and the fact that there are four circles, we write<br /> <br /> &lt;cmath&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;d&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;, where with &lt;math&gt;\pi \approx 3&lt;/math&gt;, we get &lt;math&gt;d = \frac{1}{\sqrt{6}}&lt;/math&gt;, and from here, we simplify and see that &lt;math&gt;d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}&lt;/math&gt; ~Crypthes<br /> <br /> &lt;math&gt;\textbf{Note:}&lt;/math&gt; To be more rigorous, note that &lt;math&gt;d&lt;0.5&lt;/math&gt; since if &lt;math&gt;d\geq0.5&lt;/math&gt; then clearly the probability is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This would make sure the above solution works, as if &lt;math&gt;d\geq0.5&lt;/math&gt; there is overlap with the quartercircles. &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> <br /> == Solution 2 ==<br /> As in the previous solution, we obtain the equation &lt;math&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;\pi d^2 = \frac{1}{2} = 0.5&lt;/math&gt;. Since &lt;math&gt;\pi&lt;/math&gt; is slightly more than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;d^2&lt;/math&gt; is slightly less than &lt;math&gt;\frac{0.5}{3} = 0.1\bar{6}&lt;/math&gt;. We notice that &lt;math&gt;0.1\bar{6}&lt;/math&gt; is slightly more than &lt;math&gt;0.4^2 = 0.16&lt;/math&gt;, so &lt;math&gt;d&lt;/math&gt; is roughly &lt;math&gt;\boxed{\textbf{(B) } 0.4}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> == Solution 3 (Estimating) ==<br /> <br /> As above, we find that we need to estimate &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;. <br /> <br /> Note that we can approximate &lt;math&gt;2\pi \approx 6.28318 \approx 6.25&lt;/math&gt; and so &lt;math&gt;\frac{1}{\sqrt{2\pi}}&lt;/math&gt; &lt;math&gt;\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4&lt;/math&gt;.<br /> <br /> And so our answer is &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;.<br /> <br /> ~Silverdragon<br /> <br /> ==Solution 4 (Estimating but a bit different)==<br /> We only need to figure out the probability for a unit square, as it will scale up to the &lt;math&gt;2020\times 2020&lt;/math&gt; square. Since we want to find the probability that a point inside a unit square that is &lt;math&gt;d&lt;/math&gt; units away from a lattice point (a corner of the square) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, we can find which answer will come the closest to covering &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the area. <br /> <br /> Since the closest is &lt;math&gt;0.4&lt;/math&gt; which turns out to be &lt;math&gt;(0.4)^2\times \pi = 0.16 \times \pi&lt;/math&gt; which is about &lt;math&gt;0.502&lt;/math&gt;, we find that the answer rounded to the nearest tenth is &lt;math&gt;0.4&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ~RuiyangWu<br /> <br /> ==Solution 5 (Estimating but differently again)==<br /> As per the above diagram, realize that &lt;math&gt;\pi d^2 = \frac{1}{2}&lt;/math&gt;, so &lt;math&gt;d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{2} \approx 1.4 = \frac{7}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{\pi}&lt;/math&gt; is between &lt;math&gt;1.7&lt;/math&gt; and &lt;math&gt;1.8&lt;/math&gt; &lt;math&gt;((1.7)^2 = 2.89&lt;/math&gt; and &lt;math&gt;(1.8)^2 = 3.24)&lt;/math&gt;, so we can say &lt;math&gt;\sqrt{\pi} \approx 1.75 = \frac{7}{4}&lt;/math&gt;.<br /> <br /> So &lt;math&gt;d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}&lt;/math&gt;. This is slightly above &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;, since &lt;math&gt;\frac{20}{49} \approx \frac{2}{5}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==Video Solution==<br /> <br /> Education, The Study of Everything<br /> <br /> https://youtu.be/napCkujyrac<br /> <br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141441 2017 AMC 12B Problems/Problem 18 2021-01-03T20:09:54Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate geometry==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, therefore &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and line &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=141440 2020 AMC 10A Problems/Problem 16 2021-01-03T20:08:07Z <p>Joeya: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br /> <br /> == Problem ==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> === Diagram ===<br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray);<br /> draw(arc((1,0), 0.3989, 90, 180));<br /> filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray);<br /> draw(arc((1,1), 0.3989, 180, 270));<br /> filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray);<br /> draw(arc((0,1), 0.3989, 270, 360));<br /> filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> Diagram by [[User:Mathandski|MathandSki]] Using Asymptote<br /> <br /> Note: The diagram represents each unit square of the given &lt;math&gt;2020 * 2020&lt;/math&gt; square.<br /> <br /> ===Solution===<br /> <br /> We consider an individual one-by-one block.<br /> <br /> If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius &lt;math&gt;d&lt;/math&gt;, the area covered by the circles should be &lt;math&gt;0.5&lt;/math&gt;. Because of this, and the fact that there are four circles, we write<br /> <br /> &lt;cmath&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;d&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;, where with &lt;math&gt;\pi \approx 3&lt;/math&gt;, we get &lt;math&gt;d = \frac{1}{\sqrt{6}}&lt;/math&gt;, and from here, we simplify and see that &lt;math&gt;d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}&lt;/math&gt; ~Crypthes<br /> <br /> &lt;math&gt;\textbf{Note:}&lt;/math&gt; To be more rigorous, note that &lt;math&gt;d&lt;0.5&lt;/math&gt; since if &lt;math&gt;d\geq0.5&lt;/math&gt; then clearly the probability is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This would make sure the above solution works, as if &lt;math&gt;d\geq0.5&lt;/math&gt; there is overlap with the quartercircles. &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> <br /> == Solution 2 ==<br /> As in the previous solution, we obtain the equation &lt;math&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;\pi d^2 = \frac{1}{2} = 0.5&lt;/math&gt;. Since &lt;math&gt;\pi&lt;/math&gt; is slightly more than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;d^2&lt;/math&gt; is slightly less than &lt;math&gt;\frac{0.5}{3} = 0.1\bar{6}&lt;/math&gt;. We notice that &lt;math&gt;0.1\bar{6}&lt;/math&gt; is slightly more than &lt;math&gt;0.4^2 = 0.16&lt;/math&gt;, so &lt;math&gt;d&lt;/math&gt; is roughly &lt;math&gt;\boxed{\textbf{(B) } 0.4}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> == Solution 3 (Estimating) ==<br /> <br /> As above, we find that we need to estimate &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;. <br /> <br /> Note that we can approximate &lt;math&gt;2\pi \approx 6.28318 \approx 6.25&lt;/math&gt; and so &lt;math&gt;\frac{1}{\sqrt{2\pi}}&lt;/math&gt; &lt;math&gt;\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4&lt;/math&gt;.<br /> <br /> And so our answer is &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;.<br /> <br /> ~Silverdragon<br /> <br /> ==Solution 4 (Estimating but a bit different)==<br /> We only need to figure out the probability for a unit square, as it will scale up to the &lt;math&gt;2020\times 2020&lt;/math&gt; square. Since we want to find the probability that a point inside a unit square that is &lt;math&gt;d&lt;/math&gt; units away from a lattice point (a corner of the square) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, we can find which answer will come the closest to covering &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the area. <br /> <br /> Since the closest is &lt;math&gt;0.4&lt;/math&gt; which turns out to be &lt;math&gt;(0.4)^2\times \pi = 0.16 \times \pi&lt;/math&gt; which is about &lt;math&gt;0.502&lt;/math&gt;, we find that the answer rounded to the nearest tenth is &lt;math&gt;0.4&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ~RuiyangWu<br /> <br /> ==Solution 5 (Estimating but differently again)==<br /> As per the above diagram, realize that &lt;math&gt;\pi d^2 = \frac{1}{2}&lt;/math&gt;, so &lt;math&gt;d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{2} \approx 1.4 = \frac{7}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{\pi}&lt;/math&gt; is between &lt;math&gt;1.7&lt;/math&gt; and &lt;math&gt;1.8&lt;/math&gt; &lt;math&gt;((1.7)^2 = 2.89&lt;/math&gt; and &lt;math&gt;(1.8)^2 = 3.24)&lt;/math&gt;, so we can say &lt;math&gt;\sqrt{\pi} \approx 1.75 = \frac{7}{4}&lt;/math&gt;.<br /> <br /> So &lt;math&gt;d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}&lt;/math&gt;. This is slightly above &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;, since &lt;math&gt;\frac{20}{49} \approx \frac{2}{5}&lt;/math&gt;.<br /> <br /> ~Joeya<br /> <br /> ==Video Solution==<br /> <br /> Education, The Study of Everything<br /> <br /> https://youtu.be/napCkujyrac<br /> <br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141173 2017 AMC 12B Problems/Problem 18 2020-12-31T20:10:03Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate plane method==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, therefore &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and line &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141131 2017 AMC 12B Problems/Problem 18 2020-12-31T04:09:38Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate plane method==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, thus &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and line &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141109 2017 AMC 12B Problems/Problem 18 2020-12-31T01:42:39Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate plane method==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, thus &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane, since this is the point of intersection of the circle and line &lt;math&gt;AE&lt;/math&gt;. Note that one root will yield the intersection of the circle and &lt;math&gt;AE&lt;/math&gt; at the origin, so we will ignore this root.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141108 2017 AMC 12B Problems/Problem 18 2020-12-31T01:33:04Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate plane method==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, thus &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to yield the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=141107 2017 AMC 12B Problems/Problem 18 2020-12-31T01:30:44Z <p>Joeya: /* Solution 4: Coordinate plane method */</p> <hr /> <div>==Problem==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br /> import graph; size(8.865514650638614cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br /> <br /> <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br /> /* draw figures */<br /> draw(circle((0.,0.), 2.)); <br /> draw((-2.,0.)--(5.,5.)); <br /> draw((5.,5.)--(5.,0.)); <br /> draw((5.,0.)--(-2.,0.)); <br /> draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br /> draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br /> draw((2.,0.)--(-2.,0.)); <br /> draw((2.,0.)--(5.,5.)); <br /> draw((0.,0.)--(5.,5.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$O$&quot;, (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br /> dot((-2.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br /> dot((2.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br /> dot((5.,0.),dotstyle); <br /> label(&quot;$D$&quot;, (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br /> dot((5.,5.),dotstyle); <br /> label(&quot;$E$&quot;, (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br /> dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br /> label(&quot;$C$&quot;, (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Note that &lt;math&gt;EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}&lt;/math&gt;. However, by Power of a Point, &lt;math&gt;(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}&lt;/math&gt;. Now &lt;math&gt;BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}&lt;/math&gt;. Since &lt;math&gt;\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2: Similar triangles with Pythagorean==<br /> &lt;math&gt;AB&lt;/math&gt; is the diameter of the circle, so &lt;math&gt;\angle ACB&lt;/math&gt; is a right angle, and therefore by AA similarity, &lt;math&gt;\triangle ACB \sim \triangle ADE&lt;/math&gt;.<br /> <br /> Because of this, &lt;math&gt;\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}&lt;/math&gt;, so &lt;math&gt;AC = \frac{28}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Likewise, &lt;math&gt;\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}&lt;/math&gt;, so &lt;math&gt;BC = \frac{20}{\sqrt{74}}&lt;/math&gt;.<br /> <br /> Thus the area of &lt;math&gt;\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> == Solution 2b: Area shortcut ==<br /> <br /> Because &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\sqrt{74}&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, the ratio of the sides is &lt;math&gt;\frac{\sqrt{74}}{4}&lt;/math&gt;, meaning the ratio of the areas is thus &lt;math&gt;{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}&lt;/math&gt;. We then have the proportion &lt;math&gt;\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;<br /> <br /> ==Solution 3: Similar triangles without Pythagorean==<br /> Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:<br /> <br /> Draw &lt;math&gt;BF \parallel ED&lt;/math&gt; with &lt;math&gt;F&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;. &lt;math&gt;BF=5\times\frac{4}{7}=\frac{20}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC:CB:CF=49:35:25&lt;/math&gt;. (&lt;math&gt;7:5&lt;/math&gt; ratio applied twice)<br /> <br /> &lt;math&gt;[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Coordinate plane method==<br /> Let &lt;math&gt;A&lt;/math&gt; be at the origin &lt;math&gt;(0, 0)&lt;/math&gt; of a coordinate plane, with &lt;math&gt;B&lt;/math&gt; being located at &lt;math&gt;(4, 0)&lt;/math&gt;, etc.<br /> <br /> We can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt; by finding the the altitude from line &lt;math&gt;AB&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Realize that this altitude is the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; on the coordinate plane, since the respective base of &lt;math&gt;\triangle ABC&lt;/math&gt; is on the &lt;math&gt;x&lt;/math&gt;-axis.<br /> <br /> Using the diagram in solution one, the equation for circle &lt;math&gt;O&lt;/math&gt; is &lt;math&gt;(x-2)^2+y^2 = 4&lt;/math&gt;. <br /> <br /> The equation for line &lt;math&gt;AE&lt;/math&gt; is then &lt;math&gt;y = \frac{5}{7}x&lt;/math&gt;, thus &lt;math&gt;x = \frac{7}{5}y&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;\frac{7}{5}y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the equation for circle &lt;math&gt;O&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;(\frac{7}{5}y-2)^2+y^2 = 4&lt;/math&gt;<br /> <br /> We can solve for &lt;math&gt;y&lt;/math&gt; to determine the &lt;math&gt;y&lt;/math&gt; coordinate of point &lt;math&gt;C&lt;/math&gt; in the coordinate plane.<br /> <br /> Expanding the expression and factoring, we get:<br /> <br /> &lt;math&gt;(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{74}{25}y^2-\frac{28}{5}y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;50y(37y-70) = 0&lt;/math&gt;<br /> <br /> Our non-zero root is thus &lt;math&gt;\frac{70}{37}&lt;/math&gt;. Calculating the area of &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;4&lt;/math&gt; as the length of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;\frac{70}{37}&lt;/math&gt; as the altitude, we get:<br /> <br /> &lt;math&gt;\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Joeya