https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Captainsnake&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-30T06:15:50Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_11&diff=148676 2021 AMC 12B Problems/Problem 11 2021-03-06T05:23:43Z <p>Captainsnake: /* Solution 5 (Barycentric coordinates) */</p> <hr /> <div>==Problem==<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 /> ==Diagram==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (5,12);<br /> pair B = (0,0);<br /> pair C = (14,0);<br /> pair P = 2/3*A+1/3*C;<br /> pair D = 3/2*P;<br /> pair E = 3*P;<br /> draw(A--B--C--A);<br /> draw(A--D);<br /> draw(C--E--B);<br /> <br /> dot(&quot;$A$&quot;,A,N);<br /> dot(&quot;$B$&quot;,B,W);<br /> dot(&quot;$C$&quot;,C,ESE);<br /> dot(&quot;$D$&quot;,D,N);<br /> dot(&quot;$P$&quot;,P,W);<br /> dot(&quot;$E$&quot;,E,N);<br /> <br /> defaultpen(fontsize(9pt));<br /> label(&quot;$13$&quot;, (A+B)/2, NW);<br /> label(&quot;$14$&quot;, (B+C)/2, S);<br /> label(&quot;$5$&quot;,(A+P)/2, NE);<br /> label(&quot;$10$&quot;, (C+P)/2, NE);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1 (analytic geometry)==<br /> Toss on the Cartesian plane with &lt;math&gt;A=(5, 12), B=(0, 0),&lt;/math&gt; and &lt;math&gt;C=(14, 0)&lt;/math&gt;. Then &lt;math&gt;\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}&lt;/math&gt; by the trapezoid condition, where &lt;math&gt;D, E\in\overline{BP}&lt;/math&gt;. Since &lt;math&gt;PC=10&lt;/math&gt;, point &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;\tfrac{10}{15}=\tfrac{2}{3}&lt;/math&gt; of the way from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; and is located at &lt;math&gt;(8, 8)&lt;/math&gt;. Thus line &lt;math&gt;BP&lt;/math&gt; has equation &lt;math&gt;y=x&lt;/math&gt;. Since &lt;math&gt;\overline{AD}\parallel\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; is parallel to the ground, we know &lt;math&gt;D&lt;/math&gt; has the same &lt;math&gt;y&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, except it'll also lie on the line &lt;math&gt;y=x&lt;/math&gt;. Therefore, &lt;math&gt;D=(12, 12). \, \blacksquare&lt;/math&gt;<br /> <br /> To find the location of point &lt;math&gt;E&lt;/math&gt;, we need to find the intersection of &lt;math&gt;y=x&lt;/math&gt; with a line parallel to &lt;math&gt;\overline{AB}&lt;/math&gt; passing through &lt;math&gt;C&lt;/math&gt;. The slope of this line is the same as the slope of &lt;math&gt;\overline{AB}&lt;/math&gt;, or &lt;math&gt;\tfrac{12}{5}&lt;/math&gt;, and has equation &lt;math&gt;y=\tfrac{12}{5}x-\tfrac{168}{5}&lt;/math&gt;. The intersection of this line with &lt;math&gt;y=x&lt;/math&gt; is &lt;math&gt;(24, 24)&lt;/math&gt;. Therefore point &lt;math&gt;E&lt;/math&gt; is located at &lt;math&gt;(24, 24). \, \blacksquare&lt;/math&gt;<br /> <br /> The distance &lt;math&gt;DE&lt;/math&gt; is equal to the distance between &lt;math&gt;(12, 12)&lt;/math&gt; and &lt;math&gt;(24, 24)&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(D)} ~12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Using Stewart's Theorem we find &lt;math&gt;BP = 8\sqrt{2}&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;DP = BP\cdot\frac{PC}{PA} = 2BP&lt;/cmath&gt;<br /> &lt;cmath&gt;EP = BP\cdot\frac{PA}{PC} = \frac12 BP&lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;x&lt;/math&gt; be the length &lt;math&gt;PE&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;BP = \frac{PA}{PC}x = \frac12 x&lt;/cmath&gt;<br /> &lt;cmath&gt;PD = \frac{PA}{PC}BP = \frac14 x&lt;/cmath&gt;<br /> Therefore &lt;math&gt;BD = DE = \frac{3}{4}x&lt;/math&gt;. Now extend line &lt;math&gt;CD&lt;/math&gt; to the point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;, forming parallelogram &lt;math&gt;ZABC&lt;/math&gt;. As &lt;math&gt;BD = DE&lt;/math&gt; we also have &lt;math&gt;EZ = ZC = 13&lt;/math&gt; so &lt;math&gt;EC = 26&lt;/math&gt;.<br /> <br /> We now use the Law of Cosines to find &lt;math&gt;x&lt;/math&gt; (the length of &lt;math&gt;PE&lt;/math&gt;):<br /> &lt;cmath&gt;x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)&lt;/cmath&gt;<br /> As &lt;math&gt;\angle PCE = \angle BAC&lt;/math&gt;, we have (by Law of Cosines on triangle &lt;math&gt;BAC&lt;/math&gt;)<br /> &lt;cmath&gt;\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;\begin{align*}<br /> x^2 &amp;= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\<br /> &amp;= 776 - 264\\<br /> &amp;= 512<br /> \end{align*}&lt;/cmath&gt;<br /> And &lt;math&gt;x = 16\sqrt2&lt;/math&gt;. The answer is then &lt;math&gt;\frac34x = \boxed{\textbf{(D) }12\sqrt2}&lt;/math&gt;<br /> <br /> ==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)==<br /> Let the brackets denote areas. By Heron's Formula, we have<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC]&amp;=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\<br /> &amp;=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\<br /> &amp;=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\<br /> &amp;=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\<br /> &amp;=2^2\cdot3\cdot7 \\<br /> &amp;=84.<br /> \end{align*}&lt;/cmath&gt;<br /> It follows that the height of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\frac{2[ABC]}{14}=12.&lt;/math&gt;<br /> <br /> Next, we drop the altitudes &lt;math&gt;\overline{AF}&lt;/math&gt; and &lt;math&gt;\overline{DG}&lt;/math&gt; of &lt;math&gt;ABCD.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle AFB,&lt;/math&gt; we get &lt;math&gt;BF=5.&lt;/math&gt; By the AA Similarity, &lt;math&gt;\triangle ADP\sim\triangle CBP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;AD=7.&lt;/math&gt; Since &lt;math&gt;ADGF&lt;/math&gt; is a rectangle, &lt;math&gt;FG=AD=7.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle DGB,&lt;/math&gt; we get &lt;math&gt;BD=12\sqrt2.&lt;/math&gt;<br /> <br /> By &lt;math&gt;\triangle ADP\sim\triangle CBP&lt;/math&gt; again, we have &lt;math&gt;BP=8\sqrt2&lt;/math&gt; and &lt;math&gt;DP=4\sqrt2.&lt;/math&gt; Also, by the AA Similarity, &lt;math&gt;\triangle ABP\sim\triangle CEP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;EP=16\sqrt2.&lt;/math&gt;<br /> <br /> Finally, &lt;math&gt;DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.&lt;/math&gt;<br /> <br /> &lt;b&gt;PS: If you memorize that the area of a &lt;math&gt;\mathbf{13}\textbf{-}\mathbf{14}\textbf{-}\mathbf{15}&lt;/math&gt; triangle is &lt;math&gt;\mathbf{84,}&lt;/math&gt; the Heron's Formula part will be done instantly.&lt;/b&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 5 (Barycentric coordinates)==<br /> <br /> (For those unfamiliar with barycentric coordinates, consider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions#Articles)<br /> <br /> We can find &lt;math&gt;P&lt;/math&gt; in barycentric coordinates as &lt;math&gt;\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)&lt;/math&gt;. We can then write &lt;math&gt;\overline{BP}&lt;/math&gt; as &lt;math&gt;x-2z=0&lt;/math&gt;, where &lt;math&gt;(x,y,z)&lt;/math&gt; defines a point in barycentric coordinates. We have &lt;math&gt;\overline{AD}\parallel \overline{BC}&lt;/math&gt; as &lt;math&gt;y+z=0&lt;/math&gt; and &lt;math&gt;\overline{CE}\parallel \overline{AB}&lt;/math&gt; as &lt;math&gt;x+y=0&lt;/math&gt;. We can then compute &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; by intersecting lines:<br /> <br /> &lt;cmath&gt;\begin{cases}<br /> x-2z=0\\<br /> y+z=0\\<br /> x+y+z=1<br /> \end{cases}&lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;D=(1, -\frac{1}{2}, \frac{1}{2})&lt;/math&gt;. We can get &lt;math&gt;E&lt;/math&gt; with:<br /> <br /> &lt;cmath&gt;<br /> \begin{cases}<br /> x-2z=0\\<br /> x+y=0\\<br /> x+y+z=1<br /> \end{cases}<br /> &lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;E=(2, -2, 1)&lt;/math&gt;. Then, finding the displacement vector, we have &lt;math&gt;\overrightarrow{ED}=(1,-\frac{3}{2},\frac{1}{2})&lt;/math&gt;. Using the barycentric distance formula:<br /> <br /> &lt;cmath&gt;\text{dist}(D,E)=\sqrt{-a^2yx-b^2zx-c^2xy}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sqrt{-14^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)-15^2\Bigr(1\Bigr)\Bigr(\frac{1}{2}\Bigr)-13^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)}&lt;/cmath&gt;<br /> We get &lt;math&gt;\boxed{\textbf{(D)}12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://YouTube.com/watch?v=yxt8-rUUosI&amp;t=450s<br /> <br /> == Video Solution by OmegaLearn (Using properties of 13-14-15 triangle) ==<br /> https://youtu.be/mTcdKf5-FWg<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_11&diff=148675 2021 AMC 12B Problems/Problem 11 2021-03-06T05:23:25Z <p>Captainsnake: /* Solution 5 (Barycentric coordinates) */</p> <hr /> <div>==Problem==<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 /> ==Diagram==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (5,12);<br /> pair B = (0,0);<br /> pair C = (14,0);<br /> pair P = 2/3*A+1/3*C;<br /> pair D = 3/2*P;<br /> pair E = 3*P;<br /> draw(A--B--C--A);<br /> draw(A--D);<br /> draw(C--E--B);<br /> <br /> dot(&quot;$A$&quot;,A,N);<br /> dot(&quot;$B$&quot;,B,W);<br /> dot(&quot;$C$&quot;,C,ESE);<br /> dot(&quot;$D$&quot;,D,N);<br /> dot(&quot;$P$&quot;,P,W);<br /> dot(&quot;$E$&quot;,E,N);<br /> <br /> defaultpen(fontsize(9pt));<br /> label(&quot;$13$&quot;, (A+B)/2, NW);<br /> label(&quot;$14$&quot;, (B+C)/2, S);<br /> label(&quot;$5$&quot;,(A+P)/2, NE);<br /> label(&quot;$10$&quot;, (C+P)/2, NE);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1 (analytic geometry)==<br /> Toss on the Cartesian plane with &lt;math&gt;A=(5, 12), B=(0, 0),&lt;/math&gt; and &lt;math&gt;C=(14, 0)&lt;/math&gt;. Then &lt;math&gt;\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}&lt;/math&gt; by the trapezoid condition, where &lt;math&gt;D, E\in\overline{BP}&lt;/math&gt;. Since &lt;math&gt;PC=10&lt;/math&gt;, point &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;\tfrac{10}{15}=\tfrac{2}{3}&lt;/math&gt; of the way from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; and is located at &lt;math&gt;(8, 8)&lt;/math&gt;. Thus line &lt;math&gt;BP&lt;/math&gt; has equation &lt;math&gt;y=x&lt;/math&gt;. Since &lt;math&gt;\overline{AD}\parallel\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; is parallel to the ground, we know &lt;math&gt;D&lt;/math&gt; has the same &lt;math&gt;y&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, except it'll also lie on the line &lt;math&gt;y=x&lt;/math&gt;. Therefore, &lt;math&gt;D=(12, 12). \, \blacksquare&lt;/math&gt;<br /> <br /> To find the location of point &lt;math&gt;E&lt;/math&gt;, we need to find the intersection of &lt;math&gt;y=x&lt;/math&gt; with a line parallel to &lt;math&gt;\overline{AB}&lt;/math&gt; passing through &lt;math&gt;C&lt;/math&gt;. The slope of this line is the same as the slope of &lt;math&gt;\overline{AB}&lt;/math&gt;, or &lt;math&gt;\tfrac{12}{5}&lt;/math&gt;, and has equation &lt;math&gt;y=\tfrac{12}{5}x-\tfrac{168}{5}&lt;/math&gt;. The intersection of this line with &lt;math&gt;y=x&lt;/math&gt; is &lt;math&gt;(24, 24)&lt;/math&gt;. Therefore point &lt;math&gt;E&lt;/math&gt; is located at &lt;math&gt;(24, 24). \, \blacksquare&lt;/math&gt;<br /> <br /> The distance &lt;math&gt;DE&lt;/math&gt; is equal to the distance between &lt;math&gt;(12, 12)&lt;/math&gt; and &lt;math&gt;(24, 24)&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(D)} ~12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Using Stewart's Theorem we find &lt;math&gt;BP = 8\sqrt{2}&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;DP = BP\cdot\frac{PC}{PA} = 2BP&lt;/cmath&gt;<br /> &lt;cmath&gt;EP = BP\cdot\frac{PA}{PC} = \frac12 BP&lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;x&lt;/math&gt; be the length &lt;math&gt;PE&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;BP = \frac{PA}{PC}x = \frac12 x&lt;/cmath&gt;<br /> &lt;cmath&gt;PD = \frac{PA}{PC}BP = \frac14 x&lt;/cmath&gt;<br /> Therefore &lt;math&gt;BD = DE = \frac{3}{4}x&lt;/math&gt;. Now extend line &lt;math&gt;CD&lt;/math&gt; to the point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;, forming parallelogram &lt;math&gt;ZABC&lt;/math&gt;. As &lt;math&gt;BD = DE&lt;/math&gt; we also have &lt;math&gt;EZ = ZC = 13&lt;/math&gt; so &lt;math&gt;EC = 26&lt;/math&gt;.<br /> <br /> We now use the Law of Cosines to find &lt;math&gt;x&lt;/math&gt; (the length of &lt;math&gt;PE&lt;/math&gt;):<br /> &lt;cmath&gt;x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)&lt;/cmath&gt;<br /> As &lt;math&gt;\angle PCE = \angle BAC&lt;/math&gt;, we have (by Law of Cosines on triangle &lt;math&gt;BAC&lt;/math&gt;)<br /> &lt;cmath&gt;\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;\begin{align*}<br /> x^2 &amp;= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\<br /> &amp;= 776 - 264\\<br /> &amp;= 512<br /> \end{align*}&lt;/cmath&gt;<br /> And &lt;math&gt;x = 16\sqrt2&lt;/math&gt;. The answer is then &lt;math&gt;\frac34x = \boxed{\textbf{(D) }12\sqrt2}&lt;/math&gt;<br /> <br /> ==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)==<br /> Let the brackets denote areas. By Heron's Formula, we have<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC]&amp;=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\<br /> &amp;=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\<br /> &amp;=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\<br /> &amp;=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\<br /> &amp;=2^2\cdot3\cdot7 \\<br /> &amp;=84.<br /> \end{align*}&lt;/cmath&gt;<br /> It follows that the height of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\frac{2[ABC]}{14}=12.&lt;/math&gt;<br /> <br /> Next, we drop the altitudes &lt;math&gt;\overline{AF}&lt;/math&gt; and &lt;math&gt;\overline{DG}&lt;/math&gt; of &lt;math&gt;ABCD.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle AFB,&lt;/math&gt; we get &lt;math&gt;BF=5.&lt;/math&gt; By the AA Similarity, &lt;math&gt;\triangle ADP\sim\triangle CBP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;AD=7.&lt;/math&gt; Since &lt;math&gt;ADGF&lt;/math&gt; is a rectangle, &lt;math&gt;FG=AD=7.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle DGB,&lt;/math&gt; we get &lt;math&gt;BD=12\sqrt2.&lt;/math&gt;<br /> <br /> By &lt;math&gt;\triangle ADP\sim\triangle CBP&lt;/math&gt; again, we have &lt;math&gt;BP=8\sqrt2&lt;/math&gt; and &lt;math&gt;DP=4\sqrt2.&lt;/math&gt; Also, by the AA Similarity, &lt;math&gt;\triangle ABP\sim\triangle CEP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;EP=16\sqrt2.&lt;/math&gt;<br /> <br /> Finally, &lt;math&gt;DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.&lt;/math&gt;<br /> <br /> &lt;b&gt;PS: If you memorize that the area of a &lt;math&gt;\mathbf{13}\textbf{-}\mathbf{14}\textbf{-}\mathbf{15}&lt;/math&gt; triangle is &lt;math&gt;\mathbf{84,}&lt;/math&gt; the Heron's Formula part will be done instantly.&lt;/b&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 5 (Barycentric coordinates)==<br /> <br /> (For those unfamiliar with barycentric coordinates, consider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions#Articles)<br /> <br /> We can find &lt;math&gt;P&lt;/math&gt; in barycentric coordinates as &lt;math&gt;\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)&lt;/math&gt;. We can then write &lt;math&gt;\overline{BP}&lt;/math&gt; as &lt;math&gt;x-2z=0&lt;/math&gt;, where &lt;math&gt;(x,y,z)&lt;/math&gt; defines a point in barycentric coordinates. We have &lt;math&gt;\overline{AD}\parallel \overline{BC}&lt;/math&gt; as &lt;math&gt;y+z=0&lt;/math&gt; and &lt;math&gt;\overline{CE}\parallel \overline{AB}&lt;/math&gt; as &lt;math&gt;x+y=0&lt;/math&gt;. We can then compute &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; by intersecting lines:<br /> <br /> &lt;cmath&gt;\begin{cases}<br /> x-2z=0\\<br /> y+z=0\\<br /> x+y+z=1<br /> \end{cases}&lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;D=(1, -\frac{1}{2}, \frac{1}{2})&lt;/math&gt;. We can get &lt;math&gt;E&lt;/math&gt; with:<br /> <br /> &lt;cmath&gt;<br /> \begin{cases}<br /> x-2z=0\\<br /> x+y=0\\<br /> x+y+z=1<br /> \end{cases}<br /> &lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;E=(2, -2, 1)&lt;/math&gt;. Then, finding the displacement vector, we have &lt;math&gt;\overrightarrow{ED}=(1,-\frac{3}{2},frac{1}{2})&lt;/math&gt;. Using the barycentric distance formula:<br /> <br /> &lt;cmath&gt;\text{dist}(D,E)=\sqrt{-a^2yx-b^2zx-c^2xy}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sqrt{-14^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)-15^2\Bigr(1\Bigr)\Bigr(\frac{1}{2}\Bigr)-13^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)}&lt;/cmath&gt;<br /> We get &lt;math&gt;\boxed{\textbf{(D)}12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://YouTube.com/watch?v=yxt8-rUUosI&amp;t=450s<br /> <br /> == Video Solution by OmegaLearn (Using properties of 13-14-15 triangle) ==<br /> https://youtu.be/mTcdKf5-FWg<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_11&diff=148646 2021 AMC 12B Problems/Problem 11 2021-03-06T02:39:02Z <p>Captainsnake: </p> <hr /> <div>==Problem==<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 /> ==Diagram==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A = (5,12);<br /> pair B = (0,0);<br /> pair C = (14,0);<br /> pair P = 2/3*A+1/3*C;<br /> pair D = 3/2*P;<br /> pair E = 3*P;<br /> draw(A--B--C--A);<br /> draw(A--D);<br /> draw(C--E--B);<br /> <br /> dot(&quot;$A$&quot;,A,N);<br /> dot(&quot;$B$&quot;,B,W);<br /> dot(&quot;$C$&quot;,C,ESE);<br /> dot(&quot;$D$&quot;,D,N);<br /> dot(&quot;$P$&quot;,P,W);<br /> dot(&quot;$E$&quot;,E,N);<br /> <br /> defaultpen(fontsize(9pt));<br /> label(&quot;$13$&quot;, (A+B)/2, NW);<br /> label(&quot;$14$&quot;, (B+C)/2, S);<br /> label(&quot;$5$&quot;,(A+P)/2, NE);<br /> label(&quot;$10$&quot;, (C+P)/2, NE);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1 (analytic geometry)==<br /> Toss on the Cartesian plane with &lt;math&gt;A=(5, 12), B=(0, 0),&lt;/math&gt; and &lt;math&gt;C=(14, 0)&lt;/math&gt;. Then &lt;math&gt;\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}&lt;/math&gt; by the trapezoid condition, where &lt;math&gt;D, E\in\overline{BP}&lt;/math&gt;. Since &lt;math&gt;PC=10&lt;/math&gt;, point &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;\tfrac{10}{15}=\tfrac{2}{3}&lt;/math&gt; of the way from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; and is located at &lt;math&gt;(8, 8)&lt;/math&gt;. Thus line &lt;math&gt;BP&lt;/math&gt; has equation &lt;math&gt;y=x&lt;/math&gt;. Since &lt;math&gt;\overline{AD}\parallel\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; is parallel to the ground, we know &lt;math&gt;D&lt;/math&gt; has the same &lt;math&gt;y&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, except it'll also lie on the line &lt;math&gt;y=x&lt;/math&gt;. Therefore, &lt;math&gt;D=(12, 12). \, \blacksquare&lt;/math&gt;<br /> <br /> To find the location of point &lt;math&gt;E&lt;/math&gt;, we need to find the intersection of &lt;math&gt;y=x&lt;/math&gt; with a line parallel to &lt;math&gt;\overline{AB}&lt;/math&gt; passing through &lt;math&gt;C&lt;/math&gt;. The slope of this line is the same as the slope of &lt;math&gt;\overline{AB}&lt;/math&gt;, or &lt;math&gt;\tfrac{12}{5}&lt;/math&gt;, and has equation &lt;math&gt;y=\tfrac{12}{5}x-\tfrac{168}{5}&lt;/math&gt;. The intersection of this line with &lt;math&gt;y=x&lt;/math&gt; is &lt;math&gt;(24, 24)&lt;/math&gt;. Therefore point &lt;math&gt;E&lt;/math&gt; is located at &lt;math&gt;(24, 24). \, \blacksquare&lt;/math&gt;<br /> <br /> The distance &lt;math&gt;DE&lt;/math&gt; is equal to the distance between &lt;math&gt;(12, 12)&lt;/math&gt; and &lt;math&gt;(24, 24)&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(D)} ~12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Using Stewart's Theorem we find &lt;math&gt;BP = 8\sqrt{2}&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;DP = BP\cdot\frac{PC}{PA} = 2BP&lt;/cmath&gt;<br /> &lt;cmath&gt;EP = BP\cdot\frac{PA}{PC} = \frac12 BP&lt;/cmath&gt;<br /> So<br /> &lt;cmath&gt;DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;x&lt;/math&gt; be the length &lt;math&gt;PE&lt;/math&gt;. From the similar triangles &lt;math&gt;BPA\sim DPC&lt;/math&gt; and &lt;math&gt;BPC\sim EPA&lt;/math&gt; we have<br /> &lt;cmath&gt;BP = \frac{PA}{PC}x = \frac12 x&lt;/cmath&gt;<br /> &lt;cmath&gt;PD = \frac{PA}{PC}BP = \frac14 x&lt;/cmath&gt;<br /> Therefore &lt;math&gt;BD = DE = \frac{3}{4}x&lt;/math&gt;. Now extend line &lt;math&gt;CD&lt;/math&gt; to the point &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;AE&lt;/math&gt;, forming parallelogram &lt;math&gt;ZABC&lt;/math&gt;. As &lt;math&gt;BD = DE&lt;/math&gt; we also have &lt;math&gt;EZ = ZC = 13&lt;/math&gt; so &lt;math&gt;EC = 26&lt;/math&gt;.<br /> <br /> We now use the Law of Cosines to find &lt;math&gt;x&lt;/math&gt; (the length of &lt;math&gt;PE&lt;/math&gt;):<br /> &lt;cmath&gt;x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)&lt;/cmath&gt;<br /> As &lt;math&gt;\angle PCE = \angle BAC&lt;/math&gt;, we have (by Law of Cosines on triangle &lt;math&gt;BAC&lt;/math&gt;)<br /> &lt;cmath&gt;\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.&lt;/cmath&gt;<br /> Therefore<br /> &lt;cmath&gt;\begin{align*}<br /> x^2 &amp;= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\<br /> &amp;= 776 - 264\\<br /> &amp;= 512<br /> \end{align*}&lt;/cmath&gt;<br /> And &lt;math&gt;x = 16\sqrt2&lt;/math&gt;. The answer is then &lt;math&gt;\frac34x = \boxed{\textbf{(D) }12\sqrt2}&lt;/math&gt;<br /> <br /> ==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)==<br /> Let the brackets denote areas. By Heron's Formula, we have<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC]&amp;=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\<br /> &amp;=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\<br /> &amp;=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\<br /> &amp;=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\<br /> &amp;=2^2\cdot3\cdot7 \\<br /> &amp;=84.<br /> \end{align*}&lt;/cmath&gt;<br /> It follows that the height of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\frac{2[ABC]}{14}=12.&lt;/math&gt;<br /> <br /> Next, we drop the altitudes &lt;math&gt;\overline{AF}&lt;/math&gt; and &lt;math&gt;\overline{DG}&lt;/math&gt; of &lt;math&gt;ABCD.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle AFB,&lt;/math&gt; we get &lt;math&gt;BF=5.&lt;/math&gt; By the AA Similarity, &lt;math&gt;\triangle ADP\sim\triangle CBP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;AD=7.&lt;/math&gt; Since &lt;math&gt;ADGF&lt;/math&gt; is a rectangle, &lt;math&gt;FG=AD=7.&lt;/math&gt; By the Pythagorean Theorem on &lt;math&gt;\triangle DGB,&lt;/math&gt; we get &lt;math&gt;BD=12\sqrt2.&lt;/math&gt;<br /> <br /> By &lt;math&gt;\triangle ADP\sim\triangle CBP&lt;/math&gt; again, we have &lt;math&gt;BP=8\sqrt2&lt;/math&gt; and &lt;math&gt;DP=4\sqrt2.&lt;/math&gt; Also, by the AA Similarity, &lt;math&gt;\triangle ABP\sim\triangle CEP,&lt;/math&gt; with the ratio of similitude &lt;math&gt;1:2.&lt;/math&gt; It follows that &lt;math&gt;EP=16\sqrt2.&lt;/math&gt;<br /> <br /> Finally, &lt;math&gt;DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.&lt;/math&gt;<br /> <br /> &lt;b&gt;PS: If you memorize that the area of a &lt;math&gt;\mathbf{13}\textbf{-}\mathbf{14}\textbf{-}\mathbf{15}&lt;/math&gt; triangle is &lt;math&gt;\mathbf{84,}&lt;/math&gt; the Heron's Formula part will be done instantly.&lt;/b&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 5 (Barycentric coordinates)==<br /> <br /> (For those unfamiliar with barycentric coordinates, consider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions#Articles)<br /> <br /> We can find &lt;math&gt;P&lt;/math&gt; in barycentric coordinates as &lt;math&gt;\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)&lt;/math&gt;. We can then write &lt;math&gt;\overline{BP}&lt;/math&gt; as &lt;math&gt;x-2z=0&lt;/math&gt;, where &lt;math&gt;(x,y,z)&lt;/math&gt; defines a point in barycentric coordinates. We have &lt;math&gt;\overline{AD}\parallel \overline{BC}&lt;/math&gt; as &lt;math&gt;y+z=0&lt;/math&gt; and &lt;math&gt;\overline{CE}\parallel \overline{AB}&lt;/math&gt; as &lt;math&gt;x+y=0&lt;/math&gt;. We can then compute &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; by intersecting lines:<br /> <br /> &lt;cmath&gt;\begin{cases}<br /> x-2z=0\\<br /> y+z=0\\<br /> x+y+z=1<br /> \end{cases}&lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;D=(1, -\frac{1}{2}, \frac{1}{2})&lt;/math&gt;. We can get &lt;math&gt;E&lt;/math&gt; with:<br /> <br /> &lt;cmath&gt;<br /> \begin{cases}<br /> x-2z=0\\<br /> x+y=0\\<br /> x+y+z=1<br /> \end{cases}<br /> &lt;/cmath&gt;<br /> <br /> Which gives us &lt;math&gt;E=(2, -2, 1)&lt;/math&gt;. Then, finding the displacement vector, we have \overrightarrow{ED}=(1,-\frac{3}{2},frac{1}{2}). Using the barycentric distance formula:<br /> <br /> &lt;cmath&gt;\text{dist}(D,E)=\sqrt{-a^2yx-b^2zx-c^2xy}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sqrt{-14^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)-15^2\Bigr(1\Bigr)\Bigr(\frac{1}{2}\Bigr)-13^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)}&lt;/cmath&gt;<br /> We get &lt;math&gt;\boxed{\textbf{(D)}12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://YouTube.com/watch?v=yxt8-rUUosI&amp;t=450s<br /> <br /> == Video Solution by OmegaLearn (Using properties of 13-14-15 triangle) ==<br /> https://youtu.be/mTcdKf5-FWg<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=p4iCAZRUESs<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_10&diff=147993 2000 AIME II Problems/Problem 10 2021-02-26T18:16:20Z <p>Captainsnake: Added fancier version of solution 2 that involves less bashing, also fixed some typos.</p> <hr /> <div>== Problem ==<br /> A [[circle]] is [[inscribe]]d in [[quadrilateral]] &lt;math&gt;ABCD&lt;/math&gt;, [[tangent]] to &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and to &lt;math&gt;\overline{CD}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt;. Given that &lt;math&gt;AP=19&lt;/math&gt;, &lt;math&gt;PB=26&lt;/math&gt;, &lt;math&gt;CQ=37&lt;/math&gt;, and &lt;math&gt;QD=23&lt;/math&gt;, find the [[Perfect square|square]] of the [[radius]] of the circle.<br /> <br /> == Solution 1==<br /> Call the [[center]] of the circle &lt;math&gt;O&lt;/math&gt;. By drawing the lines from &lt;math&gt;O&lt;/math&gt; tangent to the sides and from &lt;math&gt;O&lt;/math&gt; to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed.<br /> <br /> Thus, &lt;math&gt;\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180&lt;/math&gt;, or &lt;math&gt;(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180&lt;/math&gt;.<br /> <br /> Take the &lt;math&gt;\tan&lt;/math&gt; of both sides and use the identity for &lt;math&gt;\tan(A+B)&lt;/math&gt; to get &lt;cmath&gt;\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.&lt;/cmath&gt;<br /> <br /> Use the identity for &lt;math&gt;\tan(A+B)&lt;/math&gt; again to get &lt;cmath&gt;\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.&lt;/cmath&gt;<br /> <br /> Solving gives &lt;math&gt;r^2=\boxed{647}&lt;/math&gt;.<br /> <br /> Note: the equation may seem nasty at first, but once you cancel the &lt;math&gt;r&lt;/math&gt;s and other factors, you are just left with &lt;math&gt;r^2&lt;/math&gt;. That gives us &lt;math&gt;647&lt;/math&gt; quite easily.<br /> <br /> == Solution 2==<br /> Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on (&lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are the tangent lengths, not the side lengths).<br /> &lt;cmath&gt;A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}&lt;/cmath&gt;<br /> &lt;math&gt;r^2=\frac{A}{a+b+c+d}^2 = \boxed{647}&lt;/math&gt;.<br /> <br /> == Solution 3 (Smart algebra to make 2 less annoying) ==<br /> <br /> Using the formulas established in solution 2, one notices:<br /> &lt;cmath&gt;r^2=\frac{A^2}{a+b+c+d}&lt;/cmath&gt;<br /> &lt;cmath&gt;r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}&lt;/cmath&gt;<br /> &lt;cmath&gt;r^2=\boxed{647}&lt;/cmath&gt;<br /> <br /> which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_12&diff=147172 2000 AMC 12 Problems/Problem 12 2021-02-16T15:06:46Z <p>Captainsnake: /* Solution 2 (Nonrigorous) */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Let &lt;math&gt;A, M,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; be [[nonnegative integer]]s such that &lt;math&gt;A + M + C=12&lt;/math&gt;. What is the maximum value of &lt;math&gt;A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> It is not hard to see that <br /> &lt;cmath&gt;(A+1)(M+1)(C+1)=&lt;/cmath&gt;<br /> &lt;cmath&gt;AMC+AM+AC+MC+A+M+C+1&lt;/cmath&gt;<br /> Since &lt;math&gt;A+M+C=12&lt;/math&gt;, we can rewrite this as<br /> &lt;cmath&gt;(A+1)(M+1)(C+1)=&lt;/cmath&gt;<br /> &lt;cmath&gt;AMC+AM+AC+MC+13&lt;/cmath&gt;<br /> So we wish to maximize<br /> &lt;cmath&gt;(A+1)(M+1)(C+1)-13&lt;/cmath&gt;<br /> Which is largest when all the factors are equal (consequence of AM-GM). Since &lt;math&gt;A+M+C=12&lt;/math&gt;, we set &lt;math&gt;A=M=C=4&lt;/math&gt;<br /> Which gives us <br /> &lt;cmath&gt;(4+1)(4+1)(4+1)-13=112&lt;/cmath&gt;<br /> so the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;.<br /> <br /> == Solution 2 (Nonrigorous) ==<br /> <br /> If you know that to maximize your result you &lt;math&gt;\textit{usually}&lt;/math&gt; have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make &lt;math&gt;A,M&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; as close as possible. In this case, they would all be equal to &lt;math&gt;4&lt;/math&gt;, so &lt;math&gt;AMC+AM+AC+MC=64+16+16+16=112&lt;/math&gt;, giving you the answer of &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_12&diff=147171 2000 AMC 12 Problems/Problem 12 2021-02-16T15:06:27Z <p>Captainsnake: /* Solution 2 */ Clarified that this solution isn't rigorous</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Let &lt;math&gt;A, M,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; be [[nonnegative integer]]s such that &lt;math&gt;A + M + C=12&lt;/math&gt;. What is the maximum value of &lt;math&gt;A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> It is not hard to see that <br /> &lt;cmath&gt;(A+1)(M+1)(C+1)=&lt;/cmath&gt;<br /> &lt;cmath&gt;AMC+AM+AC+MC+A+M+C+1&lt;/cmath&gt;<br /> Since &lt;math&gt;A+M+C=12&lt;/math&gt;, we can rewrite this as<br /> &lt;cmath&gt;(A+1)(M+1)(C+1)=&lt;/cmath&gt;<br /> &lt;cmath&gt;AMC+AM+AC+MC+13&lt;/cmath&gt;<br /> So we wish to maximize<br /> &lt;cmath&gt;(A+1)(M+1)(C+1)-13&lt;/cmath&gt;<br /> Which is largest when all the factors are equal (consequence of AM-GM). Since &lt;math&gt;A+M+C=12&lt;/math&gt;, we set &lt;math&gt;A=M=C=4&lt;/math&gt;<br /> Which gives us <br /> &lt;cmath&gt;(4+1)(4+1)(4+1)-13=112&lt;/cmath&gt;<br /> so the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;.<br /> <br /> == Solution 2 (Nonrigorous) ==<br /> <br /> If you know that to maximize your result you \textit{usually} have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make &lt;math&gt;A,M&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; as close as possible. In this case, they would all be equal to &lt;math&gt;4&lt;/math&gt;, so &lt;math&gt;AMC+AM+AC+MC=64+16+16+16=112&lt;/math&gt;, giving you the answer of &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B&diff=144916 2021 AMC 12B 2021-02-11T19:18:38Z <p>Captainsnake: Fixed tense since the test is no longer in the future lol</p> <hr /> <div>'''2021 AMC 12B''' problems and solutions. The test was held on Wednesday, February &lt;math&gt;10&lt;/math&gt;, &lt;math&gt;2021&lt;/math&gt;.<br /> <br /> *[[2021 AMC 12B Problems]]<br /> *[[2021 AMC 12B Answer Key]]<br /> **[[2021 AMC 12B Problems/Problem 1|Problem 1]]<br /> **[[2021 AMC 12B Problems/Problem 2|Problem 2]]<br /> **[[2021 AMC 12B Problems/Problem 3|Problem 3]]<br /> **[[2021 AMC 12B Problems/Problem 4|Problem 4]]<br /> **[[2021 AMC 12B Problems/Problem 5|Problem 5]]<br /> **[[2021 AMC 12B Problems/Problem 6|Problem 6]]<br /> **[[2021 AMC 12B Problems/Problem 7|Problem 7]]<br /> **[[2021 AMC 12B Problems/Problem 8|Problem 8]]<br /> **[[2021 AMC 12B Problems/Problem 9|Problem 9]]<br /> **[[2021 AMC 12B Problems/Problem 10|Problem 10]]<br /> **[[2021 AMC 12B Problems/Problem 11|Problem 11]]<br /> **[[2021 AMC 12B Problems/Problem 12|Problem 12]]<br /> **[[2021 AMC 12B Problems/Problem 13|Problem 13]]<br /> **[[2021 AMC 12B Problems/Problem 14|Problem 14]]<br /> **[[2021 AMC 12B Problems/Problem 15|Problem 15]]<br /> **[[2021 AMC 12B Problems/Problem 16|Problem 16]]<br /> **[[2021 AMC 12B Problems/Problem 17|Problem 17]]<br /> **[[2021 AMC 12B Problems/Problem 18|Problem 18]]<br /> **[[2021 AMC 12B Problems/Problem 19|Problem 19]]<br /> **[[2021 AMC 12B Problems/Problem 20|Problem 20]]<br /> **[[2021 AMC 12B Problems/Problem 21|Problem 21]]<br /> **[[2021 AMC 12B Problems/Problem 22|Problem 22]]<br /> **[[2021 AMC 12B Problems/Problem 23|Problem 23]]<br /> **[[2021 AMC 12B Problems/Problem 24|Problem 24]]<br /> **[[2021 AMC 12B Problems/Problem 25|Problem 25]]<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|before= [[2020 AMC 12A|2020 AMC 12A]], [[2020 AMC 12B|B]], [[2021 AMC 12A]]|after=[[2022 AMC 12A|2022 AMC 12A]], [[2022 AMC 12B|B]]|ab=B}}<br /> * [[AMC 12]]<br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}<br /> --&gt;</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_7&diff=144668 2014 AMC 12B Problems/Problem 7 2021-02-02T17:05:00Z <p>Captainsnake: /* Solutions */</p> <hr /> <div>==Problem==<br /> <br /> For how many positive integers &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n}{30-n}&lt;/math&gt; also a positive integer?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> We know that &lt;math&gt;n \le 30&lt;/math&gt; or else &lt;math&gt;30-n&lt;/math&gt; will be negative, resulting in a negative fraction. We also know that &lt;math&gt;n \ge 15&lt;/math&gt; or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;15&lt;/math&gt; to &lt;math&gt;30&lt;/math&gt; gives us integer values for &lt;math&gt;n=15, 20, 24, 25, 27, 28, 29&lt;/math&gt;. Counting them up, we have &lt;math&gt;\boxed{\textbf{(D)}\ 7}&lt;/math&gt; possible values for &lt;math&gt;n&lt;/math&gt;. <br /> <br /> ===Solution 2===<br /> Let &lt;math&gt; \frac{n}{30-n}=m &lt;/math&gt;, where &lt;math&gt; m \in \mathbb{N} &lt;/math&gt;. Solving for &lt;math&gt; n &lt;/math&gt;, we find that &lt;math&gt; n=\frac{30m}{m+1} &lt;/math&gt;. Because &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; m+1 &lt;/math&gt; are relatively prime, &lt;math&gt; m+1|30 &lt;/math&gt;. Our answer is the number of proper divisors of &lt;math&gt; 2^13^15^1 &lt;/math&gt;, which is &lt;math&gt; (1+1)(1+1)(1+1)-1 = \boxed{\textbf{(D)}\ 7} &lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We know that &lt;math&gt;30-n|n&lt;/math&gt;. Then, by divisibility rules:<br /> <br /> &lt;cmath&gt;\Leftrightarrow 30-n|n+30-n&lt;/cmath&gt;<br /> &lt;cmath&gt;\Leftrightarrow 30-n|30&lt;/cmath&gt;<br /> <br /> There are &lt;math&gt;8&lt;/math&gt; divisors of &lt;math&gt;30&lt;/math&gt;, but &lt;math&gt;n&lt;/math&gt; must be positive, so &lt;math&gt;30|30&lt;/math&gt; isn't counted, meaning we have &lt;math&gt;\boxed{\textbf{(D)}\ 7} &lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_20&diff=144636 2014 AMC 10A Problems/Problem 20 2021-02-02T15:31:38Z <p>Captainsnake: added formal proof to solution 1 for those who would like a more rigorous version of it.</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #16]] and [[2014 AMC 10A Problems|2014 AMC 10A #20]]}}<br /> ==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;k&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can list the first few numbers in the form &lt;math&gt;8 \cdot (8....8)&lt;/math&gt;<br /> <br /> (Hard problem to do without the multiplication, but you can see the pattern early on)<br /> <br /> &lt;math&gt;8 \cdot 8 = 64&lt;/math&gt;<br /> <br /> &lt;math&gt;8 \cdot 88 = 704&lt;/math&gt;<br /> <br /> &lt;math&gt;8 \cdot 888 = 7104&lt;/math&gt;<br /> <br /> &lt;math&gt;8 \cdot 8888 = 71104&lt;/math&gt;<br /> <br /> &lt;math&gt;8 \cdot 88888 = 711104&lt;/math&gt;<br /> <br /> By now it's clear that the numbers will be in the form &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;k-2&lt;/math&gt; &lt;math&gt;1&lt;/math&gt;s, and &lt;math&gt;04&lt;/math&gt;. We want to make the numbers sum to 1000, so &lt;math&gt;7+4+(k-2) = 1000&lt;/math&gt;. Solving, we get &lt;math&gt;k = 991&lt;/math&gt;, meaning the answer is &lt;math&gt;\fbox{(D)}&lt;/math&gt;<br /> <br /> Another way to proceed is that we know the difference between the sum of the digits of each product and &lt;math&gt;k&lt;/math&gt; is always &lt;math&gt;9&lt;/math&gt;, so we just do &lt;math&gt;1000-9=\boxed{\textbf{(D)991}}&lt;/math&gt;.<br /> <br /> ===Proof of this solution's validity===<br /> <br /> Since this solution won't fly on a proof-based competition, here's a proof that it's valid:<br /> <br /> We will call &lt;math&gt;x_k=8(888\dots8)&lt;/math&gt; with exactly &lt;math&gt;k&lt;/math&gt; &lt;math&gt;8&lt;/math&gt;s. We then rewrite this more formally, as:<br /> <br /> &lt;cmath&gt;x_k=8\biggr(\sum_{n=0}^{k}8(10)^n\biggr)&lt;/cmath&gt;<br /> &lt;cmath&gt;=64\biggr(\sum_{n=0}^{k}(10)^n\biggr)&lt;/cmath&gt;<br /> &lt;cmath&gt;=64\frac{10^{k+1}-1}{9}&lt;/cmath&gt;<br /> <br /> Then, finding a recursive formula, we get:<br /> <br /> &lt;cmath&gt;x_{k+1}=64\times 10^{k+1}+x_k&lt;/cmath&gt;<br /> <br /> We will now use induction, Our base case will be &lt;math&gt;k=2&lt;/math&gt;. It's easy to see that this becomes &lt;math&gt;x_2=704&lt;/math&gt;. Then, the &lt;math&gt;k+1&lt;/math&gt; case: let x_k=7111\dots104 with &lt;math&gt;k-2&lt;/math&gt; &lt;math&gt;1&lt;/math&gt;s. Then &lt;math&gt;x_{k+1}=64000\dots000+7111\dots104&lt;/math&gt;. Adding these numbers, we get &lt;math&gt;x_{k+1}=71111\dots104&lt;/math&gt;. <br /> <br /> Summing these digits, we have &lt;math&gt;4+7+(k-2)=1000&lt;/math&gt;, giving us &lt;math&gt;k=991&lt;/math&gt;.<br /> <br /> ==Solution 2(Educated Guesses if you have no time)==<br /> We first note that &lt;math&gt;125 \cdot 8 = 1000&lt;/math&gt; and so we assume there are &lt;math&gt;125&lt;/math&gt; 8s.<br /> <br /> Then we note that it is asking for the second factor, so we subtract &lt;math&gt;1&lt;/math&gt;(the original &lt;math&gt;8&lt;/math&gt; in the first factor).<br /> <br /> Now we have &lt;math&gt;125-1=124.&lt;/math&gt; The second factor is obviously a multiple of &lt;math&gt;124&lt;/math&gt;.<br /> <br /> Listing the first few, we have &lt;math&gt;124, 248, 372, 496, 620, 744, 868, 992, 1116, 1240, ...&lt;/math&gt;<br /> <br /> We notice that the 4th answer choice is 1 less than a 992(a multiple of 124.)<br /> <br /> Thus we make an educated guess that it is somehow less by 1, so we get &lt;math&gt;\fbox{(D)}&lt;/math&gt;. ~mathboy282<br /> <br /> ===Note(Must Read)===<br /> We were just lucky; this method is NOT reliable. Please note that this probably will not work for other problems and is just a lucky coincidence.<br /> <br /> ==Video Solution==<br /> https://youtu.be/wQzuQZvq8sk<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2014|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Number Theory Problems]]</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_15&diff=144627 2014 AMC 12A Problems/Problem 15 2021-02-02T15:12:09Z <p>Captainsnake: Added solution</p> <hr /> <div>==Problem==<br /> <br /> A five-digit palindrome is a positive integer with respective digits &lt;math&gt;abcba&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; is non-zero. Let &lt;math&gt;S&lt;/math&gt; be the sum of all five-digit palindromes. What is the sum of the digits of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad<br /> \textbf{(B) }18\qquad<br /> \textbf{(C) }27\qquad<br /> \textbf{(D) }36\qquad<br /> \textbf{(E) }45\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For each digit &lt;math&gt;a=1,2,\ldots,9&lt;/math&gt; there are &lt;math&gt;10\cdot10&lt;/math&gt; (ways of choosing &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;) palindromes. So the &lt;math&gt;a&lt;/math&gt;s contribute &lt;math&gt;(1+2+\cdots+9)(100)(10^4+1)&lt;/math&gt; to the sum.<br /> For each digit &lt;math&gt;b=0,1,2,\ldots,9&lt;/math&gt; there are &lt;math&gt;9\cdot10&lt;/math&gt; (since &lt;math&gt;a \neq 0&lt;/math&gt;) palindromes. So the &lt;math&gt;b&lt;/math&gt;s contribute &lt;math&gt;(0+1+2+\cdots+9)(90)(10^3+10)&lt;/math&gt; to the sum.<br /> Similarly, for each &lt;math&gt;c=0,1,2,\ldots,9&lt;/math&gt; there are &lt;math&gt;9\cdot10&lt;/math&gt; palindromes, so the &lt;math&gt;c&lt;/math&gt; contributes &lt;math&gt;(0+1+2+\cdots+9)(90)(10^2)&lt;/math&gt; to the sum.<br /> <br /> It just so happens that &lt;cmath&gt; (1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000 &lt;/cmath&gt; so the sum of the digits of the sum is &lt;math&gt;\boxed{\textbf{(B)}\; 18}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Notice that &lt;math&gt;10001+ 99999 = 110000.&lt;/math&gt; In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is &lt;math&gt;110000.&lt;/math&gt; We have &lt;math&gt;9*10*10&lt;/math&gt; palindromes, or &lt;math&gt;450&lt;/math&gt; pairs of palindromes summing to &lt;math&gt;110000.&lt;/math&gt; Performing the multiplication gives &lt;math&gt;49500000&lt;/math&gt;, so the sum &lt;math&gt; \boxed{\textbf{(B)}\; 18}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> As shown above, there are a total of &lt;math&gt;900&lt;/math&gt; five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by &lt;math&gt;900&lt;/math&gt; to get our sum. The expected value for the ten-thousands and the units digit is &lt;math&gt;\frac{1+2+3+\cdots+9}{9}=5&lt;/math&gt;, and the expected value for the thousands, hundreds, and tens digit is &lt;math&gt;\frac{0+1+2+\cdots+9}{10}=4.5&lt;/math&gt;. Therefore our expected value is &lt;math&gt;5\times10^4+4.5\times10^3+4.5\times10^2+4.5\times10^1+5\times10^0=55,\!000&lt;/math&gt;. Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either &lt;math&gt;55,\!000&lt;/math&gt; or &lt;math&gt;900&lt;/math&gt;. Thus we only need to calculate &lt;math&gt;55\times9=495&lt;/math&gt;, and the desired sum is &lt;math&gt;\boxed{\textbf{(B) }18}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Variation of #2)==<br /> First, allow &lt;math&gt;a&lt;/math&gt; to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its &lt;math&gt;\textit{complement}&lt;/math&gt;. If &lt;math&gt;abcba&lt;/math&gt; is a palindrome, then its complement is &lt;math&gt;defed&lt;/math&gt; where &lt;math&gt;d=9-a&lt;/math&gt;, &lt;math&gt;e=9-b&lt;/math&gt;, &lt;math&gt;f=9-c&lt;/math&gt;. Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is &lt;math&gt;99999&lt;/math&gt;. Therefore, the sum of our palindromes is &lt;math&gt;99999\times (10^3/2)&lt;/math&gt;. (There are &lt;math&gt;10^3/2&lt;/math&gt; pairs.)<br /> <br /> However, we have overcounted, as something like &lt;math&gt;05350&lt;/math&gt; &lt;math&gt;\textit{isn't}&lt;/math&gt; a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form &lt;math&gt;nmn0&lt;/math&gt;. By the same argument as before, these sum to &lt;math&gt;9990\times (10^2/2)&lt;/math&gt;. Therefore, the sum that the problem asks for is:<br /> <br /> &lt;cmath&gt;500\times99999-50\times 9990&lt;/cmath&gt;<br /> &lt;cmath&gt;=500\times99999-500\times 999&lt;/cmath&gt;<br /> &lt;cmath&gt;=500(99999-999)&lt;/cmath&gt;<br /> &lt;cmath&gt;=500\times 99000&lt;/cmath&gt;<br /> <br /> Since all we care about is the sum of the digits, we can drop the &lt;math&gt;0&lt;/math&gt;'s.<br /> <br /> &lt;cmath&gt;5\times99&lt;/cmath&gt;<br /> &lt;cmath&gt;=5\times(100-1)&lt;/cmath&gt;<br /> &lt;cmath&gt;=495&lt;/cmath&gt;<br /> <br /> And finally, &lt;math&gt;4+9+5=\boxed{\textbf{(B)}18}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=143429 2014 AMC 10A Problems/Problem 6 2021-01-27T17:10:58Z <p>Captainsnake: /* Solution 6 */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We need to multiply &lt;math&gt;b&lt;/math&gt; by &lt;math&gt;\frac{d}{a}&lt;/math&gt; for the new cows and &lt;math&gt;\frac{e}{c}&lt;/math&gt; for the new time, so the answer is &lt;math&gt;b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A)} \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> We plug in &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=4&lt;/math&gt;, &lt;math&gt;d=5&lt;/math&gt;, and &lt;math&gt;e=6&lt;/math&gt;. Hence the question becomes &quot;2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?&quot;<br /> <br /> If 2 cows give 3 gallons of milk in 4 days, then 2 cows give &lt;math&gt;\frac{3}{4}&lt;/math&gt; gallons of milk in 1 day, so 1 cow gives &lt;math&gt;\frac{3}{4\cdot2}&lt;/math&gt; gallons in 1 day. This means that 5 cows give &lt;math&gt;\frac{5\cdot3}{4\cdot2}&lt;/math&gt; gallons of milk in 1 day. Finally, we see that 5 cows give &lt;math&gt;\frac{5\cdot3\cdot6}{4\cdot2}&lt;/math&gt; gallons of milk in 6 days. Substituting our values for the variables, this becomes &lt;math&gt;\frac{dbe}{ac}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is &lt;math&gt;\dfrac{ac}{b}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g&lt;/math&gt; be the answer to the question. We have &lt;math&gt;\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> The problem specifics &quot;rate,&quot; so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days &lt;cmath&gt;\implies\text{rate}=\dfrac{b}{ac}&lt;/cmath&gt;<br /> <br /> Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days &lt;cmath&gt;\implies\boxed{\textbf{(A)} \dfrac{bde}{ac}}&lt;/cmath&gt;<br /> ==Solution 5==<br /> If &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days, that means that one cow will give &lt;math&gt;\frac{b}{a}&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. Also, we want to find the number of gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days, so in &lt;math&gt;\frac{e}{c}&lt;/math&gt; days &lt;math&gt;d&lt;/math&gt; cows give &lt;math&gt;\frac{bd}{a}&lt;/math&gt; gallons of milk. Multiplying with the formula &lt;math&gt;d=rt&lt;/math&gt;, we get &lt;math&gt;\boxed{(A)\frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> In the right formula, plugging in &lt;math&gt;d=a&lt;/math&gt; and &lt;math&gt;e=c&lt;/math&gt; should simplify to &lt;math&gt;b&lt;/math&gt;, as if it doesn't, we'd essentially be saying &quot;&lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days, but &lt;math&gt;a&lt;/math&gt; cows don't give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days.&quot; The only one of the answer choices that simplifies like this is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/OW4rHUTPgPA<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=143428 2014 AMC 10A Problems/Problem 6 2021-01-27T17:10:37Z <p>Captainsnake: </p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We need to multiply &lt;math&gt;b&lt;/math&gt; by &lt;math&gt;\frac{d}{a}&lt;/math&gt; for the new cows and &lt;math&gt;\frac{e}{c}&lt;/math&gt; for the new time, so the answer is &lt;math&gt;b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A)} \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> We plug in &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=4&lt;/math&gt;, &lt;math&gt;d=5&lt;/math&gt;, and &lt;math&gt;e=6&lt;/math&gt;. Hence the question becomes &quot;2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?&quot;<br /> <br /> If 2 cows give 3 gallons of milk in 4 days, then 2 cows give &lt;math&gt;\frac{3}{4}&lt;/math&gt; gallons of milk in 1 day, so 1 cow gives &lt;math&gt;\frac{3}{4\cdot2}&lt;/math&gt; gallons in 1 day. This means that 5 cows give &lt;math&gt;\frac{5\cdot3}{4\cdot2}&lt;/math&gt; gallons of milk in 1 day. Finally, we see that 5 cows give &lt;math&gt;\frac{5\cdot3\cdot6}{4\cdot2}&lt;/math&gt; gallons of milk in 6 days. Substituting our values for the variables, this becomes &lt;math&gt;\frac{dbe}{ac}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is &lt;math&gt;\dfrac{ac}{b}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g&lt;/math&gt; be the answer to the question. We have &lt;math&gt;\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> The problem specifics &quot;rate,&quot; so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days &lt;cmath&gt;\implies\text{rate}=\dfrac{b}{ac}&lt;/cmath&gt;<br /> <br /> Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days &lt;cmath&gt;\implies\boxed{\textbf{(A)} \dfrac{bde}{ac}}&lt;/cmath&gt;<br /> ==Solution 5==<br /> If &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days, that means that one cow will give &lt;math&gt;\frac{b}{a}&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. Also, we want to find the number of gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days, so in &lt;math&gt;\frac{e}{c}&lt;/math&gt; days &lt;math&gt;d&lt;/math&gt; cows give &lt;math&gt;\frac{bd}{a}&lt;/math&gt; gallons of milk. Multiplying with the formula &lt;math&gt;d=rt&lt;/math&gt;, we get &lt;math&gt;\boxed{(A)\frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> In the right formula, plugging in &lt;math&gt;d=a&lt;/math&gt; and &lt;math&gt;e=c&lt;/math&gt; should simplify to &lt;math&gt;b&lt;/math&gt;, as if it doesn't, we'd essentially be saying &quot;&lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days, but &lt;math&gt;a&lt;/math&gt; cows don't give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days.&quot; The only one of the answer choices that simplifies like his is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/OW4rHUTPgPA<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_21&diff=143426 2013 AMC 12A Problems/Problem 21 2021-01-27T16:16:45Z <p>Captainsnake: /* Solutions */ added another solution</p> <hr /> <div>== Problem ==<br /> Consider &lt;math&gt;A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots ))))&lt;/math&gt;. Which of the following intervals contains &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ (\log 2016, \log 2017) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(B)} \ (\log 2017, \log 2018) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} \ (\log 2018, \log 2019) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} \ (\log 2019, \log 2020) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} \ (\log 2020, \log 2021) &lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;f(x) = \log(x + f(x-1))&lt;/math&gt; and &lt;math&gt;f(2) = \log(2)&lt;/math&gt;, and from the problem description, &lt;math&gt;A = f(2013)&lt;/math&gt;<br /> <br /> We can reason out an approximation, by ignoring the &lt;math&gt;f(x-1)&lt;/math&gt;:<br /> <br /> &lt;math&gt;f_{0}(x) \approx \log x&lt;/math&gt;<br /> <br /> And a better approximation, by plugging in our first approximation for &lt;math&gt;f(x-1)&lt;/math&gt; in our original definition for &lt;math&gt;f(x)&lt;/math&gt;:<br /> <br /> &lt;math&gt;f_{1}(x) \approx \log(x + \log(x-1))&lt;/math&gt;<br /> <br /> And an even better approximation:<br /> <br /> &lt;math&gt;f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))&lt;/math&gt;<br /> <br /> Continuing this pattern, obviously, will eventually terminate at &lt;math&gt;f_{x-1}(x)&lt;/math&gt;, in other words our original definition of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> However, at &lt;math&gt;x = 2013&lt;/math&gt;, going further than &lt;math&gt;f_{1}(x)&lt;/math&gt; will not distinguish between our answer choices. &lt;math&gt;\log(2012 + \log(2011))&lt;/math&gt; is nearly indistinguishable from &lt;math&gt;\log(2012)&lt;/math&gt;.<br /> <br /> So we take &lt;math&gt;f_{1}(x)&lt;/math&gt; and plug in.<br /> <br /> &lt;math&gt;f(2013) \approx \log(2013 + \log 2012)&lt;/math&gt;<br /> <br /> Since &lt;math&gt;1000 &lt; 2012 &lt; 10000&lt;/math&gt;, we know &lt;math&gt;3 &lt; \log(2012) &lt; 4&lt;/math&gt;. This gives us our answer range:<br /> <br /> &lt;math&gt;\log 2016 &lt; \log(2013 + \log 2012) &lt; \log(2017)&lt;/math&gt;<br /> <br /> &lt;math&gt;(\log 2016, \log 2017)&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> Suppose &lt;math&gt;A=\log(x)&lt;/math&gt;. <br /> Then &lt;math&gt;\log(2012+ \cdots)=x-2013&lt;/math&gt;. <br /> So if &lt;math&gt;x&gt;2017&lt;/math&gt;, then &lt;math&gt;\log(2012+\log(2011+\cdots))&gt;4&lt;/math&gt;. <br /> So &lt;math&gt;2012+\log(2011+\cdots)&gt;10000&lt;/math&gt;. <br /> Repeating, we then get &lt;math&gt;2011+\log(2010+\cdots)&gt;10^{7988}&lt;/math&gt;. <br /> This is clearly absurd (the RHS continues to grow more than exponentially for each iteration).<br /> So, &lt;math&gt;x&lt;/math&gt; is not greater than &lt;math&gt;2017&lt;/math&gt;. <br /> So &lt;math&gt;A&lt;\log(2017)&lt;/math&gt;. <br /> But this leaves only one answer, so we are done.<br /> <br /> === Solution 3 ===<br /> Define &lt;math&gt;f(2) = \log(2)&lt;/math&gt;, and &lt;math&gt;f(n) = \log(n+f(n-1)), \text{ for } n &gt; 2.&lt;/math&gt; We are looking for &lt;math&gt;f(2013)&lt;/math&gt;. First we show<br /> <br /> '''Lemma.''' For any integer &lt;math&gt;n&gt;2&lt;/math&gt;, if &lt;math&gt;n &lt; 10^k-k&lt;/math&gt; then &lt;math&gt;f(n) &lt; k&lt;/math&gt;.<br /> <br /> '''Proof.''' First note that &lt;math&gt;f(2) &lt; 1&lt;/math&gt;. Let &lt;math&gt;n&lt;10^k-k&lt;/math&gt;. Then &lt;math&gt;n+k&lt;10^k&lt;/math&gt;, so &lt;math&gt;\log(n+k)&lt; k&lt;/math&gt;. Suppose the claim is true for &lt;math&gt;n-1&lt;/math&gt;. Then &lt;math&gt;f(n) = \log(n+f(n-1)) &lt; \log(n + k) &lt; k&lt;/math&gt;. The Lemma is thus proved by induction.<br /> <br /> Finally, note that &lt;math&gt;2012 &lt; 10^4 - 4&lt;/math&gt; so that the Lemma implies that &lt;math&gt;f(2012) &lt; 4&lt;/math&gt;. This means that &lt;math&gt;f(2013) = \log(2013+f(2012)) &lt; \log(2017)&lt;/math&gt;, which leaves us with only one option &lt;math&gt;\boxed{\textbf{(A) } (\log 2016, \log 2017)}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Define &lt;math&gt;f(2) = \log(2)&lt;/math&gt;, and &lt;math&gt;f(n) = \log(n+f(n-1)), \text{ for } n &gt; 2.&lt;/math&gt; We start with a simple observation:<br /> <br /> '''Lemma.''' For &lt;math&gt;x,y&gt;2&lt;/math&gt;, &lt;math&gt;\log(x+\log(y)) &lt; \log(x)+\log(y)=\log(xy)&lt;/math&gt;.<br /> <br /> '''Proof.''' Since &lt;math&gt;x,y&gt;2&lt;/math&gt;, we have &lt;math&gt;xy-x-y = (x-1)(y-1) - 1 &gt; 0&lt;/math&gt;, so &lt;math&gt;xy - x-\log(y) &gt; xy - x - y &gt; 0&lt;/math&gt;.<br /> <br /> It follows that &lt;math&gt;\log(z+\log(x+\log(y))) &lt; \log(x)+\log(y)+\log(z)&lt;/math&gt;, and so on. <br /> <br /> Thus &lt;math&gt;f(2010) &lt; \log 2 + \log 3 + \cdots + \log 2010 &lt; \log 2010 + \cdots + \log 2010 &lt; 2009\cdot 4 = 8036&lt;/math&gt;. <br /> <br /> Then &lt;math&gt;f(2011) = \log(2011+f(2010)) &lt; \log(10047) &lt; 5&lt;/math&gt;. <br /> <br /> It follows that &lt;math&gt;f(2012) = \log(2012+f(2011)) &lt; \log(2017) &lt; 4&lt;/math&gt;. <br /> <br /> Finally, we get &lt;math&gt;f(2013) = \log(2013 + f(2012)) &lt; \log(2017)&lt;/math&gt;, which leaves us with only option &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> === Solution 5 (nonrigorous + abusing answer choices.) ===<br /> <br /> Intuitively, you can notice that &lt;math&gt;\log(2012+\log(2011+\cdots(\log(2)\cdots))&lt;\log(2013+\log(2012+\cdots(\log(2))\cdots))&lt;/math&gt;, therefore (by the answer choices) &lt;math&gt;\log(2012+\log(2011+\cdots(\log(2)\cdots))&lt;\log(2021)&lt;/math&gt;. We can then say:<br /> <br /> &lt;cmath&gt;x=\log(2013+\log(2012+\cdots(\log(2))\cdots))&lt;/cmath&gt;<br /> &lt;cmath&gt;\log(2013)&lt;x&lt;\log(2013+\log(2021))&lt;/cmath&gt;<br /> &lt;cmath&gt;\log(2013)&lt;x&lt;\log(2013+4)&lt;/cmath&gt;<br /> &lt;cmath&gt;\log(2013)&lt;x&lt;\log(2017)&lt;/cmath&gt;<br /> <br /> The only answer choice that is possible given this information is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> https://artofproblemsolving.com/videos/amc/2013amc12a/360<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_23&diff=139709 2001 AMC 8 Problems/Problem 23 2020-12-15T14:58:57Z <p>Captainsnake: /* Solution 3 (Multiple-choice abuse) */</p> <hr /> <div>==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Points &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; are vertices of an equilateral triangle, and points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; are midpoints of its sides. How many noncongruent triangles can be<br /> drawn using any three of these six points as vertices?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;asy&gt;<br /> pair SS,R,T,X,Y,Z;<br /> SS = (2,2*sqrt(3)); R = (0,0); T = (4,0);<br /> X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3));<br /> dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z);<br /> label(&quot;$S$&quot;,SS,N); label(&quot;$R$&quot;,R,SW); label(&quot;$T$&quot;,T,SE);<br /> label(&quot;$X$&quot;,X,S); label(&quot;$Y$&quot;,Y,NW); label(&quot;$Z$&quot;,Z,NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are &lt;math&gt; 6 &lt;/math&gt; points in the figure, and &lt;math&gt; 3 &lt;/math&gt; of them are needed to form a triangle, so there are &lt;math&gt; {6\choose{3}} =20 &lt;/math&gt; possible triples of &lt;math&gt; 3 &lt;/math&gt; of the &lt;math&gt; 6 &lt;/math&gt; points. However, some of these created congruent triangles, and some don't even make triangles at all.<br /> <br /> '''Case 1: Triangles congruent to &lt;math&gt; \triangle RST &lt;/math&gt;''' There is obviously only &lt;math&gt; 1 &lt;/math&gt; of these: &lt;math&gt; \triangle RST &lt;/math&gt; itself.<br /> <br /> '''Case 2: Triangles congruent to &lt;math&gt; \triangle SYZ &lt;/math&gt;''' There are &lt;math&gt; 4 &lt;/math&gt; of these: &lt;math&gt; \triangle SYZ, \triangle RXY, \triangle TXZ, &lt;/math&gt; and &lt;math&gt; \triangle XYZ &lt;/math&gt;.<br /> <br /> '''Case 3: Triangles congruent to &lt;math&gt; \triangle RSX &lt;/math&gt;''' There are &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ, &lt;/math&gt; and &lt;math&gt; \triangle RTZ &lt;/math&gt;.<br /> <br /> '''Case 4: Triangles congruent to &lt;math&gt; \triangle SYX &lt;/math&gt;''' There are again &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ, &lt;/math&gt; and &lt;math&gt; \triangle RYZ &lt;/math&gt;.<br /> <br /> However, if we add these up, we accounted for only &lt;math&gt; 1+4+6+6=17 &lt;/math&gt; of the &lt;math&gt; 20 &lt;/math&gt; possible triplets. We see that the remaining triplets don't even form triangles; they are &lt;math&gt; SYR, RXT, &lt;/math&gt; and &lt;math&gt; TZS &lt;/math&gt;. Adding these &lt;math&gt; 3 &lt;/math&gt; into the total yields for all of the possible triplets, so we see that there are only &lt;math&gt; 4 &lt;/math&gt; possible non-congruent, non-degenerate triangles, &lt;math&gt; \boxed{\text{D}} &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is &lt;math&gt;3+1 = \boxed{4}&lt;/math&gt;, which is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> -FIREDRAGONMATH16<br /> <br /> ==Solution 3 (Multiple-choice abuse)==<br /> <br /> Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: &lt;math&gt;\triangle RYX&lt;/math&gt;, &lt;math&gt;\triangle RYT&lt;/math&gt;, &lt;math&gt;\triangle RYZ&lt;/math&gt;, &lt;math&gt;\triangle RST&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{\text{(D) } 4}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2001|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_23&diff=139708 2001 AMC 8 Problems/Problem 23 2020-12-15T14:58:39Z <p>Captainsnake: /* Solution 3 (Multiple-choice abuse) */</p> <hr /> <div>==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Points &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; are vertices of an equilateral triangle, and points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; are midpoints of its sides. How many noncongruent triangles can be<br /> drawn using any three of these six points as vertices?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;asy&gt;<br /> pair SS,R,T,X,Y,Z;<br /> SS = (2,2*sqrt(3)); R = (0,0); T = (4,0);<br /> X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3));<br /> dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z);<br /> label(&quot;$S$&quot;,SS,N); label(&quot;$R$&quot;,R,SW); label(&quot;$T$&quot;,T,SE);<br /> label(&quot;$X$&quot;,X,S); label(&quot;$Y$&quot;,Y,NW); label(&quot;$Z$&quot;,Z,NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are &lt;math&gt; 6 &lt;/math&gt; points in the figure, and &lt;math&gt; 3 &lt;/math&gt; of them are needed to form a triangle, so there are &lt;math&gt; {6\choose{3}} =20 &lt;/math&gt; possible triples of &lt;math&gt; 3 &lt;/math&gt; of the &lt;math&gt; 6 &lt;/math&gt; points. However, some of these created congruent triangles, and some don't even make triangles at all.<br /> <br /> '''Case 1: Triangles congruent to &lt;math&gt; \triangle RST &lt;/math&gt;''' There is obviously only &lt;math&gt; 1 &lt;/math&gt; of these: &lt;math&gt; \triangle RST &lt;/math&gt; itself.<br /> <br /> '''Case 2: Triangles congruent to &lt;math&gt; \triangle SYZ &lt;/math&gt;''' There are &lt;math&gt; 4 &lt;/math&gt; of these: &lt;math&gt; \triangle SYZ, \triangle RXY, \triangle TXZ, &lt;/math&gt; and &lt;math&gt; \triangle XYZ &lt;/math&gt;.<br /> <br /> '''Case 3: Triangles congruent to &lt;math&gt; \triangle RSX &lt;/math&gt;''' There are &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ, &lt;/math&gt; and &lt;math&gt; \triangle RTZ &lt;/math&gt;.<br /> <br /> '''Case 4: Triangles congruent to &lt;math&gt; \triangle SYX &lt;/math&gt;''' There are again &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ, &lt;/math&gt; and &lt;math&gt; \triangle RYZ &lt;/math&gt;.<br /> <br /> However, if we add these up, we accounted for only &lt;math&gt; 1+4+6+6=17 &lt;/math&gt; of the &lt;math&gt; 20 &lt;/math&gt; possible triplets. We see that the remaining triplets don't even form triangles; they are &lt;math&gt; SYR, RXT, &lt;/math&gt; and &lt;math&gt; TZS &lt;/math&gt;. Adding these &lt;math&gt; 3 &lt;/math&gt; into the total yields for all of the possible triplets, so we see that there are only &lt;math&gt; 4 &lt;/math&gt; possible non-congruent, non-degenerate triangles, &lt;math&gt; \boxed{\text{D}} &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is &lt;math&gt;3+1 = \boxed{4}&lt;/math&gt;, which is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> -FIREDRAGONMATH16<br /> <br /> ==Solution 3 (Multiple-choice abuse)==<br /> <br /> Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: &lt;math&gt;\triangle RYX&lt;/math&gt;, &lt;math&gt;\triangle RYT&lt;/math&gt;, &lt;math&gt;\triangle RYZ&lt;/math&gt;, &lt;math&gt;\triangle RST&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{(D) 4}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2001|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_23&diff=139707 2001 AMC 8 Problems/Problem 23 2020-12-15T14:58:11Z <p>Captainsnake: Added another solution</p> <hr /> <div>==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Points &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; are vertices of an equilateral triangle, and points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; are midpoints of its sides. How many noncongruent triangles can be<br /> drawn using any three of these six points as vertices?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;asy&gt;<br /> pair SS,R,T,X,Y,Z;<br /> SS = (2,2*sqrt(3)); R = (0,0); T = (4,0);<br /> X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3));<br /> dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z);<br /> label(&quot;$S$&quot;,SS,N); label(&quot;$R$&quot;,R,SW); label(&quot;$T$&quot;,T,SE);<br /> label(&quot;$X$&quot;,X,S); label(&quot;$Y$&quot;,Y,NW); label(&quot;$Z$&quot;,Z,NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are &lt;math&gt; 6 &lt;/math&gt; points in the figure, and &lt;math&gt; 3 &lt;/math&gt; of them are needed to form a triangle, so there are &lt;math&gt; {6\choose{3}} =20 &lt;/math&gt; possible triples of &lt;math&gt; 3 &lt;/math&gt; of the &lt;math&gt; 6 &lt;/math&gt; points. However, some of these created congruent triangles, and some don't even make triangles at all.<br /> <br /> '''Case 1: Triangles congruent to &lt;math&gt; \triangle RST &lt;/math&gt;''' There is obviously only &lt;math&gt; 1 &lt;/math&gt; of these: &lt;math&gt; \triangle RST &lt;/math&gt; itself.<br /> <br /> '''Case 2: Triangles congruent to &lt;math&gt; \triangle SYZ &lt;/math&gt;''' There are &lt;math&gt; 4 &lt;/math&gt; of these: &lt;math&gt; \triangle SYZ, \triangle RXY, \triangle TXZ, &lt;/math&gt; and &lt;math&gt; \triangle XYZ &lt;/math&gt;.<br /> <br /> '''Case 3: Triangles congruent to &lt;math&gt; \triangle RSX &lt;/math&gt;''' There are &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ, &lt;/math&gt; and &lt;math&gt; \triangle RTZ &lt;/math&gt;.<br /> <br /> '''Case 4: Triangles congruent to &lt;math&gt; \triangle SYX &lt;/math&gt;''' There are again &lt;math&gt; 6 &lt;/math&gt; of these: &lt;math&gt; \triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ, &lt;/math&gt; and &lt;math&gt; \triangle RYZ &lt;/math&gt;.<br /> <br /> However, if we add these up, we accounted for only &lt;math&gt; 1+4+6+6=17 &lt;/math&gt; of the &lt;math&gt; 20 &lt;/math&gt; possible triplets. We see that the remaining triplets don't even form triangles; they are &lt;math&gt; SYR, RXT, &lt;/math&gt; and &lt;math&gt; TZS &lt;/math&gt;. Adding these &lt;math&gt; 3 &lt;/math&gt; into the total yields for all of the possible triplets, so we see that there are only &lt;math&gt; 4 &lt;/math&gt; possible non-congruent, non-degenerate triangles, &lt;math&gt; \boxed{\text{D}} &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is &lt;math&gt;3+1 = \boxed{4}&lt;/math&gt;, which is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> -FIREDRAGONMATH16<br /> <br /> ==Solution 3 (Multiple-choice abuse)==<br /> <br /> Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: &lt;math&gt;\triangle RYX&lt;/math&gt;, &lt;math&gt;\triangle RYT&lt;/math&gt;, &lt;math&gt;\triangle RYZ&lt;/math&gt;, &lt;math&gt;\triangle RST&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{4}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2001|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_15&diff=138962 2009 AMC 12A Problems/Problem 15 2020-12-03T20:34:10Z <p>Captainsnake: /* Solution 3 (Fast) */</p> <hr /> <div>== Problem ==<br /> For what value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i&lt;/math&gt;?<br /> <br /> Note: here &lt;math&gt;i = \sqrt { - 1}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98&lt;/math&gt;<br /> <br /> == Solution 1 == <br /> We know that &lt;math&gt;i^x&lt;/math&gt; cycles every &lt;math&gt;4&lt;/math&gt; powers so we group the sum in &lt;math&gt;4&lt;/math&gt;s. <br /> &lt;cmath&gt;i+2i^2+3i^3+4i^4=2-2i&lt;/cmath&gt;<br /> &lt;cmath&gt;5i^5+6i^6+7i^7+8i^8=2-2i&lt;/cmath&gt;<br /> <br /> We can postulate that every group of &lt;math&gt;4&lt;/math&gt; is equal to &lt;math&gt;2-2i&lt;/math&gt;.<br /> For 24 groups we thus, get &lt;math&gt;48-48i&lt;/math&gt; as our sum. <br /> We know the solution must lie near<br /> The next term is the &lt;math&gt;24*4+1=97&lt;/math&gt;th term. This term is equal to &lt;math&gt;97i&lt;/math&gt; (first in a group of &lt;math&gt;4&lt;/math&gt; so &lt;math&gt;i^{97}=i&lt;/math&gt;) and our sum is now &lt;math&gt;48+49i&lt;/math&gt; so &lt;math&gt;n=97\Rightarrow\boxed{\mathbf{D}}&lt;/math&gt; is our answer<br /> <br /> == Solution 2==<br /> <br /> Obviously, even powers of &lt;math&gt;i&lt;/math&gt; are real and odd powers of &lt;math&gt;i&lt;/math&gt; are imaginary. <br /> Hence the real part of the sum is &lt;math&gt;2i^2 + 4i^4 + 6i^6 + \ldots&lt;/math&gt;, and <br /> the imaginary part is &lt;math&gt;i + 3i^3 + 5i^5 + \cdots&lt;/math&gt;.<br /> <br /> Let's take a look at the real part first. We have &lt;math&gt;i^2=-1&lt;/math&gt;, hence the real part simplifies to &lt;math&gt;-2+4-6+8-10+\cdots&lt;/math&gt;.<br /> If there were an odd number of terms, we could pair them as follows: &lt;math&gt;-2 + (4-6) + (8-10) + \cdots&lt;/math&gt;, hence the result would be negative. As we need the real part to be &lt;math&gt;48&lt;/math&gt;, we must have an even number of terms. If we have an even number of terms, we can pair them as &lt;math&gt;(-2+4) + (-6+8) + \cdots&lt;/math&gt;. Each parenthesis is equal to &lt;math&gt;2&lt;/math&gt;, thus there are &lt;math&gt;24&lt;/math&gt; of them, and the last value used is &lt;math&gt;96&lt;/math&gt;. This happens for &lt;math&gt;n=96&lt;/math&gt; and &lt;math&gt;n=97&lt;/math&gt;. As &lt;math&gt;n=96&lt;/math&gt; is not present as an option, we may conclude that the answer is &lt;math&gt;\boxed{97}&lt;/math&gt;.<br /> <br /> In a complete solution, we should now verify which of &lt;math&gt;n=96&lt;/math&gt; and &lt;math&gt;n=97&lt;/math&gt; will give us the correct imaginary part.<br /> <br /> We can rewrite the imaginary part as follows: &lt;math&gt;i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)&lt;/math&gt;. We need to obtain &lt;math&gt;(1 - 3 + 5 - \cdots) = 49&lt;/math&gt;. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as &lt;math&gt;1 + (-3+5) + (-7+9) + \cdots&lt;/math&gt;. We need &lt;math&gt;24&lt;/math&gt; parentheses, therefore the last value used is &lt;math&gt;97&lt;/math&gt;. This happens when &lt;math&gt;n=97&lt;/math&gt; or &lt;math&gt;n=98&lt;/math&gt;, and we are done.<br /> <br /> == Solution 3 (Fast)==<br /> <br /> Some may know the equation:<br /> <br /> &lt;cmath&gt;\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}&lt;/cmath&gt;<br /> <br /> (For those curious, this comes from differentiating the equation for finite geometric sums)<br /> <br /> Using this equation, we have<br /> <br /> &lt;cmath&gt;48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\frac{1-(n+1)i^n+ni^{n+1}}{-2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}&lt;/cmath&gt;<br /> <br /> Since the imaginary and the real part must be positive, we know that &lt;math&gt;i^{n+1}=-1&lt;/math&gt; or &lt;math&gt;i^{n+1}=-i&lt;/math&gt;. By the same line of reason, &lt;math&gt;i^{n}=1,i&lt;/math&gt;. This only works for &lt;math&gt;n\equiv 1 \mod 4&lt;/math&gt;. Therefore, we have:<br /> <br /> &lt;cmath&gt;\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i&lt;/cmath&gt;<br /> <br /> Solving either the real or imaginary part gives &lt;math&gt;\boxed{\mathbf{(D) }97}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/VfgUhcw112s<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_15&diff=138961 2009 AMC 12A Problems/Problem 15 2020-12-03T20:32:58Z <p>Captainsnake: Added another solution</p> <hr /> <div>== Problem ==<br /> For what value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i&lt;/math&gt;?<br /> <br /> Note: here &lt;math&gt;i = \sqrt { - 1}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98&lt;/math&gt;<br /> <br /> == Solution 1 == <br /> We know that &lt;math&gt;i^x&lt;/math&gt; cycles every &lt;math&gt;4&lt;/math&gt; powers so we group the sum in &lt;math&gt;4&lt;/math&gt;s. <br /> &lt;cmath&gt;i+2i^2+3i^3+4i^4=2-2i&lt;/cmath&gt;<br /> &lt;cmath&gt;5i^5+6i^6+7i^7+8i^8=2-2i&lt;/cmath&gt;<br /> <br /> We can postulate that every group of &lt;math&gt;4&lt;/math&gt; is equal to &lt;math&gt;2-2i&lt;/math&gt;.<br /> For 24 groups we thus, get &lt;math&gt;48-48i&lt;/math&gt; as our sum. <br /> We know the solution must lie near<br /> The next term is the &lt;math&gt;24*4+1=97&lt;/math&gt;th term. This term is equal to &lt;math&gt;97i&lt;/math&gt; (first in a group of &lt;math&gt;4&lt;/math&gt; so &lt;math&gt;i^{97}=i&lt;/math&gt;) and our sum is now &lt;math&gt;48+49i&lt;/math&gt; so &lt;math&gt;n=97\Rightarrow\boxed{\mathbf{D}}&lt;/math&gt; is our answer<br /> <br /> == Solution 2==<br /> <br /> Obviously, even powers of &lt;math&gt;i&lt;/math&gt; are real and odd powers of &lt;math&gt;i&lt;/math&gt; are imaginary. <br /> Hence the real part of the sum is &lt;math&gt;2i^2 + 4i^4 + 6i^6 + \ldots&lt;/math&gt;, and <br /> the imaginary part is &lt;math&gt;i + 3i^3 + 5i^5 + \cdots&lt;/math&gt;.<br /> <br /> Let's take a look at the real part first. We have &lt;math&gt;i^2=-1&lt;/math&gt;, hence the real part simplifies to &lt;math&gt;-2+4-6+8-10+\cdots&lt;/math&gt;.<br /> If there were an odd number of terms, we could pair them as follows: &lt;math&gt;-2 + (4-6) + (8-10) + \cdots&lt;/math&gt;, hence the result would be negative. As we need the real part to be &lt;math&gt;48&lt;/math&gt;, we must have an even number of terms. If we have an even number of terms, we can pair them as &lt;math&gt;(-2+4) + (-6+8) + \cdots&lt;/math&gt;. Each parenthesis is equal to &lt;math&gt;2&lt;/math&gt;, thus there are &lt;math&gt;24&lt;/math&gt; of them, and the last value used is &lt;math&gt;96&lt;/math&gt;. This happens for &lt;math&gt;n=96&lt;/math&gt; and &lt;math&gt;n=97&lt;/math&gt;. As &lt;math&gt;n=96&lt;/math&gt; is not present as an option, we may conclude that the answer is &lt;math&gt;\boxed{97}&lt;/math&gt;.<br /> <br /> In a complete solution, we should now verify which of &lt;math&gt;n=96&lt;/math&gt; and &lt;math&gt;n=97&lt;/math&gt; will give us the correct imaginary part.<br /> <br /> We can rewrite the imaginary part as follows: &lt;math&gt;i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)&lt;/math&gt;. We need to obtain &lt;math&gt;(1 - 3 + 5 - \cdots) = 49&lt;/math&gt;. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as &lt;math&gt;1 + (-3+5) + (-7+9) + \cdots&lt;/math&gt;. We need &lt;math&gt;24&lt;/math&gt; parentheses, therefore the last value used is &lt;math&gt;97&lt;/math&gt;. This happens when &lt;math&gt;n=97&lt;/math&gt; or &lt;math&gt;n=98&lt;/math&gt;, and we are done.<br /> <br /> == Solution 3 (Fast)==<br /> <br /> Some may know the equation:<br /> <br /> &lt;cmath&gt;\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}&lt;/cmath&gt;<br /> <br /> (For those curious, this comes from differentiating the equation for finite geometric sums)<br /> <br /> Using this equation, we have<br /> <br /> &lt;cmath&gt;48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\frac{1-(n+1)i^n+ni^{n+1}}{-2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}&lt;/cmath&gt;<br /> <br /> Since the imaginary and the real part must be positive, we know that &lt;math&gt;i^{n+1}=-1&lt;/math&gt; or &lt;math&gt;i^{n+1}=-i&lt;/math&gt;. By the same line of reason, &lt;math&gt;i^{n}=1,i&lt;/math&gt;. This only works for &lt;math&gt;n\equiv 1 (\mod 4)&lt;/math&gt;. Therefore, we have:<br /> <br /> &lt;cmath&gt;\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i&lt;/cmath&gt;<br /> <br /> Solving either the real or imaginary part gives &lt;math&gt;\boxed{\mathbf{(D)}97}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/VfgUhcw112s<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_12&diff=136757 2020 AIME I Problems/Problem 12 2020-11-06T19:18:55Z <p>Captainsnake: Clarified the meaning of part of the solution</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer for which &lt;math&gt;149^n-2^n&lt;/math&gt; is divisible by &lt;math&gt;3^3\cdot5^5\cdot7^7.&lt;/math&gt; Find the number of positive integer divisors of &lt;math&gt;n.&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Lifting the Exponent shows that &lt;cmath&gt;v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1&lt;/cmath&gt; so thus, &lt;math&gt;3^2&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. It also shows that &lt;cmath&gt;v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2&lt;/cmath&gt; so thus, &lt;math&gt;7^5&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. <br /> <br /> <br /> Now, multiplying &lt;math&gt;n&lt;/math&gt; by &lt;math&gt;4&lt;/math&gt;, we see &lt;cmath&gt;v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})&lt;/cmath&gt; and since &lt;math&gt;149^{4} \equiv 1 \pmod{25}&lt;/math&gt; and &lt;math&gt;16^1 \equiv 16 \pmod{25}&lt;/math&gt; then &lt;math&gt;v_5(149^{4n}-2^{4n})=1+v_5(n)&lt;/math&gt; meaning that we have that by LTE, &lt;math&gt;4 \cdot 5^4&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;3^2&lt;/math&gt;, &lt;math&gt;7^5&lt;/math&gt; and &lt;math&gt;4\cdot 5^4&lt;/math&gt; all divide &lt;math&gt;n&lt;/math&gt;, the smallest value of &lt;math&gt;n&lt;/math&gt; working is their LCM, also &lt;math&gt;3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5&lt;/math&gt;. Thus the number of divisors is &lt;math&gt;(2+1)(2+1)(4+1)(5+1) = \boxed{270}&lt;/math&gt;.<br /> <br /> ~kevinmathz<br /> <br /> <br /> Lol you can get &lt;math&gt;4 \cdot 5^4&lt;/math&gt; by just Euler Totient.<br /> <br /> ~LLL2019<br /> <br /> == Solution 2 (Simpler, just basic mods and Fermat's theorem)==<br /> Note that for all &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;149^n - 2^n&lt;/math&gt; is divisible by &lt;math&gt;149-2 = 147&lt;/math&gt; by difference of &lt;math&gt;n&lt;/math&gt;th powers. That is &lt;math&gt;3\cdot7^2&lt;/math&gt;, so now we can clearly see that the smallest &lt;math&gt;n&lt;/math&gt; to make the expression divisible by &lt;math&gt;3^3&lt;/math&gt; is just &lt;math&gt;3^2&lt;/math&gt;. Similarly, we can reason that the smallest &lt;math&gt;n&lt;/math&gt; to make the expression divisible by &lt;math&gt;7^7&lt;/math&gt; is just &lt;math&gt;7^5&lt;/math&gt;. <br /> <br /> Finally, for &lt;math&gt;5^5&lt;/math&gt;, take &lt;math&gt;\pmod {5}&lt;/math&gt; and &lt;math&gt;\pmod {25}&lt;/math&gt; of each quantity (They happen to both be &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum &lt;math&gt;n&lt;/math&gt; for divisibility by &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, and other values are factors of &lt;math&gt;4&lt;/math&gt;. Testing all of them(just &lt;math&gt;1&lt;/math&gt;,&lt;math&gt;2&lt;/math&gt;,&lt;math&gt;4&lt;/math&gt; using mods-not too bad), &lt;math&gt;4&lt;/math&gt; is indeed the smallest value to make the expression divisible by &lt;math&gt;5&lt;/math&gt;, and this clearly is NOT divisible by &lt;math&gt;25&lt;/math&gt;.<br /> Therefore, the smallest &lt;math&gt;n&lt;/math&gt; to make this expression divisible by &lt;math&gt;5^5&lt;/math&gt; is &lt;math&gt;2^2 \cdot 5^4&lt;/math&gt;.<br /> <br /> Calculating the LCM of all these, one gets &lt;math&gt;2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5&lt;/math&gt;. Using the factor counting formula,<br /> the answer is &lt;math&gt;3\cdot3\cdot5\cdot6&lt;/math&gt; = &lt;math&gt;\boxed{270}&lt;/math&gt;.<br /> <br /> ~Solution by thanosaops<br /> <br /> ~formatted by MY-2<br /> <br /> ~also formatted by pandyhu2001<br /> <br /> == Solution 3 (Elementary and Thorough) ==<br /> As usual, denote &lt;math&gt;v_p(n)&lt;/math&gt; the highest power of prime &lt;math&gt;p&lt;/math&gt; that divides &lt;math&gt;n&lt;/math&gt;. For divisibility by &lt;math&gt;3^3&lt;/math&gt;, notice that &lt;math&gt;v_3(149^3 - 2^3) = 2&lt;/math&gt; as &lt;math&gt;149^3 - 2^3 =&lt;/math&gt; &lt;math&gt;(147)(149^2 + 2\cdot149 + 2^2)&lt;/math&gt;, and upon checking mods, &lt;math&gt;149^2 + 2\cdot149 + 2^2&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; but not &lt;math&gt;9&lt;/math&gt;. In addition, &lt;math&gt;149^9 - 2^9&lt;/math&gt; is divisible by &lt;math&gt;3^3&lt;/math&gt; because &lt;math&gt;149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)&lt;/math&gt;, and the rightmost factor equates to &lt;math&gt;1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}&lt;/math&gt;. In fact, &lt;math&gt;n = 9 = 3^2&lt;/math&gt; is the least possible choice to ensure divisibility by &lt;math&gt;3^3&lt;/math&gt; because if &lt;math&gt;n = a \cdot 3^b&lt;/math&gt;, with &lt;math&gt;3 \nmid a&lt;/math&gt; and &lt;math&gt;b &lt; 2&lt;/math&gt;, we write &lt;cmath&gt;149^{a \cdot 3^b} - 2^{a \cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).&lt;/cmath&gt; Then, the rightmost factor is equivalent to &lt;math&gt;\pm a \pmod{3} \not\equiv 0 \pmod{3}&lt;/math&gt;, and &lt;math&gt;v_3(149^{3^b} - 2^{3^b}) = b + 1 &lt; 3&lt;/math&gt;.<br /> <br /> For divisibility by &lt;math&gt;7^7&lt;/math&gt;, we'll induct, claiming that &lt;math&gt;v_7(149^{7^k} - 2^{7^k}) = k + 2&lt;/math&gt; for whole numbers &lt;math&gt;k&lt;/math&gt;. The base case is clear. Then, &lt;cmath&gt;v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).&lt;/cmath&gt; By the induction hypothesis, &lt;math&gt;v_7(149^{7^k} - 2^{7^k}) = k + 2&lt;/math&gt;. Then, notice that &lt;cmath&gt;S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.&lt;/cmath&gt; This tells us that &lt;math&gt;S(k)&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;, but not &lt;math&gt;49&lt;/math&gt; so that &lt;math&gt;v_7\left(S(k)\right) = 1&lt;/math&gt;, completing our induction. We can verify that &lt;math&gt;7^5&lt;/math&gt; is the least choice of &lt;math&gt;n&lt;/math&gt; to ensure divisibility by &lt;math&gt;7^7&lt;/math&gt; by arguing similarly to the &lt;math&gt;3^3&lt;/math&gt; case. <br /> <br /> Finally, for &lt;math&gt;5^5&lt;/math&gt;, we take the powers of &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; in mod &lt;math&gt;5&lt;/math&gt; and mod &lt;math&gt;25&lt;/math&gt;. Writing out these mods, we have that &lt;math&gt;149^n \equiv 2^n \pmod{5}&lt;/math&gt; if and only if &lt;math&gt;4 | n&lt;/math&gt;, in which &lt;math&gt;149^n \equiv 2^n \equiv 1 \pmod{5}&lt;/math&gt;. So here we claim that &lt;math&gt;v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1&lt;/math&gt; and perform yet another induction. The base case is true: &lt;math&gt;5 | 149^4 - 2^4&lt;/math&gt;, but &lt;math&gt;149^4 - 2^4 \equiv 1 - 16 \pmod{25}&lt;/math&gt;. Now then, assuming the induction statement to hold for some &lt;math&gt;k&lt;/math&gt;, &lt;cmath&gt;v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).&lt;/cmath&gt; Note that &lt;math&gt;S'(k) = 149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}&lt;/math&gt; equates to &lt;math&gt;S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}&lt;/math&gt; in both mod &lt;math&gt;5&lt;/math&gt; and mod &lt;math&gt;25&lt;/math&gt;. We notice that &lt;math&gt;S''(k) \equiv 0 \pmod{5}&lt;/math&gt;. Writing out the powers of &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;25&lt;/math&gt;, we have &lt;math&gt;S''(0) \equiv 5 \pmod{25}&lt;/math&gt;. Also &lt;math&gt;2^n \equiv 1 \pmod{25}&lt;/math&gt; when &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;20&lt;/math&gt;. Hence for &lt;math&gt;k &gt; 0&lt;/math&gt;, &lt;math&gt;S''(k) \equiv 5 \mod{25}&lt;/math&gt;. Thus, &lt;math&gt;v_5\left(S'(k)\right) = 1&lt;/math&gt;, completing our induction. Applying the same argument from the previous two cases, &lt;math&gt;4\cdot5^4&lt;/math&gt; is the least choice to ensure divisibility by &lt;math&gt;5^5&lt;/math&gt;.<br /> <br /> Our answer is the number of divisors of &lt;math&gt;\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5&lt;/math&gt;. It is &lt;math&gt;(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}&lt;/math&gt;.<br /> <br /> ~hnkevin42<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&diff=136595 2003 AMC 12A Problems/Problem 23 2020-11-05T15:53:01Z <p>Captainsnake: </p> <hr /> <div>How many perfect squares are divisors of the product &lt;math&gt;1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We want to find the number of perfect square factors in the product of all the factorials of numbers from &lt;math&gt;1 - 9&lt;/math&gt;. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to &lt;math&gt;2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3&lt;/math&gt;. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: &lt;math&gt;2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1&lt;/math&gt;. To find the total number of possibilities, we add &lt;math&gt;1&lt;/math&gt; to each exponent and multiply them all together. This gives us &lt;math&gt;16 \cdot 7 \cdot 3 \cdot 2 = 672&lt;/math&gt; &lt;math&gt;\Rightarrow\boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> ( Just Explanation Of 1 )<br /> <br /> We can easily find that factorials upto 9 in product lead to prime factorization &lt;math&gt;2^{30}&lt;/math&gt; &lt;math&gt;*&lt;/math&gt; &lt;math&gt;3^{13}&lt;/math&gt; &lt;math&gt;*&lt;/math&gt; &lt;math&gt;5^5&lt;/math&gt; &lt;math&gt;*&lt;/math&gt; &lt;math&gt;7^3&lt;/math&gt; So number of pairs possible = { &lt;math&gt;2^{0}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; &lt;math&gt;2^{2}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; ... &lt;math&gt;2^{30}&lt;/math&gt; }{ &lt;math&gt;3^{0}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; &lt;math&gt;3^{2}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; ... &lt;math&gt;3^{12}&lt;/math&gt; }{ &lt;math&gt;5^{0}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; &lt;math&gt;5^{2}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; ... &lt;math&gt;5^{4}&lt;/math&gt; }{ &lt;math&gt;7^{0}&lt;/math&gt; &lt;math&gt;,&lt;/math&gt; &lt;math&gt;7^{2}&lt;/math&gt; }<br /> <br /> Which leads to resulting number of pairs = &lt;math&gt;16*7*3*2&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;672&lt;/math&gt; &lt;math&gt;\Rightarrow\boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_22&diff=136594 2003 AMC 12A Problems/Problem 22 2020-11-05T15:49:36Z <p>Captainsnake: Added solution</p> <hr /> <div>== Problem ==<br /> <br /> Objects &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object &lt;math&gt;A&lt;/math&gt; starts at &lt;math&gt;(0,0)&lt;/math&gt; and each of its steps is either right or up, both equally likely. Object &lt;math&gt;B&lt;/math&gt; starts at &lt;math&gt;(5,7)&lt;/math&gt; and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)} \ 0.10 \qquad<br /> \mathrm{(B)} \ 0.15 \qquad<br /> \mathrm{(C)} \ 0.20 \qquad<br /> \mathrm{(D)} \ 0.25 \qquad<br /> \mathrm{(E)} \ 0.30 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; meet, their paths connect &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(5,7).&lt;/math&gt; There are &lt;math&gt;\binom{12}{5}=792&lt;/math&gt; such paths. Since the path is &lt;math&gt;12&lt;/math&gt; units long, they must meet after each travels &lt;math&gt;6&lt;/math&gt; units, so the probability is &lt;math&gt;\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}&lt;/math&gt;.<br /> <br /> == Solution 2 (Generating Functions) ==<br /> <br /> We know that the sum of the vertical steps must be equal to &lt;math&gt;7&lt;/math&gt;. We also know that they must take &lt;math&gt;6&lt;/math&gt; steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:<br /> <br /> &lt;cmath&gt;P(x)=(x+1)^{12}&lt;/cmath&gt;<br /> <br /> Where we need to extract the &lt;math&gt;x^5&lt;/math&gt; coefficient. By the binomial coefficient theorem, that term is &lt;math&gt;binom{12}{5}=792&lt;/math&gt; paths. Since there are also &lt;math&gt;2^{12}&lt;/math&gt; paths, we have:<br /> <br /> &lt;math&gt;\frac{792}{2^12}=\frac{792}{4096}\approx\frac{800}{4000}=\boxed{\text{(C) } 0.20}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2003|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=136140 2010 AMC 12B Problems/Problem 16 2020-10-30T17:24:46Z <p>Captainsnake: /* Solution 3 (Fancier version of Solution 1) */</p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We group this into groups of &lt;math&gt;3&lt;/math&gt;, because &lt;math&gt;3|2010&lt;/math&gt;. This means that every residue class mod 3 has an equal probability.<br /> <br /> If &lt;math&gt;3|a&lt;/math&gt;, we are done. There is a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt; that that happens.<br /> <br /> Otherwise, we have &lt;math&gt;3|bc+b+1&lt;/math&gt;, which means that &lt;math&gt;b(c+1) \equiv 2\pmod{3}&lt;/math&gt;. So either &lt;cmath&gt;b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}&lt;/cmath&gt; or &lt;cmath&gt;b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3&lt;/cmath&gt; which will lead to the property being true. There are a &lt;math&gt;\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}&lt;/math&gt; chance for each bundle of cases to be true. Thus, the total for the cases is &lt;math&gt;\frac{2}{9}&lt;/math&gt;. But we have to multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; because this only happens with a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance. So the total is actually &lt;math&gt;\frac{4}{27}&lt;/math&gt;.<br /> <br /> The grand total is &lt;cmath&gt;\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}&lt;/cmath&gt;<br /> <br /> == Solution 2 (Minor change from Solution 1) ==<br /> <br /> Just like solution 1, we see that there is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of &lt;math&gt;3|a&lt;/math&gt; and &lt;math&gt;\frac{2}{9}&lt;/math&gt; chance of &lt;math&gt;3|1+b+bc&lt;/math&gt; <br /> <br /> Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be &lt;math&gt;\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}&lt;/math&gt;<br /> <br /> -Conantwiz2023<br /> <br /> ==Solution 3 (Fancier version of Solution 1)==<br /> <br /> As with solution one, we conclude that if &lt;math&gt;a\equiv0\mod 3&lt;/math&gt; then the requirements are satisfied. We then have:<br /> &lt;cmath&gt;a(bc+c)+a\equiv0 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;a(bc+c)\equiv-a \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;c(b+1)\equiv-1 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;b+1\equiv \frac{-1}{c} \mod 3&lt;/cmath&gt;<br /> Which is true for one &lt;math&gt;b&lt;/math&gt; when &lt;math&gt;c\not\equiv 0 \mod 3&lt;/math&gt; because the integers&lt;math&gt;\mod 3&lt;/math&gt; form a field under multiplication and addition with absorbing element &lt;math&gt;0&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;P=\frac{1}{3}+\frac{4}{27}=\boxed{\text{(E) }\frac{13}{27}}&lt;/math&gt;.<br /> <br /> ~Snacc<br /> <br /> ==Video Solution==<br /> https://youtu.be/FQO-0E2zUVI?t=437<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=136139 2010 AMC 12B Problems/Problem 16 2020-10-30T17:24:21Z <p>Captainsnake: </p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We group this into groups of &lt;math&gt;3&lt;/math&gt;, because &lt;math&gt;3|2010&lt;/math&gt;. This means that every residue class mod 3 has an equal probability.<br /> <br /> If &lt;math&gt;3|a&lt;/math&gt;, we are done. There is a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt; that that happens.<br /> <br /> Otherwise, we have &lt;math&gt;3|bc+b+1&lt;/math&gt;, which means that &lt;math&gt;b(c+1) \equiv 2\pmod{3}&lt;/math&gt;. So either &lt;cmath&gt;b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}&lt;/cmath&gt; or &lt;cmath&gt;b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3&lt;/cmath&gt; which will lead to the property being true. There are a &lt;math&gt;\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}&lt;/math&gt; chance for each bundle of cases to be true. Thus, the total for the cases is &lt;math&gt;\frac{2}{9}&lt;/math&gt;. But we have to multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; because this only happens with a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance. So the total is actually &lt;math&gt;\frac{4}{27}&lt;/math&gt;.<br /> <br /> The grand total is &lt;cmath&gt;\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}&lt;/cmath&gt;<br /> <br /> == Solution 2 (Minor change from Solution 1) ==<br /> <br /> Just like solution 1, we see that there is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of &lt;math&gt;3|a&lt;/math&gt; and &lt;math&gt;\frac{2}{9}&lt;/math&gt; chance of &lt;math&gt;3|1+b+bc&lt;/math&gt; <br /> <br /> Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be &lt;math&gt;\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}&lt;/math&gt;<br /> <br /> -Conantwiz2023<br /> <br /> ==Solution 3 (Fancier version of Solution 1)==<br /> <br /> As with solution one, we conclude that if &lt;math&gt;a\equiv0\mod 3&lt;/math&gt; then the requirements are satisfied. We then have:<br /> &lt;cmath&gt;a(bc+c)+a\equiv0 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;a(bc+c)\equiv-a \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;c(b+1)\equiv-1 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;b+1\equiv \frac{-1}{c} \mod 3&lt;/cmath&gt;<br /> Which is true for one &lt;math&gt;b&lt;/math&gt; when &lt;math&gt;c\not\equiv 0 \mod 3&lt;/math&gt; because the integers &lt;math&gt;\mod 3&lt;/math&gt; form a field under multiplication and addition with absorbing element &lt;math&gt;0&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;P=\frac{1}{3}+\frac{4}{27}=\boxed{\text{(E) }\frac{13}{27}}&lt;/math&gt;.<br /> <br /> ~Snacc<br /> <br /> ==Video Solution==<br /> https://youtu.be/FQO-0E2zUVI?t=437<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=136138 2010 AMC 12B Problems/Problem 16 2020-10-30T17:23:24Z <p>Captainsnake: /* Solution 3 (Fancier version of Solution 1) */</p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We group this into groups of &lt;math&gt;3&lt;/math&gt;, because &lt;math&gt;3|2010&lt;/math&gt;. This means that every residue class mod 3 has an equal probability.<br /> <br /> If &lt;math&gt;3|a&lt;/math&gt;, we are done. There is a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt; that that happens.<br /> <br /> Otherwise, we have &lt;math&gt;3|bc+b+1&lt;/math&gt;, which means that &lt;math&gt;b(c+1) \equiv 2\pmod{3}&lt;/math&gt;. So either &lt;cmath&gt;b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}&lt;/cmath&gt; or &lt;cmath&gt;b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3&lt;/cmath&gt; which will lead to the property being true. There are a &lt;math&gt;\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}&lt;/math&gt; chance for each bundle of cases to be true. Thus, the total for the cases is &lt;math&gt;\frac{2}{9}&lt;/math&gt;. But we have to multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; because this only happens with a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance. So the total is actually &lt;math&gt;\frac{4}{27}&lt;/math&gt;.<br /> <br /> The grand total is &lt;cmath&gt;\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}&lt;/cmath&gt;<br /> <br /> == Solution 2 (Minor change from Solution 1) ==<br /> <br /> Just like solution 1, we see that there is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of &lt;math&gt;3|a&lt;/math&gt; and &lt;math&gt;\frac{2}{9}&lt;/math&gt; chance of &lt;math&gt;3|1+b+bc&lt;/math&gt; <br /> <br /> Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be &lt;math&gt;\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}&lt;/math&gt;<br /> <br /> -Conantwiz2023<br /> <br /> ==Solution 3 (Fancier version of Solution 1)==<br /> <br /> As with solution one, we conclude that if &lt;math&gt;a\equiv0\mod 3&lt;/math&gt; then the requirements are satisfied. We then have:<br /> &lt;cmath&gt;a(bc+c)+a\equiv0 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;a(bc+c)\equiv-a \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;c(b+1)\equiv-1 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;b+1\equiv \frac{-1}{c} \mod 3&lt;/cmath&gt;<br /> Which is true for one &lt;math&gt;b&lt;/math&gt; when &lt;math&gt;c\not\equiv 0 \mod 3&lt;/math&gt; because the integers &lt;math&gt;\mod 3&lt;/math&gt; form a field under multiplication and addition with absorbing element &lt;math&gt;0&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;P=\frac{1}{3}+\frac{4}{27}=(b)\frac{13}{27}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/FQO-0E2zUVI?t=437<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=136137 2010 AMC 12B Problems/Problem 16 2020-10-30T17:22:03Z <p>Captainsnake: </p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}<br /> <br /> == Problem 16 ==<br /> Positive integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are randomly and independently selected with replacement from the set &lt;math&gt;\{1, 2, 3,\dots, 2010\}&lt;/math&gt;. What is the probability that &lt;math&gt;abc + ab + a&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We group this into groups of &lt;math&gt;3&lt;/math&gt;, because &lt;math&gt;3|2010&lt;/math&gt;. This means that every residue class mod 3 has an equal probability.<br /> <br /> If &lt;math&gt;3|a&lt;/math&gt;, we are done. There is a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt; that that happens.<br /> <br /> Otherwise, we have &lt;math&gt;3|bc+b+1&lt;/math&gt;, which means that &lt;math&gt;b(c+1) \equiv 2\pmod{3}&lt;/math&gt;. So either &lt;cmath&gt;b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}&lt;/cmath&gt; or &lt;cmath&gt;b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3&lt;/cmath&gt; which will lead to the property being true. There are a &lt;math&gt;\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}&lt;/math&gt; chance for each bundle of cases to be true. Thus, the total for the cases is &lt;math&gt;\frac{2}{9}&lt;/math&gt;. But we have to multiply by &lt;math&gt;\frac{2}{3}&lt;/math&gt; because this only happens with a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance. So the total is actually &lt;math&gt;\frac{4}{27}&lt;/math&gt;.<br /> <br /> The grand total is &lt;cmath&gt;\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}&lt;/cmath&gt;<br /> <br /> == Solution 2 (Minor change from Solution 1) ==<br /> <br /> Just like solution 1, we see that there is a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of &lt;math&gt;3|a&lt;/math&gt; and &lt;math&gt;\frac{2}{9}&lt;/math&gt; chance of &lt;math&gt;3|1+b+bc&lt;/math&gt; <br /> <br /> Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be &lt;math&gt;\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}&lt;/math&gt;<br /> <br /> -Conantwiz2023<br /> <br /> ==Solution 3 (Fancier version of Solution 1)==<br /> <br /> As with solution one, we conclude that if &lt;math&gt;a\equiv0\mod 3&lt;/math&gt; then the requirements are satisfied. We then have:<br /> &lt;cmath&gt;a(bc+c)+a\equiv0 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;a(bc+c)\equiv-a \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;c(b+1)\equiv-1 \mod 3&lt;/cmath&gt;<br /> &lt;cmath&gt;b+1\equiv \frac{-1}{c} \mod 3&lt;/cmath&gt;<br /> Which is true for all &lt;math&gt;c\not\equiv 0 \mod 3&lt;/math&gt; because the integers &lt;math&gt;\mod 3&lt;/math&gt; form a field under multiplication and addition with absorbing element &lt;math&gt;0&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/FQO-0E2zUVI?t=437<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_21&diff=135429 2000 AMC 12 Problems/Problem 21 2020-10-19T23:26:02Z <p>Captainsnake: Added another solution and removed unnecesary comments</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}}<br /> <br /> == Problem ==<br /> Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is &lt;math&gt;m&lt;/math&gt; times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is <br /> <br /> &lt;math&gt;\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}\ \frac{1}{8m^2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(36);<br /> draw((0,0)--(6,0)--(0,3)--cycle);<br /> draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br /> label(&quot;$1$&quot;,(1,2),S);<br /> label(&quot;$1$&quot;,(2,1),W);<br /> label(&quot;$2m$&quot;,(4,0),S);<br /> label(&quot;$x$&quot;,(0,2.5),W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> WLOG, let a side of the square be &lt;math&gt;1&lt;/math&gt;. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since &lt;math&gt;A = \frac{1}{2}bh = \frac{h}{2}&lt;/math&gt;, the height of the triangle with area &lt;math&gt;m&lt;/math&gt; is &lt;math&gt;2m&lt;/math&gt;. Therefore &lt;math&gt;\frac{2m}{1} = \frac{1}{x}&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; is the base of the other triangle. &lt;math&gt;x = \frac{1}{2m}&lt;/math&gt;, and the area of that triangle is &lt;math&gt;\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}&lt;/math&gt;.<br /> <br /> <br /> === Solution 2 ===<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(36);<br /> draw((0,0)--(6,0)--(0,3)--cycle);<br /> draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br /> label(&quot;$b$&quot;,(2.5,0),S);<br /> label(&quot;$a$&quot;,(0,1.5),W);<br /> label(&quot;$c$&quot;,(2.5,1),W);<br /> label(&quot;$A$&quot;,(0.5,2.5),W);<br /> label(&quot;$B$&quot;,(3.5,0.75),W);<br /> label(&quot;$C$&quot;,(1,1),W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> From the diagram from the previous solution, we have &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; as the legs and &lt;math&gt;c&lt;/math&gt; as the side length of the square. WLOG, let the area of triangle &lt;math&gt;A&lt;/math&gt;<br /> be &lt;math&gt;m&lt;/math&gt; times the area of square &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Since triangle &lt;math&gt;A&lt;/math&gt; is similar to the large triangle, it has &lt;math&gt;h_A = a(\frac{c}{b}) = \frac{ac}{b}&lt;/math&gt;, &lt;math&gt;b_A = c&lt;/math&gt; and &lt;cmath&gt;[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2&lt;/cmath&gt;<br /> Thus &lt;math&gt;\frac{a}{2b} = m&lt;/math&gt;<br /> <br /> Now since triangle &lt;math&gt;B&lt;/math&gt; is similar to the large triangle, it has &lt;math&gt;h_B = c&lt;/math&gt;, &lt;math&gt;b_B = b\frac{c}{a} = \frac{bc}{a}&lt;/math&gt; and &lt;cmath&gt;[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]&lt;/cmath&gt; <br /> <br /> Thus &lt;math&gt;n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}&lt;/math&gt;. &lt;math&gt;\text{(D)}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> === Solution 3 (process of elimination) ===<br /> <br /> Simply testing specific triangles is sufficient.<br /> <br /> A triangle with legs of 1 and 2 gives a square of area &lt;math&gt;S=\frac{2}{3}\frac{2}{3}=\frac{4}{9}&lt;/math&gt;. The larger sub-triangle has area &lt;math&gt;T_1=\frac{\frac{2}{3}\frac{4}{3}}{2}=\frac{4}{9}&lt;/math&gt;, and the smaller triangle has area &lt;math&gt;T_2=\frac{\frac{2}{3}\frac{1}{3}}{2}=\frac{1}{9}&lt;/math&gt;. Computing ratios you get &lt;math&gt;\frac{T_1}{S}=1&lt;/math&gt; and &lt;math&gt;\frac{T_2}{S}=\frac{1}{4}&lt;/math&gt;. Plugging &lt;math&gt;m=1&lt;/math&gt; in shows that the only possible answer is &lt;math&gt;\text{(D)}&lt;/math&gt;<br /> <br /> ~ Snacc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=20|num-a=22}}<br /> {{AMC10 box|year=2000|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Shortlist_Problems/N4&diff=135154 2001 IMO Shortlist Problems/N4 2020-10-17T02:27:00Z <p>Captainsnake: /* Solution (Elementary Group Theory) */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p \geq 5&lt;/math&gt; be a prime number. Prove that there exists an integer &lt;math&gt;a&lt;/math&gt; with &lt;math&gt;1 \leq a \leq p - 2&lt;/math&gt; such that neither &lt;math&gt;a^{p - 1} - 1&lt;/math&gt; nor &lt;math&gt;(a + 1)^{p - 1} - 1&lt;/math&gt; is divisible by &lt;math&gt;p^2&lt;/math&gt;.<br /> <br /> == Solution (Elementary Group Theory) ==<br /> Let us work in the group &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt;.<br /> <br /> Note that the order of this group is &lt;math&gt;\phi({p^2})=p^2-p&lt;/math&gt;. Since &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt; is cyclic, we know that it is isomorphic to &lt;math&gt;(\mathbb{Z}/(p^2-p)\mathbb{Z})^{+}&lt;/math&gt;. <br /> <br /> We can then conclude how many residues &lt;math&gt;n&lt;/math&gt; there are such that &lt;math&gt;(p-1)n\equiv0\text{ (mod }p^2-p\text{)}&lt;/math&gt;. For some arbitrary natural number &lt;math&gt;k&lt;/math&gt;, one has:<br /> &lt;cmath&gt;(p-1)n=k(p^2-p)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=kp&lt;/cmath&gt;<br /> Therefore there are only &lt;math&gt;p&lt;/math&gt; possible values of &lt;math&gt;n&lt;/math&gt;. It's also notable that if this is true for &lt;math&gt;n_0&lt;/math&gt;, then it is also true for &lt;math&gt;p^2-p-n_0&lt;/math&gt; and &lt;math&gt;2n_0&lt;/math&gt;. This implies that there are also &lt;math&gt;p&lt;/math&gt; different values of a such that &lt;math&gt;a^{p-1}\equiv 1\text{ (mod }p^2\text{)}&lt;/math&gt;. It also implies that if and only if &lt;math&gt;a_0&lt;/math&gt; has this property, so does &lt;math&gt;\frac{1}{a_0}&lt;/math&gt;, &lt;math&gt;a_0^2&lt;/math&gt; and &lt;math&gt;\frac{1}{a_0^2}&lt;/math&gt; when working in &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt;. The squares also have their inverses which are guaranteed to have the property.<br /> <br /> There exists no numbers &lt;math&gt;1&lt;m_1,m_2\leq p-2&lt;/math&gt; such that &lt;math&gt;m_1m_2\equiv1\text{ (mod }p^2\text{)}&lt;/math&gt; as &lt;math&gt;(p-2)(p-2)&lt;p^2&lt;/math&gt; and &lt;math&gt;2^2&gt;1&lt;/math&gt;. Therefore, at most half of the values where &lt;math&gt;a_0^{p-1}\equiv 1\text{ (mod }p^2\text{)}&lt;/math&gt; are in range &lt;math&gt;1&lt;a_0\leq p-2&lt;/math&gt;. We can again remove some of the potential values by squaring all &lt;math&gt;a_0\geq\sqrt{p}&lt;/math&gt; and taking their inverse. As long as no more than half of the values in that range can have the required property, there must be a pair &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a+1&lt;/math&gt; that satisfy the requirements by the pigeonhole principle.<br /> &lt;cmath&gt;\frac{p-\sqrt(p-2)-1}{2}\leq\frac{p-3}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;2\leq(\sqrt(p-2))&lt;/cmath&gt;<br /> &lt;cmath&gt;p\geq6&lt;/cmath&gt;<br /> <br /> Lastly, you must show that there is a pair for &lt;math&gt;p=5&lt;/math&gt;. This is satisfied by &lt;math&gt;a=2&lt;/math&gt;.<br /> &lt;cmath&gt;2^{4}\equiv16\text{ (mod }25\text{)}&lt;/cmath&gt;<br /> &lt;cmath&gt;3^{4}\equiv6\text{ (mod }25\text{)}&lt;/cmath&gt;<br /> <br /> == Resources ==<br /> <br /> * [[2001 IMO Shortlist Problems]]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17473 Discussion on AOPS/MathLinks]<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Captainsnake https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Shortlist_Problems/N4&diff=135153 2001 IMO Shortlist Problems/N4 2020-10-17T02:23:31Z <p>Captainsnake: Added solution. My solution uses group theory, I suspect there is a very similar line of reasoning that can be written in the language of modular arithmetic, but I couldn't think of one.</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p \geq 5&lt;/math&gt; be a prime number. Prove that there exists an integer &lt;math&gt;a&lt;/math&gt; with &lt;math&gt;1 \leq a \leq p - 2&lt;/math&gt; such that neither &lt;math&gt;a^{p - 1} - 1&lt;/math&gt; nor &lt;math&gt;(a + 1)^{p - 1} - 1&lt;/math&gt; is divisible by &lt;math&gt;p^2&lt;/math&gt;.<br /> <br /> == Solution (Elementary Group Theory) ==<br /> Let us work in the group &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt;.<br /> <br /> Note that the order of this group is &lt;math&gt;\phi({p^2})=p^2-p&lt;/math&gt;. Since &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt; is cyclic, we know that it is isomorphic to &lt;math&gt;(\mathbb{Z}/(p^2-p)\mathbb{Z})^{+}&lt;/math&gt;. <br /> <br /> We can then conclude how many residues &lt;math&gt;n&lt;/math&gt; there are such that &lt;math&gt;(p-1)n\equiv0\text{ (mod }p^2-p\text{)}&lt;/math&gt;. For some arbitrary natural number &lt;math&gt;k&lt;/math&gt;, one has:<br /> &lt;cmath&gt;(p-1)n=k(p^2-p)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=kp&lt;/cmath&gt;<br /> Therefore there are only &lt;math&gt;p&lt;/math&gt; possible values of &lt;math&gt;n&lt;/math&gt;. It's also notable that if this is true for &lt;math&gt;n_0&lt;/math&gt;, then it is also true for &lt;math&gt;p^2-p-n_0&lt;/math&gt;, and &lt;math&gt;2n_0&lt;/math&gt;. This implies that there are also &lt;math&gt;p&lt;/math&gt; different values of a such that &lt;math&gt;a^{p-1}\equiv 1\text{ (mod }p^2\text{)}&lt;/math&gt;, more specifically, it implies that if and only if &lt;math&gt;a_0&lt;/math&gt; has this property, so does &lt;math&gt;\frac{1}{a_0}&lt;/math&gt; and &lt;math&gt;\frac{1}{a^2}&lt;/math&gt; when working in &lt;math&gt;(\mathbb{Z}/p^2\mathbb{Z})^{\times}&lt;/math&gt;. The squares also have their inverses which are guaranteed to have the propert.<br /> <br /> There exists no numbers &lt;math&gt;1&lt;m_1,m_2\leq p-2&lt;/math&gt; such that &lt;math&gt;m_1m_2\equiv1\text{ (mod }p^2\text{)}&lt;/math&gt; as &lt;math&gt;(p-2)(p-2)&lt;p^2&lt;/math&gt; and &lt;math&gt;2^2&gt;1&lt;/math&gt;. Therefore, at most half of the values where &lt;math&gt;a_0^{p-1}\equiv 1\text{ (mod }p^2\text{)}&lt;/math&gt; are in range &lt;math&gt;1&lt;a_0\leq p-2&lt;/math&gt;. We can again remove some of the potential values by squaring all &lt;math&gt;a_0\geq\sqrt{p}&lt;/math&gt; and taking their inverse. As long as no more than half of the values in that range can have the required property, there must be a pair &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;a+1&lt;/math&gt; that satisfy the requirements by the pigeonhole principle.<br /> &lt;cmath&gt;\frac{p-\sqrt(p-2)-1}{2}\leq\frac{p-3}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;2\leq(\sqrt(p-2))&lt;/cmath&gt;<br /> &lt;cmath&gt;p\geq6&lt;/cmath&gt;<br /> <br /> Lastly, you must show that there is a pair for &lt;math&gt;p=5&lt;/math&gt;.<br /> &lt;cmath&gt;2^{4}\equiv16\text{ (mod }25\text{)}&lt;/cmath&gt;<br /> &lt;cmath&gt;3^{4}\equiv6\text{ (mod }25\text{)}&lt;/cmath&gt;<br /> == Resources ==<br /> <br /> * [[2001 IMO Shortlist Problems]]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17473 Discussion on AOPS/MathLinks]<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Captainsnake