https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Xxu110&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-20T07:46:46Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=148842 2000 AIME I Problems/Problem 6 2021-03-07T22:25:35Z <p>Xxu110: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> For how many [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of [[integer]]s is it true that &lt;math&gt;0 &lt; x &lt; y &lt; 10^6&lt;/math&gt; and that the [[arithmetic mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is exactly &lt;math&gt;2&lt;/math&gt; more than the [[geometric mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;?<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{x+y}{2} &amp;=&amp; \sqrt{xy} + 2\\<br /> x+y-4 &amp;=&amp; 2\sqrt{xy}\\<br /> y - 2\sqrt{xy} + x &amp;=&amp; 4\\<br /> \sqrt{y} - \sqrt{x} &amp;=&amp; \pm 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;y &gt; x&lt;/math&gt;, we only consider &lt;math&gt;+2&lt;/math&gt;.<br /> <br /> For simplicity, we can count how many valid pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt; that satisfy our equation.<br /> <br /> The maximum that &lt;math&gt;\sqrt{y}&lt;/math&gt; can be is &lt;math&gt;\sqrt{10^6} - 1 = 999&lt;/math&gt; because &lt;math&gt;\sqrt{y}&lt;/math&gt; must be an integer (this is because &lt;math&gt;\sqrt{y} - \sqrt{x} = 2&lt;/math&gt;, an integer). Then &lt;math&gt;\sqrt{x} = 997&lt;/math&gt;, and we continue this downward until &lt;math&gt;\sqrt{y} = 3&lt;/math&gt;, in which case &lt;math&gt;\sqrt{x} = 1&lt;/math&gt;. The number of pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt;, and so &lt;math&gt;(x,y)&lt;/math&gt; is then &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> &lt;!-- solution lost in edit conflict - azjps -<br /> Since &lt;math&gt;y&gt;x&lt;/math&gt;, it follows that each ordered pair &lt;math&gt;(x,y) = (n^2, (n+2)^2)&lt;/math&gt; satisfies this equation. The minimum value of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; and the maximum value of &lt;math&gt;y = 999^2&lt;/math&gt; which would make &lt;math&gt;x = 997^2&lt;/math&gt;. Thus &lt;math&gt;x&lt;/math&gt; can be any of the squares between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;997^2&lt;/math&gt; inclusive and the answer is &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> --&gt;<br /> <br /> === Solution 2 ===<br /> <br /> <br /> Let &lt;math&gt;a^2&lt;/math&gt; = &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;b^2&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive.<br /> <br /> Then &lt;cmath&gt;\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2 + b^2 = 2ab + 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b)^2 = 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b) = \pm 2&lt;/cmath&gt;<br /> <br /> This makes counting a lot easier since now we just have to find all pairs &lt;math&gt;(a,b)&lt;/math&gt; that differ by 2.<br /> <br /> <br /> Because &lt;math&gt;\sqrt{10^6} = 10^3&lt;/math&gt;, then we can use all positive integers less than 1000 for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> <br /> We know that because &lt;math&gt;x &lt; y&lt;/math&gt;, we get &lt;math&gt;a &lt; b&lt;/math&gt;.<br /> <br /> <br /> We can count even and odd pairs separately to make things easier*:<br /> <br /> <br /> Odd: &lt;cmath&gt;(1,3) , (3,5) , (5,7) . . . (997,999)&lt;/cmath&gt;<br /> <br /> <br /> Even: &lt;cmath&gt;(2,4) , (4,6) , (6,8) . . . (996,998)&lt;/cmath&gt;<br /> <br /> <br /> This makes 499 odd pairs and 498 even pairs, for a total of &lt;math&gt;\boxed{997}&lt;/math&gt; pairs.<br /> <br /> <br /> <br /> &lt;math&gt;*&lt;/math&gt;Note: We are counting the pairs for the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, which, when squared, translate to the pairs of &lt;math&gt;(x,y)&lt;/math&gt; we are trying to find.<br /> <br /> === Solution 3 ===<br /> Since the arithmetic mean is 2 more than the geometric mean, &lt;math&gt;\frac{x+y}{2} = 2 + \sqrt{xy}&lt;/math&gt;. We can multiply by 2 to get &lt;math&gt;x + y = 4 + 2\sqrt{xy}&lt;/math&gt;. Subtracting 4 and squaring gives <br /> &lt;cmath&gt;((x+y)-4)^2 = 4xy&lt;/cmath&gt;<br /> &lt;cmath&gt;((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2 - 2xy + y^2 + 16 - 8x - 8y = 0&lt;/cmath&gt;<br /> <br /> Notice that &lt;math&gt;((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y&lt;/math&gt;, so the problem asks for solutions of <br /> &lt;cmath&gt;(x-y-4)^2 = 16y&lt;/cmath&gt;<br /> Since the left hand side is a perfect square, and 16 is a perfect square, &lt;math&gt;y&lt;/math&gt; must also be a perfect square. Since &lt;math&gt;0 &lt; y &lt; (1000)^2&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; must be from &lt;math&gt;1^2&lt;/math&gt; to &lt;math&gt;999^2&lt;/math&gt;, giving at most 999 options for &lt;math&gt;y&lt;/math&gt;.<br /> <br /> However if &lt;math&gt;y = 1^2&lt;/math&gt;, you get &lt;math&gt;(x-5)^2 = 16&lt;/math&gt;, which has solutions &lt;math&gt;x = 9&lt;/math&gt; and &lt;math&gt;x = 1&lt;/math&gt;. Both of those solutions are not less than &lt;math&gt;y&lt;/math&gt;, so &lt;math&gt;y&lt;/math&gt; cannot be equal to 1. If &lt;math&gt;y = 2^2 = 4&lt;/math&gt;, you get &lt;math&gt;(x - 8)^2 = 64&lt;/math&gt;, which has 2 solutions, &lt;math&gt;x = 16&lt;/math&gt;, and &lt;math&gt;x = 0&lt;/math&gt;. 16 is not less than 4, and &lt;math&gt;x&lt;/math&gt; cannot be 0, so &lt;math&gt;y&lt;/math&gt; cannot be 4. However, for all other &lt;math&gt;y&lt;/math&gt;, you get exactly 1 solution for &lt;math&gt;x&lt;/math&gt;, and that gives a total of &lt;math&gt;999 - 2 = \boxed{997}&lt;/math&gt; pairs.<br /> <br /> - asbodke<br /> <br /> <br /> === Solution 4 (Similar to Solution 3) ===<br /> Rearranging our conditions to <br /> <br /> &lt;cmath&gt;x^2-2xy+y^2+16-8x-8y=0 \implies&lt;/cmath&gt;<br /> &lt;cmath&gt;(y-x)^2=8(x+y-2).&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;4|y-x.&lt;/math&gt;<br /> <br /> Now, let &lt;math&gt;y = 4k+x.&lt;/math&gt; Plugging this back into our expression, we get<br /> <br /> &lt;cmath&gt;(k-1)^2=x-1.&lt;/cmath&gt;<br /> <br /> There, a unique value of &lt;math&gt;x, y&lt;/math&gt; is formed for every value of &lt;math&gt;k&lt;/math&gt;. However, we must have <br /> <br /> &lt;cmath&gt;y&lt;10^6 \implies (k+1)^2&lt; 10^6-1&lt;/cmath&gt;<br /> <br /> and <br /> <br /> &lt;cmath&gt;x=(k-1)^2+1&gt;0.&lt;/cmath&gt;<br /> <br /> Therefore, there are only &lt;math&gt;997&lt;/math&gt; pairs of &lt;math&gt;(x,y).&lt;/math&gt;<br /> <br /> Solution by Williamgolly<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_15&diff=135207 2020 AIME II Problems/Problem 15 2020-10-17T23:20:31Z <p>Xxu110: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an acute scalene triangle with circumcircle &lt;math&gt;\omega&lt;/math&gt;. The tangents to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; be the projections of &lt;math&gt;T&lt;/math&gt; onto lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. Suppose &lt;math&gt;BT = CT = 16&lt;/math&gt;, &lt;math&gt;BC = 22&lt;/math&gt;, and &lt;math&gt;TX^2 + TY^2 + XY^2 = 1143&lt;/math&gt;. Find &lt;math&gt;XY^2&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Assume &lt;math&gt;O&lt;/math&gt; to be the center of triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;OT&lt;/math&gt; cross &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;, link &lt;math&gt;XM&lt;/math&gt;, &lt;math&gt;YM&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the middle point of &lt;math&gt;BT&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the middle point of &lt;math&gt;CT&lt;/math&gt;, so we have &lt;math&gt;MT=3\sqrt{15}&lt;/math&gt;. Since &lt;math&gt;\angle A=\angle CBT=\angle BCT&lt;/math&gt;, we have &lt;math&gt;\cos A=\frac{11}{16}&lt;/math&gt;. Notice that &lt;math&gt;\angle XTY=180^{\circ}-A&lt;/math&gt;, so &lt;math&gt;\cos XYT=-\cos A&lt;/math&gt;, and this gives us &lt;math&gt;1143-2XY^2=\frac{-11}{8}XT\cdot YT&lt;/math&gt;. Since &lt;math&gt;TM&lt;/math&gt; is perpendicular to &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;BXTM&lt;/math&gt; and &lt;math&gt;CYTM&lt;/math&gt; cocycle (respectively), so &lt;math&gt;\theta_1=\angle ABC=\angle MTX&lt;/math&gt; and &lt;math&gt;\theta_2=\angle ACB=\angle YTM&lt;/math&gt;. So &lt;math&gt;\angle XPM=2\theta_1&lt;/math&gt;, so &lt;cmath&gt;\frac{\frac{XM}{2}}{XP}=\sin \theta_1&lt;/cmath&gt;, which yields &lt;math&gt;XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.&lt;/math&gt; So same we have &lt;math&gt;YM=XT&lt;/math&gt;. Apply Ptolemy theorem in &lt;math&gt;BXTM&lt;/math&gt; we have &lt;math&gt;16TY=11TX+3\sqrt{15}BX&lt;/math&gt;, and use Pythagoras theorem we have &lt;math&gt;BX^2+XT^2=16^2&lt;/math&gt;. Same in &lt;math&gt;YTMC&lt;/math&gt; and triangle &lt;math&gt;CYT&lt;/math&gt; we have &lt;math&gt;16TX=11TY+3\sqrt{15}CY&lt;/math&gt; and &lt;math&gt;CY^2+YT^2=16^2&lt;/math&gt;. Solve this for &lt;math&gt;XT&lt;/math&gt; and &lt;math&gt;TY&lt;/math&gt; and submit into the equation about &lt;math&gt;\cos XYT&lt;/math&gt;, we can obtain the result &lt;math&gt;XY^2=\boxed{717}&lt;/math&gt;.<br /> <br /> (Notice that &lt;math&gt;MXTY&lt;/math&gt; is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)<br /> <br /> -Fanyuchen20020715<br /> <br /> ==Solution 2 (Official MAA)==<br /> Let &lt;math&gt;M&lt;/math&gt; denote the midpoint of &lt;math&gt;\overline{BC}&lt;/math&gt;. The critical claim is that &lt;math&gt;M&lt;/math&gt; is the orthocenter of &lt;math&gt;\triangle AXY&lt;/math&gt;, which has the circle with diameter &lt;math&gt;\overline{AT}&lt;/math&gt; as its circumcircle. To see this, note that because &lt;math&gt;\angle BXT = \angle BMT = 90^\circ&lt;/math&gt;, the quadrilateral &lt;math&gt;MBXT&lt;/math&gt; is cyclic, it follows that<br /> &lt;cmath&gt;\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,&lt;/cmath&gt; implying that &lt;math&gt;\overline{MX} \perp \overline{AC}&lt;/math&gt;. Similarly, &lt;math&gt;\overline{MY} \perp \overline{AB}&lt;/math&gt;. In particular, &lt;math&gt;MXTY&lt;/math&gt; is a parallelogram.<br /> &lt;asy&gt;<br /> defaultpen(fontsize(8pt));<br /> unitsize(0.8cm);<br /> <br /> pair A = (0,0); <br /> pair B = (-1.26,-4.43);<br /> pair C = (-1.26+3.89, -4.43);<br /> pair M = (B+C)/2; <br /> pair O = circumcenter(A,B,C); <br /> pair T = (0.68, -6.49);<br /> pair X = foot(T,A,B); <br /> pair Y = foot(T,A,C);<br /> path omega = circumcircle(A,B,C);<br /> real rad = circumradius(A,B,C);<br /> <br /> <br /> <br /> filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69));<br /> label(&quot;$\omega$&quot;, O + rad*dir(45), SW);<br /> filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255));<br /> draw(M--T); <br /> draw(X--Y);<br /> draw(B--T--C);<br /> draw(A--X--Y--cycle);<br /> draw(omega);<br /> dot(&quot;$X$&quot;, X, W); <br /> dot(&quot;$Y$&quot;, Y, E);<br /> dot(&quot;$O$&quot;, O, W);<br /> dot(&quot;$T$&quot;, T, S); <br /> dot(&quot;$A$&quot;, A, N); <br /> dot(&quot;$B$&quot;, B, W); <br /> dot(&quot;$C$&quot;, C, E); <br /> dot(&quot;$M$&quot;, M, N);<br /> <br /> <br /> &lt;/asy&gt;<br /> Hence, by the Parallelogram Law,<br /> &lt;cmath&gt; TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).&lt;/cmath&gt; But &lt;math&gt;TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135&lt;/math&gt;. Therefore &lt;cmath&gt;XY^2 = \frac13(2 \cdot 1143-135) = 717.&lt;/cmath&gt;<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/bz5N-jI2e0U?t=710<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/zXGhABDIANY<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=14|after=Last Problem}}<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_21&diff=131441 2006 AMC 10A Problems/Problem 21 2020-08-11T01:54:24Z <p>Xxu110: /* Solution(Complementary Counting) */</p> <hr /> <div>== Problem ==<br /> How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad&lt;/math&gt;<br /> <br /> == Solution (Complementary Counting) ==<br /> Since we are asked for the number of positive 4-digit [[integer]]s with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. <br /> <br /> The total number of 4-digit integers is &lt;math&gt;9 \cdot 10 \cdot 10 \cdot 10 = 9000&lt;/math&gt;, since we have 10 choices for each digit except the first (which can't be 0).<br /> <br /> Similarly, the total number of 4-digit integers without any 2 or 3 is &lt;math&gt;7 \cdot 8 \cdot 8 \cdot 8 ={3584}&lt;/math&gt;.<br /> <br /> Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is &lt;math&gt;9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)} &lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_21&diff=131440 2006 AMC 10A Problems/Problem 21 2020-08-11T01:54:08Z <p>Xxu110: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad&lt;/math&gt;<br /> <br /> == Solution(Complementary Counting) ==<br /> Since we are asked for the number of positive 4-digit [[integer]]s with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. <br /> <br /> The total number of 4-digit integers is &lt;math&gt;9 \cdot 10 \cdot 10 \cdot 10 = 9000&lt;/math&gt;, since we have 10 choices for each digit except the first (which can't be 0).<br /> <br /> Similarly, the total number of 4-digit integers without any 2 or 3 is &lt;math&gt;7 \cdot 8 \cdot 8 \cdot 8 ={3584}&lt;/math&gt;.<br /> <br /> Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is &lt;math&gt;9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)} &lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_9&diff=131439 2002 AMC 10B Problems/Problem 9 2020-08-11T01:52:07Z <p>Xxu110: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Using the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt;, and &lt;math&gt;U&lt;/math&gt;, we can form five-letter &quot;words&quot;. If these &quot;words&quot; are arranged in alphabetical order, then the &quot;word&quot; &lt;math&gt;USAMO&lt;/math&gt; occupies position<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> There are &lt;math&gt;4!\cdot 4&lt;/math&gt; &quot;words&quot; beginning with each of the first four letters alphabetically. From there, there are &lt;math&gt;3!\cdot 3&lt;/math&gt; with &lt;math&gt;U&lt;/math&gt; as the first letter and each of the first three letters alphabetically. After that, the next &quot;word&quot; is &lt;math&gt;USAMO&lt;/math&gt;, hence our answer is &lt;math&gt;4\cdot 4!+3\cdot 3!+1=\boxed{115\Rightarrow\text{(D)}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;A = 1&lt;/math&gt;, &lt;math&gt;M = 2&lt;/math&gt;, &lt;math&gt;O = 3&lt;/math&gt;, &lt;math&gt;S = 4&lt;/math&gt;, and &lt;math&gt;U = 5&lt;/math&gt;. Then counting backwards, &lt;math&gt;54321, 54312, 54231, 54213, 54132, 54123&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{115\Rightarrow\text{(D)}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=B|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_15&diff=131432 2016 AIME I Problems/Problem 15 2020-08-11T01:07:31Z <p>Xxu110: /* Solution 5 (not too different) */</p> <hr /> <div>==Problem ==<br /> <br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; intersect at points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. Line &lt;math&gt;\ell&lt;/math&gt; is tangent to &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, respectively, with line &lt;math&gt;AB&lt;/math&gt; closer to point &lt;math&gt;X&lt;/math&gt; than to &lt;math&gt;Y&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; passes through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersecting &lt;math&gt;\omega_1&lt;/math&gt; again at &lt;math&gt;D \neq A&lt;/math&gt; and intersecting &lt;math&gt;\omega_2&lt;/math&gt; again at &lt;math&gt;C \neq B&lt;/math&gt;. The three points &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; are collinear, &lt;math&gt;XC = 67&lt;/math&gt;, &lt;math&gt;XY = 47&lt;/math&gt;, and &lt;math&gt;XD = 37&lt;/math&gt;. Find &lt;math&gt;AB^2&lt;/math&gt;.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> Let &lt;math&gt;Z = XY \cap AB&lt;/math&gt;. By the Radical Axis Theorem &lt;math&gt;AD, XY, BC&lt;/math&gt; concur at point &lt;math&gt;E&lt;/math&gt;. Furthermore, by simple angle chasing, &lt;math&gt;\triangle DXE \sim \triangle EXC&lt;/math&gt;. Let &lt;math&gt;y = EX, x = XZ&lt;/math&gt;. Then &lt;math&gt;\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67&lt;/math&gt;. Now, by Power of a Point, &lt;math&gt;AZ^2 = \frac{AB^2}{4}&lt;/math&gt;, &lt;math&gt;(y-x)x = \frac{AB^2}{4}&lt;/math&gt;, and &lt;math&gt;x(47+x) = \frac{AB^2}{4}&lt;/math&gt;. Solving, we get &lt;math&gt;\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> By the Radical Axis Theorem &lt;math&gt;AD, XY, BC&lt;/math&gt; concur at point &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;EY&lt;/math&gt; intersect at &lt;math&gt;S&lt;/math&gt;. Note that because &lt;math&gt;AXDY&lt;/math&gt; and &lt;math&gt;CYXB&lt;/math&gt; are cyclic, by Miquel's Theorem &lt;math&gt;AXBE&lt;/math&gt; is cyclic as well. Thus<br /> &lt;cmath&gt;\angle AEX = \angle ABX = \angle XCB = \angle XYB&lt;/cmath&gt;and<br /> &lt;cmath&gt;\angle XEB = \angle XAB = \angle XDA = \angle XYA.&lt;/cmath&gt;Thus &lt;math&gt;AY \parallel EB&lt;/math&gt; and &lt;math&gt;YB \parallel EA&lt;/math&gt;, so &lt;math&gt;AEBY&lt;/math&gt; is a parallelogram. Hence &lt;math&gt;AS = SB&lt;/math&gt; and &lt;math&gt;SE = SY&lt;/math&gt;. But notice that &lt;math&gt;DXE&lt;/math&gt; and &lt;math&gt;EXC&lt;/math&gt; are similar by &lt;math&gt;AA&lt;/math&gt; Similarity, so &lt;math&gt;XE^2 = XD \cdot XC = 37 \cdot 67&lt;/math&gt;. But<br /> &lt;cmath&gt;XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.&lt;/cmath&gt;Hence &lt;math&gt;AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> First, we note that as &lt;math&gt;\triangle XDY&lt;/math&gt; and &lt;math&gt;\triangle XYC&lt;/math&gt; have bases along the same line, &lt;math&gt;\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}&lt;/math&gt;. We can also find the ratio of their areas using the circumradius area formula. If &lt;math&gt;R_1&lt;/math&gt; is the radius of &lt;math&gt;\omega_1&lt;/math&gt; and if &lt;math&gt;R_2&lt;/math&gt; is the radius of &lt;math&gt;\omega_2&lt;/math&gt;, then <br /> &lt;cmath&gt;\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.&lt;/cmath&gt;<br /> Since we showed this to be &lt;math&gt;\frac{DY}{YC}&lt;/math&gt;, we see that &lt;math&gt;\frac{R_2}{R_1}=\frac{67}{37}&lt;/math&gt;.<br /> <br /> We extend &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; to meet at point &lt;math&gt;P&lt;/math&gt;, and we extend &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; to meet at point &lt;math&gt;Q&lt;/math&gt; as shown below.<br /> &lt;asy&gt;<br /> size(200);<br /> import olympiad;<br /> real R1=45,R2=67*R1/37;<br /> real m1=sqrt(R1^2-23.5^2);<br /> real m2=sqrt(R2^2-23.5^2);<br /> pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);<br /> draw(circle(o1,R1));<br /> draw(circle(o2,R2));<br /> pair q=(-R1/(R2-R1)*o2.x,0);<br /> pair a=tangent(q,o1,R1,2);<br /> pair b=tangent(q,o2,R2,2);<br /> pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q));<br /> pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q));<br /> pair p=extension(a,d,b,c);<br /> dot(q^^a^^b^^x^^y^^c^^d^^p);<br /> draw(q--b^^q--c);<br /> draw(p--d^^p--c^^x--y);<br /> draw(a--y^^b--y);<br /> draw(d--x--c);<br /> label(&quot;$A$&quot;,a,NW,fontsize(8));<br /> label(&quot;$B$&quot;,b,NE,fontsize(8));<br /> label(&quot;$C$&quot;,c,SE,fontsize(8));<br /> label(&quot;$D$&quot;,d,SW,fontsize(8));<br /> label(&quot;$X$&quot;,x,2*WNW,fontsize(8));<br /> label(&quot;$Y$&quot;,y,3*S,fontsize(8));<br /> label(&quot;$P$&quot;,p,N,fontsize(8));<br /> label(&quot;$Q$&quot;,q,W,fontsize(8));<br /> &lt;/asy&gt;<br /> As &lt;math&gt;ABCD&lt;/math&gt; is cyclic, we know that &lt;math&gt;\angle BCD=180-\angle DAB=\angle BAP&lt;/math&gt;. But then as &lt;math&gt;AB&lt;/math&gt; is tangent to &lt;math&gt;\omega_2&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;\angle BCD=\angle ABY&lt;/math&gt;. Therefore, &lt;math&gt;\angle ABY=\angle BAP&lt;/math&gt;, and &lt;math&gt;BY\parallel PD&lt;/math&gt;. A similar argument shows &lt;math&gt;AY\parallel PC&lt;/math&gt;. These parallel lines show &lt;math&gt;\triangle PDC\sim\triangle ADY\sim\triangle BYC&lt;/math&gt;. Also, we showed that &lt;math&gt;\frac{R_2}{R_1}=\frac{67}{37}&lt;/math&gt;, so the ratio of similarity between &lt;math&gt;\triangle ADY&lt;/math&gt; and &lt;math&gt;\triangle BYC&lt;/math&gt; is &lt;math&gt;\frac{37}{67}&lt;/math&gt;, or rather<br /> &lt;cmath&gt;\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.&lt;/cmath&gt;<br /> We can now use the parallel lines to find more similar triangles. As &lt;math&gt;\triangle AQD\sim \triangle BQY&lt;/math&gt;, we know that<br /> &lt;cmath&gt;\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.&lt;/cmath&gt;<br /> Setting &lt;math&gt;QA=37x&lt;/math&gt;, we see that &lt;math&gt;QB=67x&lt;/math&gt;, hence &lt;math&gt;AB=30x&lt;/math&gt;, and the problem simplifies to finding &lt;math&gt;30^2x^2&lt;/math&gt;. Setting &lt;math&gt;QD=37^2y&lt;/math&gt;, we also see that &lt;math&gt;QY=37\cdot 67y&lt;/math&gt;, hence &lt;math&gt;DY=37\cdot 30y&lt;/math&gt;. Also, as &lt;math&gt;\triangle AQY\sim \triangle BQC&lt;/math&gt;, we find that<br /> &lt;cmath&gt;\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.&lt;/cmath&gt;<br /> As &lt;math&gt;QY=37\cdot 67y&lt;/math&gt;, we see that &lt;math&gt;QC=67^2y&lt;/math&gt;, hence &lt;math&gt;YC=67\cdot30y&lt;/math&gt;.<br /> <br /> Applying Power of a Point to point &lt;math&gt;Q&lt;/math&gt; with respect to &lt;math&gt;\omega_2&lt;/math&gt;, we find<br /> &lt;cmath&gt;67^2x^2=37\cdot 67^3 y^2,&lt;/cmath&gt;<br /> or &lt;math&gt;x^2=37\cdot 67 y^2&lt;/math&gt;. We wish to find &lt;math&gt;AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2&lt;/math&gt;.<br /> <br /> Applying Stewart's Theorem to &lt;math&gt;\triangle XDC&lt;/math&gt;, we find<br /> &lt;cmath&gt;37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).&lt;/cmath&gt;<br /> We can cancel &lt;math&gt;30\cdot 104\cdot y&lt;/math&gt; from both sides, finding &lt;math&gt;37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2&lt;/math&gt;. Therefore, <br /> &lt;cmath&gt;AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.&lt;/cmath&gt;<br /> <br /> ===Solution 4===<br /> &lt;asy&gt;<br /> size(9cm);<br /> import olympiad;<br /> real R1=45,R2=67*R1/37;<br /> real m1=sqrt(R1^2-23.5^2);<br /> real m2=sqrt(R2^2-23.5^2);<br /> pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);<br /> draw(circle(o1,R1));<br /> draw(circle(o2,R2));<br /> pair q=(-R1/(R2-R1)*o2.x,0);<br /> pair a=tangent(q,o1,R1,2);<br /> pair b=tangent(q,o2,R2,2);<br /> pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q));<br /> pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q));<br /> dot(a^^b^^x^^y^^c^^d);<br /> draw(x--y);<br /> draw(a--y^^b--y);<br /> draw(d--x--c);<br /> draw(a--b--c--d--cycle);<br /> draw(x--a^^x--b);<br /> label(&quot;$A$&quot;,a,NW,fontsize(9));<br /> label(&quot;$B$&quot;,b,NE,fontsize(9));<br /> label(&quot;$C$&quot;,c,SE,fontsize(9));<br /> label(&quot;$D$&quot;,d,SW,fontsize(9));<br /> label(&quot;$X$&quot;,x,2*N,fontsize(9));<br /> label(&quot;$Y$&quot;,y,3*S,fontsize(9));<br /> &lt;/asy&gt;<br /> First of all, since quadrilaterals &lt;math&gt;ADYX&lt;/math&gt; and &lt;math&gt;XYCB&lt;/math&gt; are cyclic, we can let &lt;math&gt;\angle DAX = \angle XYC = \theta&lt;/math&gt;, and &lt;math&gt;\angle XYD = \angle CBX = 180 - \theta&lt;/math&gt;, due to the properties of cyclic quadrilaterals. In addition, let &lt;math&gt;\angle BAX = x&lt;/math&gt; and &lt;math&gt;\angle ABX = y&lt;/math&gt;. Thus, &lt;math&gt;\angle ADX = \angle AYX = x&lt;/math&gt; and &lt;math&gt;\angle XYB = \angle XCB = y&lt;/math&gt;. Then, since quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cyclic as well, we have the following sums:<br /> &lt;cmath&gt;\theta + x +\angle XCY + y = 180^{\circ}&lt;/cmath&gt;<br /> &lt;cmath&gt;180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}&lt;/cmath&gt;<br /> Cancelling out &lt;math&gt;180^{\circ}&lt;/math&gt; in the second equation and isolating &lt;math&gt;\theta&lt;/math&gt; yields &lt;math&gt;\theta = y + \angle XDY + x&lt;/math&gt;. Substituting &lt;math&gt;\theta&lt;/math&gt; back into the first equation, we obtain<br /> &lt;cmath&gt;2x + 2y + \angle XCY + \angle XDY = 180^{\circ}&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}&lt;/cmath&gt;<br /> &lt;cmath&gt;x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}&lt;/cmath&gt;<br /> we can then imply that &lt;math&gt;\angle DAY = x + y&lt;/math&gt;. Similarly, &lt;math&gt;\angle YBC = x + y&lt;/math&gt;. So then &lt;math&gt;\angle DXY = \angle YXC = x + y&lt;/math&gt;, so since we know that &lt;math&gt;XY&lt;/math&gt; bisects &lt;math&gt;\angle DXC&lt;/math&gt;, we can solve for &lt;math&gt;DY&lt;/math&gt; and &lt;math&gt;YC&lt;/math&gt; with Stewart’s Theorem. Let &lt;math&gt;DY = 37n&lt;/math&gt; and &lt;math&gt;YC = 67n&lt;/math&gt;. Then<br /> &lt;cmath&gt;37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n&lt;/cmath&gt;<br /> &lt;cmath&gt;37n \cdot 67n + 47^2 = 37 \cdot 67&lt;/cmath&gt;<br /> &lt;cmath&gt;n^2 = \frac{270}{2479}&lt;/cmath&gt;<br /> Now, since &lt;math&gt;\angle AYX = x&lt;/math&gt; and &lt;math&gt;\angle BYX = y&lt;/math&gt;, &lt;math&gt;\angle AYB = x + y&lt;/math&gt;. From there, let &lt;math&gt;\angle AYD = \alpha&lt;/math&gt; and &lt;math&gt;\angle BYC = \beta&lt;/math&gt;. From angle chasing we can derive that &lt;math&gt;\angle YDX = \angle YAX = \beta - x&lt;/math&gt; and &lt;math&gt;\angle YCX = \angle YBX = \alpha - y&lt;/math&gt;. From there, since &lt;math&gt;\angle ADX = x&lt;/math&gt;, it is quite clear that &lt;math&gt;\angle ADY = \beta&lt;/math&gt;, and &lt;math&gt;\angle YAB = \beta&lt;/math&gt; can be found similarly. From there, since &lt;math&gt;\angle ADY = \angle YAB = \angle BYC = \beta&lt;/math&gt; and &lt;math&gt;\angle DAY = \angle AYB = \angle YBC = x + y&lt;/math&gt;, we have &lt;math&gt;AA&lt;/math&gt; similarity between &lt;math&gt;\triangle DAY&lt;/math&gt;, &lt;math&gt;\triangle AYB&lt;/math&gt;, and &lt;math&gt;\triangle YBC&lt;/math&gt;. Therefore the length of &lt;math&gt;AY&lt;/math&gt; is the geometric mean of the lengths of &lt;math&gt;DA&lt;/math&gt; and &lt;math&gt;YB&lt;/math&gt; (from &lt;math&gt;\triangle DAY \sim \triangle AYB&lt;/math&gt;). However, &lt;math&gt;\triangle DAY \sim \triangle AYB \sim \triangle YBC&lt;/math&gt; yields the proportion &lt;math&gt;\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}&lt;/math&gt;; hence, the length of &lt;math&gt;AB&lt;/math&gt; is the geometric mean of the lengths of &lt;math&gt;DY&lt;/math&gt; and &lt;math&gt;YC&lt;/math&gt;. <br /> We can now simply use arithmetic to calculate &lt;math&gt;AB^2&lt;/math&gt;. <br /> &lt;cmath&gt;AB^2 = DY \cdot YC&lt;/cmath&gt;<br /> &lt;cmath&gt;AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}&lt;/cmath&gt;<br /> &lt;cmath&gt;AB^2 = \boxed{270}&lt;/cmath&gt;<br /> <br /> '''-Solution by TheBoomBox77'''<br /> ===Solution 5 (not too different)===<br /> Let &lt;math&gt;E = DA \cap CB&lt;/math&gt;. By Radical Axes, &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;XY&lt;/math&gt;. Note that &lt;math&gt;EAXB&lt;/math&gt; is cyclic as &lt;math&gt;X&lt;/math&gt; is the Miquel point of &lt;math&gt;\triangle EDC&lt;/math&gt; in this configuration.<br /> <br /> Claim. &lt;math&gt;\triangle DXE \sim \triangle EXC&lt;/math&gt;<br /> Proof. We angle chase. &lt;cmath&gt;\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE&lt;/cmath&gt;and&lt;cmath&gt;\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;F = EX \cap AB&lt;/math&gt;. Note &lt;cmath&gt;FA^2 = FX \cdot FY = FB^2&lt;/cmath&gt;and&lt;cmath&gt;EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY&lt;/cmath&gt;By our claim, &lt;cmath&gt;\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}&lt;/cmath&gt;and&lt;cmath&gt;FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}&lt;/cmath&gt;Finally, &lt;cmath&gt;AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare&lt;/cmath&gt;~Mathscienceclass<br /> <br /> ==See Also==<br /> {{AIME box|year=2016|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_9&diff=130986 2020 AMC 10B Problems/Problem 9 2020-08-07T17:20:38Z <p>Xxu110: /* Video Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #9]] and [[2020 AMC 12B Problems|2020 AMC 12B #8]]}}<br /> <br /> ==Problem==<br /> <br /> How many ordered pairs of integers &lt;math&gt;(x, y)&lt;/math&gt; satisfy the equation &lt;cmath&gt;x^{2020}+y^2=2y?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Rearranging the terms and and completing the square for &lt;math&gt;y&lt;/math&gt; yields the result &lt;math&gt;x^{2020}+(y-1)^2=1&lt;/math&gt;. Then, notice that &lt;math&gt;x&lt;/math&gt; can only be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;-1&lt;/math&gt; because any value of &lt;math&gt;x^{2020}&lt;/math&gt; that is greater than 1 will cause the term &lt;math&gt;(y-1)^2&lt;/math&gt; to be less than &lt;math&gt;0&lt;/math&gt;, which is impossible as &lt;math&gt;y&lt;/math&gt; must be real. Therefore, plugging in the above values for &lt;math&gt;x&lt;/math&gt; gives the ordered pairs &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(1,1)&lt;/math&gt;, &lt;math&gt;(-1,1)&lt;/math&gt;, and &lt;math&gt;(0,2)&lt;/math&gt; gives a total of &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; ordered pairs.<br /> <br /> ==Solution 2==<br /> <br /> Bringing all of the terms to the LHS, we see a quadratic equation &lt;cmath&gt;y^2 - 2y + x^{2020} = 0&lt;/cmath&gt; in terms of &lt;math&gt;y&lt;/math&gt;. Applying the quadratic formula, we get &lt;cmath&gt;y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.&lt;/cmath&gt; In order for &lt;math&gt;y&lt;/math&gt; to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, &lt;math&gt;4(1-x^{2020})&lt;/math&gt; must be nonnegative. Therefore, &lt;cmath&gt;4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.&lt;/cmath&gt; Here, we see that we must split the inequality into a compound, resulting in &lt;math&gt;-1 \leq x \leq 1&lt;/math&gt;. <br /> <br /> The only integers that satisfy this are &lt;math&gt;x \in \{-1,0,1\}&lt;/math&gt;. Plugging these values back into the quadratic equation, we see that &lt;math&gt;x = \{-1,1\}&lt;/math&gt; both produce a discriminant of &lt;math&gt;0&lt;/math&gt;, meaning that there is only 1 solution for &lt;math&gt;y&lt;/math&gt;. If &lt;math&gt;x = \{0\}&lt;/math&gt;, then the discriminant is nonzero, therefore resulting in two solutions for &lt;math&gt;y&lt;/math&gt;. <br /> <br /> Thus, the answer is &lt;math&gt;2 \cdot 1 + 1 \cdot 2 = \boxed{\textbf{(D) }4}&lt;/math&gt;.<br /> <br /> ~Tiblis<br /> <br /> <br /> ==Solution 3, x first==<br /> Set it up as a quadratic in terms of y:<br /> &lt;cmath&gt;y^2-2y+x^{2020}=0&lt;/cmath&gt;<br /> Then the discriminant is<br /> &lt;cmath&gt;\Delta = 4-4x^{2020}&lt;/cmath&gt;<br /> This will clearly only yield real solutions when &lt;math&gt;x^{2020} \leq 1&lt;/math&gt;, because it is always positive.<br /> Then &lt;math&gt;x=-1,0,1&lt;/math&gt;. Checking each one:<br /> &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; are the same when raised to the 2020th power:<br /> &lt;cmath&gt;y^2-2y+1=(y-1)^2=0&lt;/cmath&gt;<br /> This has only has solutions &lt;math&gt;1&lt;/math&gt;, so &lt;math&gt;(\pm 1,1)&lt;/math&gt; are solutions.<br /> Next, if &lt;math&gt;x=0&lt;/math&gt;:<br /> &lt;cmath&gt;y^2-2y=0&lt;/cmath&gt;<br /> Which has 2 solutions, so &lt;math&gt;(0,2)&lt;/math&gt; and &lt;math&gt;(0,0)&lt;/math&gt;<br /> <br /> These are the only 4 solutions, so &lt;math&gt;\boxed{\textbf{(D) } 4}&lt;/math&gt;<br /> <br /> ==Solution 4, y first==<br /> <br /> Move the &lt;math&gt;y^2&lt;/math&gt; term to the other side to get &lt;math&gt;x^{2020}=2y-y^2 = y(2-y)&lt;/math&gt;. Because &lt;math&gt;x^{2020} \geq 0&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt;, then &lt;math&gt;y(2-y) \geq 0 \Rightarrow y = 0,1,2&lt;/math&gt;. If &lt;math&gt;y=0&lt;/math&gt; or &lt;math&gt;y=2&lt;/math&gt;, the right side is &lt;math&gt;0&lt;/math&gt; and therefore &lt;math&gt;x=0&lt;/math&gt;. When &lt;math&gt;y=1&lt;/math&gt;, the right side become &lt;math&gt;1&lt;/math&gt;, therefore &lt;math&gt;x=1,-1&lt;/math&gt;. Our solutions are &lt;math&gt;(0,2)&lt;/math&gt;, &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(1,1)&lt;/math&gt;, &lt;math&gt;(-1,1)&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; solutions, so the answer is &lt;math&gt;\boxed{\textbf{(D) } 4}&lt;/math&gt; - wwt7535<br /> <br /> ==Video Solutions==<br /> https://youtu.be/6ujfjGLzVoE<br /> <br /> ~IceMatrix<br /> <br /> https://youtu.be/7dQ423hhgac<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=8|num-a=10}}<br /> {{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}<br /> <br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=130298 2017 AIME I Problems/Problem 15 2020-08-03T02:48:21Z <p>Xxu110: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_13&diff=129445 2017 AIME I Problems/Problem 13 2020-07-27T04:25:06Z <p>Xxu110: /* Solution 2 */</p> <hr /> <div>==Problem 13==<br /> For every &lt;math&gt;m \geq 2&lt;/math&gt;, let &lt;math&gt;Q(m)&lt;/math&gt; be the least positive integer with the following property: For every &lt;math&gt;n \geq Q(m)&lt;/math&gt;, there is always a perfect cube &lt;math&gt;k^3&lt;/math&gt; in the range &lt;math&gt;n &lt; k^3 \leq m \cdot n&lt;/math&gt;. Find the remainder when &lt;cmath&gt;\sum_{m = 2}^{2017} Q(m)&lt;/cmath&gt;is divided by 1000.<br /> <br /> ==Solution 1==<br /> Lemma 1: The ratio between &lt;math&gt;k^3&lt;/math&gt; and &lt;math&gt;(k+1)^3&lt;/math&gt; decreases as &lt;math&gt;k&lt;/math&gt; increases.<br /> <br /> Lemma 2: If the range &lt;math&gt;(n,mn]&lt;/math&gt; includes &lt;math&gt;y&lt;/math&gt; cubes, &lt;math&gt;(p,mp]&lt;/math&gt; will always contain at least &lt;math&gt;y-1&lt;/math&gt; cubes for all &lt;math&gt;p&lt;/math&gt; in &lt;math&gt;[n,+\infty)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;m=14&lt;/math&gt;, the range &lt;math&gt;(1,14]&lt;/math&gt; includes one cube. The range &lt;math&gt;(2,28]&lt;/math&gt; includes 2 cubes, which fulfills the Lemma. Since &lt;math&gt;n=1&lt;/math&gt; also included a cube, we can assume that &lt;math&gt;Q(m)=1&lt;/math&gt; for all &lt;math&gt;m&gt;14&lt;/math&gt;. Two groups of 1000 are included in the sum modulo 1000. They do not count since &lt;math&gt;Q(m)=1&lt;/math&gt; for all of them, therefore &lt;cmath&gt;\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000&lt;/cmath&gt;<br /> <br /> Now that we know this we will find the smallest &lt;math&gt;n&lt;/math&gt; that causes &lt;math&gt;(n,mn]&lt;/math&gt; to contain two cubes and work backwards (recursion) until there is no cube in &lt;math&gt;(n,mn]&lt;/math&gt;.<br /> <br /> For &lt;math&gt;m=2&lt;/math&gt; there are two cubes in &lt;math&gt;(n,2n]&lt;/math&gt; for &lt;math&gt;n=63&lt;/math&gt;. There are no cubes in &lt;math&gt;(31,62]&lt;/math&gt; but there is one in &lt;math&gt;(32,64]&lt;/math&gt;. Therefore &lt;math&gt;Q(2)=32&lt;/math&gt;.<br /> <br /> For &lt;math&gt;m=3&lt;/math&gt; there are two cubes in &lt;math&gt;(n,3n]&lt;/math&gt; for &lt;math&gt;n=22&lt;/math&gt;. There are no cubes in &lt;math&gt;(8,24]&lt;/math&gt; but there is one in &lt;math&gt;(9,27]&lt;/math&gt;. Therefore &lt;math&gt;Q(3)=9&lt;/math&gt;.<br /> <br /> For &lt;math&gt;m&lt;/math&gt; in &lt;math&gt;\{4,5,6,7\}&lt;/math&gt; there are two cubes in &lt;math&gt;(n,4n]&lt;/math&gt; for &lt;math&gt;n=7&lt;/math&gt;. There are no cubes in &lt;math&gt;(1,4]&lt;/math&gt; but there is one in &lt;math&gt;(2,8]&lt;/math&gt;. Therefore &lt;math&gt;Q(4)=2&lt;/math&gt;, and the same for &lt;math&gt;Q(5)&lt;/math&gt;, &lt;math&gt;Q(6)&lt;/math&gt;, and &lt;math&gt;Q(7)&lt;/math&gt; for a sum of &lt;math&gt;8&lt;/math&gt;.<br /> <br /> For all other &lt;math&gt;m&lt;/math&gt; there is one cube in &lt;math&gt;(1,8]&lt;/math&gt;, &lt;math&gt;(2,16]&lt;/math&gt;, &lt;math&gt;(3,24]&lt;/math&gt;, and there are two in &lt;math&gt;(4,32]&lt;/math&gt;. Therefore, since there are 10 values of &lt;math&gt;m&lt;/math&gt; in the sum, this part sums to &lt;math&gt;10&lt;/math&gt;.<br /> <br /> When the partial sums are added, we get &lt;math&gt;\boxed{059}\hspace{2 mm}QED\hspace{2 mm} \blacksquare&lt;/math&gt;<br /> <br /> This solution is brought to you by [[User:a1b2|a1b2]]<br /> <br /> ==Solution 2==<br /> <br /> We claim that &lt;math&gt;Q(m) = 1&lt;/math&gt; when &lt;math&gt;m \ge 8&lt;/math&gt;. <br /> <br /> When &lt;math&gt;m \ge 8&lt;/math&gt;, for every &lt;math&gt;n \ge Q(m) = 1&lt;/math&gt;, we need to prove there exists an integer &lt;math&gt;k&lt;/math&gt;, such that &lt;math&gt;n &lt; k^3 \le m*n&lt;/math&gt;.<br /> <br /> That because &lt;math&gt;\sqrt{m*n} - \sqrt{n} \ge 2\sqrt{n} - \sqrt{n} = \sqrt{n} \ge 1&lt;/math&gt;, so k exists between &lt;math&gt;\sqrt{m*n}&lt;/math&gt; and &lt;math&gt;\sqrt{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{n} &lt; k \le \sqrt{m*n}&lt;/math&gt;.<br /> <br /> We can then hand evaluate &lt;math&gt;Q(m)&lt;/math&gt; for &lt;math&gt;m = 2,3,4,5,6,7&lt;/math&gt;, and get &lt;math&gt;Q(2) = 32&lt;/math&gt;, &lt;math&gt;Q(3) = 9&lt;/math&gt;, and all the others equal 2.<br /> <br /> There are a total of 2010 integers from 8 to 2017.<br /> <br /> &lt;cmath&gt;\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{7} Q(m) + 2010 \equiv 32+9+2+2+2+2+10 = \boxed{059} \mod 1000&lt;/cmath&gt;<br /> <br /> -AlexLikeMath<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=129330 2017 AIME I Problems/Problem 15 2020-07-26T14:37:57Z <p>Xxu110: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=129315 2017 AIME I Problems/Problem 15 2020-07-26T04:11:09Z <p>Xxu110: /* Solution 1 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_12&diff=129310 2018 AIME I Problems/Problem 12 2020-07-26T02:36:49Z <p>Xxu110: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Rewrite the set after mod 3<br /> <br /> 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0<br /> <br /> All 0s can be omitted <br /> <br /> Case 1<br /> No 1 No 2<br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> Case 2<br /> &lt;math&gt;222&lt;/math&gt;<br /> &lt;math&gt;20&lt;/math&gt;<br /> <br /> Case 3<br /> &lt;math&gt;222222&lt;/math&gt;<br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> Case 4<br /> &lt;math&gt;12&lt;/math&gt;<br /> &lt;math&gt;6*6=36&lt;/math&gt;<br /> <br /> Case 5<br /> &lt;math&gt;12222&lt;/math&gt;<br /> &lt;math&gt;6*15=90&lt;/math&gt;<br /> <br /> Case 6<br /> &lt;math&gt;1122&lt;/math&gt;<br /> &lt;math&gt;15*15=225&lt;/math&gt;<br /> <br /> Case 7<br /> &lt;math&gt;1122222&lt;/math&gt;<br /> &lt;math&gt;15*6=90&lt;/math&gt;<br /> <br /> Case 8<br /> &lt;math&gt;111&lt;/math&gt;<br /> &lt;math&gt;20&lt;/math&gt;<br /> <br /> Case 9<br /> &lt;math&gt;111222&lt;/math&gt;<br /> &lt;math&gt;20*20=400&lt;/math&gt;<br /> <br /> Case 10<br /> &lt;math&gt;111222222&lt;/math&gt;<br /> &lt;math&gt;20&lt;/math&gt;<br /> <br /> Case 11<br /> &lt;math&gt;11112&lt;/math&gt;<br /> &lt;math&gt;15*6=90&lt;/math&gt;<br /> <br /> Case 12<br /> &lt;math&gt;11112222&lt;/math&gt;<br /> &lt;math&gt;15*15=225&lt;/math&gt;<br /> <br /> Case 13<br /> &lt;math&gt;1111122&lt;/math&gt;<br /> &lt;math&gt;6*15=90&lt;/math&gt;<br /> <br /> Case 14<br /> &lt;math&gt;1111122222&lt;/math&gt;<br /> &lt;math&gt;6*6=36&lt;/math&gt;<br /> <br /> Case 15<br /> &lt;math&gt;111111&lt;/math&gt;<br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> Case 16<br /> &lt;math&gt;111111222&lt;/math&gt;<br /> &lt;math&gt;20&lt;/math&gt;<br /> <br /> Case 17<br /> &lt;math&gt;111111222222&lt;/math&gt;<br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> Total &lt;math&gt;1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366&lt;/math&gt;<br /> <br /> &lt;math&gt;P=\frac{1366}{2^{12}}=\frac{683}{2^{11}}&lt;/math&gt;<br /> <br /> ANS=&lt;math&gt;\boxed{683}&lt;/math&gt;<br /> <br /> By SpecialBeing2017<br /> <br /> ==Solution 2==<br /> Consider the numbers &lt;math&gt;\{1,4,7,10,13,16\}&lt;/math&gt;. Each of those are congruent to &lt;math&gt;1 \pmod 3&lt;/math&gt;. There is &lt;math&gt;{6 \choose 0}=1&lt;/math&gt; way to choose zero numbers &lt;math&gt;{6 \choose 1}=6&lt;/math&gt; ways to choose &lt;math&gt;1&lt;/math&gt; and so on. There ends up being &lt;math&gt;{6 \choose 0}+{6 \choose 3}+{6 \choose 6} = 22&lt;/math&gt; possible subsets congruent to &lt;math&gt;0\pmod 3&lt;/math&gt;. There are &lt;math&gt;2^6=64&lt;/math&gt; possible subsets of these numbers. By symmetry there are &lt;math&gt;21&lt;/math&gt; subsets each for &lt;math&gt;1 \pmod 3&lt;/math&gt; and &lt;math&gt;2 \pmod3&lt;/math&gt;.<br /> <br /> We get the same numbers for the subsets of &lt;math&gt;\{2,5,8,11,14,17\}&lt;/math&gt;.<br /> <br /> For &lt;math&gt;\{3,6,9,12,15,18\}&lt;/math&gt;, all &lt;math&gt;2^6&lt;/math&gt; subsets are &lt;math&gt;0 \pmod3&lt;/math&gt;.<br /> <br /> So the probability is: &lt;math&gt;\frac{(22\cdot22+2\cdot21\cdot21)\cdot2^6}{2^{18}}=\frac{683}{2^{11}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Notice that six numbers are &lt;math&gt;0\pmod3&lt;/math&gt;, six are &lt;math&gt;1\pmod3&lt;/math&gt;, and six are &lt;math&gt;2\pmod3&lt;/math&gt;. Having numbers &lt;math&gt;0\pmod3&lt;/math&gt; will not change the remainder when &lt;math&gt;s(T)&lt;/math&gt; is divided by &lt;math&gt;3&lt;/math&gt;, so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are &lt;math&gt;1\pmod3&lt;/math&gt;, minus the number of numbers that are &lt;math&gt;2\pmod3&lt;/math&gt;, must be a multiple of &lt;math&gt;3&lt;/math&gt;, possibly zero or negative. We can now split into cases based on how many numbers that are &lt;math&gt;1\pmod3&lt;/math&gt; are in the set.<br /> <br /> Case 1- &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers: There can be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484&lt;/math&gt; ways.<br /> <br /> Case 2- &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers: There can be &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441&lt;/math&gt; ways.<br /> <br /> Case 3- &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers: There can be &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441&lt;/math&gt; ways.<br /> <br /> Adding these up, we get that there are &lt;math&gt;1366&lt;/math&gt; ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of &lt;math&gt;3&lt;/math&gt; in our set, we have that there are &lt;math&gt;1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66\right)=1366\cdot2^6&lt;/math&gt; subsets &lt;math&gt;T&lt;/math&gt; with a sum that is a multiple of &lt;math&gt;3&lt;/math&gt;. Since there are &lt;math&gt;2^{18}&lt;/math&gt; total subsets, the probability is &lt;math&gt;\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{683}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We use generating functions. Each element of &lt;math&gt;U&lt;/math&gt; has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given &lt;math&gt;n\in U&lt;/math&gt;, the probability generating function is<br /> &lt;cmath&gt;\frac{1}{2}+\frac{1}{2}x^n.&lt;/cmath&gt;<br /> Therefore, in the generating function<br /> &lt;cmath&gt;\frac{1}{2^{18}}(1+x)(1+x^2)(1+x^3)\cdots (1+x^{18}),&lt;/cmath&gt;<br /> the coefficient of &lt;math&gt;x^k&lt;/math&gt; represents the probability of obtaining a sum of &lt;math&gt;k&lt;/math&gt;. We wish to find the sum of the coefficients of all terms of the form &lt;math&gt;x^{3k}&lt;/math&gt;. If &lt;math&gt;\omega=e^{2\pi i/3}&lt;/math&gt; is a cube root of unity, then it is well know that for a polynomial &lt;math&gt;P(x)&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{P(1)+P(\omega)+P(\omega^2)}{3}&lt;/cmath&gt;<br /> will yield the sum of the coefficients of the terms of the form &lt;math&gt;x^{3k}&lt;/math&gt;. Then we find<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{1}{2^{18}}(1+1)(1+1^2)(1+1^3)\cdots (1+1^{18})&amp;=1\\\frac{1}{2^{18}}(1+\omega)(1+\omega^2)(1+\omega^3)\cdots (1+\omega^{18})&amp;=\frac{1}{2^{12}}\\\frac{1}{2^{18}}(1+\omega^2)(1+\omega^4)(1+\omega^6)\cdots (1+\omega^{36})&amp;=\frac{1}{2^{12}}.<br /> \end{align*}&lt;/cmath&gt;<br /> To evaluate the last two products, we utilized the facts that &lt;math&gt;\omega^3=1&lt;/math&gt; and &lt;math&gt;1+\omega+\omega^2=0&lt;/math&gt;. Therefore, the desired probability is<br /> &lt;cmath&gt;\frac{1+1/2^{12}+1/2^{12}}{3}=\frac{683}{2^{11}}.&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;\boxed{683}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Define a set that &quot;works&quot; to be a set for which the sum of the terms is &lt;math&gt;0&lt;/math&gt; mod &lt;math&gt;3&lt;/math&gt;. The given set mod &lt;math&gt;3&lt;/math&gt; is<br /> &lt;cmath&gt;\{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}.&lt;/cmath&gt;<br /> Let &lt;math&gt;w(N)&lt;/math&gt; be the number of subsets of the first &lt;math&gt;N&lt;/math&gt; terms of this set that &quot;work.&quot;<br /> Consider just the first &lt;math&gt;3&lt;/math&gt; terms:<br /> &lt;cmath&gt;\{1,2,0\}.&lt;/cmath&gt;<br /> There are &lt;math&gt;2^3=8&lt;/math&gt; total subsets, and &lt;math&gt;w(3)=4&lt;/math&gt; (the subsets are &lt;math&gt;\emptyset, \{0\}, \{1,2\},&lt;/math&gt; and &lt;math&gt;\{1,2,0\}&lt;/math&gt;). Now consider the first &lt;math&gt;6&lt;/math&gt; terms:<br /> &lt;cmath&gt;\{1,2,0,1,2,0\}.&lt;/cmath&gt;<br /> To find &lt;math&gt;w(6)&lt;/math&gt;, we set aside the last &lt;math&gt;3&lt;/math&gt; terms as a &quot;reserve&quot; that we can pair with subsets of the first &lt;math&gt;3&lt;/math&gt; terms (which we looked at earlier). <br /> <br /> First, all &lt;math&gt;2^3&lt;/math&gt; subsets of the first &lt;math&gt;3&lt;/math&gt; terms can either be paired with a &lt;math&gt;1&lt;/math&gt; or a &lt;math&gt;2&lt;/math&gt; (or nothing) from the &quot;reserve&quot; terms so that they &quot;work,&quot; creating &lt;math&gt;2^3&lt;/math&gt; subsets that &quot;work&quot; already. But for each of these, we have the option to add a &lt;math&gt;0&lt;/math&gt; from the reserve, so we now have &lt;math&gt;2\cdot2^3&lt;/math&gt; subsets that &quot;work.&quot; For each of the &lt;math&gt;w(3)=4&lt;/math&gt; subsets of the first &lt;math&gt;3&lt;/math&gt; terms that &quot;work&quot;, we can also add on a &lt;math&gt;\{1,2\}&lt;/math&gt; or a &lt;math&gt;\{1,2,0\}&lt;/math&gt; from the reserves, creating &lt;math&gt;2w(3)&lt;/math&gt; new subsets that &quot;work.&quot; And that covers all of them. With all of this information, we can write &lt;math&gt;w(6)&lt;/math&gt; as<br /> &lt;cmath&gt;w(6)=2\cdot2^3+2w(3)=2(2^3+w(3)).&lt;/cmath&gt;<br /> The same process can be used to calculate &lt;math&gt;w(9)&lt;/math&gt; then &lt;math&gt;w(12)&lt;/math&gt; etc. so we can generalize it to<br /> &lt;cmath&gt;w(N)=2(2^{N-3}+w(N-3)).&lt;/cmath&gt;<br /> Thus, we calculate &lt;math&gt;w(18)&lt;/math&gt; with this formula:<br /> &lt;cmath&gt;w(18)=2( 2^{15} + 2( 2^{12} + 2( 2^9 + 2( 2^6 + 2( 2^3 + 4 ) ) ) ) )=87424.&lt;/cmath&gt;<br /> Because there are &lt;math&gt;2^{18}&lt;/math&gt; total possible subsets, the desired probability is<br /> &lt;cmath&gt;\frac{w(3)}{2^{18}}=\frac{87424}{2^{18}}=\frac{683}{2048},&lt;/cmath&gt;<br /> so the answer is &lt;math&gt;\boxed{683}.&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if &lt;math&gt;U&lt;/math&gt; is of a small size. <br /> <br /> If &lt;math&gt;U = \{ 1,2,0\} \pmod 3&lt;/math&gt;, then &lt;math&gt;4&lt;/math&gt; out of &lt;math&gt;8&lt;/math&gt; subsets work, for a probability of &lt;math&gt;\frac12&lt;/math&gt;.<br /> <br /> If &lt;math&gt;U = \{ 1,2,0,1,2,0\} \pmod 3&lt;/math&gt;, then &lt;math&gt;24&lt;/math&gt; out of &lt;math&gt;64&lt;/math&gt; subsets work, for a probability of &lt;math&gt;\frac38&lt;/math&gt;.<br /> <br /> If &lt;math&gt;U = \{ 1,2,0,1,2,0,1,2,0\} \pmod 3&lt;/math&gt;, then &lt;math&gt;176&lt;/math&gt; out of &lt;math&gt;512&lt;/math&gt; subsets work, for a probability of &lt;math&gt;\frac{11}{32}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the numerator of the desired probability if &lt;math&gt;U&lt;/math&gt; is of size &lt;math&gt;3n&lt;/math&gt;. Then &lt;math&gt;a_1 = 1, a_2 = 3,&lt;/math&gt; and &lt;math&gt;a_3 = 11&lt;/math&gt;. Noticing that the denominators are multiplied by &lt;math&gt;4&lt;/math&gt; each time, we can conclude that the pattern of the numerators is &lt;math&gt;a_n = 4a_{n-1} - 1&lt;/math&gt;, so &lt;math&gt;a_6 = \boxed{683}&lt;/math&gt;.<br /> <br /> ==Solution 7 (Quick guesswork for about 2 minutes remaining)==<br /> <br /> <br /> We conjecture that the difference from the probability will be as small as possible from 1/3<br /> <br /> (The value approached as n--&gt;infinity, where n is the number of terms in the largest subset.)<br /> <br /> <br /> We also see that there are 2^18 subsets and know the denominator will be a power of 2<br /> (since the numerator is an integer).<br /> <br /> We essentially want to guess that the greatest integer n satisfying (2^n/3)-1 &lt;1000 can be plugged in to get the solution of round(2^n/3)<br /> <br /> <br /> We see that this occurs at n=11, and get round(2^11/3)=round(682.66...)= 683.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_3&diff=127334 2020 AIME II Problems/Problem 3 2020-07-03T05:38:04Z <p>Xxu110: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n&lt;/math&gt;. Based on the equation, we get &lt;math&gt;(2^x)^n=3^{20}&lt;/math&gt; and &lt;math&gt;(2^{x+3})^n=3^{2020}&lt;/math&gt;. Expanding the second equation, we get &lt;math&gt;8^n\cdot2^{xn}=3^{2020}&lt;/math&gt;. Substituting the first equation in, we get &lt;math&gt;8^n\cdot3^{20}=3^{2020}&lt;/math&gt;, so &lt;math&gt;8^n=3^{2000}&lt;/math&gt;. Taking the 100th root, we get &lt;math&gt;8^{\frac{n}{100}}=3^{20}&lt;/math&gt;. Therefore, &lt;math&gt;(2^{\frac{3}{100}})^n=3^{20}&lt;/math&gt;, and using the our first equation(&lt;math&gt;2^{xn}=3^{20}&lt;/math&gt;), we get &lt;math&gt;x=\frac{3}{100}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{103}&lt;/math&gt;.<br /> ~rayfish<br /> <br /> ==Easiest Solution==<br /> Recall the identity &lt;math&gt;\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b &lt;/math&gt; (which is easily proven using exponents or change of base).<br /> Then this problem turns into &lt;cmath&gt;\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3&lt;/cmath&gt;<br /> Divide &lt;math&gt;\log_{2} 3&lt;/math&gt; from both sides. And we are left with &lt;math&gt;\frac{20}{x}=\frac{2020}{x+3}&lt;/math&gt;.Solving this simple equation we get &lt;cmath&gt;x = \tfrac{3}{100} \Rightarrow \boxed{103}&lt;/cmath&gt;<br /> ~mlgjeffdoge21<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;\log_a{b^c}=c\log_a{b},&lt;/math&gt; we have that &lt;math&gt;20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,&lt;/math&gt; or &lt;math&gt;\log_{2^x} 3 = 101\log_{2^{x+3}} 3.&lt;/math&gt; Since &lt;math&gt;\log_a{b}=\dfrac{1}{\log_b{a}},&lt;/math&gt; &lt;math&gt;\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},&lt;/math&gt; and &lt;math&gt;101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},&lt;/math&gt; thus resulting in &lt;math&gt;\log_{3}2^{x+3}=101\log_{3} 2^x,&lt;/math&gt; or &lt;math&gt;\log_{3}2^{x+3}=\log_{3} 2^{101x}.&lt;/math&gt; We remove the base 3 logarithm and the power of 2 to yield &lt;math&gt;x+3=101x,&lt;/math&gt; or &lt;math&gt;x=\dfrac{3}{100}.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{3+100=103}.&lt;/math&gt;<br /> ~ OreoChocolate<br /> <br /> ==Solution 3 (Official MAA)==<br /> Using the Change of Base Formula to convert the logarithms in the given equation to base &lt;math&gt;2&lt;/math&gt; yields<br /> &lt;cmath&gt;\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~}<br /> \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.&lt;/cmath&gt;Canceling the logarithm factors then yields&lt;cmath&gt;\frac{20}x = \frac{2020}{x+3},&lt;/cmath&gt;which has solution &lt;math&gt;x = \frac3{100}.&lt;/math&gt; The requested sum is &lt;math&gt;3 + 100 = 103&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/lPr4fYEoXi0 ~ CNCM<br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=x0QznvXcwHY?t=528<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution 3==<br /> <br /> https://youtu.be/-CkEF5nWOaI <br /> <br /> ~avn<br /> <br /> ==Video Solution 4==<br /> <br /> https://www.youtube.com/watch?v=2TSNY2DDUbQ&amp;t=3s ~ MathEx<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=2|num-a=4}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xxu110 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_3&diff=127333 2020 AIME II Problems/Problem 3 2020-07-03T05:34:45Z <p>Xxu110: /* Easiest Solution */</p> <hr /> <div>==Problem==<br /> <br /> The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n&lt;/math&gt;. Based on the equation, we get &lt;math&gt;(2^x)^n=3^{20}&lt;/math&gt; and &lt;math&gt;(2^{x+3})^n=3^{2020}&lt;/math&gt;. Expanding the second equation, we get &lt;math&gt;8^n\cdot2^{xn}=3^{2020}&lt;/math&gt;. Substituting the first equation in, we get &lt;math&gt;8^n\cdot3^{20}=3^{2020}&lt;/math&gt;, so &lt;math&gt;8^n=3^{2000}&lt;/math&gt;. Taking the 100th root, we get &lt;math&gt;8^{\frac{n}{100}}=3^{20}&lt;/math&gt;. Therefore, &lt;math&gt;(2^{\frac{3}{100}})^n=3^{20}&lt;/math&gt;, so &lt;math&gt;n=\frac{3}{100}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{103}&lt;/math&gt;.<br /> ~rayfish<br /> <br /> ==Easiest Solution==<br /> Recall the identity &lt;math&gt;\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b &lt;/math&gt; (which is easily proven using exponents or change of base).<br /> Then this problem turns into &lt;cmath&gt;\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3&lt;/cmath&gt;<br /> Divide &lt;math&gt;\log_{2} 3&lt;/math&gt; from both sides. And we are left with &lt;math&gt;\frac{20}{x}=\frac{2020}{x+3}&lt;/math&gt;.Solving this simple equation we get &lt;cmath&gt;x = \tfrac{3}{100} \Rightarrow \boxed{103}&lt;/cmath&gt;<br /> ~mlgjeffdoge21<br /> <br /> ==Solution 2==<br /> Because &lt;math&gt;\log_a{b^c}=c\log_a{b},&lt;/math&gt; we have that &lt;math&gt;20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,&lt;/math&gt; or &lt;math&gt;\log_{2^x} 3 = 101\log_{2^{x+3}} 3.&lt;/math&gt; Since &lt;math&gt;\log_a{b}=\dfrac{1}{\log_b{a}},&lt;/math&gt; &lt;math&gt;\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},&lt;/math&gt; and &lt;math&gt;101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},&lt;/math&gt; thus resulting in &lt;math&gt;\log_{3}2^{x+3}=101\log_{3} 2^x,&lt;/math&gt; or &lt;math&gt;\log_{3}2^{x+3}=\log_{3} 2^{101x}.&lt;/math&gt; We remove the base 3 logarithm and the power of 2 to yield &lt;math&gt;x+3=101x,&lt;/math&gt; or &lt;math&gt;x=\dfrac{3}{100}.&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;\boxed{3+100=103}.&lt;/math&gt;<br /> ~ OreoChocolate<br /> <br /> ==Solution 3 (Official MAA)==<br /> Using the Change of Base Formula to convert the logarithms in the given equation to base &lt;math&gt;2&lt;/math&gt; yields<br /> &lt;cmath&gt;\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~}<br /> \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.&lt;/cmath&gt;Canceling the logarithm factors then yields&lt;cmath&gt;\frac{20}x = \frac{2020}{x+3},&lt;/cmath&gt;which has solution &lt;math&gt;x = \frac3{100}.&lt;/math&gt; The requested sum is &lt;math&gt;3 + 100 = 103&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/lPr4fYEoXi0 ~ CNCM<br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=x0QznvXcwHY?t=528<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution 3==<br /> <br /> https://youtu.be/-CkEF5nWOaI <br /> <br /> ~avn<br /> <br /> ==Video Solution 4==<br /> <br /> https://www.youtube.com/watch?v=2TSNY2DDUbQ&amp;t=3s ~ MathEx<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=2|num-a=4}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xxu110